MECHANOPHYSICS (I)
SOME VALUES GIVEN IN THIS WORK
Gravity microcorpuscle: mg = 7.811 x 10—58 Kg.
Light corpuscle: mc = 1.47236 x 10—50 Kg.
Propeller particle, 7 orbit m7 = 4.8324 x 10—38 Kg.
Mass of the electron me = 9.1091 x 10—31 Kg
¨ ¨ a proton: mp = 1.6725 x 10--27 Kg.
¨ ¨ ¨ Earth = 5.975 x 1024 Kgs.
¨ ¨ ¨ Sun = 1.985 x 1030 ¨
Radius ¨ proton: r+ = 1.7143 x 10—14 m
¨ ¨ ¨ electron: re = 1.4 x 10—15 m
¨ ¨ ¨ seven shell: r7 = 1.13763 x 10—10 m
¨ ¨ ¨ Earth: r = 6.375 x 106 m
¨ ¨ ¨ Sun: rs = 7 x 108 m
¨ ¨ ¨ Earth orbit: r = 1.48 x 1011 m
Length of a side of polygonal orbit: lv = .2.55 x 10—11 m
Length of a multimolecule: lm = 5.68815 x 10—4 m
Radius of a virtual orbit: rv = 2.06 x 10 --4 m (maximum orbit)
Radius (maximum) of a Coulomb´s orbit: rc = 0.895 m
Time of return distance of light corpuscles: tr =1.58 x 1018 Sec
Time of deflection (electron 7 orbit): t>7 = 4.47 x 10—20 Sec.
Time of orbit ( ¨ ¨ ¨ ): to = 2.3084 x 10—10 Sec.
Time of vertex : ( ¨ ¨ ¨ ) tv7 = 8.27 x 10—17 Sec.
Time of jump ( ¨ ¨ ¨ ) tj7 = 1.12 x 10—11 Sec. = 135,632 tv7
Time of multimolecule: tm = 2.8 x 10—5 Sec.
1 year = 31,557,600 seconds
Deflection velocity of sever orbit electron: v_l_7 = 69,098 m / Sec.
Velocity of the seven orbit electron: ve7 = 308,570 m / Sec.
Velocity of the seven nushel positive particle: v+7 = 672.27 m / Sec.
h = 6.6256 x 10—34 joule = 1 quantum.
1 coulomb = 1 joule
Electron charge: qe = 1 eV. = 1.6 x 10—19 joule
m = 9.2754 x 10—28 joule = 1 magnetron of Bohr.
K = 9 x 109 newton m2 / coulomb2 = coefficient of Coulomb´s law
k = 1.38 x 10--23. joule / molecule oK = coefficient of Boltzmann´s law
G = 6.673 x 10—11 m3 / (Kg Sec2) = constant of universal gravitation.
135,632 t> = 1 tj
135,632 t>7 = 1 tj
L = 0.895 m = maximum disance at which works the Coulomb´s formula-
MODEL OF ELECTRIC CURRENT AND ITS APPLICATION TO MEDICINE
Before the experiment of Griffith, it was suppose that with certain heat a microorganism could be destroyed. Grifflith proved that with certain heat not all the structure of the microorganism were destroyed, and that at least its DNA was not affected. The structure of the DNA has many short circuits that nullify many effects of secondary currents that are produced in the cells and in the microorganisms. If are placed the extremes of two electrodes spaced one millimeter, when an electric current (10 volts, 0.008 Amp.) is circulating, it will produce a continuous spark between them. This spark produces heat, whose intensity destroy some parts of the microorganisms, and this have been proved in the practice.
In mechanophysics is not accepted the model of the free electrons. It is considered that the electrons move along a chain of multimolecules that are formed in the conductor when is produced a difference of electric potential. The electrons of the current move from the atom of less potential toward the atom of more potential (here we consider that from the less positive toward the more positive); this movement is around each atom of the multimolecule, during a time equal to: 1.12 x 10—11 second / atom, as will be explained later. When the electron arrives to the atom of higher potential of the multimolecule, it jumps to the atom of less potential of the adjacent multimolecule, in which is repeated the same movement. When the electron arrives to the atom of higher potential of the multimolecule (last atom), jumps to the adjacent multimolecule, that has more potential.
I am not sure, exactly, when began my investigation on physics. First I was interested in divulgate the theories of my father, so I studied them and afterward those of relativity; in these last ones I found many fails; so, also I got interested in study some modern physics, and also found many fails in it, and determine they had not good roots; that is, they were not founded in the structure of the atoms. I remember. as in 1965, doctor Rodolfo Castillo Bahena was director of the department of physics in the ISTESM. that told me that the real progress of physics was conditioned to a good model of atom; immediately I accepted this because always was in accordance with it. I began to study all the models of atoms existing them, but none satisfied me. Finally I conceived that of the polygonal orbits. and tried to make a detailed one of it. Afterward. and due that there are many alternatives in some details, I decided that this would be obtained when we advanced in our studies, and that is better to give the most simple model in first place.
The correct study of physics must be based in the structure of the atoms, and for to understand this we ought to know many things of a correct physics; Nevertheless physics science has not been developed parting from its roots, but from the macroscopic and more evident bodies and particles. For develop mechanophysics I parted from a model of atom that consider was the most logic; for understand this there are many details that here will be omitted, but afterward will be given in other themes. Exist the idea that physics ( and all sciences) have developed because all the discoveries made of it have been proved; but here there is a confusion; one thing is to prove something and other is to give an explanation of it (correct or wrong) of it. Newton explained with his fields of gravity forces, the movement of the planets around the Sun, and this explanation has prevailed till the present time; but with this has not been proved how works the gravity.
In his electromagnetic theories Maxwell supposed that any electric particles, from the most elemental one, that in this case would be an electron with negative charge, or a proton or a positron with positive charge; they would emit their electric fluid in all directions, forming an electric field with their corresponding sign. I am not in accordance with this, because this would mean that neither the electrons nor the protons have only one side in which they can emit (and receive) their radiations. An electron or a proton have only one side in which they emit their radiation. An electron or a proton, in a given instant only can emit their radiation in only one way; in many instant they can emit many radiation in many direction forming a field. A particle formed by many atoms and that has an electric positive charge, or a negative one, can form an electric field with many atoms, positive, or negative (in accordance with the excess of protons or electrons); this because in the charged particle each atom will try to emit its radiation at the same time than the others, in all directions (in all angles...).
As said before, I conceived a model of atom, taking in account many facts, but without pretending give the proves of it; these will be obtained with the progress of physics.. In some way I imagine an atom similar to a solar system; in other aspects have to take in account the behavior of the atoms; had to take in account the limitation of the laws of the classical dynamics; the behavior of the electromagnetic and the gravity radiations; of the black body radiations, etc. Avoiding many explanations that will be understand when we advance in our study, was conceived this model with the polygonal orbits, corresponding to the shells of the atom. Not all the atoms have the same radii; and these could vary in accordance with the model it is consider; mathematically can give a great approximation to the atomic values, in order that fix very well with a model (but not with all). It was consider an average radius: r7 = 1.13763 x 10—10 m., that is equal to the most external orbit or shell. There are seven shells (n = 7). spaced at equal distance one from each other. In the first orbit (square one) the electron moves at v1 = 2160 Kms./ Sec; in the (n) shell, formed by a regular polygon of 4 n sides, the orbital electron moves at a velocity: vn = v1 /n = 2160 / n Kms./ Sec.. In our model the orbit of the electron is not produced in a flat plane, because this plane has a spin, as will be explained afterward. Considering the rotation of the electron and the spin of its orbit, the electron will move around its shell in a time we will call time of jump: tj = 1.12 x 10—11 Sec.
Till here, only have been consider the orbital electrons of the atoms but have not talked of the positive particles that could be supposed as a unitary particle that control an orbital electron of the atom. In our model have consider that each orbital electron has its exclusive positive particle formed by one proton in the H atom, and by one proton plus one, or two neutrons in the other atoms. To an electron plus one positive particle, working as was say, we will call: binary system. Here both particles rotate around a point between them; for instance he electron of the seven shell, at a distance: r7 » 1.13763 x 10—10 m; and the positive particle: r7´ = 4 r7 / 1836 = 4 x 1.13763 x 10—10 / 1836 = 2.988 x 10—13 m; etc.
Had not been explained how are formed and how work the binary systems. In our model have been consider a structure for the positive and for the negative particles; an axis, as a tube in the diameter of them in which are received and emitted some propeller particles formed by corpuscles as those of light. For to be attraction between both particles they must be oriented, co lineal and in the same way, their axis. All the binary system of an atom fulfill the fore condition; and as the positive particles are very much near one to each other in the nucleus of the atom, there is no rejection between them, because their diameter, always are oriented to their negative particles (never to the other positive particles…).
The two particles of a binary system are spaced a distance equal to rn = (n / 7) r7. The propeller particle m^n; produces the deflection of the orbits of both particles, moving along the diameter of one of them, and after that, moves jumping to the other particle in which is repeated the fore mentioned movement and interior propeller action.. When this propeller particle is into the electron, it is produced a deflection in the trajectory of it (in a vertex); similar thing happens when is into the positive one, in its trajectory. All say here will be understand better with the numerical examples. The propeller particle m^n; is formed by corpuscle as those of light. The mass of a corpuscle is: mc = 1.47236 x 10—50 Kg.; its kinetic energy when moving at light velocity is equal to one quantum, h = 6.6256 x 10—34 joule. A propeller particle of the orbital electron of the seven shell, deflects it with a kinetic energy equal to: K>^7 = 0.5 m^7 c2 = 0.5 m^7 (3 x 108)2 = 4.5 x 1016 m^7
K>^7 = 0.5 me v^72 = 0.5 x 9.1091 x 10—31 x 69,1002 = 2.1747 x 10—21 joule
v^7 = radial velocity of the orbital electron of the seven shell
m^7 = 2.1747 x 10—21 / 4.5 x 1016 = 4.8327 x 10—38 Kg.
Modern physics acceptsy as model of electric current that of the free electrons. This model is too much deficient, as have been explained many times. When it was conceived and accepted did not exist a good structure of the atoms in which could flow the electrons of the current. They avoided this problem considering that the electrons could be free to flow by themselves. With a logical structure of the atoms we have conceived a model in which the electron of the electric current move along chains of atoms we name: multimolecules. The electric current electrons advance from an atom to the adjacent one, moving in each one a time of jump: tj = 1.12 x 10—11 Sec. = 135,632 t>7 = 135,632 x 8.27 x 10—17 Sec.; t>7 = time of vertex (7 shell); equal to the time in which the electron moves from a vertex to the adjacent one. The fundamental idea of our model of electric current can be obtained if we consider a conductor wire in which are formed a bunch of multimolecules. If it is applied a difference of electric potential the electrons that are in one extreme of it will try to move to the opposite extreme spinning around each atom as was say, in a time of jump; after than jump to the adjacent and forward atom, repeating this process, and so on, till they reach the highest potential extreme of the conductor wire. In this process they can advance limited distances, we call multimolecules, along the conductor. In the higher potential atom of each multimolecule they get over saturated; with an interior propeller fluid, they get free of the fore condition and can jump to the forward multimolecule, their excess fluid is absorbed by the last atom of the fore multimolecule; in this way can jump to the lower potential atom of the adjacent multimolecule and advance along it, as did in the fore multimolecule, and so on.. A multimolecule is formed by 2.5 x 106 atoms. The time employed by each current electron along a multimolecule is: tm = 2.5 x 106 tj = 2.5 x 106 x 1.12 x 10—11 = 2.8 x 10--5 Sec. The fundamental times of our model of electric current are: t>7, tj, tm , etc.
In figure (3), with (O) is indicated the nucleus of an atom, around it spin all the positive particles of ùuu++
the binary systems that form the nucleus of the atom, also spin the orbital electrons. The regular polygons concentric to the nucleus, have from the seven orbit toward the first the following quantity of sides: 7 x 4 = 28, 6 x 4 = 24, ... 1 x 4 = 4. The fore are the orbits of the nucleus formed by the positive particles of the binary systems; the negative particles or electrons have similar orbits, but much bigger. Practically all the sides of the same sign have the same length. In figure (2) is indicated in a schematic way a piece of trajectory of a seven orbit electront: y, z, e, a; (o) is the center of rotation. The 7 orbital electron moves around the nucleus (o) with a velocity: v7 = 309 Kms../ Sec.. The polygonal orbit also will spin, but in a perpendicular way around the nucleus (o), this due to the Bohr´s magnetron, that acts normally to the electron trajectory, as is indicated by the displacement e e´. Here the distance (ze) is proportional to the translation velocity v7 of the electron and ee´ is proportional to the perpendicular velocity of the vertex of the orbit; this due to the kinetic energy produced by the Bohr´s magnetron, and is equal to the angular velocity a of the plane of the orbit. Considering the plane of the orbit. in figure (2´) the distances: zy, ze, are equal each one to a side of the polygonal orbit; e´ e is the radial deflection of the orbital electron in a vertex.
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The effect of “energy” of the magnetron of Bohr is equal to: m. = 9.2754 x 10—28 joule; acting n accordance with the left hand rule; but this action is inverted when the orbital electron is moving in opposite direction than the advance of the electrons current, this in accordance with the explanations given in the theme: Rejecting Velocity and Admision Velocity, when two bodies that produce some field move in opposite side, there is a rejection effect between them. The action of the magnetic field produced by the moving electron, will be in the same way than the hands of the clock, for an observer that is in front of the advancing electron,when it moves in the same way of the flux of the electric current (of arrow F); and in the opposite way when the flux is inverted, as is indicated by the arrow (F), as can be appreciated in figure (2) (experiment of Stern & Gerlach). In this figure, schematically is represented the effective advance of the electron with the arrow F (equal to the direction that has the conductor). While the electron moves from the vertex (z) to (e), at a velocity v7, its orbit spins an angle (a); in such way that the vertex (e) is displaced to (e´), because the action of the magnetron of Bohr
Kinetic energy of the orbital electron: (7ª orbit) : Ke7 = 0.5 x 9.11 x 10—28 (3.09x 107 )2 = 4.35 x 10—13 erg Each time that the velocity of Bohr acts in the orbital electron of the electric current in each vertex, the orbit spin the following angle
a = (m / 2 Ke7)0.5 p / 14 = (9.2754 x 10-- 21 / {2 x 4.35 x 10-- 13})0.5 p / 14 = 2.317 x 10-- 5 radians. 14 = 28 sides of orbit / 2
(m / 2) acts in the positive particle (nucleus), and .(m / 2) acts in the orbital electron. For the orbit to spins p radians it is required: np = p / a = 3.1416 / 2.317 x 10--5 » 135,632 Bohr´s velocities. the orbital electron will move from one vertex to the next one in a time of vertex:
t>7 = 2 p r7 / (28 ve) = 2.p (1.13763 x 10—10) / (28 x 3.085 x 105) = 8.27 x 10—17 Sec.
Each time the orbit of the electron spins an angle = 180º (the electron has moved around all the surface of its shell), the electron will jump to the adjacent atom of the multimolecule, in a time of jump: tj = 8.27 x 10—17 x 135,632 = 1.12 x 10—11 Sec. = time of jump
In the seven shell of an atom one of its molecable electrons, or electron that can form a molecule, has certain saturation, In the H atom, its binary system when ionized can gets its electron moving in the first orbit, before it get ionized; then the ionization potential of the electron of the hydrogen would be: i = 0.5 me v12 = 0,5 x 9.1091 x 10—31 x 2,160,0002 = 2.125 x 10—18 joule = 2.125 x 10—18 / 1.6 x 10—19 = 13.3 eV; 1 eV = 1.6 x 10—19 joule
Since was known the existence of the electrons and of the nuclei (with our theory, the negative particles and the positive ones of the binary systems...) of the atoms, have been considered that an atom gets ionized when it gets free of an electron, because this the electron emits a propeller fluid, formed by corpuscles, as those of light, that have all the energetic capacity to realize this process. But this is not the only way an atom gets free of an electron, and jumps toward the first orbit. In a molecule formed by two atoms, in a given instant (time of jump) one of the atoms gets free of the molecular electron (molecable electron), because this moves to the other atom; in the following instant, the process is inverted; an so on; and all these without the production of any external propeller fluid. In our multimolecule the current electron moves along all the atoms without producing external propeller fluids, except in the extreme atoms of the chain of multimolecules…. It is possible (?) that the multimolecules are formed when is produced a difference of potential. For produce an ionization, this is accompany of a propeller fluid in the ionizer electron. Some atoms are ionized easier than others because they have a higher degree of saturation of corpuscles, such that with less corpuscles that are added, is produced the ionization fluid in the electrons that are liberated. In the molecules there is a continuous interchange of internal propeller fluid, that is of energy. In the electron of the electric current and the intermediate atoms of the multimolecule happens something similar. When an electron of the electric current of a multimolecule jumps to another one, only the electron gets free of its over saturation, that will be absorbed by the adjacent and last atom of the kuti,olrcule..
In my model of electric current, that could be named model of ionization or of multimolecules, is considered that the higher electric potential, is that that have the most positive particles, and the less potential that that have the less positive particles, or more negative ones. Here is considered the movement of the electric current equal to the movement of advance of the electrons. These move along a chain of atoms that we call: multimolecules, of nearly 2.5 x 106 atoms. The electron of the current moves along of each multimolecule, spinning around each atom, jumping from the less potential toward the most potential of the multimolecule; and next jumps to the atom of less potential of the adjacent multimolecule, and of higher potential than the fore one. Each time that the electron of the current moves along a multimolecule, as was say, gets a degree of over saturation of corpuscles that are yieded by the atom od the multimolecule (as those of light). When the electron of the current gets the mentioned degree of over saturation, can not advance, because it need gets free of this over saturation;, in accordance with the model, in order can do this and jumps to the adjacent multimolecule, were is repeated the fore process . When the electron jumps to the atoms of less potential of the adjacent multimolecule, before this, has get free of its over saturation, is produced a fluid is not manifested in an exterior way, because is transmitted directly between the electron to its adjacent atom, except in the case is produced what we have call an ionization effect in which is manifested in an exterior way the propeller fluid. In this way, the electron, being free of the mentioned over saturation, will be in condition to move freely in the less potential atom of the forward multimolecule, next will move toward the atom of higher potential of this multimolecule. Here is repeated the jumps process mentioned before, and so on.
In the same way that the electron (negative particle of the binary system....) gets over saturated when interacts with positive particles 2.5 x 106 times; the positive particle that absorbs the over saturations of the negative particles, also gets over saturated, in accordance with the capacity of the positive particle here will be produced an ionization....) when it is affected by an equal quantity (2.5 x 106) of the current electrons . The number mentioned before is equal to the atoms that has a multimolecule. The positive particle, that in the given moment gets over saturated and ionized correspond to the atom of higher potential of 2.5 x 106 multimolecules. Such ionization is manifested with a propeller fluid that emits the ionizer electron with the energy and temperature required for this process. In order to effectuate this ionization it is required a time equal to that that an electron of the current moves along (2.5 x 106)2 atoms of the conductor = 2.5 x 106 multimolecules. The ionization that is produced is equivalent to a high temperature.
In all the movements in which intervene the electron and the atom (more correct to say: the negative particle and the positive one in the binary systems....) there is an interchange of energy, manifested by the corpuscles acting as propeller fluids of the ionizer electron, as was say in other paragraphs. The ionization is produced when the electron has moved along nm = 2.5 x 106 multimolecules, in an electric current.. With our theories has been defined that the electrons of the electric current move along the chain of atoms of the multimolecules. Each is formed by about 2.5 x 106 atoms, as have been justified in other themes, when have obtained some concordances between some effects and phenomena. An electron of the electric current is able to move without problem from the atom of lower potential toward the atom of higher potential of the multimolecule. This can be explained considering that the electron in all this movement, around 2.5 x 106 atoms (binary system in the atoms....) gets a degree of saturation of corpuscles (as those of light). When the electron arrives to the atom of higher potential of the multimolecule it gets all its saturation (gets over saturated), so can not advance freely, so that in order to jumps to the atom of less potential of the adjacent and more potential multimolecule, the electron will have to gets free with an inerior fluid, of all its over saturation, that will be absorbed by the atom of higher potential of the multimolecule.
In the same way that the electron of the electric current requires to move around 2.5 x 106 atoms of the multimolecule, in order to gets over saturated and afterward gets free of this over saturation, that will be absorbed by the adjacent atom, that gets an over saturation, when electrons in an electric current have afected (2.5 x 106)2 times to such atom, so that can be be uinized. The effective distance that moves an electron in the fore process will be: L i = 2 r7 (2.5 x 106)2 It was say that the ionization of an atom of the conductor is equal to: i = nm io,---------------------------------------------- nm = quantity of atoms that has a multimolecule; io = energy of over saturation of the current electron. But in the same conductor can circulate currents of different intensity, depending of the different electric potential at which they are working. For produce an ionization in an atom, in all cases, this will be affected for the same quantity of interactions with the electrons of the current. It was say that between the electrons of the current and the atoms of the multimolecules there is an interchange of saturation, , by nm saturations produced by the electrons in the atom is going to be ionized, as just have been say As in these explanations there are many details that complicate the explanations; next will be given numerical examples; one with a conductor as mercury Hg, and other with a conductor of cooper Cu. In medicine, electric currents with about: V » 10 volts and: I » 0.008 Amps, are not harmful to the organism; because this, we are interested to know in which conditions all the conductors work in order to fulfill the fore requirement
In the table of radii of atoms, for Hg we have: rHg = 1.44 m; so we have a length of multimolecule for Hg: Lm = (2 r7) nm = (2 x 1,44 x 10—10) 2.5 x 106 = 0.00072 m.
In nm multimolecules is produced an ionization in an atom.
Time of multimolecule: tm = 2.8 x 10—5 Sec.
Time of ionization of an atom: ti = nm tm = 2.5 x 106 x 2.8 x 10—5 = 70 Sec.
As we are working with I acting in one second, this distance will be:
L1 = Lm nm / 70 = 0.00072 x 2,5 x 106 / 70 = 1800 / 70 = 25.714 m. in one second
r7 = radius of an atom.; nm = quantity of atoms that has a multimolecule. nm2 = quantity of atoms that travel the electron between an ionized atom and the following one.
Numerical example of a Hg conductor
(Hg): I = 0.008 amperes; r = 0.0000958 ohm / cm3; i = 10.4 eV. = potential of ionization; A = 0.01 cm2 = section of the conductor
R = r L1 / A = =.0.0000958 x 2571.4 / 0.01 = 24.634 ohm / L1 = resistance. V = R I = 24.634 x 0.008 = 0.197 volts / L1 I = intensity of the current expressed in amperes = 0.008 Amps . Ie = intensity of the current expressed in electrons / seconds = 0.008 x 6.3 x 1018 = 5,04 x 1016 Energy of the potential of ionization: i = 10.4 x 1.6 x 10—19 joule = 1.664 x 10—18 joule.
Time of ionization: ti = 2.5 x 106 tm = 2.5 x 106 x 2.8 x 10—5 = 70 Secs.
Wr = R I2 = 24.634 x 0.0082 = 0,0015766 watt
Wi = i Ie / 70 = 1.664 x 10—18 x 5.04 x 1016 / 70 = 0.0011981 joule / Sec.
Wr / Wi = 0.0015766 / 0.0011981 = 1.3159
Wi´ = 1.3159 x 0.0011981 = 0.0015766 joule / Sec..
i´ = 1.3159 x 1.664 x 19—18 = 2.1897 x 10—18 coulomb
Temperature produced every 70 Sec. (formula of Boltzmann) :
Ti = 2 I´ / 3 k = 2 x 2.1897 x 10—18 / 3 x 1.38 x 110--23= 105,780o K
An atom (a binary system). in every (2.5 x 106) multimolecules, gets ionized with an exterior fluid produced by its electron. This fluid is equivalent to a high temperature (Ti), which value is given by the Boltzmann´s formula. The pathogen microorganisms have a solid cover (glicoproteins) that protect them of the humidity of our organism, but not of the heat produced by the ionization. In this is based the electrotherapy. Our organism is not affected in a no reversible way by this heat, because its humidity that dissipate the heat; also the ionization effects of the electrotherapy are very well distributed in the space and in the time in all the organism
For a copper wire we have:
(Cu) I = 0.008 Amps. i = 7.7 eV; A = 0.01 cm2 ; r = 0.0000017; Ie = 5.04 x 1016 electrons / Sec.; Energy of the potential of ionization:
i = 7.7 x 1.6 x 10—19 = 1,232 x 10—18 coulomb
Lm = (2 r7) nm = (2 x 1.17 x 10—10) 2.5 x 106 = 0.000585 m
Space advanced by the electron in one second:
L1 = Lm nm / 70 = 0.000585 x 2.5 x 106 / 70 = 20.893 m R = r L1 / A = 0.0000017 x 2089.3 / 0.01 = 0.35518 ohms / L1 V = R I = 0.35518 x 0.008 = 0.00284 volts / L1 Ie = 0.008 x 6.3 x 1018 = 5.04 x 1016 electrons / Sec.
Wr = R´I 2 = 0.35518 (0.008)2 = 0.00002273 joule / Sec. Wi = Ie i / 70 = 5.04 x 1016 x 1.232 x 10—18/ 70 = 0.0008870 joule / Sec. Wr / Wi = 0.00002273 / 0.000887 = 0.02563 i´ = 0.02563 i = 0.02563 x 1.232 x 10--18 = 3.1576 x 10-20 Wi = i´ Ie / 70= 3.1576 x 10—20 x 5.04 x 1016 / 70= 0.00002273
Ti = 2 i´ / 3 k = 2 x 3.1576 x 10—20 / (3 x 1.38 x 10—23) = 1,525o K
In our organism Ti is bigger than in Cu and smaller tha in Hg. In any book of chemistry there are tables that give the energy of ionization (electron potential of ionization: i) of all the atomic elements; but these tables are obtained with the elements at certain temperature, pressure, etc.. With some higher temperature some atoms of a sample get ionized without been produced a difference of electrical potential: .To our model of electric current of multimolecules, also we can call: model of electric current by ionization.
Monterrey, México; 1965
August / 2003
If I were a protected investigator of a prestige university I would win a Nobel price of medicine by a work like this one.
MULTIMOPLECULES
A very much important value is that of the multimolecule. Since a long time ago I am aware of this; for obtain it had to consider several things. For all people that had read my work, it is not necessary to explain what is a multimolecule. In my investigation work of an electric current had to meditate in many facts. In first place concluded that a model of electric current as that of the free electrons was with too many deficiencies; I have been commenting many times in other theories, here will not repeat them: With respect to the resistance of the current, the practical experience for us is very much valuable and the first useful experience we can take for structure our model. As the current electrons do not get free so easy as is considered in the theory of the free electrons, we had to take in account that they moved along a chain of atoms, that we called multimolecule. This means that in all the conductors the current electrons move at the same velocity on their conductor wires, along the multimolecules; some with less resistance (good conductors), and other with more resistance. Our first problem was to determine how move the current electrons in every one of the atoms of the multimolecule. Based in many other phenomena, this solution was relative easy, as could be understand in the numerical problems. Other problem was to determine how many atoms has a multimolecule. Really this is not an easy problem. Since the beginning we got the idea that each current electron jumps from an atom to the forward one of the multimolecule in some way, similar as an electron of a molecule. How many jumps could make in this way, in order to move along the multimolecule? To answer the fore question a logic and scientific way, is not easy for us; so we studied it considering the behavior of some conductors: For this we choose the mercury atom, that practically is the atom most resistant of all conductors, and with its liquid structure there is less risk that its resistance can be altered by irregular distribution of its atoms. Some kind of atoms, as iron are more resistant to the flow of electrons, but they could be magnetized, etc., etc. Depending of the length and the cross section of a mercury conductor, can give to it any one resistance. By first instance this does not affect us. Here will be accepted the value given in a conventional way; and from this value will determine an electric intensity that will be a balanced one; that is, that will give: Wr = Wi (see the fore problem). If the resistance R is reduced to its half, its intensity I is duplicated, and the energies Wr and Wi get incremented in the same magnitude than I. In few words, the maximum variation could have the mercury conductor could not affect to the other conductors that are better ones. For instance, in the fore numerical problems we had: RHg = 1,362.3 ohms; RCu = 24.1747. We can give equal values to both conductors if: LHg = 142,204 cm.; LCu´ = 1,342.3 LHg / 24.1747 = 59.8774 LHg = 8,514,807 cm.
For instance, we have for Hg: L = 100 m: A = 1 cm2; I = 120 Amps.; r = 0.0000968; i = 10.4 eV.
RHg = r L / A = 9.58 x 10—5 x 10,000 / 1 = 0.958 ohms.
ik´ = 10.4 x 1.6 x 10—19 = 1.664 x 10—18 joule / atom
I = 120 Amps.; Ie = 120 x 6.3 x 1018 = 7.56 x 1020 electrons / Sec
Wr = R I2 = 0.958 x 1202 = 13,705 joule / Sec.
Wi = ik´Ie = 1.664 x 10—18 x 7.56 x 1020 = 1,260 joule / Sec.
Wr / Wi = 13,795 / 1,260 = 10.95
If: I´ = 120 / 10.95 = 10.96 Amps. = balanced intensity.
Wr´ = 0.958 x 10.962 = 115 joule.
Ie´ = 10.96 x 6.3 x 1018 = 6.9048 x 1019 electrons / Sec.
Wi´ = 1.664 x 10—18 x 6.9048 x 1019 = 115 joules.
For a copper conductor we have: A = 1 cm2.; I = 120 Amps.; i = 7.7 eV.;
r = 0.000001725.
We can have equal resistance than the Hg conductor if it is of a length:
LCu = rHg LHg / rCu = 0.0000958 x 100 / 0.000001725 = 5,553.62 m.
RCu´ = rCu LCu / A = 0.000001725 x 555,362 /1.0 = 0.958 ohm.
For copper we have: r = 0.000001725; i = 7.7 eV.
Determination of: nm = quantity of atoms that has a multimolecule.
In the Hg conductor we have: r = 1.13763 x 10—8 cm.; 2 r = 2.2752 x 10—8 cm.
A = 1.0 cm2 = 1.0 / /2 r)2 = 1.0 / (2.27526 x 10—8)2 = 1.931692245 x 1015 atoms.
Ie = 6.3 x 1018 I´ = 6.3 z 1018 x 10.96 = 6.905 x 1019 = quantity of current electrons that flow in one second.
6.905 x 1019 tj = 6.905 x 1019 x 1.12 x 10—11 = 7.733 x 108 electrons / tj.
1.9317 x 1015 = quantity of multimolecules there are in a volume: A nm of atoms.
-- - - - - - -- - - - - - - -- - - - -
Originally, when I made the model of the polygonal orbits, I considered that the magnetron of m: m = 9.2754 x 10—21 was an energy value; afterward, I saw that could work as an energy if it was affected by a term: Q, that is expressed in (gr. / erg). The formula, expressed in e,m, units of the magnetron of Bohr, has an energy would be: M = Q e h / 4 p m. Here the term (e / m) only gives a mathematical proportion, not a physical one… I did not pay much attention to this problem because I was working in other problems of physics; but now that I am studding the flux of the atomic particles have find many interpretations in the problem of the magnetron of Bohr…. Because this, I decided to search other energy that could explain the perpendicular deflection of the orbit of the electron. With this modification will vary the time of jump (tj) and the time of multimolecule (tm): But this will not affect our theories, only will be some quantitative variations in some problems, that in the future, and without precipitation will be corrected.
-- - - - -- - - - - - - - - -- -
Considering that a current electron moves along a multimolecule in a time: tm = nm tj, we have:
1.9317 x 1015 = 7.733 x 108 nm;
nm = 1.9317 x 1015 / 7.733 x 108 = 2.498 x 106 » 2.5 x 106 atoms = number of atoms that has a multimolecule. Also is very much important to point out that the fore number is equal to the quantity of atoms there are in the cross section of the conductor for each active atom in any given moment. -
Monterrey, México; September / 2003 Manuel de Hoyos Robles - --)
TWO HOLES EXPERIMENT
In the XIX century the undulated theory of light was imposed because with waves could explain many phenomena, that they could not do with a corpuscular theory. In the XX century, also happened the same thing; and also, afterward were found some phenomena they could not be explained by an undulated theory; so they considered that matter (corpuscles) could be transformed in energy (waves). The partisan of the undulated theory considered that the experiment of the two holes proved that light is produced by waves. They used two instruments, one that produces waves, and a slug gun. They had two parallel walls; one with two holes, and in front of it other in which they measure the intensity of the waves and of the slug particles. In the opposite side are the instruments that produce the two mentioned effects. When only one hole is opened in the middle , and the experiment is made with both instruments, the maximum intensity is appreciated similar in the testing wall, as could be seen with a symmetric curve with one maximum values. When both holes are opened 
THE RETURN DISTANCE (NO GROWING OF ENTR0PY)
Since the epoch of Newton, Coulomb, etc. the fields of forces have got a fundamental importance ; nevertheless if it is considered the infinite time of existence of the Universe, in this time all the fields of the Universe would be exhausted because the bodies that produce them would not be able to continue producing the microatomic particles that form such fields, but this is not so, because the particles that produce such fields, when they are expelled, recede from such bodies at light velocity, move during millions of years and after that time they return to the place were they were emitted, to this process we call return distance in such way that the fields never get exhausted ; to these fields we call mechanized fields for to differentiate of the structured ones that were accepted since the Newton’s epoch, and in which there were not considered as of corpuscular nature. With the mechanized fields of mechanophysics these problems are solved. Mechanized fields are formed by ultramicroscopic particles emitted at light velocity by the bodies that form such fields in all directions, when these field particles reach another body they act in this one, making the body to acts in accordance with the kind of field is considered, and as there is reciprocity between causes and effects, the second body affects to the first one in a similar way. As can be appreciated the mechanized fields by far are more simple than the structured ones. The first step of structure of the mechanized field has been given.
In the infinite time of existence of the Universe all the bodies producer of mechanized fields would be exhausted of their field fluids they emit, unless we consider other property of such fluids that we call: return distance, this return distance can vary in considerable magnitudes of time and distance in accordance with many facts; when a cycle of return distance is effectuated the fluid that original was emitted return toward the emitter body and is “recuperated”. Not all the corpuscles or particles emitted by a body return to the place they were emitted, because are absorbed by other celestial bodies. This, and other celestial bodies also emit particles that are absorbed by the first mentioned body. The second step of the mechanized fields has been made, that is, of no exhausted fields and with this solution is obtained other solution of a problem that have been of preponderant importance since the Boltzmann’s epoch, that is, the growing of entropy. It has been considered that the energy moves always from a higher level toward a lower one. In some way this is a simple criterion that with the mechanophysics is seen in a more subtle way, in accordance with this all particles (entropy particles) of the Universe always had have a determined energy that remain in them in the infinite time in the past and in the future, this energy acts in such particles in accordance with many facts in any determined process and in accordance with one of the fore mentioned facts the particle lose energy, but in accordance with other not evident facts the particle win energy (law of conservation of energy) in such way that all the particles of the Universe have not lost nor win energy in the infinite time of their existence in the Universe; forward it is going to see a subtle case, in this (returning distance) there is not the problem given by the growing of entropy, because the energy (in form of mechanized fields) that is emitted away in all directions, because the returning effect return to the opposite way (maybe in millions years).
There is a fundamental difference between the structured fields and the mechanized ones. It is considered that the first ones have an innate existence and that a body is affected by them in proportion to the magnitude of the physical body properties. The mechanized fields can not have an existence if there not exist the interacting bodies, this means that this bodies are the producer of such fields, sc the theory of the mechanized fields is very much simplified. How many mechanized fields we can have? We can answer the fore question if we observe all kinds of radiation that can emits a body in different circumstances: a gravity field, an electromagnetic field,, a light field, etc. With the mechanized fields we return the acceptance of the Newton’s law of gravity attraction. That was considered was not true because they fail in the precession of the perihelion of Mercury and in the deflection of a light ray moving tangential and near of a heavy celestial body as the Sun? I am not in accordance with the fore trace because in it was not considered that the gravity velocity was incremented (aberrated) by the velocity of the affected or receptive bodies: (Mercury planet and the light corpuscles mentioned); the effect that produces the fore increment of receiving gravity velocity is proportional to the square of such increment. The structured fields depart from the amplitude of the Universe affecting to the interacting bodies: the mechanized fields depart from the interacting bodies affecting the space of the Universe in which they interact.
The problem of the red shift can be explained with a numerical example and at the same time will be given a logical explanation of what has been interpreted as the expansion of the Universe given by modern physics. It has been observed that while farther recede a celestial body, its light radiation (gamma,.....radio radiation) move toward the red frequency, this is considering the constancy of light velocity, about this model (expansion Universe) there are some contradictory points we are not going to mention because the lack of space. In mechanophysics are considered the light radiation formed by material corpuscles (corpuscular nature). A light ray is formed by a straight line of light corpuscles spaced one from each other at equidistant we call: corpuscular distance (wave length in the undulate theory of light). In our model of mechanophysics the frequency of radiation is given by the corpuscular distance, and for distant celestial bodies diminishes not because the corpuscular distance grows, but because the light velocity diminishes due to the action of the remanent gravity (as we will see forward) that acts at equal intervals of time we call: braking time in each corpuscle of the ray of light At the present time the advance of the technology permit us to measure the velocity of light affected by the mentioned braking effect, who does this must not be influenced with the idea that light velocity is a constant value..
With respect to the action of remanent gravity in the light radiation’s we can make more than one model with a structure criterion. Of them and with a fundamentalist criterion we will choose one on which is considered that the electrons that emit the light are not affected by gravity, only their “corresponding” proton in the nucleus of the atom. Each corpuscle of a ray of light emitted, has a quantity of remanent gravity (nugravity) enough to acts at each time of brake during all the returning distance trajectory in such a way that the ray of light emitted by the electron in a given point of space, return to the same point after millions years, as could be appreciated by the numerical example will be given forward. In other themes, forward will be seen that particles with inherent energy can multiply this effect, with multiple interchanges of this energy. Here the kinetic energy produced by the remanent gravity is double than the kinetic energy with which the light corpuscle was emitted; this in accordance with the entropy particles of mechanophysics that consider that all the particles of the Universe have a fixed energy that is not lost nor win in any process we consider In accordance with our model the corpuscles of light are integrated with the nugravity in the surface of the “corresponding” proton with the quantity of nugravity in each corpuscle, enough to act producing the braking effect during all the returning time, with a force can be determined with the law of Newton acting in a corpuscle placed in the surface of the proton, by the proton exclusively. If the rays of the light are produced by atomic particles that are in a heavy celestial body, the remanent gravity in each light corpuscle will grow, so is required to sum the remanent gravity of the proton (nugravity) plus the remanent gravity produced by the heavy celestial body (astrogravity) all these in such way that the return distance trajectory will be smaller than that produced by a body (in vacuum space) emitted with a velocity little bad smaller than that of escape velocity as would be in the case of a celestial body with a mass that approximates to that of a black hole.
An emitted corpuscle of light will return to the point it was emitted due to the return effect produced by the nugravity, the force produced by nugravity, in accordance with our model is:
Fp = G M m / r2 = 6.673 x 10-- 11 x 1.6725 x 10-- 27 x 1.47236 x 10-- 50 /(1.7142951 x 10-- 14)2 =
5.5915227 x 10-- 60 Kg m / Sec2
G = 6.673 x 10-- 11 m3 / Kg Sec2. = constant of universal gravitation.
M = mass of proton (Kg)
m = mass of a corpuscle (Kg)
r = radius of the proton (m)
0.5 m c2 = 0.5 x 1.47236 x 10-- 50 (3 x 108)2 = 6.62562 x 10-- 34 joule = 1 quantum of energy
As the positive particle of the binary system is affected directly by the remanent gravity (nugravity), we suspect that the electrons are not affected directly by the gravity. The time required for the corpuscle affected by the remanent gravity to return due to the corresponding force F would be:
2 tp = 2 m c / Fp = 2 x 1.47236 x 10-- 50 x 3 x 108 / 5.5915227 x 10-- 60 = 1.57992 x 1018 Sec.
In accordance with the theories of expanding Universe, they consider that at a distance equal to:
4 x 1010 to 6 x 1010 light years there are galaxies that recede to a velocity equal to that of light
5 x 1010 x 31,557,600 = 1.57788 x 1018 Sec. = 2 tp
31,557,600 Sec. = 1 year.
About the return distance in a black hole it could be from a limited distance to practically zero distance, so the light corpuscles do not escape from it. In accordance with the concept of continuous action, if an electron emits (n) corpuscles (as those of light) they get light velocity with a kinetic energy equal to n h (n quantum) and the electron gets a kinetic energy equal to the fore one, but in the opposite way (action = reaction). In accordance with this and with mechanophysics explained the theory of the polygonal orbits of the orbital electrons around the nucleus of the atom. In each vertex of the polygonal orbit of an orbital electron are acting (n) light corpuscles that deflect the trajectory of the electron in order to follow its corresponding polygonal orbit. If no spectrums are seen of the acting light corpuscles;: How can be explained the action of them? If we consider that in each vertex of the polygonal orbit the return distance of the corpuscles is equal to the diameter of the electron, when they move from the interior face to the exterior one of the electron, they do not escape from this last face, but begin their return trajectory, from zero velocity to c velocity, toward the interior face of the electron and from here they escape toward the interior face of the positive particle (see binary systems, theme : Hydrogenated Model of Atom).
Monterrey, Mexico, August 12, 1996
HYDROGENIZED MODEL OF ATOM
In the fore theme of Return Distance was simplified in a great deal the concept of fields, because they were considered in their structure, not as something characteristic of all the space of the Universe, but only characteristic of a limited space of the mentioned Universe, corresponding to the interacting bodies (or particles), in such limited space.
Now that we are working in the electrotherapy have had all kind of problems, from that produced by impertinent persons, to technical ones, because this we developed a very much satisfactory model of electric current, but this was not enough, due some persons are too much skeptic. We are not skeptics, but too much exigent with our work, so we began to work trying to make a model of atom that fulfill a series of required conditions, with much more success than that we expected. In doing this I remember a good friend of mine, that since I began to work with mechanophysics, he was very much cooperative with me, encouraging and stimulating my work, all this in spite he was supporter of modern physics, I refer to Dr. Rodolfo Castillo Bahena, he was director and professor of the Physics Department in the Instituto Tecnológico y de Estudios Superiores de Monterrey: his indications, commentaries, and critics were very much useful, I remember in one occasion he commented me that in order that the mechanophysics could be accepted, it would be necessary to give a model of atom with certain characteristics. Because then I had many ideas in my head, I considered this as a simple conversation, but now I consider it as a prophesy (now also have many ideas, but try to be more organized).
With the structure of the atoms it is possible to make a great simplification, and we believe that with this it is possible to make great progress in this field. Till now it has been considered the nucleus of an atom formed by a determined quantity of protons and neutrons, making a unitary structure, affecting in this way to all the orbital electrons, and viceversa. With this criterion have been given different models of nuclei affected with many complexities and deficiencies. Here will be given a model by far more simple that will simplify its investigation and understanding; in this model will be considered that each orbital electron is only affected in a direct way by its corresponding proton and neutron; to this structure we will call: binary system. The intervention of a binary system in the other ones of the atoms will be minimum in all the normal conditions and only could be appreciated in an evident way in phenomena of radioactivity or half life of atoms; in which is manifested the interference of a binary system with another one. With respect to the spectrums produced by the excited atoms, they could be explained by indirect or secondary interaction of a binary system with other ones.
The medullar point of a binary system is that the electron and the positive particle work in a synchronized way emitting jn an interior way their propeller fluids every time of vertex, but this is because other effects also are synchronized; a car can move because has a motor, but its movement is not arbitrary, but in accordance with a driver that controls it, next are going to see some facts that contribute in the control of the behavior of the binary systems. In order that a binary system is not affected by other (s) one and vice-versa, as was say, it will be considered that in a general way, either the electrons, as the protons of each binary system, have a determined side or face were their propeller fluids are emitted, in the corresponding side are received and absorbed the microcorpuscles of the electric fields that excite the atomic particles (their propeller fluids); the amplitude of this side permits a small tolerance given by a small solid angle, in such way that can exist variations (in the orbital electrons) in the spectral lines (s, p, d, f). If the atomic particles (electron or proton) receive the microcorpuscles of the electric field in a different side than the fore mentioned, the microcorpuscles are not absorbed, not exciting the atomic particle. Another property of the microcorpuscles of the electric fields, in our model of binary system is that they have a return distance equal to the distance that there is between the orbital electron and the nuclear proton (or positive particle), of the binary system. Continue with the required conditions, will be considered that the electric fields formed by emitted microcorpuscles, are not formed in an immediate way, but in a progressive one, from a null (maybe small) intensity to its maximum value; the variation of intensity of the electric fields emitted by the orbital electron and the positive particle coincide with the angular rotation or spin, that is manifested in each vertex of their polygonal orbits, here the intensity is null (or small) when the deflection start, and gets the maximum and final value when the deflection finish, When the charged (-) particle has turned an angle of 360o. In accordance with the binary system, as will be appreciated with the numerical example will be given later.. Let see a binary system formed with an orbital electron as negative particle, and a proton as a positive one. The orbital electron is excited in each vertex (with a deflection angle = a) of its orbit.
The positive particle (proton) is excited in each vertex by a fields. Although the fields act fast in the vertices of the orbits, and with the same magnitude in the positive particles as in the negative particles, this does not mean that the positive particle get equal quantity of charge than the negative one, If the positive particle got its charge in one instant. In the binary system the orbital electron and the corresponding proton, always have their corresponding side of absorbing microcorpuscle of the electric fields in front one from each other, this because in this way are oriented the particles in the binary system. A binary system in the hydrogen atom is formed by one orbital electron and one nuclear proton; in a deuteron atom the binary system is formed by one orbital electron and one positive particle formed with one proton + one neutron; in the triton atom by one orbital electron and a positive particle; formed by one proton and two neutrons. From the point of view of the mass of the positive particle, there are three types of binary systems, as just have seen. From the point of view of the orbital electrons moving in any one of the 7 shells of the atom, obviously there are 7 types of binary systems. In the fore cases the negative particle (orbital electron) has the same mass. In the way can be supposed how is formed the positive particle can be imagined one sphere (the proton) two or three spheres are fusion, forming one bigger sphere,
In previous investigation works in mechanophysics, and based in some experimental data, and considering that all the elementary particles have the same density, we obtained that the radius of the electron is: re = 1.4 x 10-- 13 cm and of the proton: rp = 1.7 x 10-- 12 cm. About the nucleus of an atom, there are many objections, for instance, if we consider the nucleus of hydrogen atom as a proton or as a nushell (see forward), will be a great difference of density. In the value of: r7 = 1.1376 x 10-- 8 cm. = radius of the 7 orbit = radius of the atom, we can confide more because the Avogadro’s number, also in the mass of electron: me = 9.1091 x 10-- 28 gr. and the mass of proton: mp = 1.6725 x 10-- 24 gr., because the fields effects. Some one could object that can not confide in the binary systems, but if with two complete different ways can arrive to the same results, the fore objection is demerit.
We can continue without impediment, so for a heavy atom as gold: Au, for example: Z = 79; A = 197, we can imagine a structure formed by 79 binary systems, of them 39 have the positive particle formed with one proton + one neutron; and 39 formed with one proton + two neutrons. in the 7 shell (?) 2 with one proton: 39 x 3 + 39 x 2 + 2 x 1 = 197.
In accordance with the characteristics of a binary system, the orbital electron and the positive particle spin around the center of the nucleus of atom in a synchronized way, the orbital electron at a distance equal to the radius of the shell; the positive particle to the fore distance divided by 1836 / 4 = 459; . 459 me = m1+ / 4. The positive particle could be triple mass = 1 proton + 2 neutrons, double mass = 1proton + 1 neutron; single mass = 1 proton. Although we are ignorant of chemistry knowledge, also we believe in the future could be determined different chemical affinities in function of the mass of the positive particles of the binary systems of the seven shell . Both particles (+ -) describe their corresponding polygonal orbit of a similar shape, around the center of the nucleus, but of course of different radius of gyration, so that it is considered that the orbital electrons move in shells, the positive particles will move in çnushells (forming the nucleus of the atom). The size of the nushells, by far are smaller that their corresponding shells, also the size of the orbital electron compared with that of the positive particles, in the first or most interior nushell hardly is space for two moving positive particles, so it can be explained why in the first shell only can spin two orbital electrons. In Fig.(1), second orbit, with O is represented the center of the nucleus of the atom, that also is the center of gyration of the binary system; PN is the positive particle; E the orbital electron; rN the radius of gyration of the positive particle; Rn the radius of gyration of the negative particle (orbital electron)
When the orbital electron moves in the: n = 7 shell (most external shell);
Rn = R7 = 1.1376 x 10—8 cm = radius of the atom.
The radius of gyration of the positive particle:
rN = r7 = 4 Rn / 1,836 = 4 R7 / 1,836 = 4 x 1.1376 x 10—8 / 1,836 = 2.484 x 10—11 cm = radius of the nucleus (nushell)
For: n = 1 orbit :r1 = 2.484 x 10—11 / 7 = 3.5486 x 10-- 12 cm.(*about two radiuses of proton, (see Fig. 4)
The electron E moves in a polygonal orbit, for correlated reasons the proton PN moves in a
synchronized way in another polygonal orbit, smaller than that of E. For the orbital electron to deflect in a vertex of its orbit (7) it is required an energy to produce the velocity of deflection v_l_7 (see Fig.2)
v_l_7 = 2 x 309 Sin 6o 43 = 69.1 Km./ Sec. (7a orbit); 6º.43 x 28 = 180o
The energy of deflection of the electron is: E7 = 0.5 me v_l_72 = 0.5 x 9.1091 x 10-- 28 (6.91 x 106 )2 = 2.175 x 10-- 14 erg.
In a vertex of the 7 polygonal orbit is formed a field affecting the orbital electron of the binary system and other similar one affecting the positive particle (this in accordance with a model).
Considering the length of each side of the polygonal orbit: d = 2.55 x 10-- 9 cm, and also considering in Fig.(2) there is a proportion between velocities and distances, is obtained the deflection distance:
d_l_7 = 69.1 d / 308 = 69.1 x 2.55 x 10-- 9 / 308.6 = 5.73 x 10-- 10 cm.
Next is going to see a binary system corresponding to the 2 orbit. From Fig.(3), corresponding to two sides of the second orbit, is obtained the deflection velocity: v_l_2 = 826.58 Km./ Sec. The deflected distance: d_l_2 -= 1.96 x 10-- 9 cm. = 2.55 x 10-- 9 x 2 Sin 22o.5
The energy of deflection is:
E2 = 0.5 x 9.1091 x 10-- 28 (8.2658 x 107)2 = 3.112 x 10-- 12 erg.
In Fig.(4) is represented the orbit of a positive particle in the first nushell, from this can be deduced that in the first shell can move as much as two binary systems. Next are given some relations in which are harmonized geometrical and kinetic conditions.
v_l_n = 2 vn Sin(45o / n) . . . . ( c) 
SUPERFLUID HELIUM 3
In a newspaper dated October 10, 1996, I read that 3 Americans: David Lee, Douglas Osheroff and Robert Richardson won the Nobel Prize of Physics 1996, because their discovery of the fluidity of helium 3, a progress in the low temperature physics that was realized at the beginning of the decade 1970 in the University of Cornell in New York. In accordance with the data of the news paper they discovered that the helium 3 can became superfluid at a temperature of 0o.002 K.
With respect to the report of the newspaper, I do not know if the names of the Nobel Prize winners are correct, what is wrong is datum of temperature, because the so called lambda temperature is 2o.2 K. In the lambda point the helium pass from the gaseous state to the liquid one, as seen with the helium 3. In He4 the two binary systems are equilibrated; for this reason the helium 4 evaporates toward the upper atmosphere in form of separated atoms, this is because the helium is lighter than the air. In the helium 3, the 2 binary systems are different. The helium in liquid state, because its viscosity, does not evaporate, and a multimolecule, because its relative great mass, behaves as a liquid; it does not evaporate, but as is lighter than air tries to climb up the wall of the recipient that contain it, this because its adhesive property characteristic of a liquid state is stronger than the tendency to evaporate toward the upper atmosphere.
In our theme: Model of Electric Current, was obtained the electric currents, that move along multimolecules, each one formed by a line of atoms, forming a chain with: n = 2,497,996 atoms / multimolecule. In a transversal section of a conductor, behind each atom of the section could be formed the multimolecules were the current’s electrons move from the lower potential toward the higher one of the conductor. In our case of current, in every one of the multimolecules is moving one current’s electron, as was specified in the mentioned theme; can be considered that not all the multimolecules required by the current in a given time are working at the same time with a moving electron of the current, but only one in (n) of the mentioned multimolecules, and we say this, because it is considered that for a multimolecule to work, it is required that the orbit of the current’s electron behaves as was specified with the spin velocity, the time of jump and of multimolecule given, etc. With this criterion, as could be appreciated, the most logical seems the maximum simplification. In an atom of helium 4 there are two binary systems formed each one with a negative particle with one orbital electron and a positive particle with one proton + one neutron; if both binary systems move in opposite sides, the helium 4 will have equal spin. Now it is going to see an atom of helium 3, in this case, one binary system is formed by a negative particle with one orbital electron and one positive particle with one proton + one neutron; and the other binary system is formed by a negative particle with one orbital electron, and one positive particle with one proton.
Every binary system spin in a synchronized way, for to attain this the orbit of the shell must be similar (equal number of sides and vertices) than the orbit of the nushell, the only difference will be the dimension of them; the dimension of the shell must be proportional to the velocity of the orbital electron and the dimension of the nushell must be proportional to the velocity of the positive particle. As the positive particle can be formed with three different masses, that is: one orbital proton (single mass); one orbital proton + one neutron (double mass), and one orbital proton + two neutrons (triple mass). When a sample of an element almost get 0o K temperature could be imagined that all the movement ceases to act, this could mean that the binary systems would stop moving, this, because at 0o K the atoms are destroyed, but this is not so, in contradiction with the theory that considers that heat is produced by the kinetic energy of the atomic particles. At practically 0o K temperature, the binary systems continue moving into the atoms as have been specified in the models and maybe because this have never reached a real 0o K temperature. In a binary system the negative particle or electron has the maximum energy that oppose to reach the 0o K temperature, as was say, but also the positive particle contribute to this.
The fore conclusions that have been given from a point of view of deduction by observation, can explain the difficulties for to obtain the mentioned temperature. To K be in accordance with the experimental results, it is going to be considered that the no null temperature effect is produced by the kinetic energy and the electric charge of the orbital electron and proton. The electric charge is equivalent to a kinetic energy and viceversa, the kinetic energy is equivalent to an electric charge; all this in accordance with the structured model we will give. As the neutrons have no electric charge, they do not contribute to produce the no null temperature effect.
No null temperature effect produced by the negative particle of a binary system in accordance with our model:
Ke = 4 x 0.5 me ve2 = 4 x 0.5 x 9.1091 x 10-- 31 (3.09 x 105)2 = 1.7395 x 10-- 19 joule.
Idem. for the positive particle v+ = 4 x 308570 / 1836 = 672.27 m / Sec.:
K+ = 0.5 m+ v+2 = 0.5 x 1.6725 x 10-- 27 (672.27)2 = 3.7794 x 10-- 22 joule ; so:
Ke = 459 K+
From the book: The Science and the Life of Albert Einstein, by Abraham Pais, are the following paragraphs:
In December 1924 Einstein wrote to Ehrenfest from a certain temperature on the molecules “condense” without attractive forces, that is, they accumulate at zero velocity. Until 1938, the Bose-Einstein condensation had the reputation of having only a purely imaginary character. This phase transition was not discovered until 1928 by Willem Hendrik Keesib. In 1938 Flitz London proposed interpreting the helium transition as a Bose-Einstein condensation. Experimentally, the transition point lies at: T = 2o 19 K. In accordance with their theory: T = 3o.1 K. It is generally believed, but not proved that the difference between this two values is due to the neglecting of intermolecular forces in the theoretical derivation.
The concept of temperature is related with that of movement, in such way that some physicists consider that at zero Kelvin temperature all kind of movement ceases to act; this definition is ambiguous, because at that temperature the orbital electrons move with their great velocity around the nuclei of the atoms. We can imagine a macroscopic body moving at great velocity in the outer space at temperature almost zero degree Kelvin. From the most elementary particles, as the light corpuscles, to the atoms or molecules, can move at great velocity at zero Kelvin temperature while they are not affected directly by other particles, or by themselves. In all these cases the particles have a potential energy that is manifested when there is an intersection with them and the medium in which they move, this interaction could be by absorption, reflection, collision, etc.
In the XIX century the physicists believed that all the laws of physics were known and that progress of the science will be obtained by more detailed knowledge; with the discovery of the atomic particles and with modern physics the fore concept had a radical change, in such a way that the physics’ world was considered too complex. Now with mechanophysics we are returning to the concepts that the physicists had of classic physics before modern physics born; nevertheless it is not prudent to be too optimist, because if it is seen with ampler criterion the fields in which physics acts or intervene, can be applied in chemistry, in biology, etc.
As the objective of a mechanophysics is to find true facts with true and objective theories or laws, in all these papers have been proved that this is feasible and that it is not required to employ theories with false structure, as those of modern physics, for to find real facts. From the mentioned book of Abraham Pais is the following paragraph.
Physicists -good physicists- enjoy scientific speculations in private but tends to frown upon it when done in public. They are conservative revolutionaries resisting innovations as long as possible and at all intellectual cost, but embracing it when the evidence is incontrovertible. If they do not, physics tends to pass them by.
I would amplify the fore paragraph by saying: that ,more important than the conservative spirit, are the resulting facts obtained by the recognition of the new ideas. In our case is more important the life of the sick people (see electrotherapy) than the conservative spirit of modern physics.
It was say that at a temperature near 0o K a binary system of an atom can moves without incrementing such temperature; if the fore happens with the orbital electron of one atom, also could happens with a multimolecule. Till now have been considered the multimolecules for to explain the movement of the current´s electrons, later will be seen that the sound can move along them. From all say before it is not find any trouble for the electron of the multimolecule to move along this, without affecting the near 0o K temperature. It is well known that when the temperature diminishes, the gaseous substances try to became to a liquid state and the liquid substances try to became solid substances; the fore one only could be explained considering that the multimolecules of the substance try to unite one to each other, and for to do this, the nultimolecule electron of one multimolecule will jump to the adjacent multimolecule, and for to do this it is forceful to be produced an ionization effect. This means that if by one side, with the fore effect at the same time will be produce an increment of the temperature. The ionization of helium is 24.6 eV.
Energy of ionization: K = 24.6 x 1.6 x 10—19 = 3.936 x 10-- 18 joule / atom.
The fore energy is equivalent to a temperature, in accordance with Boltzmann´s law is:
T = 2K / 3k = 2 x 3.936 x 10-- 18 / 3 x 1.38 x 10—23 = 190,145o K.
Apparently the fore result is in contradiction with what is say in the theme: The Maximum Temperature Could Exist; that consider this temperature is equal to: T1 = 102,656º K. But there is not any contradiction, because: K = 2.1245 x 10—18 + 1.8105 x 10—18. Here the first value is employed in ionize one electron; and the second value is the opposition of the second electron to the ionization of the first one.
The fore mentioned ionization, affecting the ionized atom gives a great temperature, that afterward is distributed in all the atoms of the sample. The minimum part of the sample in which are produced all the fore mentioned effects is in a cube which volume is: nm3 (isometric distribution of atoms): nm = 2,497,996 atoms = number of atoms that has a multimolecule. In the fore volume there are as much multimolecules as: nm2 . Next it is going to be made some considerations: although almost all the mass of an atom correspond to the positive particles of the binary system; from the point of view of kinetic energy these particles, in the sample will produce the increment of heat; because the fore increment is produced by vibration, or collision, or impact with other particles, then, because action = reaction, in the formula: Ke = 459 K+ , will be consider a double mass of the positive particles of the binary systems of the atoms acting (getting ionized) in nm multimolecules, or: (459 x 2)0.5 = 9180.5 atoms acting in nm multimolecules.
T = 190,145 x 9180.5 / 2,497,996 = 2o.306 K
With the fore increment of temperature of the sample, the ionization effect ceases till the temperature diminishes the following value: (30 - 1) 2.306 / 30 = 2o.22 K
+ - 1 minimum variation of one atom for produce ionization or / and decrement of temperature.
Monterrey, México, October 15, 1996; March 28, 1998
BINARY SYSTEMS WITH POSITIVE PARTICLES HEAVIER THAN ONE PROTON.
In all the problems of mechanophysics dealing with atoms have been considered them with a radius equal to: r7 = 1.13763 x 10-- 8 cm. In many cases this does not affect our final result, because there are some compensate effects, in such way that if some values grow because the approximate data, others diminish by the same reason and when act simultaneously there is a reduction of errors instead of a sum or accumulation of them. But not always happens so, it has been observed that not all the atoms have the same diameter. In some cases it could be explained this variation considering that the diameter of the atoms grow due to the increment of temperature; could be other increments or decrements of the diameter of the atoms due to other effects, for instance if they are compressed, if they combine with other atoms in a chemical way, etc., but all these variations are not due to the structure of the binary systems, but to different physical ,chemical, geometrical distribution of the atoms, etc., and they do not contribute in a really evident way in the variation of the diameter of the atoms in our everyday normal conditions; of course, at very high temperatures and pression, the structure of the atoms is affected.
Considering the structure of the atoms as a sphere has been employed a method for to determine the diameter of the atoms (see book: Chemistry by Michell D. Sienko and Robert A. Plane; Cornell University teachers), for instance: if there is a sample of atoms with a unitary volume, and it is known the atomic mass of each atom, it would be easy to determine the volume occupied by each atom, knowing the weight of the sample. But we can not confide in this method because in the sample the atoms can be distributed in different ways, by different circumstances, and this could give us different densities; all this, if it is considered the atoms occupying fixed spheres; but imagine hydrogen atoms, with only one orbital electron each one; from a mathematical point of view the effective volume of a hydrogen atom is as a disc with a diameter equal to: 2 r7 and in the volume of a sphere with the fore diameter there is space for many discs, as the fore mentioned.
Due to the fore indeterminations, it would not be easy to determine how much press it is required to give to gaseous sample for to determine the real dimension of the atoms of the sample. The more trust way to find the real volume of the atom would be by a more direct way and this is not so easy problem as the fore mentioned method. In our model made in accordance with mechanophysics have been made some special considerations for to simplify some explanations, as for example: to consider an isometric distribution of the atoms, a radius of the seven shell: r7 = 1.13763 x 10-- 10 m, that the atoms of a multimolecule are aligned in a straight line, that the 7 orbit of one electron of an atom does not penetrate in the seventh shell of an adjacent atom, in a given moment and circumstance; that the velocities of the orbital electrons have fixed values, etc. Because all these divergence have been tried to make models in such way to obtain compensate effects and at the same time to orient the theory for to obtain ,more precise models with data more fondness with reality. In this theme will not be given a direct method to determine the real diameter of an atom in the practice, here will limit to determine it in a theoretical way, and in this way will be considered such diameter determined by the dimension of the binary systems of the seventh shell; when the binary system was formed by one proton as positive particle, it was considered the diameter: 2r7 = 2 x 1.13763 x 10-- 10 m. But if the positive particle is formed by a proton + one or two neutrons, the fore value would not be exact as will be seen forward.
For to analyze the problem of a binary system, forward is seen with a positive particle with one proton and two neutrons. First will be seen the binary system with a positive particle formed with one proton, in this binary system, its orbital electron has a translation velocity equal to: v7 = 308,571 m / Sec. and the positive particle has its translation velocity equal to: v7’ = 4 x 308,571 / 1836 = 672.268 m / Sec. and a radius of gyration: r+7 = 4 x 1.13763 x 10-- 10 / 1836 = 2.4785 x 10-- 13 m.
The kinetic energy of the orbital electron is:
Ke = 0.5 x 9.1091 x 10-- 31 x 308,5712 = 4.3367 x 10—20 joule
The kinetic energy of the positive particle:
K+ = 0.5 x 1.6725 x 10-- 27 x 672.2682 = 3.77938 x 10—22 joule.
Now let see from a geometrical point of view a binary system with a positive particle formed with: 1 proton + 2 neutrons. If it is considered a gyration axis as the fore seen binary system, then the orbit of the electron will grow: re7’ = (3)1/3 re7 = 1.4422495 x 1.13763 x 10-- 10 m = 1.6407463 x 10-- 10 m its velocity: ve7’
1.44225 x 308,571 = 445,036 m / Sec.
Its kinetic energy: Ke´ = 0.5 x 9.1091 x 10-- 31 x 445,0362 = 9.0206422 x 10—20 joule.
Incremented axial or radial velocity of the orbital electron: v_l_ = 69,098 x 31/3 = 99,657 m / Sec.
In accordance with our model the velocity of the positive particle will grow:
v7‘ = 1.4422 x 672.26 = 969.5 m / Sec;
r+7‘ = 1.4422 x 2.4785 x 10—13 = 3.5744 x 10-- 13 m
Its kinetic energy: K+ = 0.5 x 3 x 1.6725 x 10-- 27 x 969.52 = 2.36 x 10-- 21 joule
For a binary system formed with a positive particle with one proton + one neutron, the radius of gyration of the orbital electron is:
re7’ = (2)1/3 r7 = 1.26 x 1.13763 x 10-- 10 = 1.4334 x 10-- 10 m
ve7’ = 1.26 x 308,571 = 388,800 m / Sec. The radius of the positive particle: r+‘ = 1.26 x 2.4785 x 10-- 13 = 3.123 x 10-- 13 m.
Monterrey, México, December 10, 1997
Manuel de Hoyos Robles
INTERPRETATION OF THE EXPERIMENT OF MICHELSON AND MORLEY WITH THE ABERRALOGY
The following papers were take from a book I made in 1972 of the investigation work of my father Estanislao R de Hoyos, in the aberralogy and in an objective physics, since before the second world war. Practically all the work is presented as he conceived it, except some commentaries and amplifications I made to actualize it in the mentioned year 1972 Before interpreting the results of this experiment, it is going to be given an explanation of such experiment. From the book: La Teoria de la Relatividad by Alfred Wulf, Tacubaya, México, 1924, is the following translation:
In Fig.(32), where details have been omitted to simplify the understanding, is shown the interferometer of the physicist of the University of Chicago, Michelson. It consist of two fixed arms, perpendicular one to each other, with the same length and with the mirrors E1 and E2 in their extremities. In the crossing of both arms there is a semitransparent mirror W, inclined an angle of 45o, that lets pass part of the light that proceeds from the luminous point L, and the other part of this light is reflected. So that there are obtained two rays that propagate perpendicularly one to each other toward the mirrors E1 and E2 were they are reflected, returning to W, from here they continue united as far as the telescope of observation T. As the Earth’s movement has different influence in both rays the light need different times, according with the classic theory, to move over the equal paths WE1 and WE2. Due to this, it is necessary to prove that the returning rays suffer in W the reciprocal effect known as interference The telescope T is to observe the interference stripes produced by both rays. All the instrument is installed so that it could be turned around, that is ,either the arm WE1 or the arm WE2 could be oriented in the direction of the Earth’s movement. If the instrument is turned around 90 o, so that WE2 be in the direction mentioned before, it could be expected that the same interference stripes appear as they looked in the direction WE1, having as only difference their movement to the opposite side, but being their displacement equal to those due to the position WE1. This instrument was perfected by Michelson by continuous reforms, until obtaining an instrument of precision with a usefulness without precedent till now; thanks to it, it is possible to obtain a precision of one hundredth of the effect calculated in advance, but as was explained before, the experiment has given negative results, because the instrument did not show the influence of the Earth’s movement....
The interpretation of this experiment by the aberralogy can be resumed to the following: Since the observer and the luminous fountain are in the same system (the Earth), both are affected by the same movements, velocities, magnitudes, etc., that the Earth imposes on both, in the same magnitude, so that the observer can not notice any movement of the Earth by observing the luminous focus, he needs to observe another body outside the system; although the problem is too clear, it is going to be subdivided in all the parts that are necessary to study them, with the aberralogy; this, beside a more complete proof, will show us that the experiment is only a combination of cases of aberralogy and not a different phenomenon, as it has been interpreted.
Suppose an observer expecting to appreciate the movement of the Earth with the Michelson & Morley’s experiment; when obtaining negative results, was not satisfied with this, and tried to find the “error” running over the trajectory of the light, observing space by space. Starting from the luminous point L, will arrive to W, from this point he will observe the light coming from L, in this case the body L is approaching with velocity v’, to the observer in W, simultaneously, the observer recedes from W with velocity: v = v’, because both velocities are equal to the velocity of the Earth, the formula which will give the aberrated time due to the movement of the Earth is:
tr = ta (1 - v / c) / (1 - v’ / c) = ta . . . . (2,3)
from here it is deduced that the movement of the Earth does not modify the time that the light expends to run over this space.
Now the observer has moved to E1, and from here he will see the light coming from W (in this case W will represent the luminous point); this case is exactly equal to the fore one (space LW) and the movement of the Earth will not affect the time that light spends to run over this other space, as it would be appreciated by any observer that would be displacing in the same system.
If the observer now returns to W, following the route as the ray of light mentioned did, he will see the light reflected from the mirror in E1 arriving to him as if E1 were the luminous point. The observer in W approaches with velocity v, and the luminous point (here E1 is equivalent to a luminous point) E1 recedes with velocity v’, the formula of the time aberrated by the movement of the Earth, will be:
tr = ta (1 + v / c) / (1 + v’ / c) = ta . . . . (1,4)
here: v = v’ = Earth’s velocity
Here the movement of the Earth does not affect the time that light expends to run over the trajectory. It just has been seen before, that the movement of the Earth does not affect the time expended by the light to run over the trajectory given before (trajectory parallel to the Earth’s movement)
Let us see now the case of a trajectory of the light perpendicular to the movement of the Earth. Now the observer moves to E2, and from here he sees the light ray coming from W, that in this case, W represents the luminous point. To be in accordance with the nomenclature of the formulas of time and frequency aberrated, we are going to express with O to the observer in E2, and with On the displaced observer in A, to the luminous point W, by F (see Fig. 33, 34).
As in this case, either the observer O as the luminous body F, move parallel to each other and with a velocity: v = v’ = Earth’s velocity. It is convenient to analyze the problem in two phases.
1) Supposing the observer is moving and the luminous body motionless.
2) Supposing the resultant of the first phase combined with the luminous body moving.
First phase: Observer moving, luminous body motionless. In this condition the observer will appreciate the luminous body in an aberrated time tp (preberrated) in accordance with the following formula:
tr c = [(tp c)2 + (tp v)2 - 2 tp2 c v Cos On]0.5 . . . . (6); tr = real time.
From the fore formula (6), is discovered the preberrated time: tp
tp2 = [ (tr c )2 / ( c2 + v2 - 2c v Cos On )] . . . . (6’)
This time that is aberrated (preberrated) for the first phase, that is, the resultant originated by the combination of the movements of the observer and the light; for the second phase will be the real time, and starting with this for the second phase we find the transberrated time due to the movement of the luminous body combined with the resultant of the first phase. In synthesis, the transberrated time of the second phase, is equivalent to the aberrated time of both phases’ system, due to the movement of the observer (first phase) plus the movement of the luminous body (second phase).
Applying the formula (8) of frequency, or time transberrated tt and taking in account the fore indications, it is obtained:
tr c = [ (tt c)2 + (tt v’)2 - 2 tt2 c v’ Cos qn ]0.5 . . . . (8)
We are going to substitute these values in the fore formula (8), so as to be in accordance with the literal of formula (6’):
tt c = [(tr c)2 / (c2 – v2 - 2 c v´ Cos qn]0.5 . . . . (8’)
Substituting formula (6’) in this, we have:
(c2 + v’2 - 2 c v’ Cos qn) / (c2 + v2 - 2 c v Cos On) = 1
FIRST LAW OF THE DOUBLE FLUID THEORY
If we study more widely the theory of the double fluid, we could see that the emitted-received fluid can have different velocities, so we refer ourselves to the interaction between atomic particles or between bodies perceived directly by our sensorial organs, etc., For the case of the gravity emitted-received fluid, we will consider it, equal to light velocity. If the body that emits the gravity fluid, is fixed, and the body that receives it, is moving, this could be appreciated as the fluid having light velocity plus the vectoriall sum of the velocity of the moving (and receiving) body; if the emitting body is moving and the receiving one is fixed; or if both bodies are moving, we can make identical considerations, all in accordance with the aberralogy.
As we have just seen, the aberralogy permits us to appreciate one type of emitted-received fluid that can have different velocities. This variation of velocities has a notable influence; based on the fact that the energy with which the emitted-received fluid is received, is proportional to the square of the reception velocity divided by ( c),as seen this by a series of observations of different phenomena, was deduced the following law, that is going to be called: First law of the double fluid theory, that says: The energy of the propeller fluid is proportional to the square of the velocity with which the emitted-received fluid is incremented. The limitation of this law could have by interference with other phenomenon, are not going to be mentioned here, due to the lack of space.
It is important to make clear that the wearing out of masses produced by the gravity fluids, is so insignificant that the changes (additions or reductions) are due mainly to other effects that are much bigger.
DEFLECTION OF A RAY OF LIGHT IN A GRAVITATORY FIELD.
Before we study this problem I am going to reproduce some paragraphs of the book: “Electromagnetism & Relativity” by Edmund P. Ney, page 107.
Deflection of Light in the Gravitational Field of a Star.
In 1801 a German mathematician named Soldner, calculated the deflection of light in the gravitational field of the Sun. His result which we reproduce here leads to the prediction of a deflection of 0”.87 seconds of arc. Shortly after developing the general theory, Einstein calculated the deflection and got the same result. however, he later developed the theory and found that general relativity actually predicts a value twice as large, or 1”.75 of arc. The experiment was first performed in 1919, and the result reported was 1.”7 of arc. The experiment is difficult, and more recent experiments have not agreed with this theory as the 1919 experiment, This prediction of the theory can only be considered to be confirmed to an accuracy of perhaps 20 %. The procedure of the experiment, is to photograph a star field around the Sun during a total eclipse (so that the stars are visible). Six months later, when the same stars are visible at night, the star field is photographed again. The displacement of the apparent positions of the stars can be measured by comparing the two photographs.
Soldner’s derivation.
The transverse momentum P_l_, imparted to an object of mass m can be shown to be given at point P (see figure 68) by:
dP_l_ = G M m CosQ d Q / c R
The total transversal momentum is:
P_l_ = dP_l_ = -G M m Cos Q dQ / cR = 2 G M m / c R . . . . (42)
The angle : f = P_l_ / P = (2 G M m /c R) / m c = 2 G M / c2 R radians . . . . (43’)
Substitution in this formula for the case of the Sun, leads to the “classical” value: f = 0”.87 of arc. Note that although the mass of the photon cancels itself out in the derivation, the mass must be finite, i.e. m =/= 0, and this was not established in prerelativity physics. It could also be emphasized that Soldner’s value is only half the correct value given by the general theory.
The determination of the angle f by the relativity theory is not going to be effectuate here, because it is not necessary in this study. In the determination of the angle f by Soldner, he gave the gravity force an instantaneous action, but this is not the case, because the gravity has light velocity, and we have to take in account the aberralogy.
First we consider the light corpuscles moving from P to Q as in Fig.68. When the corpuscle that we take passes by point P with c velocity, it meets a gravity ray with practically the same velocity that moves in the OP direction, the velocity that meet both rays is equal to the vectorial sum of their velocities. In Fig. 69 we have represented by cx the velocity of the corpuscle of light, and by c the velocity of the gravity that meets the light ray (or the corpuscle of light), and by ca the velocity of reception of the emitted-received fluid by light corpuscle.
Distance OP = R / Cos Q
Distance PP’ = dS = OP d Q / Cos Q = R d Q / Cos2 Q
Gravity force on a corpuscle at point P, according with Newton’s law, is:
F = G M m / (R / Cos Q)2; here: m = mass of a light corpuscle; M = mass of the Sun.
The resultant of this gravity force perpendicular to trajectory PQT (but in the same plane PQO) is:
F_l_ = G M m Cos Q / (R / Cos Q)2 = G M m Cos3 Q / R2 = perpendicular force
The momentum of the corpuscle in the point we are considering or any other point of the trajectory PQT is practically constant (= mcx) and in the same way of the trajectory. The momentum normally the trajectory is obtained multiplying the normal component of the gravity force by dt (time differential).
With both magnitudes we make the triangle of Fig. 70 and obtain the deviation angle df for the chosen point. The sum of all these angles df will give us the total angular deviation f , and because this angle is very small, we consider the normal in the trajectory equal to the normal of the straight line PQT, without appreciable error.
d f = G M m Cos3 Q dt / m cx R2
We solve the problem by the principle of impulse and momentum.
dS = PP’ = (PO) d Q / Cos Q = R d Q / Cos2 Q = cx dt
cx = velocity of the corpuscle
dt = differential of time = dS / cx = R d Q / cx Cos2 Q
dF_l_ = F_l_ dt = (G M m Cos3 Q / R2) (R d Q / cx Cos2 Q) = G M m Cos Q d Q / cxR
1) We are going to consider the corpuscle moving as indicated by the arrow. When the corpuscle is in P, the gravity fluid is moving from O to P with velocity c, equal to light velocity. The corpuscle moves with a velocity equal to cx in the direction PP’Q, so that this corpuscle receives the gravitatory ray with an aberrated velocity:
ca = (cx2 + c2 + 2 c cx Sen Q)0.5 (69)
In accordance with the first law of the double fluid theory of the aberralogy, the increment of energy that produce the radiation (of gravity) is proportional to the square of the increment of velocity with which is received such radiation. So we have that the increment of the action of gravity is equal to: ca / c. Here ca = incremented velocity of gravity (see fugyre 69). As velocity of gravity is equal to velocity of light c = cx; In accordance with the principle of impulse and momentum and considering the increment (ca / c), of the gravity force by aberration, and the first law of the double fluid theory, we have:
Ca2 / c2 = (cx2 +c2 + 2 cx c Sinq )´/ c2 = 2 + 2 Sinq
The angle of deflection is: df = d P_l_ / P ; dP = dF P = mcx
f = P_l_ / P ; Considering the increment of the action of gravity:
f = P_l_ dt ca2 / c2 m c = (- G M m / c R) Cos q dq (ca2 / c2 m c) =
(- G M / c2 R) (2 + 2 Sin q) Cos q dq = 2 G M / c2 R Sin q
+ 2 G M / c2 R Sin2q / 2 = 4 G M / c2 R + 0 Radians =
4 x 6.673 x 10—11 x 1.983 x 1030 / [ (3 x108)2 7 x 108 ] = 8.4 x 10—6 Rad. =
206,265 x 8.4 x 10—6 = 1.¨73. The experiment of the eclipse of the Sun is not very good, because the rays of light that pass near the Sun are affected by the refraction of its atmospnere.
DETERMINATION OF THE PRECESSION OF THE PERIHELIUM OF A SATELLITE WITH AN ELLIPTICAL ORBIT.
Suppose a satellite that moves in an orbit as the one shown in Fig. 71. In this figure, O is the attraction center, and corresponds to one of the foci of the ellipse. We are going to represent by the distance OP the radius, and it will change it magnitude in function of the angle Q.
r = a (1 - E2) / (1 + E Cos Q) . . . . (44) dr = a (1 - E2) E Sin Q d Q / (1 + E Cos Q)2 . . (45)
E = eccentricity
In the first place we are going to consider the satellite moving without any aberralogic effect; in this condition the velocity of the satellite could be obtained using the second Kepler’s law, which considers the radius covering equals areas in equal times. To fulfill this condition, the velocity of the satellite must be in a proportion inversely to the radius of the orbit. If (v) is the velocity, this velocity too, will be a function of the angle for the same reason we mentioned before. Considering the aberralogic effects, and giving gravity light velocity, then for a given time, the satellite will move from P to P’ (distance dS), with the velocity (v) specified, this velocity can be divided in two:
1) one radial: vr = v dr / [(r dq)2 + (dr)2]0.5 . . . . (46);
2) one normal: v n = v r d q / [ (r dq)2 + (dr)2]0.5 . . . (47)
Now we are going to determine the velocity (just aberrated) with which the satellite receives the gravity rays. Fig. (71), show us in a bigger scale one piece of trajectory (Fig. 71´). the piece of trajectory that we are studying is the PP’, that has a differential magnitude of first order. To point P corresponds a radius PO, of which only appears one superior piece in the figure. To point P’ correspond a radius P’O. If we prolong the two pieces of radius showed in the figure they will meet at point O; the radius P’O is smaller than PO in a magnitude equal to CP’ (= d r), so that if we rotate the radius PO an angle dq to the right, the point P will coincide with the C. If in our analysis we approximate to the differential of first order, the arc PC (= r dq), is confounded with the straight line that joint both points and the angles in P and C can be considered right angles For our analysis it is not enough such approximation, as we will see below, so the trajectory PC will be an arc of circle, as indicated with the pointed line; here the angle in P and in C are really right angles because they are measured between the tangents of the trajectory PC (in the points P and C) and the radiuses. In this last case the angles have grown a quantity equal to dq / 2 each one. The trajectory PP’ of the satellite, approximated to the differential of first order, also it could be confused with the straight line that join both points; for a better approximation we have to take in account the curvature in that differential piece of curve and make an identical analysis with the piece PC, the angle in P (triangle OPP’ in Fig. 71) will be increased in a quantity equal to dq / 2 , for this second case this angle in P’ also will increase dq / 2, for identical reasons. When the satellite arrives at the point P’, this has a trajectory that makes an angle in this point, with the straight line PP’ equal to d q / 2. Due to the aberralogic effects, the gravity rays seem to come toward the satellite in the direction O’P’, so the angle of the trajectory of the satellite in that point, with the line P’O’ will be equal to an angle; PP’O + 0.5 dq = 1.5 dq
The projection of the velocity (in the point P’) of the satellite in the line P’O’ is:
(v)po = vr + 1.5 dq v [(r dq)2 + (dr)2]0.5 / r dq . . . . (48’)
c = velocity of the gravity fluid; v = velocity of the satellite.
We can make the following proportion:
[ (r dq)2 + (dr)2 ]0.5 : v :: : r : c ; {(r dq)2 + (dr)2]0.5 = v r / c
if we substitute this value and equation (46) in the former equation, we obtain:
(v)po = v dr [(r dq)2 + (dr)2]--0.5 + 1.5 v2 / c . . . . (48)
In accordance with the Kepler´s law:
v r = vm rm ; v = vm rm / r . . . . (52)
For the planet Mercury we have the following:
t = one year of Mercury (= 88 Earth’s days)
vm = 47.9 Kms./ Sec. = average translation velocity of Mercury.
rm = 57.86 x 106 Kms. = medium radius of the orbit of Mercury
E = eccentricity of the orbit = 0.206.
a = 58.4872 x 106 Kms. = mayor semiaxis of the elliptic orbit.
C = 2 E / (1 + E2) = 0.395228 . . . . (53)
In the fore paragraph have been defined the characteristics of an elliptic orbit of a planet, as Mercury. Of course these characteristics were obtained by observations and applying the Newton and Kepler´s laws, in which were consider the masses of the Sun, of the planet, the distances, velocities, momemtums or impulses, etc. In accordance with all the fore data was obtained the orbit of the planet Mercury with a great approximation; nevertheless, in 1859, by observations made by Le Verrier, he discovered that the planet Mercury had an acceleration of its perihelion of 43”.8 / century, with respect to the calculations made before with the Newton and Kepler´s laws. The fore discrepancy was interpreted then, due to the action of unknown masses... With the general theory of relativity, Einstein gave a more convincing interpretation, that my father did not accept: He considered that such discrepancy was due to some aberralogic effects, but not only geometrical ones, but energetic ones. The effect of a body or radiation moving at certain velocity is manifested not only in a geometrical way; but at the square of their velocity: This was the first law of the double fluid theory conceived by him, and that has not been recognized till now by the scientific world.
In the fore theme: Deflection of a Ray of Light in a Gravitatory Field was solved the problem mentioned by the tittle of the theme, considering all the physical properties of the interacting elements. Here, with the physical data, and the first law of the double fluid theory, was determine all the require data of the elliptic orbit of Mercury, so our problem will be limited to geometrical and mathematical solutions. In accordance with this we can obtain the time of orbit of the planet without considering the aberratories effects; and also can obtain the increment of the acceleration produced by the aberatories effects, in accordance with the first law of the double fluid theory..
From Figs.(71) can be obtained the following formulas:
dt = dS / v = [(r dq)2 + (dr)2]0.5 (r / vm rm)
F_l_ dt ; F_l_ = (G M m p / r2) r dq [(r dq)2 + (dr)2]--0.5
F_l_ dt = (G M m p / r2) r dq [(r dq)2 + (dr)2]—0.5 [(r dq)2 + (dr)2]0.5 (r / vm rm) =
G M m p dq / (vm rm) = perpendicular impulse produced by the Sun in the planet in a time (dt).
Average impulse that has the planet Mercury: m vm rm = average radius of the orbit;
F_l_ = perpendicular force produced by the Sun on the planet.
If we account for the small dimension of the angle dq, practically the aberrated velocity of the received gravity ray by the satellite in the point P’, will be:
ca = c + (v p dr) [(r dq)2 + (dr)2]--0.5 + 1.5 v2 / c . . . . (49)
ca / c = 1 + (v p dr / c)[(r dq)2 + (dr)2]—0.5 + 1.5 v2 / c2
Considering the kinetic aberralogic effects we have
ca2 / c2 = 1 + (2 v p dr / c) [(r dq)2 + (dr)2]—0.5 + (v p dr / c)2 [(r dq)2 + (dr)2]—1 + 3 v2 / c2 + 3 (v3 p dr / c3) [(r dq)2 + (dr)2]—0.5 + 2.25 v4 / c4
The last two terms are very much small, affected by (1 / c3) and (1 / c4), so we can disregard them.
ca2 / c2 = 1 + (2 v p dr / c) [(r dq)2 + (dr)2]—0.5 + (v p dr / c)2 [(r dq)2 + (dr)2]—1 + 3 v2 / c2 . . . (50´)
In accordance with Kepler´s law, we have: v r = vm rm ; v = vm rm / r . . . . (52)
ca2 / c2 = 1 + (2 vm rm p / c) (dr / r) [ (r dq)2 + (dr)2]—0.5 + (vm2 rm2 p2 / c2) (dr / r)2 [(r dq)2 + (dr)2]—1 + (3 vm2 rm2) / (r2 c2) . . . . (50)
dr / r = [a (1 – E2) E Sin.q dq (1 + E Cos.q)—2] [a (1 – E2) (1 + E Cos.q)—1]--1 = E Sin.q dq (1+ E Cos.q)—1
(r dq)2 + (dr)2 = a2 (1 – E2)2 (1 + E Cos.q)—2 [1 + E2 Sin.2q (1 + E Cos.q)—2 ] (dq)2 ;
1 + E2 Sin.2q (1 + E Cos.q)—2 = [(1 + E Cos.q)2 + E2 Sin.2q)] (1 + E Cos.q)—2 = (1 + 2 E Cos.q + E2) (1 + E Cos.q)—2
C = 2 E / (1 + E2)
(1 + 2 E Cos.q + E2) (1 + E Cos.q)—2 = (2 E / C + 2 E Cos.q) ( 1 + E Cos.q)—2 =
(2 E / C) ( 1 + C Cos.q) (1 + E Cos.q)—2
[(r dq)2 + (dr)2]—0.5 = [a2 (1 + E2)2 (1 + E Cos.q)--2]—0.5 [(2 E / C) (1 + C Cos.q) ]—0.5 (! + E Cos.q)--1]—0.5 (dq)—1 = a—1 (1- E2)—1 (1 + E Cos.q)2 (2 E / C)—0.5 (1 + C Cos.q)--0.5 (dq)—1
(dr / r) [(r dq)2 + (dr)2]—0.5 = E Sin.q (1 + E Cos.q)—1 (dq) a—1 (1 – E2)—1 (1 + E Cos.q)2 (2 E / C)—0.5 (1 + C Cos.q)—0.5 (dq)—1
= A Sin.q (1 + E Cos.q) (1 + C Cos.q)—0.5 . . . . (50A)
A = E a--1 (1 – E2)—1 (2 E / C)—0.5 =
0.206 x 5.84872 x 10—10 (1 –0.042436)—1 .x 0.206 x 2 / 0.395228)—0.5 = 0.206 / 5.7181135 x 10--10 = 3.60257 x 10—12 m--1
(dr / r)2 [(r dq)2 + (dr)2]–1 = A2 Sin.2q (1 + E Cos.q)2 (1 + C Cos.q)—1 . . . . (50´A)
2 B = A (2 vm rm) = 3.60257 x 10—12 (2 x 4.79 x 104 x 5.786 x 1010 =
3.60257 x 10—12 x 5.542988 x 1015 = 19,969 m / Sec.
Substituting equations (50A) and (50´A) in equation (50)
ca2 / c2 = [ 1 + (2 B p / c) Sin.q (1 + E Cos.q) (1 + C Cos.q)—0.5 + (B p / c)2 Sin.2q (1 + E Cos.q)2 (1 + C Cos.q)—1 + (3 vm2 rm2) / (r2 c2) ] dq . . . .(51)
In accordance with the second aw of the double fluid theory (not well study yet) The fore formula (51) gives a satisfactory mathematical variation of the increment of the velocity of gravity in a planet. The first term (= 1) is proportional to the normal velocity of gravity without any aberration. The second tern gives the lineal aberration produced by the movement of the planet; by now we ccan not affirm that the gravity can affects the planet in a lineal way; it would be interesting study this effect; not only from a mathematical point of view. The third and fourth terms give the aberration gravity not only from a mathematical point of view, but from a physical one, that in accordance with the first law of the double fluid theory increment the effect of gravity in a proportion equal to the square of its velocity. In these last terms, beside the increment of velocity we have to consider the intensity of gravity; this in accordance with a second law of the double fluid theory (not well study yet); for instance, if the intensity is one, by this reason the effect will be proportional to one; if the intensity is two the corresponding effect would be equal to two… In the third term, it is affected by the coefficient (B p / c )2, in which: B = 19,969 / 2 m / Sec.; but really with the intensity of gravity produced by the Sun, the velocity of the planet is: vm = 4.79 x 104 m / Sec.; also the value vm is an average one in a lineal way; but is bigger considering a second powder one (v2).
Considering no aberrated effects, the first term (1) into the angular parenthesis will give the angle that produce the planet moving from the aphelion point (O) toward the perihelion point (P), and vice versa. Now considering aberrated effects, the second term into the parenthesis, affected by (1 / c) will give a delay of the perihelion when moves from the perihelion toward the aphelion, and when moves in the opposite direction will give an advance effect. Finally, the third and fourth terms into the angular parenthesis will give a definitive angular deviation of advance of second order..
Trajectory of the planet from the aphelion to the perihelion without considering aberratories effects: dq = p radians = 180o
From the perihelion to the aphelion: dq = -- 180º
The trajectories considering the oscillate aberration of first order: :
(ca / c)2 dq = (2 B p / c) [Sin.q (1 + E Cos.q) (1 + C Cos.q )--0.5 dq =
(2 B p / c) Sin q (1 + 0.206 Cos.q) (1 + 0.395228 Cos.q)—0.5 dq
In the integral tables is not found any formula as the fore one, for direct integration; but with the Newton´s binomial we obtain the following formula:
(1 + C Cos.q)—0.5 = 1 – 0.5 C Cos.q + 0.5 x 1.5 C2 Cos.2q / 2¡ - 0.5 x 1.5 x 2.5 C3 Cos.3q / 3¡ + . . . =
1 – 0.197614 Cos.q + 0.0585769 Cos.2q – 0.0192927 Cos.3q + 0.00667189 Cos.4q – . . . .
When the planet moves from the perihelion toward the apjelion, or in the ipposite way, we have:
Sin q Cosn q dq = -- [ Sin2 q Cosn+1 q / / (n + 1) = 0; there is no aberation effect. But when the planet moves from an intermediate point between the perihelion and the aphelion, to them is produced an ooscillate value.
But when the planet moves from the intermediate point: q = 0o, we have:
Sin2 q Cosn q d q = --Sin q Cos n+1 q / (n + 1) = Cosn+1q]
Cos.nq Sin.q dq = -- Cos.n+1q / (n + 1)
Sin.q dq = --Cos.q = 1
Cos.q Sin.q dq = --Cos.2q/ 2 = + 1/2
Cos.2q Sin.q dq = --Cos.3q/ 3 = +1/3
Cos.3q Sin.q dq = --Cos.4q/ 4 = +1/4 .
Cos.4q Sin.q dq = --Cos.5q/ 5 = + 1/5
Cos.5q Sin.q dq = --Cos.6q/ 6 = + 1/6
Cos.6q Sin.q dq = --Cos.7q/ 7 = + 1/7
+ 1 D = + Sin.q (1 + 0.206 Cos.q) dq = 1+ 0.206 (1/2) = + 1.103
-0.197614 D = - 0.197614 Sin.q (1 + 0-206 Cos.q) Cos.q dq = - (0.197614 (1/2) - 0.197614 x 0.206 (1/ 3) = - 0.053092
+ 0.0585769 D = + 0.0585769Sin q (1 + 0.206 Cos.q) Cos.2q dq = + 0.0585769 (1/3) + 0.0585769 x 0.206 (1/4) = + 0.0225423
-0.0192927 D = - 0.0192927 Sin q (1 + 0.206 Cos.q) Cos.3q dq = = - 0.0192927 (1/4) – 0.0192927 x 0.206 (1/5) = - 0.0086817
+ 0.00667189 D = + 0.00667189 Sin q (1 + 0.206 Cos.q) Cos.4q dq = +0.00667189 (1/5) + 0.00667189 x 0.206 (1/6) = 0.00156344
S DL = 1.04504
Angular oscillatory aberration (first order):
q2 = (2 B p / c) Sin.q (1 + Cos.q) (1 + C Cos.q)—0.5 dq = (2 B p / c ) S D´ =
(19,969 p / 3 x 108) 1.04504 = 2.185 x 10—7 radians = 45.”0 / half orbit
With the aberralogy my father proved that the theories of relativity are wrong. Before my father, the ideas of aberration were study in very elementary way, and only were seen them from geometrical point of view, as did Bradley, or Doppler in the study of the frequency of sound and other radiation; my father did this, studding not only geometrical aberrations, but he considered energetic variations; so he discovered the first law of the double fluid theory; in this way he was able to solve the deflection of a ray of light in a gravitatory field (see the fore problem with the same name).
Way he was able to solve the deflection of a ray of light in a gravitatory field (see the fore problemwith the same name).
In this theme the problems of aberration are seen in a more profound way. In the fore mentioned theme was only considered the aberration effects produced by direct variation of the radiation (gravity) that produce the aberration. In this way the aberration grows in a lineal proportion of the increment of such radiation, and diminish in a lineal proportion of the decrement of such radiation To this we call aberration of the first order. But there are other kind of aberrations, that are manifested only as positive ones. These other aberrations are manifested when exists a radiation that produce such aberration, it does not matter the radiations grow or diminish. In both cases the radiations produce an energy of aberration; in the first case such energy grows when the radiation grows; and diminish when the radiation diminishes, in the second case, also there is an energy, although it decreases. For the production of the aberration of second order it is only required there is a quantity of gravity that increment the deflection of the planet, so the second term of the equation (51), affected by /1 / c) can produce second order aberrations affected by (vm p / c)2; this beside the first order aberration affected by (2 B p / c)2; this beside the first order aberration affected by: (2 B p / c), that was given before. With the numerical problems will be seen forward will be understand better all this.
Next it is going to see the aberration (affected by 1 / c2) of second order. . In the third and fourth terms (affected by 1 / c2) the aberration is of second order. As was say before, the second order aberrations always are of advance, although the planet moves from the perihelion toward the aphelion; that is when the intensity of gravity diminishes..
First it is going to see the aberration of second order produced by the second term (affected by 1 / c) of the formula (51). The aberration of second order produced by the second term of equation (51) is considering that: S D´2 = 1.04504 is produced in 1 / 2 of orbit and in accordance with the second law of the double fluid theory:
q2 = S D´2 x (vm p / 2 c)2 = 1.04504 x (47900 p / 2 x 3 x 108) = 6.573 x 10—8 Rad.
Now we are going to solve the problem of aberration of advance, the third and fourth terms of equation (51) will give this effect. When the planet is moving from the perihelion toward the aphelion, always is affected by the gravity force; it does not matter the gravity diminishes. In a second order process always is produced advance effect, because always there is gravity, although it is reduced when the planet advances from the perihelion toward the aphelion;, this happens in the third and fourth terms that are affected by (1 / c2). Next it is going to see this problem.
Cosn q Sin.2q dq = Sin.q Cos,n+1q / (2 + n) + [ 1 / (2 + 1)] Cosnq dq = 0 + [1 / (2 + n)] Cosnq dq
Sin2q (1 + 2 E Cos.q + E2 Cos.2q) (1 + C Cos.q)--1 dq =
Sen2q (1 + 2 x 0.206 Cos.q + 0.2062 Cos2q) (1 + 0.395828 Cos.q)—1 dq =
Sin2q (1 + 0.412 Cos.q + 0.042435 Cos2q) (1 + 0.395828 Cos.q) –1 dq
(1 + C Cos.q)--1 = 1 – C Cos.q + C2 Cos.2q - C3 Cos.3q + C4 Cos.4q - . . . . = 1 – 0.395228 Cos.q + 0.156205 Cos.2q - 0.0617367 Cos.3q + 0.0244 Cos.4q - .... . .
Cosn q Sin.2q dq = Sin.q Cos,n+1q / (2 + n) = + [ 1 / (2 + 1)] Cosnq dq = 0 + [1 / (2 + n)] Cosnq dq
Sin.2q dq = q = 1.5708
Cos.q Sin.2q dq = (1 / 3) Sin.q = 0
os.2q Sin.2q dq = (1 / 4) q = 0.3927
Cos.3q Sin.2q dq = (2 / 15) Sin.q = 0
Cos.4q Sin2q dq = (3 / 48) q = =.0.19635
Cos.5q Sin.2q dq = (8 / 105) Sin.q = 0
Cos.6q Sin.2q dq = (15 / 384) q= 0.1227
+ 1 D = + Sin.2q (1 + 0.412 Cos.q + 0.042436 Cos.2q) dq =
1.571 + 0.412 x 0 + 0.042436 x 0.3927 = + 1.5877
- 0.395228 D = - 0.395228 Sin.2q (1 + 0.412 Cos.q + 0.042436 Cos.2q) Cos.q dq =
- 0.395228 x 0 - 0.395228 X 0.412 x 0.3927 – 0.395228 x 0.042436 x 0 = - 0.0639
+ 0.156205 D = + 0.156205 Sin2q (1 + 0.412 Cos.q + 0.042436 Cos2q) Cos2q dq = + 0.156205 x 0.3927 + 0.156205 x 0.412 x 0 + 0.156205 x 0.042436 x 0.19635 = + 0.0626
- 0.0617367 D = - 0.0617367 Sin2q (1 + 0.412 Cos.q + 0.042436 Cos2q) Cos3q dq = 0.0617367 x 0 – 0.0617367 x 0.412 x0.19635 – 0.0617367 x 0.042436 x 0.0 = - 0.0050
+ 0.0244 D = + 0.244 Sin2q (1 + 0.412 Cos.q + 0.042436 Cos2q) Cos4q dq = + 0.0244 x 0.19635 + 0.0244 x 0.412 x 0.0 + 0.0244 x 0.042436 x 0.12272 = + 0.0049
S D = 1.59
In accordance with the numerical data given before we have the aberrated (of advance) angle q3´ in one Mercury half orbit, equal to: q3´ = S D (vm p / 2 c)2 =
1.59 (4.79 x 104 p / (2 x3 x 108)2 = 1.00015 x 10--7 Rad. / half orbit
The radiation produced by the fourth term of equation (51) = q4” = 3 (vm / c)2 =
3 (4.79 x 104 / 3 x 108) = 7.648 x 10—8
In accordance with the second term of formula (51), we have:
q2´ = S D2´(vm p / 2 c)2 = 1.04504 (4.79 x 104 p / 2 x 3 x108)2 = 6.574 x 10--8
q2´+ q3´+ q4´ = q´ = 6.574 x 10—8 + 1.00015 x 10—7 + 7.648 x 10—8 = 2.42 x 10—7 Rad. / half orbit
2.42 x 10—7 Rad./ half orbit. Since Mercury’s years are of 88 terrestrial days, in one terrestrial century the planet Mercury will rotate around the Sun: :365.25 x 100 / 88 = 415 revolutions. A radian has 206,264”, so the angular aberratory deviation of advance (precession of the perihelion) will be: 0.000000242 x 206,264 x 415 x 2 = 41”.7 / century. This value is practically equal to the real 43”.8, the observed one-
Cananea, Sonora, México, año 1945 Estanislao R. de Hoyos
SOME VALUES GIVEN IN THIS WORK
Gravity microcorpuscle: mg = 7.811 x 10—58 Kg.
Light corpuscle: mc = 1.47236 x 10—50 Kg.
Propeller particle, 7 orbit m7 = 4.832Manuel 4 x Hoyos Robles
Químicos 224-1; Col.Tecnológico
64700, Monterrey, N. L, Monterrey, N. L: México,
MechanophysIa
SOME VALUES GIVEN IN THIS WORK
Gravity microcorpuscle: mg = 7.811 x 10—58 Kg.
Light corpuscle: mc = 1.47236 x 10—50 Kg.
Propeller particle, 7 orbit m7 = 4.8324 x 10—38 Kg.
Mass of the electron me = 9.1091 x 10—31 Kg
¨ ¨ a proton: mp = 1.6725 x 10--27 Kg.
¨ ¨ ¨ Earth = 5.975 x 1024 Kgs.
¨ ¨ ¨ Sun = 1.985 x 1030 ¨
¨ ¨ ¨ proton: r+ = 1.7143 x 10—14 m
Radius of the electron: re = 1.4 x 10—15 m
¨ ¨ ¨ seven shell: r7 = 1.13763 x 10—10 m
¨ ¨ ¨ Earth: r = 6.375 x 106 m
¨ ¨ ¨ Sun: rs = 7 x 108 m
¨ ¨ ¨ Earth orbit: r = 1.48 x 1011 m
Length of a side of polygonal orbit: lv = .2.55 x 10—11 m
Length of a multimolecule: lm = 5.68815 x 19—4 m
Radius of a virtual orbit: rv = 2.06 x 10 --4 m (maximum orbit)
Radius (maximum) of a Coulomb´s orbit: rc = 0.895 m
Time of return distance of light corpuscles: tr =1.58 x 1018 Sec
Time of deflection (electron 7 orbit): t>7 = 4.47 x 10—20 Sec.
Time of orbit ( ¨ ¨ ¨ ): to = 2.3084 x 10—10 Sec.
Time of vertex : ( ¨ ¨ ¨ ) tv7 = 8.27 x 10—17 Sec.
Time of jump ( ¨ ¨ ¨ ) tj7 = 1.12 x 10—11 Sec. = 135,632 tv7
Time of multimolecule: tm = 2.8 x 10—5 Sec.
1 year = 31,557,600 seconds
Deflection velocity of sever orbit electron: v_l_7 = 69,098 m / Sec.
Velocity of the seven orbit electron: ve7 = 308,570 m / Sec.
Velocity of the seven nushel positive particle: v+7 = 672.27 m / Sec.
h = 6.6256 x 10—34 joule = 1 quantum.
1 coulomb = 1 joule
Electron charge: qe = 1 eV. = 1.6 x 10—19 joule
μ = 9.2754 x 10—28 joule = 1 magnetron of Bohr.
K = 9 x 109 newton m2 / coulomb2 = coefficient of Coulomb´s law
k = 1.38 x 10--23. joule / molecule oK = coefficient of Boltzmann´s law
G = 6.673 x 10—11 m3 / (Kg Sec2) = constant of universal gravitation.
135,632 t>7 = 1 tj
L = 0.895 m = maximum disance at which works the Coulomb´s formula
. MODEL OF ELECTRIC CURRENT AND ITS APPLICATION TO MEDICINE (checked)
Before the experiment of Griffith, it was suppose that with certain heat a microorganism could be destroyed. Grifflith proved that with certain heat not all the structure of the microorganism were destroyed, and that at least its DNA was not affected. The structure of the DNA has many short circuits that nullify many effects of secondary currents that are produced in the cells and in the microorganisms. If are placed the extremes of two electrodes spaced one millimeter, when an electric current (10 volts, 0.008 Amp.) is circulating, it will produce a continuous spark between them. This spark produces heat, whose intensity destroy some parts of the microorganisms, and this have been proved in the practice.
In mechanophysics is not accepted the model of the free electrons. It is considered that the electrons move along a chain of multimolecules that are formed in the conductor when is produced a difference of electric potential. The electrons of the current move from the atom of less potential toward the atom of more potential (here we consider that from the less positive toward the more positive); this movement is around each atom of the multimolecule (a multimolecule is a lineal chain of atoms, as will be seen forward), during a time equal to: 1.12 x 10—11 second / atom, as will be explained later. When the electron arrives to the atom of higher potential of the multimolecule, it jumps to the atom of less potential of the adjacent multimolecule, in which is repeated the same movement. When the electron arrives to the atom of higher potential of the multimolecule (last atom), jumps to the adjacent multimolecule, that has more potential.
I am not sure when began my investigation on physics. First I tried to divulgate the theories of my father, so I studied them and afterward those of relativity; in these last ones I found many fails; so, also I got interested in study some modern physics, and also found many fails in it, and determine they had not good roots; that is, they were not founded in the structure of the atoms. I remember. as in 1965, doctor Rodolfo Castillo Bahena was director of the department of physics in the ISTESM. that told me that the real progress of physics was conditioned to a good model of atom; immediately I accepted this because always was in accordance with it. I began to study all the models of atoms existing them, but none satisfied me. Finally I conceived that of the polygonal orbits. and tried to make a detailed one of it. Afterward. and due that there are many alternatives in some details, I decided that this would be obtained when we advanced in our studies, and that is better to give the most simple model in first place.
The correct study of physics must be based in the structure of the atoms, and for to understand this we ought to know many things of a correct physics; Nevertheless physics science has not been developed parting from its roots, but from the macroscopic and more evident bodies and particles. For develop mechanophysics I parted from a model of atom that consider was the most logic; for understand this there are many details that here will be omitted, but afterward will be given in other themes. Exist the idea that physics ( and all sciences) have developed because all the discoveries made of it have been proved; but here there is a confusion; one thing is to prove something and other is to give an explanation of it (correct or wrong) of it. Newton explained with his fields of gravity forces, the movement of the planets around the Sun, and this explanation has prevailed till the present time; but with this has not been proved how works the gravity.
In his electromagnetic theories Maxwell supposed that any electric particles, from the most elemental one, that in this case would be an electron with negative charge, or a proton or a positron with positive charge; they would emit their electric fluid in all directions, forming an electric field with their corresponding sign. I am not in accordance with this, because this would mean that neither the electrons nor the protons have only one side in which they can emit (and receive) their radiations. An electron or a proton have only one side in which they emit their radiation. An electron or a proton, in a given instant only can emit their radiation in only one way; in many instant they can emit many radiation in many direction forming a field. A particle formed by many atoms and that has an electric positive charge, or a negative one, can form an electric field with many atoms, positive, or negative (in accordance with the excess of protons or electrons); this because in the charged particle each atom will try to emit its radiation at the same time than the others, in all directions (in all angles...).
As said before, I conceived a model of atom, taking in account many facts, but without pretending give the proves of it; these will be obtained with the progress of physics.. In some way I imagine an atom similar to a solar system; in other aspects have to take in account the behavior of the atoms; had to take in account the limitation of the laws of the classical dynamics; the behavior of the electromagnetic and the gravity radiations; of the black body radiations, etc. Avoiding many explanations that will be understand when we advance in our study, was conceived this model with the polygonal orbits, corresponding to the shells of the atom. Not all the atoms have the same radii; and these could vary in accordance with the model it is consider; mathematically can give a great approximation to the atomic values, in order that fix very well with a model (but not with all). It was consider an average radius: r7 = 1.13763 x 10—10 m., that is equal to the most external orbit or shell. There are seven shells (n = 7). spaced at equal distance one from each other. In the first orbit (square one) the electron moves at v1 = 2160 Kms./ Sec; in the (n) shell, formed by a regular polygon of 4 n sides, the orbital electron moves at a velocity: vn = v1 /n = 2160 / n Kms./ Sec.. In our model the orbit of the electron is not produced in a flat plane, because this plane has a spin, as will be explained afterward. Considering the rotation of the electron and the spin of its orbit, the electron will move around its shell in a time we will call time of jump: tj = 1.12 x 10—11 Sec.
Till here, only have been consider the orbital electrons of the atoms but have not talked of the positive particles that could be supposed as a unitary particle that control an orbital electron of the atom. In our model have consider that each orbital electron has its exclusive positive particle formed by one proton in the H atom, and by one proton plus one, or two neutrons in the other atoms. To an electron plus one positive particle, working as was say, we will call: binary system. Here both particles rotate around a point between them; for instance the electron of the seven shell, at a distance: r7 ≈ 1.13763 x 10—10 m; and the positive particle: r7´ = 4 r7 / 1836 = 4 x 1.13763 x 10—10 / 1836 =
2.988 x 10—13 m; etc.
Had not been explained how are formed and how work the binary systems. In our model have been consider a structure for the positive and for the negative particles; an axis, as a tube in the diameter of them in which are received and emitted some propeller particles formed by corpuscles as those of light. For to be attraction between both particles they must be oriented, co lineal and in the same way, their axis. All the binary system of an atom fulfill the fore condition; and as the positive particles are very much near one to each other in the nucleus of the atom, there is no rejection between them, because their diameter, always are oriented to their negative particles (never to the other positive particles…).
Particle m⊥n; produces the deflection of the orbits of both particles, moving along the diameter of one of them, The two particles of a binary system are spaced a distance equal to rn = (n / 7) r7. The propeller particle after that, moves jumping to the other particle in which is repeated the fore mentioned movement and interior propeller action.. When this propeller particle is into the electron, it is
produced a deflection in the trajectory of it (in a vertex); similar thing happens when is into the positive one, in its trajectory. All say here will be understand better with the numerical examples. The propeller particle m⊥n; is formed by corpuscle as those of light. The mass of a corpuscle is: mc = 1.47236 x 10—50 Kg.; its kinetic energy when moving at light velocity is equal to one quantum, h = 6.6256 x 10—34 joule. A propeller particle of the orbital electron of the seven shell, deflects it with a kinetic energy equal to: K>⊥7 = 0.5 m⊥7 c2 = 0.5 m⊥7 (3 x 108)2 = 4.5 x 1016 m⊥7 K>⊥7 = 0.5 me v⊥72 = 0.5 x 9.1091 x 10—31 x 69,1002 = 2.1747 x 10—21 joule v⊥7 = radial velocity of the orbital electron of the seven shell m⊥7 = 2.1747 x 10—21 / 4.5 x 1016 = 4.8327 x 10—38 Kg.
Modern physics accepts as model of electric current that of the free electrons. This model is too much deficient, as have been explained many times. When it was conceived and accepted did not exist a good structure of the atoms in which could flow the electrons of the current. They avoided this problem considering that the electrons could be free to flow by themselves. With a logical structure of the atoms we have conceived a model in which the electron of the electric current move along chains of atoms we name: multimolecules. The electric current electrons advance from an atom to the adjacent one, moving in each one a time of jump: tj = 1.12 x 10—11 Sec. = 135,632 t>7 = 135,632 x 8.27 x 10—17 Sec.; t>7 = time of vertex (7 shell); equal to the time in which the electron moves from a vertex to the adjacent one. The fundamental idea of our model of electric current can be obtained if we consider a conductor wire in which are formed a bunch of multimolecules. If it is applied a difference of electric potential the electrons that are in one extreme of it will try to move to the opposite extreme spinning around each atom as was say, in a time of jump; after than jump to the adjacent and forward atom, repeating this process, and so on, till they reach the highest potential extreme of the conductor wire. In this process they can advance limited distances, we call multimolecules, along the conductor. In the higher potential atom of each multimolecule they get over saturated; with an interior propeller fluid, they get free of the fore condition and can jump to the forward multimolecule, their excess fluid is absorbed by the last atom of the fore multimolecule; in this way can jump to the lower potential atom of the adjacent multimolecule and advance along it, as did in the fore multimolecule, and so on.. A multimolecule is formed by 2.5 x 106 atoms. The time employed by each current electron along a multimolecule is: tm = 2.5 x 106 x tj = 2.5 x 106 x 1.12 x 10—11 = 2.8 x 10--5 Sec. The fundamental times of our model of electric current are: t>7, tj, tm , etc.
In figure (3), with (O) is indicated the nucleus of an atom, around it spin all the positive particles
o the binary systems that form the nucleus of the atom, also spin the orbital electrons. The regular polygons concentric to the nucleus, have from the seven orbit toward the first the following quantity of sides: 7 x 4 = 28, 6 x 4 = 24, ... 1 x 4 = 4. The fore are the orbits of the nucleus formed by the positive particles of the binary systems; the negative particles or electrons have similar orbits, but much bigger. Practically all the sides of the same sign have the same length. In figure (2) is indicated in a schematic way a piece of trajectory of a seven orbit electront: y, z, e, a; (o) is the center of rotation. The 7 orbital electron moves around the nucleus (o) with a velocity: v7 = 309 Kms../ Sec.. The polygonal orbit also will spin, but in a perpendicular way around the nucleus (o), this due to the Bohr´s magnetron, that acts normally to the electron trajectory, as is indicated by the displacement e e´. Here the distance (z e) is proportional to the translation velocity v7 of the electron and ee´ is proportional to the perpendicular velocity of the vertex of the orbit; this due to the kinetic energy produced by the Bohr´s magnetron, and is equal to the angular velocity α of the plane of the orbit. Considering the plane of the orbit. in figure (2´) the distances: zy, ze, are equal each one to a side of the polygonal orbit; e´ e is the radial deflection of the orbital electron in a vertex.
The effect of “energy” of the magnetron of Bohr is equal to: μ. = 9.2754 x 10—28 joule; acting n accordance with the left hand rule; when two bodies that produce some field move in opposite side, there is a rejection effect between them. The action of the magnetic field produced by the moving electron,
will be in the same way than the hands of the clock, for an observer that is in front of the advancing electron,. In this figure, schematically is represented the effective advance of the electron with the arrow F (equal to the direction that has the conductor). While the electron moves from the vertex (z) to (e), at a velocity v7, its orbit spins an angle (α); in such way that the vertex (e) is displaced to (e´), because the action of the magnetron of Bohr
Kinetic energy of the orbital electron: (7ª orbit) : Ke7 = 0.5 x 9.11 x 10—28 (3.09x 107 )2 = 4.35 x 10—13 erg Each time that the velocity of Bohr acts in the orbital electron of the electric current in each vertex, the orbit spin the following angle α = (μ / 2 Ke7)0.5 π / 14 = (9.2754 x 10-- 21 / {2 x 4.35 x 10-- 13})0.5 π / 14 = 2.317 x 10-- 5 radians. 14 = 28 sides of orbit / 2
(μ / 2) acts in the positive particle (nucleus), and .(μ / 2) acts in the orbital electron.
For the orbit to spins π radians it is required:
nπ = π / α = 3.1416 / 2.317 x 10−−5 ≈ 135,632 Bohr´s velocities.
the orbital electron will move from one vertex to the next one in a time of vertex:
t>7 = 2 π r7 / (28 ve) = 2.π= 8.27 x 10—17 Sec.
(1.13763 x 10—10) / (28 x 3.085 x 105) Each time the orbit of the electron spins an angle = 180º (the electron has moved around all the surface of its shell), the electron will jump to the adjacent atom of the multimolecule, in a time of jump: tj = 8.27 x 10—17 x 135,632 = 1.12 x 10—11 Sec. = time of jump
In the seven shell of an atom one of its molecable electrons, or electron that can form a molecule, has certain saturation, In the H atom, its binary system when ionized can gets its electron moving in the first orbit, before it get ionized; then the ionization potential of the electron of the hydrogen would be: i = 0.5 me v12 = 0,5 x 9.1091 x 10—31 x 2,160,0002 = 2.125 x 10—18 joule = 2.125 x 10—18 / 1.6 x 10—19 = 13.3 eV; 1 eV = 1.6 x 10—19 joule
Since was known the existence of the electrons and of the nuclei (with our theory, the negative particles and the positive ones of the binary systems...) of the atoms, have been considered that an atom gets ionized when it gets free of an electron, because this the electron emits a propeller fluid, formed by corpuscles, as those of light, that have all the energetic capacity to realize this process. But this is not the only way an atom gets free of an electron, and jumps toward the first orbit. In a molecule formed by two atoms, in a given instant (time of jump) one of the atoms gets free of the molecular electron (molecable electron), because this moves to the other atom; in the following instant, the process is inverted; an so on; and all these without the production of any external propeller fluid. In our multimolecule the current electron moves along all the atoms without producing external propeller fluids, except in the extreme atoms of the chain of multimolecules…. It is possible (?) that the multimolecules are formed when is produced a difference of potential. For produce an ionization, this is accompany of a propeller fluid in the ionizer electron. Some atoms are ionized easier than others because they have a higher degree of saturation of corpuscles, such that with less corpuscles that are added, is produced the ionization fluid in the electrons that are liberated. In the molecules there is a continuous interchange of internal propeller fluid, that is of energy. In the electron of the electric current and the intermediate atoms of the multimolecule happens something similar. When an electron of the electric current of a multimolecule jumps to another one, only the electron gets free of its over saturation, that will be absorbed by the adjacent and last atom of the multimolecule..
In my model of electric current, that could be named model of ionization or of multimolecules, is considered that the higher electric potential, is that that have the most positive particles, and the less potential that that have the less positive particles, or more negative ones. Here is considered the movement of the electric current equal to the movement of advance of the electrons. These move along a chain of atoms that we call: multimolecules, of nearly 2.5 x 106 atoms. The electron of the current moves along of each multimolecule, spinning around each atom, jumping from the less potential toward the most potential of the multimolecule; and next jumps to the atom of less potential of the adjacent multimolecule, and of higher potential than the fore one. Each time that the electron of the current moves along a multimolecule, as was say, gets a degree of over saturation of corpuscles that are yieded by the atom of the multimolecule (as those of light). When the electron of the current gets the mentioned degree of over saturation, can not advance, because it need gets free of this over saturation;, in accordance with the model, in order can do this and jumps to the adjacent multimolecule, were is repeated the fore process . When the electron jumps to the atoms of less potential of the adjacent multimolecule, before this, has get free of its over saturation, is produced a fluid is not manifested in an exterior way, because is transmitted directly between the electron to its adjacent atom, except in the case is produced what we have call an ionization effect in which is manifested in an exterior way the propeller fluid. In this way, the electron, being free of the mentioned over saturation, will be in condition to move freely in the less potential atom of the forward multimolecule, next will move toward the atom of higher potential of this multimolecule. Here is repeated the jumps process mentioned before, and so on.
In the same way that the electron (negative particle of the binary system....) gets over saturated when interacts with positive particles 2.5 x 106 times; the positive particle that absorbs the over saturations of the negative particles, also gets over saturated, in accordance with the capacity of the positive particle here will be produced an ionization....) when it is affected by an equal quantity (2.5 x 106) of the current electrons . The number mentioned before is equal to the atoms that has a multimolecule. The positive particle, that in the given moment gets over saturated and ionized correspond to the atom of higher potential of 2.5 x 106 multimolecules. Such ionization is manifested with a propeller fluid that emits the ionizer electron with the energy and temperature required for this process. In order to effectuate this ionization it is required a time equal to that that an electron of the current moves along (2.5 x 106)2 atoms of the conductor = 2.5 x 106 multimolecules. The ionization that is produced is equivalent to a high temperature.
In all the movements in which intervene the electron and the atom (more correct to say: the negative particle and the positive one in the binary systems....) there is an interchange of energy, manifested by the corpuscles acting as propeller fluids of the ionizer electron, as was say in other paragraphs. The ionization is produced when the electron has moved along nm = 2.5 x 106 multimolecules, in an electric current.. With our theories has been defined that the electrons of the electric current move along the chain of atoms of the multimolecules. Each is formed by about 2.5 x 106 atoms, as have been justified in other themes, when have obtained some concordances between some effects and phenomena. An electron of the electric current is able to move without problem from the atom of lower potential toward the atom of higher potential of the multimolecule. This can be explained considering that the electron in all this movement, around 2.5 x 106 atoms (binary system in the atoms....) gets a degree of saturation of corpuscles (as those of light). When the electron arrives to the atom of higher potential of the multimolecule it gets all its saturation (gets over saturated), so can not advance freely, so that in order to jumps to the atom of less potential of the adjacent and more potential multimolecule, the electron will have to gets free with an inerior fluid, of all its over saturation, that will be absorbed by the atom of higher potential of the multimolecule.
In the same way that the electron of the electric current requires to move around 2.5 x 106 atoms of the multimolecule, in order to gets over saturated and afterward gets free of this over saturation, that will be absorbed by the adjacent atom, that gets an over saturation, when electrons in an electric current have afected (2.5 x 106) times to such atom, so that can be uinized. The effective distance that moves an electron in the fore process will be: Li = 2 r7 (2.5 x 106)2 It was say that the ionization of an atom of the conductor is equal to: i = nm io; nm = quantity of atoms that has a multimolecule; io = energy of over saturation of the current electron. But in the same conductor can circulate currents of different intensity, depending of the different electric potential at which they are working. For produce an ionization in an atom, in all cases, this will be affected for the same quantity of interactions with the electrons of the current. It was say that between the electrons of the current and the atoms of the multimolecules there is an interchange of saturation, by nm saturations produced by the electrons in the atom is going to be ionized, as just have been say As in these explanations there are many details that complicate the explanations; next will be given numerical examples; one with a conductor as mercury Hg, and other with a conductor of copper Cu. In medicine, electric currents with about: V ≈ 10 volts and: I ≈ 0.008 Amps, are not harmful to the organism; because this, we are interested to know in which conditions all the conductors work in order to fulfill the fore requirement
In the table of radii of atoms, for Hg we have: rHg = 1.44 μ; so we have a length of multimolecule for Hg: Lm = (2 r7) nm = (2 x 1.44 x 10—10) 2.5 x 106 = 0.00072 m. In nm multimolecules is produced an ionization in an atom. Time of multimolecule: tm = 2.8 x 10—5 Sec. Time of ionization of an atom: ti = nm tm = 2.5 x 106 x 2.8 x 10—5 = 70 Sec. As we are working with I acting in one second, this distance will be:
L1 = Lm nm / 70 = 0.00072 x 2,5 x 106 / 70 = 1800 / 70 = 25.714 m. in one second r7 = radius of an atom.; nm = quantity of atoms that has a multimolecule. nm2 = quantity of atoms that travel the electron between an ionized atom and the following one.
Numerical example of a Hg conductor
(Hg): I = 0.008 amperes; ρ = 0.0000958 ohm / cm3; i = 10.4 eV. = potential of
ionization; A = 0.01 cm2 = section of the conductor
R = ρ L1 / A = =.0.0000958 x 2571.4 / 0.01 = 24.634 ohm / L1 = resistance.
V = R I = 24.634 x 0.008 = 0.197 volts / L1
I = intensity of the current expressed in amperes = 0.008 Amps .
Ie = intensity of the current expressed in electrons / seconds = 0.008 x 6.3 x 1018 = 5,04 x 1016
Energy of the potential of ionization: i = 10.4 x 1.6 x 10—19 joule = 1.664 x 10—18 joule.
The ionization in the electric current is produced every 70 Sec.
Time of ionization: ti = 2.5 x 106 tm = 2.5 x 106 x 2.8 x 10—5 = 70 Secs.
Wr = R I2 = 24.634 x 0.0082 = 0,0015766 watt Wi = i Ie / 70 = 1.664 x 10—18 x 5.04 x 1016 / 70 = 0.0011981 joule / Sec. Wr / Wi = 0.0015766 / 0.0011981 = 1.3159 Wi´ = 1.3159 x 0.0011981 = 0.0015766 joule / Sec.. i´ = 1.3159 x 1.664 x 19—18 = 2.1897 x 10—18 coulomb
Heath accumulated every 70 Sec. (formula of Boltzmann) : ΣTi = 2 I´ / 3 k = 2 x 2.1897 x 10—18 / 3 x 1.38 x 110--23= 105,780o K …
An atom (a binary system). in every (2.5 x 106) multimolecules, gets ionized with an exterior fluid produced by its electron. This fluid is equivalent to a high temperature (Ti), which value is given by the Boltzmann´s formula. The pathogen microorganisms have a solid cover (glicoproteins) that protect them of the humidity of our organism, but not of the heat produced by the ionization. In this is based the electrotherapy. Our organism is not affected in a no reversible way by this heat, because its humidity that dissipate the heat; also the ionization effects of the electrotherapy are very well distributed in the space and in the time in all the organism
For a copper wire we have: (Cu) I = 0.008 Amps. i = 7.7 eV; A = 0.01 cm2 ; ρ = 0.0000017; Ie = 5.04 x 1016 electrons / Sec.; Energy of the potential of ionization: i = 7.7 x 1.6 x 10—19 = 1,232 x 10—18 coulomb Lm = (2 r7) nm = (2 x 1.17 x 10—10) 2.5 x 106 = 0.000585 m Space advanced by the electron in one second: L1 = Lm nm / 70 = 0.000585 x 2.5 x 106 / 70 = 20.893 m R = ρ L1 / A = 0.0000017 x 2089.3 / 0.01 = 0.35518 ohms / L1 V = R I = 0.35518 x 0.008 = 0.00284 volts / L1 Ie = 0.008 x 6.3 x 1018 = 5.04 x 1016 electrons / Sec. Wr = R´I 2 = 0.35518 (0.008)2 = 0.00002273 joule / Sec. Wi = Ie i / 70 = 5.04 x 1016 x 1.232 x 10—18/ 70 = 0.0008870 joule / Sec. Wr / Wi = 0.00002273 / 0.000887 = 0.02563 i´ = 0.02563 i = 0.02563 x 1.232 x 10--18 = 3.1576 x 10-20 Wi = i´ Ie / 70= 3.1576 x 10—20 x 5.04 x 1016 / 70= 0.00002273 Ti = 2 i´ / 3 k = 2 x 3.1576 x 10—20 / (3 x 1.38 x 10—23) = 1,525o K
In our organism Ti is bigger than in Cu and smaller tha in Hg. In any book of chemistry there are tables that give the energy of ionization (electron potential of ionization: i) of all the atomic elements; but these tables are obtained with the elements at certain temperature, pressure, etc.. With some higher temperature some atoms of a sample get ionized without been produced a difference of electrical potential: .To our model of electric current of multimolecules, also we can call: model of electric current by ionization.
Monterrey, México; 1065; August / 2003 November 2
. MULTIMOLECULES
A very much important value is that of the multimolecule. Since a long time ago I am aware of this; for obtain it had to consider several things. For all people that had read my work, it is not necessary to explain what is a multimolecule. In my investigation work of an electric current had to meditate in many facts. In first place concluded that a model of electric current as that of the free electrons was with too many deficiencies; I have been commenting many times in other theories, here will not repeat them: With respect to the resistance of the current, the practical experience for us is very much valuable and the first useful experience we can take for structure our model. As the current electrons do not get free so easy as is considered in the theory of the free electrons, we had to take in account that they moved along a chain of atoms, that we called multimolecule. This means that in all the conductors the current electrons move at the same velocity on their conductor wires, along the multimolecules; some with less resistance (good conductors), and other with more resistance. Our first problem was to determine how move the current electrons in every one of the atoms of the multimolecule. Based in many other phenomena, this solution was relative easy, as could be understand in the numerical problems. Other problem was to determine how many atoms has a multimolecule. Really this is not an easy problem. Since the beginning we got the idea that each current electron jumps from an atom to the forward one of the multimolecule in some way, similar as an electron of a molecule. How many jumps could make in this way, in order to move along the multimolecule? To answer the fore question a logic and scientific way, is not easy for us; so we studied it considering the behavior of some conductors: For this we choose the mercury atom, that practically is the atom most resistant of all conductors, and with its liquid structure there is less risk that its resistance can be altered by irregular distribution of its atoms. Some kind of atoms, as iron are more resistant to the flow of electrons, but they could be magnetized, etc., etc. Depending of the length and the cross section of a mercury conductor, can give to it any one resistance. By first instance this does not affect us. Here will be accepted the value given in a conventional way; and from this value will determine an electric intensity that will be a balanced one; that is, that will give: Wr = Wi (see the fore problem). If the resistance R is reduced to its half, its intensity I is duplicated, and the energies Wr and Wi get incremented in the same magnitude than I. In few words, the maximum variation could have the mercury conductor could not affect to the other conductors that are better ones. For instance, in the fore numerical problems we had: RHg = 1,362.3 ohms; RCu = 24.1747. We can give equal values to both conductors if: LHg = 142,204 cm.; LCu´ = 1,342.3 LHg / 24.1747 = 59.8774 LHg = 8,514,807 cm.
For instance, we have for Hg: L = 100 m: A = 1 cm2; I = 120 Amps.; ρ = 0.0000968; i = 10.4 eV.
RHg = ρ L / A = 9.58 x 10—5 x 10,000 / 1 = 0.958 ohms.
ik´ = 10.4 x 1.6 x 10—19 = 1.664 x 10—18 joule / atom
I = 120 Amps.; Ie = 120 x 6.3 x 1018 = 7.56 x 1020 electrons / Sec
Wr = R I2 = 0.958 x 1202 = 13,705 joule / Sec.
Wi = ik´Ie = 1.664 x 10—18 x 7.56 x 1020 = 1,260 joule / Sec.
Wr / Wi = 13,795 / 1,260 = 10.95
If: / I´ = 120 / 10.95 = 10.96 Amps. = balanced intensity.
Wr´ = 0.958 x 10.962 = 115 joule.
Ie´ = 10.96 x 6.3 x 1018 = 6.9048 x 1019 electrons / Sec.
Wi´ = 1.664 x 10—18 x 6.9048 x 1019 = 115 joules.
For a copper conductor we have: A = 1 cm2.; I = 120 Amps.; i = 7.7 eV.;
ρ = 0.000001725.
We can have equal resistance than the Hg conductor if it is of a length: LCu = ρHg LHg / ρCu = 0.0000958 x 100 / 0.000001725 = 5,553.62 m.
RCu´ = ρCu LCu / A = 0.000001725 x 555,362 /1.0 = 0.958 ohm.
Determination of: nm = quantity of atoms that has a multimolecule.
In the Hg conductor we have: r = 1.13763 x 10—8 cm.; 2 r = 2.2752 x 10—8 cm.
A = 1.0 cm2 = 1.0 / (2 r)2 = 1.0 / (2.27526 x 10—8)2 = 1.931692245 x 1015 atoms.
Ie = 6.3 x 1018 I´ = 6.3 z 1018 x 10.96 = 6.905 x 1019 = quantity of current electrons that flow in one second.
6.905 x 1019 tj = 6.905 x 1019 x 1.12 x 10—11 = 7.733 x 108 electrons / tj.
1.9317 x 1015 = quantity of multimolecules there are in a volume: A nm of atoms.
Originally, when I made the model of the polygonal orbits, I considered that the magnetron of Bohr: μ = 9.2754 x 10—21 was an energy value; afterward, I saw that could work as an energy if it was affected by a term: Q, that is expressed in (gr. / erg). The formula, expressed in e,m, units of the magnetron of Bohr, has an energy would be: M = Q e h / 4 π m. Here the term (e / m) only gives a mathematical proportion, not a physical one… I did not pay much attention to this problem because I was working in other problems of physics; but now that I am studding the flux of the atomic particles have find many interpretations in the problem of the magnetron of Bohr…. Because this, I decided to search other energy that could explain the perpendicular deflection of the orbit of the electron. With this modification will vary the time of jump (tj) and the time of multimolecule ™: But this will not affect our theories, only will be some quantitative variations in some problems, that in the future, and without precipitation will be corrected.
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Considering that a current electron moves along a multimolecule in a time: tm = nm tj, we have: 1.9317 x 1015/ 7.733 x 108 = ;nm2 /nm = 1.9317 x 1015 / 7.733 x 108 = 2.498 x 106 ≈ 2.5 x 106 atoms = number of atoms that has a multimolecule. Also is very much important to point out that the fore number is equal to the quantity of atoms there are in the cross section of the conductor for each active atom in any given moment. – Monterrey, México; September / 2003 Manuel de Hoyos Robles
.THE TWO HOLES EXPERIMENT (chequed)
n the XIX century the undulated theory of light was imposed because with waves could explain many phenomena, that they could not do with a corpuscular theory. In the XX century, also happened the same thing; and also, afterward were found some phenomena they could not be explained by an undulated theory; so they considered that matter (corpuscles) could be transformed in energy (waves). The partisan of the undulated theory considered that the experiment of the two holes proved that light is produced by waves. They used two instruments, one that produces waves, and a slug gun. They had two parallel walls; one with two holes, and in front of it other in which they measure the intensity of the waves and of the slug particles. In the opposite side are the instruments that produce the two mentioned effects. When only one hole is opened in the middle wall, and the experiment is made with both instruments, the maximum intensity is appreciated similar in the testing wall, as could be seen with a symmetric curve with one maximum values. When both holes are opened in the practice and in the experiment, with waves appeared two more intense radiations spaced by one less intensity one; adjacent to the two most intensity, were others two maximum radiations, spaced by two minimum ones; adjacent to the second maximum ones, appeared other maximum spaced by two minimum ones and so on. The experiment of the two holes with the slung gun is not a good one, because the slings are too big compared with the light corpuscles, and with this is limited the interaction of the slings in a great deal; this by one side; by other side, the velocity of the slings, by far is smaller than that of the corpuscles; so the time for produce intersection is very much slow. With the slung gun only can be produced two maximum intensities when the two holes are opened- ntensities when the two holes are opened- . . . . . . -. . . -. . . . . . -. . . . . .
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THE RETURN DISTANCE (NO GROWING OF ENTR0PY) (checked)
Since the epoch of Newton, Coulomb, etc. the fields of forces have got a fundamental importance ; nevertheless if it is considered the infinite time of existence of the Universe, in this time all the fields of the Universe would be exhausted because the bodies that produce them would not be able to continue producing the microatomic particles that form such fields, but this is not so, because the particles that produce such fields, when they are expelled, recede from such bodies at light velocity, move during millions of years and after that time they return to the place were they were emitted, to this process we call return distance in such way that the fields never get exhausted ; to these fields we call mechanized fields for to differentiate of the structured ones that were accepted since the Newton’s epoch, and in which there were not considered as of corpuscular nature. With the mechanized fields of mechanophysics these problems are solved. Mechanized fields are formed by ultramicroscopic particles emitted at light velocity by the bodies that form such fields in all directions, when these field particles reach another body they act in this one, making the body to acts in accordance with the kind of field is considered, and as there is reciprocity between causes and effects, the second body affects to the first one in a similar way. As can be appreciated the mechanized fields by far are more simple than the structured ones. The first step of structure of the mechanized field has been given.
In the infinite time of existence of the Universe all the bodies producer of mechanized fields would be exhausted of their field fluids they emit, unless we consider other property of such fluids that we call: return distance, this return distance can vary in considerable magnitudes of time and distance in accordance with many facts; when a cycle of return distance is effectuated the fluid that original was emitted return toward the emitter body and is “recuperated”. Not all the corpuscles or particles emitted by a body return to the place they were emitted, because are absorbed by other celestial bodies. This, and other celestial bodies also emit particles that are absorbed by the first mentioned body. The second step of the mechanized fields has been made, that is, of no exhausted fields and with this solution is obtained other solution of a problem that have been of preponderant importance since the Boltzmann’s epoch, that is, the growing of entropy. It has been considered that the energy moves always from a higher level toward a lower one. In some way this is a simple criterion that with the mechanophysics is seen in a more subtle way, in accordance with this all particles (entropy particles) of the Universe always had have a determined energy that remain in them in the infinite time in the past and in the future, this energy acts in such particles in accordance with many facts in any determined process and in accordance with one of the fore mentioned facts the particle lose energy, but in accordance with other not evident facts the particle win energy (law of conservation of energy) in such way that all the particles of the Universe have not lost nor win energy in the infinite time of their existence in the Universe; forward it is going to see a subtle case, in this (returning distance) there is not the problem given by the growing of entropy, because the energy (in form of mechanized fields) that is emitted away in all directions, because the returning effect return to the opposite way (maybe in millions years).
There is a fundamental difference between the structured fields and the mechanized ones. It is considered that the first ones have an innate existence and that a body is affected by them in proportion to the magnitude of the physical body properties. The mechanized fields can not have an existence if there not exist the interacting bodies, this means that this bodies are the producer of such fields, sc the theory of the mechanized fields is very much simplified. How many mechanized fields we can have? We can answer the fore question if we observe all kinds of radiation that can emits a body in different circumstances: a gravity field, an electromagnetic field,, a light field, etc. With the mechanized fields we return the acceptance of the Newton’s law of gravity attraction. That was considered was not true because they fail in the precession of the perihelion of Mercury and in the deflection of a light ray moving tangential and near of a heavy celestial body as the Sun? I am not in accordance with the fore trace because in it was not considered that the gravity velocity was incremented (aberrated) by the velocity of the affected or receptive bodies: (Mercury planet and the light corpuscles mentioned); the effect that produces the fore increment of receiving gravity velocity is proportional to the square of such increment. The structured fields depart from the amplitude of the Universe affecting to the interacting bodies: the mechanized fields depart from the interacting bodies affecting the space of the Universe in which they interact.
The problem of the red shift can be explained with a numerical example and at the same time will be given a logical explanation of what has been interpreted as the expansion of the Universe given by modern physics. It has been observed that while farther recede a celestial body, its light radiation (gamma,.....radio radiation) move toward the red frequency, this is considering the constancy of light velocity, about this model (expansion Universe) there are some contradictory points we are not going to mention because the lack of space. In mechanophysics are considered the light radiation formed by material corpuscles (corpuscular nature). A light ray is formed by a straight line of light corpuscles spaced one from each other at equidistant we call: corpuscular distance (wave length in the undulate theory of light). In our model of mechanophysics the frequency of radiation is given by the corpuscular distance, and for distant celestial bodies diminishes not because the corpuscular distance grows, but because the light velocity diminishes due to the action of the remanent gravity (as we will see forward) that acts at equal intervals of time we call: braking time in each corpuscle of the ray of light At the present time the advance of the technology permit us to measure the velocity of light affected by the mentioned braking effect, who does this must not be influenced with the idea that light velocity is a constant value..
With respect to the action of remanent gravity in the light radiation’s we can make more than one model with a structure criterion. Of them and with a fundamentalist criterion we will choose one on which is considered that the electrons that emit the light are not affected by gravity, only their “corresponding” proton in the nucleus of the atom. Each corpuscle of a ray of light emitted, has a quantity of remanent gravity (nugravity) enough to acts at each time of brake during all the returning distance trajectory in such a way that the ray of light emitted by the electron in a given point of space, return to the same point after millions years, as could be appreciated by the numerical example will be given forward. In other themes, forward will be seen that particles with inherent energy can multiply this effect, with multiple interchanges of this energy. Here the kinetic energy produced by the remanent gravity is double than the kinetic energy with which the light corpuscle was emitted; this in accordance with the entropy particles of mechanophysics that consider that all the particles of the Universe have a fixed energy that is not lost nor win in any process we consider In accordance with our model the corpuscles of light are integrated with the nugravity in the surface of the “corresponding” proton with the quantity of nugravity in each corpuscle, enough to act producing the braking effect during all the returning time, with a force can be determined with the law of Newton acting in a corpuscle placed in the surface of the proton, by the proton exclusively. If the rays of the light are produced by atomic particles that are in a heavy celestial body, the remanent gravity in each light corpuscle will grow, so is required to sum the remanent gravity of the proton (nugravity) plus the remanent gravity produced by the heavy celestial body (astrogravity) all these in such way that the return distance trajectory will be smaller than that produced by a body (in vacuum space) emitted with a velocity little bad smaller than that of escape velocity as would be in the case of a celestial body with a mass that approximates to that of a black hole.
An emitted corpuscle of light will return to the point it was emitted due to the return effect produced by the nugravity, the force produced by nugravity, in accordance with our model is:
Fp = G M m / r2 = 6.673 x 10-- 11 x 1.6725 x 10-- 27 x 1.47236 x 10-- 50 /(1.7142951 x 10-- 14)2 = 5.5915227 x 10-- 60 Kg m / Sec2 G = 6.673 x 10-- 11 m3 / Kg Sec2. = constant of universal gravitation. M = mass of proton (Kg) m = mass of a corpuscle (Kg) r = radius of the proton (m)
0.5 m c2 = 0.5 x 1.47236 x 10-- 50 (3 x 108)2 = 6.62562 x 10-- 34 joule = 1 quantum of energy
As the positive particle of the binary system is affected directly by the remanent gravity (nugravity), we suspect that the electrons are not affected directly by the gravity. The time required for the corpuscle affected by the remanent gravity to return due to the corresponding force F would be:
2 tp = 2 m c / Fp = 2 x 1.47236 x 10-- 50 x 3 x 108 / 5.5915227 x 10-- 60 = 1.57992 x 1018 Sec.
In accordance with the theories of expanding Universe, they consider that at a distance equal to: 4 x 1010 to 6 x 1010 light years there are galaxies that recede to a velocity equal to that of light 5 x 1010 x 31,557,600 = 1.57788 x 1018 Sec. = 2 tp 31,557,600 Sec. = 1 year.
About the return distance in a black hole it could be from a limited distance to practically zero distance, so the light corpuscles do not escape from it. In accordance with the concept of continuous action, if an electron emits (n) corpuscles (as those of light) they get light velocity with a kinetic energy equal to n h (n quantum) and the electron, with less velocity, gets a kinetic energy equal to the fore one, but in the opposite way (action = reaction). In accordance with this and with mechanophysics explained the theory of the polygonal orbits of the orbital electrons around the nucleus of the atom. In each vertex of the polygonal orbit of an orbital electron is produced a deflection of it, due to an interior impulse produced by a propeller particle m⊥n , with (n) light corpuscles that deflect the trajectory of the electron in order to follow its corresponding polygonal orbit. If the deflection is bigger than that corresponding to its orbit, the propulsion effect is manifested in an exterior way, with a radiation. If we consider that in each vertex of the polygonal orbit the particle moves along the diameter of the electron, from the in-side to the out-side, interchange its kinetic energy with the rlectron, in order to produce the deflection when the impulse is an interior one; and also to produce an exterior radiation if the impulse is bigger.
Monterrey, Mexico, August 12, 1996 Manuel de Hoyos Robles
April /2007
HYDROGENIZED MODEL OF ATOM (corrected theme)
In the fore theme of Return Distance was simplified in a great deal the concept of fields, because they were considered in their structure, not as something characteristic of all the space of the Universe, but only characteristic of a limited space of the mentioned Universe, corresponding to the interacting bodies (or particles), in such limited space.
Now that we are working in the electrotherapy have had all kind of problems, from that produced by impertinent persons, to technical ones, because this we developed a very much satisfactory model of electric current, but this was not enough, due some persons are too much skeptic. We are not skeptics, but too much exigent with our work, so we began to work trying to make a model of atom that fulfill a series of required conditions, with much more success than that we expected. In doing this I remember a good friend of mine, that since I began to work with mechanophysics, he was very much cooperative with me, encouraging and stimulating my work, all this in spite he was supporter of modern physics, I refer to Dr. Rodolfo Castillo Bahena, he was director and professor of the Physics Department in the Instituto Tecnológico y de Estudios Superiores de Monterrey: his indications, commentaries, and critics were very much useful, I remember in one occasion he commented me that in order that the mechanophysics could be accepted, it would be necessary to give a model of atom with certain characteristics. Because then I had many ideas in my head, I considered this as a simple conversation, but now I consider it as a prophesy (now also have many ideas, but try to be more organized).
With the structure of the atoms it is possible to make a great simplification, and we believe that with this it is possible to make great progress in this field. Till now it has been considered the nucleus of an atom formed by a determined quantity of protons and neutrons, making a unitary structure, affecting in this way to all the orbital electrons, and viceversa. With this criterion have been given different models of nuclei affected with many complexities and deficiencies. Here will be given a model by far more simple that will simplify its investigation and understanding; in this model will be considered that each orbital electron is only affected in a direct way by its corresponding proton and neutron; to this structure we will call: binary system. The intervention of a binary system in the other ones of the atoms will be minimum in all the normal conditions and only could be appreciated in an evident way in phenomena of radioactivity or half life of atoms; in which is manifested the interference of a binary system with another one. With respect to the spectrums produced by the excited atoms, they could be explained by indirect or secondary interaction of a binary system with other ones.
The medullar point of a binary system is that the electron and the positive particle work in a synchronized way emitting jn an interior way their propeller fluids every time of vertex, but this is because other effects also are synchronized; a car can move because has a motor, but its movement is not arbitrary, but in accordance with a driver that controls it, next are going to see some facts that contribute in the control of the behavior of the binary systems. In order that a binary system is not affected by other (s) one and vice-versa, as was say, it will be considered that in a general way, either the electrons, as the protons of each binary system, have a determined side or face were their propeller fluids are emitted, in the corresponding side are received and absorbed the microcorpuscles of the electric fields that excite the atomic particles (their propeller fluids); the amplitude of this side permits a small tolerance given by a small solid angle, in such way that can exist variations (in the orbital electrons) in the spectral lines (s, p, d, f). If the atomic particles (electron or proton) receive the corpuscles of the electric field in a different side than the fore mentioned, the microcorpuscles are not absorbed, not exciting the atomic particle; they can move to avoid this problem.binary systems, but if with two complete different ways can arrive to the same results, the fore objection is demerit. We can continue without impediment, so for a heavy atom as gold: Au, for example: Z = 79; A = 197, we can imagine a structure formed by 79 binary systems, of them 39 have the positive particle formed with one proton + one neutron; and 39 formed with one proton + two neutrons. in the 7 shell (?) 2 with one proton: 39 x 3 + 39 x 2 + 2 x 1 = 197.
In accordance with the characteristics of a binary system, the orbital electron and the positive particle spin around the center of the nucleus of atom in a synchronized way, the orbital electron at a distance equal to the radius of the shell; the positive particle to the fore distance divided by 1836 / 4 = 459; . 459 me = m1+ / 4. The positive particle could be triple mass = 1 proton + 2 neutrons, double mass = 1proton + 1 neutron; single mass = 1 proton. Although we are ignorant of chemistry knowledge, also we believe in the future could be determined different chemical affinities in function of the mass of the positive particles of the binary systems of the seven shell . Both particles (+ -) describe their corresponding polygonal orbit of a similar shape, around the center of the nucleus, but of course of different radius of gyration, so that it is considered that the orbital electrons move in shells, the positive particles will move in a nushells (forming the nucleus of the atom). The size of the nushells, by far are smaller that their corresponding shells, also the size of the orbital electron compared with that of the positive particles, in the first or most interior nushell hardly is space for two moving positive particles, so it can be explained why in the first shell only can spin two orbital electrons. In Fig.(1), second orbit, with O is represented the center of the nucleus of the atom, that also is the center of gyration of the binary system; PN is the positive particle; E the orbital electron; rN the radius of gyration of the positive particle; Rn the radius of gyration of the negative particle (orbital electron)
When the orbital electron moves in the: n = 7 shell (most external shell); Rn = R7 = 1.1376 x 10—8 cm = radius of the atom. The radius of gyration of the positive particle: rN = r7 = 4 Rn / 1,836 = 4 R7 / 1,836 = 4 x 1.1376 x 10—8 / 1,836 = 2.484 x 10—11 cm = radius of the nucleus (nushell) For: n = 1 orbit :r1 = 2.484 x 10—11 / 7 = 3.5486 x 10-- 12 cm.(*about two radiuses of proton, (see Fig. 4)
The electron E moves in a polygonal orbit, for correlated reasons the proton PN moves in asynchronized way in another polygonal orbit, smaller than that of E. For the orbital electron to deflect in a vertex of its orbit (7) it is required an energy to produce the velocity of deflection v (see Fig.2)
v⊥7 = 2 x 309 Sin 6o 43 = 69.1 Km./ Sec. (7a orbit); 6º.43 x 28 = 180º
The energy of deflection of the electron is : E7 = 0.5 me v⊥72 = 0.5 x 9.1091 x 10-- 28 (6.91 x 106 )2 = 2.175 x 10-- 14 erg. In a vertex of the 7 polygonal orbit is formed a field affecting the orbital electron of the binary system and other similar one affecting the positive particle (this in accordance with a model). d = 2 r Sin (180 / 28) = 2 x 1.13673 x 10..10 = 2.51 x 10—11 m ≈ 2.55 x 10--11 m
Considering the length of each side of the polygonal orbit: d = 2.55 x 10-- 9 cm, and also considering in Fig.(2) there is a proportion between velocities and distances, is obtained the deflection distance:..
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21
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SUPERFLUID HELIUM 3 (corrected theme)
In a newspaper dated October 10, 1996, I read that 3 Americans: David Lee, Douglas Osheroff and Robert Richardson won the Nobel Prize of Physics 1996, because their discovery of the fluidity of helium 3, a progress in the low temperature physics that was realized at the beginning of the decade 1970 in the University of Cornell in New York. In accordance with the data of the news paper they discovered that the helium 3 can became superfluid at a temperature of 0o.002 K.
With respect to the report of the newspaper, I do not know if the names of the Nobel Prize winners are correct, what is wrong is datum of temperature, because the so called lambda temperature is 2o.2 K. In the lambda point the helium pass from the gaseous state to the liquid one, as seen with the helium 3. In He4 the two binary systems are equilibrated; for this reason the helium 4 evaporates toward the upper atmosphere in form of separated atoms, this is because the helium is lighter than the air. In the helium 3, the 2 binary systems are different. The helium in liquid state, because its viscosity, does not evaporate, and a multimolecule, because its relative great mass, behaves as a liquid; it does not evaporate, but as is lighter than air tries to climb up the wall of the recipient that contain it, this because its adhesive property characteristic of a liquid state is stronger than the tendency to evaporate toward the upper atmosphere.
In our theme: Model of Electric Current, was obtained the electric currents, that move along multimolecules, each one formed by a line of atoms, forming a chain with: n = 2,497,996 atoms / multimolecule. In a transversal section of a conductor, behind each atom of the section could be formed the multimolecules were the current’s electrons move from the lower potential toward the higher one of the conductor. In our case of current, in every one of the multimolecules is moving one current’s electron, as was specified in the mentioned theme; can be considered that not all the multimolecules required by the current in a given time are working at the same time with a moving electron of the current, but only one in (n) of the mentioned multimolecules, and we say this, because it is considered that for a multimolecule to work, it is required that the orbit of the current’s electron behaves as was specified with the spin velocity, the time of jump and of multimolecule given, etc. With this criterion, as could be appreciated, the most logical seems the maximum simplification. In an atom of helium 4 there are two binary systems formed each one with a negative particle with one orbital electron and a positive particle with one proton + one neutron; if both binary systems move in opposite sides, the helium 4 will have equal spin. Now it is going to see an atom of helium 3, in this case, one binary system is formed by a negative particle with one orbital electron and one positive particle with one proton + one neutron; and the other binary system is formed by a negative particle with one orbital electron, and one positive particle with one proton.
Every binary system spin in a synchronized way, for to attain this the orbit of the shell must be similar (equal number of sides and vertices) than the orbit of the nushell, the only difference will be the dimension of them; the dimension of the shell must be proportional to the velocity of the orbital electron and the dimension of the nushell must be proportional to the velocity of the positive particle. As the positive particle can be formed with three different masses, that is: one orbital proton (single mass); one orbital proton + one neutron (double mass), and one orbital proton + two neutrons (triple mass). When a sample of an element almost get 0o K temperature could be imagined that all the movement ceases to act, this could mean that the binary systems would stop moving, this, because at 0o K the atoms are destroyed, but this is not so, in contradiction with the theory that considers that heat is produced by the kinetic energy of the atomic particles. At practically 0o K temperature, the binary systems continue moving into the atoms as have been specified in the models and maybe because this have never reached a real 0o K temperature. In a binary system the negative particle or electron has the maximum energy that oppose to reach the 0o K temperature, as was say, but also the positive particle contribute to this.
From the book: The Science and the Life of Albert Einstein, by Abraham Pais, are the following paragraphs: In December 1924 Einstein wrote to Ehrenfest from a certain temperature on the molecules “condense” without attractive forces, that is, they accumulate at zero velocity. Until 1938, the Bose-Einstein condensation had the reputation of having only a purely imaginary character. This phase transition was not discovered until 1928 by Willem Hendrik Keesib. In 1938 Flitz London proposed interpreting the helium transition as a Bose-Einstein condensation. Experimentally, the transition point lies at: T = 2o 19 K. In accordance with their theory: T = 3o.1 K. It is generally believed, but not proved that the difference between this two values is due to the neglecting of intermolecular forces in the theoretical derivation.
The concept of temperature is related with that of movement, in such way that some physicists consider that at zero Kelvin temperature all kind of movement ceases to act; this definition is ambiguous, because at that temperature the orbital electrons move with their great velocity around the nuclei of the atoms. We can imagine a macroscopic body moving at great velocity in the outer space at temperature almost zero degree Kelvin. From the most elementary particles, as the light corpuscles, to the atoms or molecules, can move at great velocity at zero Kelvin temperature while they are not affected directly by other particles, or by themselves. In all these cases the particles have a potential energy that is manifested when there is an intersection with them and the medium in which they move, this interaction could be by absorption, reflection, collision, etc.
In the XIX century the physicists believed that all the laws of physics were known and that progress of the science will be obtained by more detailed knowledge; with the discovery of the atomic particles and with modern physics the fore concept had a radical change, in such a way that the physics’ world was considered too complex. Now with mechanophysics we are returning to the concepts that the physicists had of classic physics before modern physics born; nevertheless it is not prudent to be too optimist, because if it is seen with ampler criterion the fields in which physics acts or intervene, can be applied in chemistry, in biology, etc.
As the objective of a mechanophysics is to find true facts with true and objective theories or laws, in all these papers have been proved that this is feasible and that it is not required to employ theories with false structure, as those of modern physics, for to find real facts. From the mentioned book of Abraham Pais is the following paragraph.
Physicists -good physicists- enjoy scientific speculations in private but tends to frown upon it when done in public. They are conservative revolutionaries resisting innovations as long as possible and at all intellectual cost, but embracing it when the evidence is incontrovertible. If they do not, physics tends to pass them by. I would amplify the fore paragraph by saying: that ,more important than the conservative spirit, are the resulting facts obtained by the recognition of the new ideas. In our case is more important the life of the sick people (see electrotherapy) than the conservative spirit of modern physics.
It was say that at a temperature near 0o K a binary system of an atom can moves without incrementing such temperature; if the fore happens with the orbital electron of one atom, also could happens with a multimolecule. Till now have been considered the multimolecules for to explain the movement of the current´s electrons, later will be seen that the sound can move along them. From all say before it is not find any trouble for the electron of the multimolecule to move along this, without affecting the near 0o K temperature. It is well known that when the temperature diminishes, the gaseous substances try to became to a liquid state and the liquid substances try to became solid substances; the fore one only could be explained considering that the multimolecules of the substance try to unite one to each other, and for to do this, the nultimolecule electron of one multimolecule will jump to the adjacent multimolecule, and for to do this it is forceful to be produced an ionization effect. This means that if by one side, with the fore effect at the same time will be produce an increment of the temperature. The ionization of helium is 24.6 eV.
In a gas its particles or atoms have not enough cohesion, so they can move freely; this happens with helium at normal temperature. At very low one (lambda temperature: 2o2 K) the helium is transformed in a liquid substance; this means that its atoms can form multimolecules. The fore mentioned is an external effect, that in a great deal reduces its kinetic effects. But into the atoms their orbital electrons have great velocities: v7 = 308,570 m / Sec. In a binary system formed with a positive particle, equal to one proton, this will moves into its nushell with a velocity: v7´ = 4 v7 / 1,836 = 672.27 m / Sec. With the fore values can obtain the kinetic inherent energy of the atom; this energy of the electron is:
Ke = 0.5 me v72 = 0.5 x 9.1091 x 10—31 x 308,5702 = 4.3366 x 10—20 joule For the positive particle formed with one proton we have: K+ = 0.5 m+ v´72 = 0.5 x 1.673 x 10—27 x 672.272 = 3.7805 x 10—22 joule. The inherent kinetic energy, due to one orbital electron and the positive particle is: Ke + 1.5 K+ = 4.3366 x 10—20 + 1.5 x 3.7805 x 10—22 = 4.3933 x 10—20 joule.
Energy of ionization: Ki = 24.6 x 1.6 x 10—19 = 3.936 x 10-- 18 joule /2 electrons. = 1.968 x 10—18 joule / 1 orbital electron. The two orbital electrons practically get ionized simultaneously, and with the same magnitude; so the ionization energy for one orbital electron is the half of the indicated in the tables; so the temperature produced by an orbital electron is:, in accordance with Boltzmann´s law: TΣ = 2K / 3 k = 2 x 1.963 x 10-- 18 / 3 x 1.38 x 10—23 = 95,072o.5 K.
The inherent energy of the atom is produced, mainly by two orbital electrons, acting in three AMU., so each AMU is affected by 2 Ke / 3. The temperature produced by the ionization by one atom is distributed in all the atoms of a multimolecule; and also, as was say in (3 AMU / 2) of the ionized atom; and this lambda temperature is proportional to the ionization potential of the atom and inverse proportional to the inherent kinetic energy of the atom; so we have:
Tλ= TΣ (Ki / Ke) / nm = 95,072.5 (2 / 3) (3.926 x 10—18 / 4.3933 x 10—20) / (2.5 x 106) =
2.o27 ≈ 2.o2 oK
Monterrey, México, October 15, 1996; March 28, 1998; March /07
Manuel de Hoyos Robles BINARY SYSTEMS WITH POSITIVE PARTICLES HEAVIER THAN ONE PROTON. (corrected theme) Å---- pendiente
In all the problems of mechanophysics dealing with atoms have been considered them with a radius equal to: r7 = 1.13763 x 10-- 8 cm. In many cases this does not affect our final result, because there are some compensate effects, in such way that if some values grow because the approximate data, others diminish by the same reason and when act simultaneously there is a reduction of errors instead of a sum or accumulation of them. But not always happens so, it has been observed that not all the atoms have the same diameter. In some cases it could be explained this variation considering that the diameter of the atoms grow due to the increment of temperature; could be other increments or decrements of the diameter of the atoms due to other effects, for instance if they are compressed, if they combine with other atoms in a chemical way, etc., but all these variations are not due to the structure of the binary systems, but to different physical ,chemical, geometrical distribution of the atoms, etc., and they do not contribute in a really evident way in the variation of the diameter of the atoms in our everyday normal conditions; of course, at very high temperatures and pression, the structure of the atoms is affected.
Considering the structure of the atoms as a sphere has been employed a method for to determine the diameter of the atoms (see book: Chemistry by Michell D. Sienko and Robert A. Plane; Cornell University teachers), for instance: if there is a sample of atoms with a unitary volume, and it is known the atomic mass of each atom, it would be easy to determine the volume occupied by each atom, knowing the weight of the sample. But we can not confide in this method because in the sample the atoms can be distributed in different ways, by different circumstances, and this could give us different densities; all this, if it is considered the atoms occupying fixed spheres; but imagine hydrogen atoms, with only one orbital electron each one; from a mathematical point of view the effective volume of a hydrogen atom is as a disc with a diameter equal to: 2 r7 and in the volume of a sphere with the fore diameter there is space for many discs, as the fore mentioned.
Due to the fore indeterminations, it would not be easy to determine how much press it is required to give to gaseous sample for to determine the real dimension of the atoms of the sample. The more trust way to find the real volume of the atom would be by a more direct way and this is not so easy problem as the fore mentioned method. In our model made in accordance with mechanophysics have been made some special considerations for to simplify some explanations, as for example: to consider an isometric distribution of the atoms, a radius of the seven shell: r7 = 1.13763 x 10-- 10 m, that the atoms of a multimolecule are aligned in a straight line, that the 7 orbit of one electron of an atom does not penetrate in the seventh shell of an adjacent atom, in a given moment and circumstance; that the velocities of the orbital electrons have fixed values, etc. Because all these divergence have been tried to make models in such way to obtain compensate effects and at the same time to orient the theory for to obtain ,more precise models with data more fondness with reality. In this theme will not be given a direct method to determine the real diameter of an atom in the practice, here will limit to determine it in a theoretical way, and in this way will be considered such diameter determined by the dimension of the binary systems of the seventh shell; when the binary system was formed by one proton as positive particle, it was considered the diameter: 2r7 = 2 x 1.13763 x 10-- 10 m. But if the positive particle is formed by a proton + one or two neutrons, the fore value would not be exact as will be seen forward. For to analyze the problem of a binary system, forward is seen with a positive particle with one proton and two neutrons. First will be seen the binary system with a positive particle formed with one proton, in this binary system, its orbital electron has a translation velocity equal to: v7 = 308,571 m / Sec. and the positive particle has its translation velocity equal to: v7’ = 4 x 308,571 / 1836 = 672.268
m / Sec. and a radius of gyration: r+7 = 4 x 1.13763 x 10-- 10 / 1836 = 2.4785 x 10-- 13 m.
The kinetic energy of the orbital electron is: Ke = 0.5 x 9.1091 x 10-- 31 x 308,5712 = 4.3367 x 10—20 joule The kinetic energy of the positive particle: K+ = 0.5 x 1.6725 x 10-- 27 x 672.2682 = 3.77938 x 10—22 joule.
Now let see a binary system with a positive particle formed with: 1 proton + 2 neutrons. If it is considered a gyration axis as the fore seen binary system, then the orbit of the electron will be: re7’ = re7 = 1.13763 x 10-- 10 m. its velocity: ve7’ = 308,571 m / Sec. Its kinetic energy: Ke´ = 0.5 x 9.1091 x 10-- 31 x 308,6702 = 4,3367 x 10—20 joule. v7‘ = 672.26 m / Sec; r+7‘ = 2.4785 x 10—13 m Its kinetic energy: K+ = 0.5 x 3 x 1.6725 x 10-- 27 x 672.2682 = 1.13887 x 10-- 21 joule The time of deflection of the positive particle will be in the nushell: t>7 = 3 x 4.47 x 10—20 Sec.
For a binary system formed with a positive particle with one proton + one neutron, the radius of gyration of the orbital electron is: re7’ = 1.13763 x 10-- 10 m ve7’ = 308,570 m / Sec. The radius of the positive particle: r+‘ = 2.4785 x 10-- 13 = m.; etc, etc.
Monterrey, México, Dicember 10, 01 Manuel de Hoyos Robles INTERPRETATION OF THE EXPERIMENT OF MICHELSON AND MORLEY WITH THE ABERRALOGY
The following papers were take from a book I made in 1972 of the investigation work of my father Estanislao R de Hoyos, in the aberralogy and in an objective physics, since before the second world war. Practically all the work is presented as he conceived it, except some commentaries and amplifications I made to actualize it in the mentioned year 1972 Before interpreting the results of this experiment, it is going to be given an explanation of such experiment. From the book: La Teoria de la Relatividad by Alfred Wulf, Tacubaya, México, 1924, is the following translation:
In Fig.(32), where details have been omitted to simplify the understanding, is shown the interferometer of the physicist of the University of Chicago, Michelson. It consist of two fixed arms, perpendicular one to each other, with the same length and with the mirrors E1 and E2 in their extremities. In the crossing of both arms there is a semitransparent mirror W, inclined an angle of 45o, that lets pass part of the light that proceeds from the luminous point L, and the other part of this light is reflected. So that there are obtained two rays that propagate perpendicularly one to each other toward the mirrors E1 and E2 were they are reflected, returning to W, from here they continue united as far as the telescope of observation T. As the Earth’s movement has different influence in both rays the light need different times, according with the classic theory, to move over the equal paths WE1 and WE2. Due to this, it is necessary to prove that the returning rays suffer in W the reciprocal effect known as interference The telescope T is to observe the interference stripes produced by both rays. All the instrument is installed so that it could be turned around, that is ,either the arm WE1 or the arm WE2 could be oriented in the direction of the Earth’s movement. If the instrument is turned around 90o, so that WE2 be in the direction mentioned before, it could be expected that the same interference stripes appear as they looked in the direction WE1, having as only difference their movement to the opposite side, but being their displacement equal to those due to the position WE1. This instrument was perfected by Michelson by continuous reforms, until obtaining an instrument of precision with a usefulness without precedent till now; thanks to it, it is possible to obtain a precision of one hundredth of the effect calculated in advance, but as was explained before, the experiment has given negative results, because the instrument did not show the influence of the Earth’s movement....
The interpretation of this experiment by the aberralogy can be resumed to the following: Since the observer and the luminous fountain are in the same system (the Earth), both are affected by the same movements, velocities, magnitudes, etc., that the Earth imposes on both, in the same magnitude, so that the observer can not notice any movement of the Earth by observing the luminous focus, he needs to observe another body outside the system; although the problem is too clear, it is going to be subdivided in all the parts that are necessary to study them, with the aberralogy; this, beside a more complete proof, will show us that the experiment is only a combination of cases of aberralogy and not a different phenomenon, as it has been interpreted.
Suppose an observer expecting to appreciate the movement of the Earth with the Michelson & Morley’s experiment; when obtaining negative results, was not satisfied with this, and tried to find the “error” running over the trajectory of the light, observing space by space. Starting from the luminous point L, will arrive to W, from this point he will observe the light coming from L, in this case the body L is approaching with velocity v’, to the observer in W, simultaneously, the observer recedes from W with velocity: v = v’, because both velocities are equal to the velocity of the Earth, the formula which will give the aberrated time due to the movement of the Earth is:
tr = ta (1 - v / c) / (1- v’ / c) = ta . . . . (2,3)
from here it is deduced that the movement of the Earth does not modify the time that the light expends to run over this space.
Now the observer has moved to E1, and from here he will see the light coming from W (in this case W will represent the luminous point); this case is exactly equal to the fore one (space LW) and the movement of the Earth will not affect the time that light spends to run over this other space, as it would be appreciated by any observer that would be displacing in the same system.
If the observer now returns to W, following the route as the ray of light mentioned did, he will see the light reflected from the mirror in E1 arriving to him as if E1 were the luminous point. The observer in W approaches with velocity v, and the luminous point (here E1 is equivalent to a luminous point) E1 recedes with velocity v’, the formula of the time aberrated by the movement of the Earth, will be:
tr = ta (1 + v / c) / (1 + v’ / c) = ta . . . . (1,4)
here: v = v’ = Earth’s velocity
Here the movement of the Earth does not affect the time that light expends to run over the trajectory. It just has been seen before, that the movement of the Earth does not affect the time expended by the light to run over the trajectory given before (trajectory parallel to the Earth’s movement)
Let us see now the case of a trajectory of the light perpendicular to the movement of the Earth. Now the observer moves to E2, and from here he sees the light ray coming from W, that in this case, W represents the luminous point. To be in accordance with the nomenclature of the formulas of time and frequency aberrated, we are going to express with O to the observer in E2, and with On the displaced observer in A, to the luminous point W, by F (see Fig. 33, 34).
As in this case, either the observer O as the luminous body F, move parallel to each other and with a velocity: v = v’ = Earth’s velocity. It is convenient to analyze the problem in two phases. 1) Supposing the observer is moving and the luminous body motionless.
2) Supposing the resultant of the first phase combined with the luminous body moving.
First phase: Observer moving, luminous body motionless. In this condition the observer will appreciate the luminous body in an aberrated time tp (preberrated) in accordance with the following formula: tr c = [(tp c)2 + (tp v)2 - 2 tp2 c v Cos θn]0.5 . . . . (6); tr = real time.
From the fore formula (6), is discovered the preberrated time: tp tp2 = [ (tr c )2 / ( c2 + v2 - 2c v Cos �n )] . . . . (6’)
This time that is aberrated (preberrated) for the first phase, that is, the resultant originated by the combination of the movements of the observer and the light; for the second phase will be the real time, and starting with this for the second phase we find the transberrated time due to the movement of the luminous body combined with the resultant of the first phase. In synthesis, the transberrated time of the second phase, is equivalent to the aberrated time of both phases’ system, due to the movement of the observer (first phase) plus the movement of the luminous body (second phase).
Applying the formula (8) of frequency, or time transberrated tt and taking in account the fore indications, it is obtained:
tr c = [ (tt c)2 + (tt v’)2 - 2 tt2 c v’ Cos θn ]0.5 . . . . (8)
We are going to substitute these values in the fore formula (8), so as to be in accordance with the literal of formula (6’):
tt c = [(tr c)2 / (c2 – v2 - 2 c v´ Cos θn]0.5 . . . . (8’)
Substituting formula (6’) in this, we have: (c2 + v’2 - 2 c v’ Cos θn) / (c2 + v2 - 2 c v Cos Οn) = 1
FIRST LAW OF THE DOUBLE FLUID THEORY
If we study more widely the theory of the double fluid, we could see that the emitted-received fluid can have different velocities, so we refer ourselves to the interaction between atomic particles or
between bodies perceived directly by our sensorial organs, etc., For the case of the gravity emitted- received fluid, we will consider it, equal to light velocity. If the body that emits the gravity fluid, is fixed, and the body that receives it, is moving, this could be appreciated as the fluid having light velocity plus the vectoriall sum of the velocity of the moving (and receiving) body; if the emitting
body is moving and the receiving one is fixed; or if both bodies are moving, we can make identical considerations, all in accordance with the aberralogy.
As we have just seen, the aberralogy permits us to appreciate one type of emitted-received fluid that can have different velocities. This variation of velocities has a notable influence; based on the fact that the energy with which the emitted-received fluid is received, is proportional to the square of the reception velocity divided by ( c),as seen this by a series of observations of different phenomena, was deduced the following law, that is going to be called: First law of the double fluid theory, that says: The energy of the propeller fluid is proportional to the square of the velocity with which the emitted-received fluid is incremented. The limitation of this law could have by interference with other phenomenon, are not going to be mentioned here, due to the lack of space.It is important to make clear that the wearing out of masses produced by the gravity fluids, is so insignificant that the changes (additions or reductions) are due mainly to other effects that are much bigger.
DEFLECTION OF A RAY OF LIGHT IN A GRAVITATORY FIELD.
Before we study this problem I am going to reproduce some paragraphs of the book: “Electromagnetism & Relativity” by Edmund P. Ney, page 107.
Deflection of Light in the Gravitational Field of a Star. In 1801 a German mathematician named Soldner, calculated the deflection of light in the gravitational field of the Sun. His result which we reproduce here leads to the prediction of a deflection of 0”.87 seconds of arc. Shortly after developing the general theory, Einstein calculated the deflection and got the same result. however, he later developed the theory and found that general relativity actually predicts a value twice as large, or 1”.75 of arc. The experiment was first performed in 1919, and the result reported was 1.”7 of arc. The experiment is difficult, and more recent experiments have not agreed with this theory as the 1919 experiment, This prediction of the theory can only be considered to be confirmed to an accuracy of perhaps 20 %. The procedure of the experiment, is to photograph a star field around the Sun during a total eclipse (so that the stars are visible). Six months later, when the same stars are visible at night, the star field is photographed again. The displacement of the apparent positions of the stars can be measured by comparing the two photographs.
Soldner’s derivation. The transverse momentum P⊥, imparted to an object of mass m can be shown to be given at point P (see figure 68) by: dP⊥ =
G M m CosΘ dΘ / c R The total transversal momentum is:
P⊥ =
dP⊥ =
-G M m Cos Θ dΘ / cR = 2 G M m / c R . . . . (42)
The angle : φ = P⊥ / P = (2 G M m /c R) / m c = 2 G M / c2 R radians . . . . (43’)
Substitution in this formula for the case of the Sun, leads to the “classical” value: φ = 0”.87 of arc. Note that although the mass of the photon cancels itself out in the derivation, the mass must be finite, i.e. m ≠ 0, and this was not established in prerelativity physics. It could also be emphasized that Soldner’s value is only half the correct value given by the general theory.
The determination of the angle φ by the relativity theory is not going to be effectuate here, because it is not necessary in this study. In the determination of the angle φ by Soldner, he gave the gravity force an instantaneous action, but this is not the case, because the gravity has light velocity, and we have to take in account the aberralogy. First we consider the light corpuscles moving from P to Q as in Fig.68. When the corpuscle that we take passes by point P with c velocity, it meets a gravity ray with practically the same velocity that moves in the OP direction, the velocity that meet both rays is equal to the vectorial sum of their velocities. In Fig. 69 we have represented by cx the velocity of the corpuscle of light, and by c the velocity of the gravity that meets the light ray (or the corpuscle of light), and by ca the velocity of reception of the emitted-received fluid by light corpuscle.
Distance OP = R / Cos Θ
Distance PP’ = dS = R dΘ/ Cos2 Θ
OP dΘ / Cos Θ =
Gravity force on a corpuscle at point P, according with Newton’s law, is: F = G M m / (R / Cos Θ)2; here: m = 1.47236 x 10—50 Kg. mass of a light corpuscle; M = 1.983 x 1030 Kgs. mass of the Sun; R = 7 x 108 m The resultant of this gravity force perpendicular to trajectory PQT (but in the same plane PQO) is: F⊥ = G M m Cos Θ / (R / Cos Θ)2 = G M m Cos3 Θ / R2 = perpendicular force
The momentum of the corpuscle in the point we are considering or any other point of the trajectory PQT is practically constant (= mcx) and in the same way of the trajectory. The momentum normally the trajectory is obtained multiplying the normal component of the gravity force by dt (time differential). With both magnitudes we make the triangle of Fig. 70 and obtain the deviation angle dφ for the chosen point. The sum of all these angles dφ will give us the total angular deviation φ , and because this angle is very small, we consider the normal in the trajectory equal to the normal of the straight line PQT, without appreciable error.
dφ= G M m Cos3 Θ dt / m cx R2
We solve the problem by the principle of impulse and momentum. dS = PP’ = (PO) dΘ / Cos Θ = R dΘ / Cos2 Θ = cx dt
cx = velocity of the corpuscle dt = = = R d
differential of timedS / cxΘ / cx Cos2 Θ
dF⊥ = F⊥ dt = (G M m Cos3 Θ / RW2) (R dΘ / cx Cos2 Θ) = G M m CosΘ dΘ / cxR
We are going to consider the corpuscle moving as indicated by the arrow. When the corpuscle is in P, the gravity fluid is moving from O to P with velocity c, equal to light velocity. The corpuscle moves with a velocity equal to cx in the direction PP’Q, so that this corpuscle receives the gravitatory ray with an aberrated velocity:
ca = (cx2 + c2 + 2 c cx Sen Θ)0.5 (69)
In accordance with the first law of the double fluid theory of the aberralogy, the increment of energy that produce the radiation (of gravity) is proportional to the square of the increment of velocity with which is received such radiation. So we have that the increment of the action of gravity is equal to: ca / c. Here ca = incremented velocity of gravity (see fugure 69). As velocity of gravity is equal to velocity of light c = cx; In accordance with the principle of impulse and momentum and considering the increment (ca / c), of the gravity force by aberration, and the first law of the double fluid theory, we have:
ca2 / c2 = (cx2 +c2 + 2 cx c Sinθ ) / c2 = 2 + 2 Sinθ The angle of deflection is: dφ=d P⊥ / P ; dP = dF P = mcx φ = dP⊥ / P ; Considering the increment of the action of gravity:
φ =
P⊥ dt ca2 / (c2 m c) =
(-G M m / c R) Cos θ dθ (ca2 / c2 m c) =
(-G M / c2 R) (2 + 2 Sin θ) Cos θ dθ = 2 G M / c2 R
Sin θ 2 G M / c2 R
Sin2θ / 2
= 4 G M / c2 R + 0 Radians =
4 x 6.673 x 10—11 x 1.983 x 1030 / [ (3 x108)2 x 1.4 x 108 ] = 8.4 x 10—6 Rad. = 206,265 x 8.4 x 10—6 = 1.¨73.
The experiment of the eclipse of the Sun is not very good, because the rays of light that pass near the Sun are affected by the refraction of its atmospnere.
DETERMINATION OF THE PRECESSION OF THE PERIHELIUM OF A SATELLITE WITH AN ELLIPTICAL ORBIT. (change from logarithms calculation to scientific calcuculator)
Suppose a satellite that moves in an orbit as the one shown in Fig. 71. In this figure, O is the attraction center, and corresponds to one of the foci of the ellipse. We are going to represent by the distance OP the radius, and it will change it magnitude in function of the angle Θ. r = a (1 - E2) / (1 + E Cos Θ) . . . . (44) dr =
a (1 - E2) E Sin Θ dΘ / (1 + E Cos Θ)2 . . (45) E = eccentricity
In the first place we are going to consider the satellite moving without any aberralogic effect; in this condition the velocity of the satellite could be obtained using the second Kepler’s law, which considers the radius covering equals areas in equal times. To fulfill this condition, the velocity of the satellite must be in a proportion inversely to the radius of the orbit. If (v) is the velocity, this velocity too, will be a function of the angle for the same reason we mentioned before. Considering the aberralogic effects, and giving gravity light velocity, then for a given time, the satellite will move from P to P’ (distance dS), with the velocity (v) specified, this velocity can be divided in two:
1) one radial: vr = v dr / [(r dθ)2 + (dr)2]0.5 . . . . (46);
2) one normal: v n = v r dθ / [ (r dθ)2 + (dr)2]0.5 . . . (47)
Now we are going to determine the velocity (just aberrated) with which the satellite receives the gravity rays. Fig. (71), show us in a bigger scale one piece of trajectory (Fig. 71´). the piece of trajectory that we are studying is the PP’, that has a differential magnitude of first order. To point P corresponds a radius PO, of which only appears one superior piece in the figure. To point P’ correspond a radius P’O. If we prolong the two pieces of radius showed in the figure they will meet at point O; the radius P’O is smaller than PO in a magnitude equal to CP’ (= d r), so that if we rotate the radius PO an angle dθ to the right, the point P will coincide with the C. If in our analysis we approximate to the differential of first order, the arc PC (= r dθ), is confounded with the straight line that joint both points and the angles in P and C can be considered right angles For our analysis it is not enough such approximation, as we will see below, so the trajectory PC will be an arc of circle, as indicated with the pointed line; here the angle in P and in C are really right angles because they are measured between the tangents of the trajectory PC (in the points P and C) and the radiuses. In this last case the angles have grown a quantity equal to dθ / 2 each one. The trajectory PP’ of the satellite, approximated to the differential of first order, also it could be confused with the straight line that join both points; for a better approximation we have to take in account the curvature in that differential piece of curve and make an identical analysis with the piece PC, the angle in P (triangle OPP’ in Fig. 71) will be increased in a quantity equal to dθ / 2 , for this second case this angle in P’ also will increase dθ / 2, for identical reasons. When the satellite arrives at the point P’, this has a trajectory that makes an angle in this point, with the straight line PP’ equal to dθ / 2. Due to the aberralogic effects, the gravity rays seem to come toward the satellite in the direction O’P’, so the angle of the trajectory of the satellite in that point, with the line P’O’ will be equal to an angle; PP’O + 0.5 dθ =
1.5 dθ
The projection of the velocity (in the point P’) of the satellite in the line P’O’ is: (v)po = vr + 1.5 dθ v [(r dθ)2 + (dr)2]0.5 / r dθ . . . . (48’)
c = velocity of the gravity fluid; v = velocity of the satellite.
We can make the following proportion: [ (r dθ)2 + (dr)2 ]0.5 : v :: : r : c; {(r dθ)2 + (dr)2]0.5 = v r / c if we substitute this value and equation (46) in the former equation, we obtain: (v)po = v dr [(r dθ)2 + (dr)2]--0.5 + 1.5 v2 / c . . . . (48) In accordance with the Kepler´s law: v r = vm rm ; v = vm rm / r . . . . (52)
For the planet Mercury we have the following: t = one year of Mercury (= 88 Earth’s days)
vm = 47.9 Kms./ Sec. = average translation velocity of Mercury. rm = 57.86 x 106 Kms. = medium radius of the orbit of Mercury E = eccentricity of the orbit = 0.206. a = 58.4872 x 106 Kms. = mayor semiaxis of the elliptic orbit.
C = 2 E / (1 + E2) = 0.395228 . . . . (53)
In the fore paragraph have been defined the characteristics of an elliptic orbit of a planet, as Mercury. Of course these characteristics were obtained by observations and applying the Newton and Kepler´s laws, in which were consider the masses of the Sun, of the planet, the distances, velocities, momemtums or impulses, etc. In accordance with all the fore data was obtained the orbit of the planet Mercury with a great approximation; nevertheless, in 1859, by observations made by Le Verrier, he discovered that the planet Mercury had an acceleration of its perihelion of 43”.8 / century, with respect to the calculations made before with the Newton and Kepler´s laws. The fore discrepancy was interpreted then, due to the action of unknown masses... With the general theory of relativity, Einstein gave a more convincing interpretation, that my father did not accept: He considered that such discrepancy was due to some aberralogic effects, but not only geometrical ones, but energetic ones. The effect of a body or radiation moving at certain velocity is manifested not only in a geometrical way; but at the square of their velocity: This was the first law of the double fluid theory conceived by him, and that has not been recognized till now by the scientific world.
In the fore theme: Deflection of a Ray of Light in a Gravitatory Field was solved the problem mentioned by the tittle of the theme, considering all the physical properties of the interacting elements. Here, with the physical data, and the first law of the double fluid theory, was determine all the require data of the elliptic orbit of Mercury, so our problem will be limited to geometrical and mathematical solutions. In accordance with this we can obtain the time of orbit of the planet without considering the aberratories effects; and also can obtain the increment of the acceleration produced by the aberatories effects, in accordance with the first law of the double fluid theory..
From Figs.(71) can be obtained the following formulas: dt = dS / v = [(r dθ)2 + (dr)2]0.5 (r / vm rm)
F⊥ dt ; F⊥ = (G M m π / r2) r dθ [(r dθ)2 + (dr)2]--0.5
F⊥ dt =
(G M m π / r2) r dθ [(r dθ)2 + (dr)2]—0.5 [(r dθ)2 + (dr)2]0.5 (r / vm rm) = G M m π dθ / (vm rm) = perpendicular impulse produced by the Sun in the planet in a time (dt). Average impulse that has the planet Mercury: m vm rm = average radius of the orbit; F⊥ = perpendicular force produced by the Sun on the planet. If we account for the small dimension of the angle dθ, practically the aberrated velocity of the received gravity ray by the satellite in the point P’, will be: ca = c + (v π dr) [(r dθ)2 + (dr)2]--0.5 + 1.5 v2hanophy / c . . . . (49) ca / c = 1 + (v π dr / c)[(r dθ)2 + (dr)2]—0.5 + 1.5 v2 / c2 Considering the kinetic aberralogic effects we have ca2 / c2 = 1 + (2 v π dr / c) [(r dθ)2 + (dr)2]—0.5 + (v π dr / c)2 [(r dθ)2 + (dr)2]—1 + 3 v2 / c2 + 3 (v3 π dr / c3) [(r dθ)2 + (dr)2]—0.5 + 2.25 v4 / c4 The last two terms are very much small, affected by (1 / c3) and (1 / c4), so we can disregard them. ca2 / c2 = 1 + (2 v π dr / c) [(r dθ)2 + (dr)2]—0.5 + (v π dr / c)2 [(r dθ)2 + (dr)2]—1 + 3 v2 / c2 . . . (50´)
In accordance with Kepler´s law, we have: v r = vm rm ; v = vm rm / r . . . . (52) ca2 / c2 = 1 + (2 vm rm π / c) (dr / r) [ (r dθ)2 + (dr)2]—0.5 + (vm2 rm2 π2 / c2) (dr / r)2 [(r dθ)2 + (dr)2]—
12
+ (3 vm rm2) / (r2 c2) . . . . (50)
dr / r = [a (1 – E2) E Sin.θ dθ (1 + E Cos.θ)—2] [a (1 – E2) (1 + E Cos.θ)—1]--1 =
E Sin.θ dθ (1+ E Cos.θ)—1
(r dθ)2 + (dr)2 = a2 (1 – E2)2 (1 + E Cos.θ)—2 [1 + E2 Sin.2θ (1 + E Cos.θ)—2 ] (dθ)2 ;
1 + E2 Sin.2θ ( + E Cos.θ)—2 = [(1 + E Cos.θ)2 + E2 Sin.2θ)] (1 + E Cos.θ)—2 =
(1 + 2 E Cos.θ + E2) (1 + E Cos.θ)—2
C = 2 E / (1 + E2)
(1 + 2 E Cos.θ + E2) (1 + E Cos.θ)—2 = (2 E / C + 2 E Cos.θ) ( 1 + E Cos.θ)—2 =
(2 E / C) ( 1 + C Cos.θ) (1 + E Cos.θ)—2
[r dθ)2 + (dr)2]—0.5 = [a2 (1 + E2)2 (1 + E Cos.θ)--2]—0.5 [(2 E / C) (1 + C Cos.θ) ] —0.5 (! +
E Cos.θ)--1]—0.5 (dθ)—1 = a—1 (1- E2)—1 (1 + E Cos.θ)2 (2 E / C)—0.5 (1 + C Cos.θ)--0.5 (dθ)—1
(dr / r) [(r dθ)2 + (dr)2]—0.5 =
E Sin.θ (1 + E Cos.θ)—1 (dθ) a—1 (1 – E2)—1 (1 + E Cos.θ)2 (2 E / C)—0.5 (1 + C Cos.θ)—0.5 (dθ)—1
= A Sin.θ (1 + E Cos.θ) (1 + C Cos.θ)—0.5 . . . . (50ª)
A = E a--1 (1 – E2)—1 (2 E / C)—0.5 =
0.206 x 5.84872 x 10—10 (1 –0.042436)—1 .x 0.206 x 2 / 0.395228)—0.5 = 0.206 / 5.7181135 x 10--10 = 3.60257 x 10—12 m—1
(dr / r)2 [(r dθ)2 + (dr)2]–1 = A2 Sin.2θ (1 + E Cos.θ)2 (1 + C Cos.θ)—1 . . . . (50´A)
2 B = A (2 vm rm) = 3.60257 x 10—12 (2 x 4.79 x 104 x 5.786 x 1010 =
3.60257 x 10—12 x 5.542988 x 1015 = 19,969 m / Sec.
Substituting equations (50A) and (50´A) in equation (50) ca2 / c2 = [ 1 + (2 B π / c) Sin.θ (1 + E Cos.θ) (1 + C Cos.θ)—0.5 + (B π / c)2 Sin.2θ (1 + E Cos.θ)2 (1
+ C Cos.θ)—1 + (3 vm2 rm2) / (r2 c2) ] dθ . . . .(51)
In accordance with the second aw of the double fluid theory (not well study yet) The fore formula (51) gives a satisfactory mathematical variation of the increment of the velocity of gravity in a planet. The first term (= 1) is proportional to the normal velocity of gravity without any aberration. The second tern gives the lineal aberration produced by the movement of the planet; by now we ccan not affirm that the gravity can affects the planet in a lineal way; it would be interesting study this effect; not only from a mathematical point of view. The third and fourth terms give the aberration gravity not only from a mathematical point of view, but from a physical one, that in accordance with the first law of the double fluid theory increment the effect of gravity in a proportion equal to the square of its velocity. In these last terms, beside the increment of velocity we have to consider the intensity of gravity; this in accordance with a second law of the double fluid theory (not well study yet); for instance, if the intensity is one, by this reason the effect will be proportional to one; if the intensity is two the corresponding effect would be equal to two… In the third term, it is affected by the coefficient (B π / c )2, in which: B = 19,969 / 2 m / Sec.; but really with the intensity of gravity produced by the Sun, the velocity of the planet is: vm = 4.79 x 104 m / Sec.; also the value vm is an average one in a lineal way; but is bigger considering a second powder one (v2).
Considering no aberrated effects, the first term (1) into the angular parenthesis will give the angle that produce the planet moving from the aphelion point (O) toward the perihelion point (P), and vice versa. Now considering aberrated effects, the second term into the parenthesis, affected by (1 / c) will give a delay of the perihelion when moves from the perihelion toward the aphelion, and when moves in the opposite direction will give an advance effect. Finally, the third and fourth terms into the angular parenthesis will give a definitive angular deviation of advance of second order..
Trajectory of the planet from the aphelion to the perihelion without considering aberratories
effects: dθ = π radians = 180o
From the perihelion to the aphelion: dθ = -- 180º
The trajectories considering the oscillate aberration of first order: :
0.5 dθ
(ca / c)2 dθ = (2 B π / c) [Sin.θ (1 + E Cos.θ) (1 + C Cos θ =
(2 B π / c) Sin θ (1 + 0.206 Cos.θ) (1 + 0.395228 Cos.θ)—0.5 dθ
In the integral tables is not found any formula as the fore one, for direct integration; but with the Newton´s binomial we obtain the following formula: (1 + C Cos.θ)—0.5 = 1 – 0.5 C Cos.θ + 0.5 x 1.5 C2 Cos.2θ / 2¡ - 0.5 x 1.5 x 2.5 C3 Cos.3θ / 3¡ + . . . =
1 – 0.197614 Cos.θ + 0.0585769 Cos.2θ – 0.0192927 Cos.3θ + 0.00667189 Cos.4θ – . . . .
When the planet moves from the perihelion toward the apjelion, or in the ipposite way, we have:
Sin θ Cosn θ dθ = --[ Sin2 θ Cosn+1 θ / / (n + 1)
= 0; there is no aberation effect. But
when the planet moves from an intermediate point between the perihelion and the aphelion, to them is
produced an ooscillate value.
But when the planet moves from the intermediate point: q = 0o, we have:
Sin2 θ Cosn θ d θ = --Sin θ Cos n+1 θ / (n + 1)
=
Cosn+1 θ]
Cos.nθ Sin.θ dθ = --
Cos.n+1θ / (n + 1)
Sin.θ dθ = --Cos.θ = 1
Cos.θ Sin.θ dθ = --Cos.2θ / 2 = + 1 /2
Cos.2θ Sin.θ dθ = --Cos.3θ / 3 = +1/3
Cos.3θ Sin.θ dθ = --Cos.4θ / 4 = +1/4 .
Cos.4θ Sin.θ dθ = --Cos.5θ / 5 = + 1/5
Cos.5θ Sin.θ dθ = --Cos.6θ / 6 = + 1/6
Cos.6θ Sin.θ dθ = --Cos.7θ / 7 = + 1/7
+1 D = + Sin.θ= 1+ 0.206 (1/2) = + 1.103
(1 + 0.206 Cos.θ) dθ
-0.197614 D = - 0.197614 Sin.θ (1 + 0-206 Cos.θ) Cos.θ dθ = - (0.197614 (1/2) -
0.197614 x 0.206 (1/ 3) =
+0.0585769 D = = + 0.0585769 (1/3) +
+ 0.0585769 Sin θ (1 + 0.206 Cos.θ) Cos.2θ dθ0.0585769 x 0.206 (1/4) = + 0.0225423
-0.0192927 D = - 0.0192927 Sin θ (1 + 0.206 Cos.θ) Cos.3θ dθ = = - 0.0192927 (1/4) – 0.0192927 x 0.206 (1/5) = - 0.0086817
+0.00667189 D = + 0.00667189 =
Sin θ (1 + 0.206 Cos.θ) Cos.4θ dθ+0.00667189 (1/5) + 0.00667189 x 0.206 (1/6) = 0.00156344
Σ DL = 1.04504
Angular oscillatory aberration (first order):
θ2 = (2 B π / c) Sin.θ (1 + Cos.θ) (1 + C Cos.θ)—0.5 dθ = (2 B π / c ) Σ D´ =
(19,969 π = 2.185 x 10—4 radians = 45.”0 / half orbit
3 x 108) 1.04504
With the aberralogy my father proved that the theories of relativity are wrong. Before my father, the ideas of aberration were study in very elementary way, and only were seen them from geometrical point of view, as did Bradley, or Doppler in the study of the frequency of sound and other radiation; my father did this, studding not only geometrical aberrations, but he considered energetic variations; so he discovered the first law of the double fluid theory; in this way he was able to solve the deflection of a ray of light in a gravitatory field (see the fore problem with the same name). Way he was able to solve the deflection of a ray of light in a gravitatory field (see the fore problem with the same name).
In this theme the problems of aberration are seen in a more profound way. In the fore mentioned theme was only considered the aberration effects produced by direct variation of the radiation (gravity) that produce the aberration. In this way the aberration grows in a lineal proportion of the increment of such radiation, and diminish in a lineal proportion of the decrement of such radiation To this we call aberration of the first order. But there are other kind of aberrations, that are manifested only as positive ones. These other aberrations are manifested when exists a radiation that produce such aberration, it does not matter the radiations grow or diminish. In both cases the radiations produce an energy of aberration; in the first case such energy grows when the radiation grows; and diminish when the radiation diminishes, in the second case, also there is an energy, although it decreases. For the production of the aberration of second order it is only required there is a quantity of gravity that increment the deflection of the planet, so the second term of the equation (51), affected by /1 / c) can produce second order aberrations affected by (vm π / c)2; this beside the first order aberration affected by (2 B π / c)2; this beside the first order aberration affected by: (2 B π / c), that was given before. With the numerical problems will be seen forward will be understand better all this.
Next it is going to see the aberration (affected by 1 / c2) of second order. . In the third and fourth terms (affected by 1 / c2) the aberration is of second order. As was say before, the second order aberrations always are of advance, although the planet moves from the perihelion toward the aphelion; that is when the intensity of gravity diminishes..
First it is going to see the aberration of second order produced by the second term (affected by 1 / c) of the formula (51). The aberration of second order produced by the second term of equation
(51) is considering that: Σ D´2 = 1.04504 is produced in 1 / 2 of orbit and in accordance with the second law of the double fluid theory: θ2 = Σ D´2 x (vm π / 2 c)2 = 1.04504 x (47900 π / 2 x 3 x 108) = 6.573 x 10—8 Rad.
Now we are going to solve the problem of aberration of advance, the third and fourth terms of equation (51) will give this effect. When the planet is moving from the perihelion toward the aphelion, always is affected by the gravity force; it does not matter the gravity diminishes. In a second order process always is produced advance effect, because always there is gravity, although it is reduced when the planet advances from the perihelion toward the aphelion;, this happens in the third and fourth terms that are affected by (1 / c2). Next it is going to see this problem. (1 + C Cos θ)—1 = 1 - C Cos θ + C2 Cos2θ - C3 Cos3θ + C4 Cos4θ -. . . = 1 - 0.395228 Cos θ + 0.156205 Cos2θ – 0.617367 Cos3θ +0.0244 Cos4θ - . . . .
Cosn θ Sin.2θ dθ = Sin.θ Cos,n+1θ / (2 + n) + [ 1 / (2 + 1)]
Cosnθdθ =
0 + [1 / (2 + n)] Cosnθ dθ
Sin2θ(1 + 2 E Cos.θ + E2 Cos.2θ) (1 + C Cos.θ)--1 dθ =
Sen2θ (1 + 2 x 0.206 Cos.θ + 0.2062 Cos2θ) (1 + 0.395828 Cos.θ)—1 dθ =
Sin2θ (1 + 0.412 Cos.θ + 0.042435 Cos2θ) (1 + 0.395828 Cos.θ) –1 dθ
(1 + C Cos.θ)--1 = 1 – C Cos.θ + C2 Cos.2θ - C3 Cos.3θ + C4 Cos.4θ - . . . . =
1 – 0.395228 Cos.θ + 0.156205 Cos.2θ - 0.0617367 Cos.3θ + 0.0244 Cos.4θ - .... . .
Cosn θ Sin.2θ dθ = Sin.θ Cos,n+1θ / (2 + n) = + [ 1 / (2 + 1)] Cosnθdθ =
0 + [1 / (2 + n)] Cosnθ dθ
:
Cos nq Sin2q dq = 0 + {(n-1) / (2 + n)} Cosn-2 dq
Cos q dq = Sin θ
Cos2q dq = q / 2
Cos3 q dq = {(3 –1)/ (2+3)} = 2
Sin q / 5 Cos5 q Sin2 q dq = {(5 –1) / (2+ 5)}
Cos3 q dq = 2 Cos3 q
Important remark: The next data do not give proportion of distances; but the propotions of theincrement of the precession of the satellite by unitary haf orbit.
Sin.2θ dθ = θ = 1.5708 Cos.θ Sin.2θ dθ = (1 / 3) Sin.θ
= 0 Cos2θ Sin2θ dθ = (1 / 4) θ = 0.3927 Cos3θ Sin.2θ dθ = (2 / 15) Sin.θ
= 0 Cos4θ Sin2θ dθ = (3 / 48) θ = =.0.19635
Cos.5θ Sin.2θ dθ = (8 / 35) Sin.θ
= 0
Cos.6θ Sin.2θ dθ = (15 / 384) θ = 0.1227
+ 1 D = + Sin2θ (1 + 0.412 Cos.θ + 0.042436 Cos2θ) dθ =
+
1.571 + 0.412 x 0 + 0.042436 x 0.3927 = + 1.5877
- 0.395228 D = - 0.395228 Sin2θ (1 + 0.412 Cos.θ + 0.042436 Cos2θ) Cos.θ dθ =
-0.395228 x 0 - 0.395228 X 0.412 x 0.3927 – 0.395228 x 0.042436 x 0 = - 0.0639
+ 0.156205 D = + 0.156205 Sin2θ (1 + 0.412 Cos.θ + 0.042436 Cos2θ) Cos2θ dθ = +
+ 0.156205 x 0.3927 + 0.156205 x 0.412 x 0 + 0.156205 x 0.042436 x 0.19635 = + 0.0626
-0.0617367 D = - 0.0617367 Sin2θ (1 + 0.412 Cos.θ + 0.042436 Cos2θ) Cos3θ dθ =
-
0.0617367 x 0 – 0.0617367 x 0.412 x0.19635 – 0.0617367 x 0.042436 x 0.0 = - 0.0050
+ 0.0244 D = + 0.244 Sin2θ (1 + 0.412 Cos.θ + 0.042436 Cos2θ) Cos4θ dθ = +0.0244 x 0.19635 + 0.0244 x 0.412 x 0.0 + 0.0244 x 0.042436 x 0.12272 = + 0.0049
Σ D = 1.59
In accordance with the second term of formula (51), we have: In accordance with the numerical data given before we have the aberrated (of advance) angle θ3´ in one Mercury half orbit, equal to: θ3´ = Σ D (vm π / 2 c)2 =
1.59 (4.79 x 104 π / (2 x3 x 108)2 = 1.00015 x 10--7 Rad. / half orbit
The radiation produced by the fourth term of equation (51) = θ4” = 3 (vm / c)2 = 3 (4.79 x 104 / 3 x 108) = 7.648 x 10—8 In accordance with the second term of formula (51), we have:
θ2´ = Σ D2´(vm π / 2 c)2 = 1.04504 (4.79 x 104 π / 2 x 3 x108)2 = 6.574 x 10--8
θ2´+ θ3´+ θ4´ = θ´ = 6.574 x 10—8 + 1.00015 x 10—7 + 7.648 x 10—8 = 2.42 x 10—7 Rad. / half orbit ..
.2.42 x 10—7 Rad./ half orbit. Since Mercury’s years are of 88 terrestrial days, in one terrestrial century the planet Mercury will rotate around the Sun: :365.25 x 100 / 88 = 415 revolutions. A radian has 206,264”, so the angular aberratory deviation of advance (precession of the perihelion) will be: 0.000000242 x 206,264 x 415 x 2 = 41”.7 / century. This value is practically equal to the real 43”.8, the observed one-
Cananea, Sonora, México, año 1945 Estanislao R. de Hoyos
Unified Fields
UNIFFIED FIELDS. (corrected theme)
In other themes we have talked of the electromagnetic corpuscles and the gravity microcorpuscles, of their masses and the property they have to get light velocity. Of course we ignore how they behave so; but by observation we have prove of such behavior; because all these, the best way to explain the theory is with a numerical example. For to study the physics in first place, it is necessary to study the structure of the atom; in some way we have done this. We have eliminated the concept of waves as fundamental ones in our physics; we only consider material particles. Other fundamental point of a good physics is the relation there is between all kinds of forces that act in nature; ; first we will see the action of the electromagnetic forces.
The atoms and the celestial bodies move during infinite time, as if there were no lost of energy in them, and this is due to the property of the corpuscles and the microcorpuscles; in all them only there is an interchange of movement; here we do not say that there is an interchange of energy. In my investigation work, by preference try to apply known concepts, rather than look for new ones. In our numerical example, for to simplify our explanations, we will consider the interaction between two particles (two binary systems) with one AMU = 1.6725 x 10-27 kg., spaced at the maximum distance could acts the Coulomb´s fields: L = 0.895 m.; see theme: Polygonal, Virtual and Couulomb´s Orbits. In the theme: 21 cm. Radiation of Interstellar Hydrogen, it is consider that the positive particle of H with mass of a proton can emits a particle with mass mL = 2.054 x 10—41 kg. as an excess matter, forming the λ21 = 21 cm.. In the solution of our problem, will see this is correct if we consider a time of action equal than tj (time of jump). In our numerical example of the ionized particles, they can emit such excess mass in one action. But also can be emitted in a uniform way during every time of vertex there are in the time of jump. Here we are giving the final results by advance, as will be seen in a theoretical way in our numerical examples.
In other themes related with the so called fields of forces, we have consider them as induced fields. The induced gravity fields are formed by microcorpuscles, so all bodies are transparent to them. It has supposed that a particle or body emits ultramicroscopic particles forming such fields, that when reach other particles or bodies, these are affected in such way that produce external propeller fluids that make them to approach or recede (in accordance with their sign) one particle or body from the other one. The emitter particle produces the inducive fluids, when this fluid reaches the affected or receiving particle; this one produces a reaction, making it to approach to the first one (or recede, in accordance with their sign), as will be explained forward. Here we have say that in the atomic magnitude the particles can get different velocities with respect to their adjacent particles. If it is consider to these effects as energy, then can say that the induced fields have a small energy compared with the energy is produced in the particles they reach, in which are produced an incremented force or effect that affects the particle that receives it with a bigger force of impulse.
In this theme will be study the Coulomb´s and the gravity fields, whose formulas are as follow: Coulomb´s field: fc = K q1a qb / L2 newton; qa and qb are equal to the interacting charges; L is equal to the distance between the two acting bodies. Originally we considered that K was equal to the incremented effect, and: (qa qb /L2) was equal to the inducive fluid; but if it is consider they are an effect produced by the atoms of the interacting particles or bodies, the fore consideration is not so clear, as will be understand better in the numerical examples.
Coulomb´s field: interacting particles, important remark: in them the masses are not the real ones of the interacting particles, but those of the ionized one of them; mass: m1 = m2 = 1.6725 x 10—27 kg., radius of the particles: r7 = 1.13763 x 10—10 m; space between them: L = 0.895 m; mass of a corpuscle: mc = 1.47236 x 19—50 kg.; mass of a microcorpuscle: mg = 7.811 x 10—58 Kg.; mass of the particle that emits the field’s particles: m1 = 1.673 x 10—27 Kg., that form the inductive field, with mass: mL = mq = 2.054 x 10—41 Kg / tj. = mass of corpuscles that emits the interstellar H (see theme: 21 cm. Radiation of Interstellar Hydrogen); time of jump: tj = 1.12 x 10—11 Sec.; time of vertex: tv7 = 8.27 x 10—17 Sec.; tj = 135,632 tv7. The emitted particle mq, here is not acted in a single way, but in uniform ones, every times of vertex, during 135,632 tv7.
The particle m1 = 1.6725 x 10—27 Kg. emits an inductive field formed by: mL = mq = 2.054 x10—41 Kg. in a time of jump: tj = 135,632 tv7; part of this fluid (expressed in intensity) is received by the other particle: m2 = 1.6725 x 10—27 Kg. in the fore mentioned time (tj), and vice versa, the same happens with particle mb with respect to particle m1.
Quantity of corpuscles of the inducive field / tj7: Nq = mq / mc = 2.054 x 10—41 / 1.47236 x 10—50 = 1.39506 x 109 corpuscles / tj7 mc = weight of a corpuscle. Quantity of corpuscles that are emitted in a time of vertex (tv7): nq = Nq / 135,632 = 1.39506 x 109 / 135,632 = 10,285.5 corpuscles / tv7 Mass of the corpuscles emitted in each vertex:
Fq = nq mc = 10,285.5 x 1.47236 x 10—50 = 1.5144 x 10—46 kg. / vertex (tv7) Mass of the corpuscles received by the induced particle (mb) FL = u Fq = 1.61568 x 10--20 x 1.5144 x 10--46 = 2.44679 x 10—66 Kg. / tv7
Force produced by FL (induced) moving at © velocity in the received particle (mb). Here the action is of the induced corpuscles with the corpuscles that has the affected particle (mb), and that produces the impulse on it: FL c = 2.44679 x 10—66 x 3 x 108 = 7.34036 x 10—58 = Kg. m / tv7 =
7.34036 x 10—58 / 8.27 x 10—17 = 8.87589 x 10—42 kg. m / Sec.
Applying the Coulomb´s formula:
Fc = K (qa qb) / L2 = 9 x 109 (1.6 x 10—19)2 / 0.8952 = 9 x 109 x 3.19598 x 10—38 = 2.87631 x 10—28 newton
Incremented value: K´ = Fc / FL c = 2.87631 x 10—28 / 8.87589 x 10—42 = 3.24059 x 1013 (Sec. / m) = 9 x 109 x 3600
From here we have important results:
9 x 109 / 3600 = 2.5 x 106 atoms, that has a multimolecule; this means that the energy induced by an ionization can produce an electric current.
In all numerical problems we have made in other themes have consider the values of the time of vertex tv7 (= 8.27 x 10—17 Sec.); the time of jump tj (= 1.12 x 10—11 Sec.); considering values of the seven shell. Here have done the same thing; mathematically this is correct, but can affect the clarity of these problems, because in each shell the value of tvn have some variation that does not affect, because: (nq)n / tvn = (nq)7 / tv7. This means that the quantity of excess matter emitted is proportional to the lapse of time it is emitted.
Before was given a model explaining how work the Coulomb´s fields; for to have a better idea would be convenient to make a practical numerical example, considering the interaction of two charges: qa = 1.5 x 10—5 coulombs; qb = 6 x 10—6 coulombs; they are spaced at a distance: L =
0.05 m. In accordance with the Coulomb´s law, they are affected by a force: Fc = K (qa qb) / L2 = 9 x 109 x 1.5 x 10—5 x 6 x 10—6 / 0.052 = 324 newtons.
Employing our model of induced fields, we have that one AMU has a charge equal to: 1.6 x 10—19 coulombs; then a charge q1 is produced by the following quantity of AMU: n1 = q1 / 1.6 x 10—19 = 1.5 x 10—5 / 1.6 x 10—19 = 9.375 x 1013 AMU ´s n2 = 6 x 10—6 / 1.6 x 10—19 = 3.75 x 1013 .
If we consider that one AMU has a mass: m = 1.6725 x 10—27 Kg.: The emitted (inducive field) produced by particle ma in a time of jump (tj) is: mq1 = n1 mq = 9.375 x 1013 x 2.054 x 10—41 = 1.92563 x 10—27 Kg. /tj mq2 = n2 mq = 3.75 x 1013 x 2.054 x 10—41 = 7.7025 x 10—28 Kg. / tj
Quantity of corpuscles emitted in a time of jump (tj) by body ma. Nq = mq1 / mc = 1.9256 x 10—27 / 1.47236 x 10—50 = 1.30785 x 1023 corpuscles / tj nq = Nq / 135,632 = 1.30783 x 1023 /136,532 = 9.64266 x 1017 corpuscles / tv7
Mass of the corpuscles emitted in a time of vertex: Fq = nq mc = 9.6266 x 1017 x 1.47236 x 10—50 = 1.41975 x 10-32 Kg / tv7 u = (r7 / L)2 = (1.13763 x 10—10 /0.05)2 = 5.17681 x 10—48
It is supposed that a more correct formula for determine the attraction (and rejection...) of two bodies or particles, producing an induced field, it is required to take in account the size, the form, the time and all the physical properties of the two interacting particles. In the solution of the gravity and of the Coulomb´s formulas, was take only in account the mass and the distance of such particles or bodies, in them was disregarded the time or instant in which the particle gets saturated; the only thing that matter, was the capacity to saturate the particles, only in function of the distances: (r / L)2 = u = unitary decrement, But also can obtain the same results, considering the variation of the volume of the field: r3 / L2 = L (r / L)3 = u. In both cases the variation of the intensity is given in accordance with the model we choose. FL = u Fq n2 = 5.17681 x 10--18 x 1.4975 x 10—32 x 3.75 x 1013 = 2.75616 x 10—36 Kg / tv7
Induced force produced by FL on body mb FLc = 2.75616 x 10—36 x 3 x 108 = 8.26848 x 10—28 Kg m / tv7 = 8.26848 x 10—28 / 8.27 x 10—17 = 9.99815 x 10—12 Kg m /Sec.
K´ = Fc / FLc = 324 / 9.99815 x 10—12 = 3.2406 x 1013 = 3600 x 9 x 109 Sec.--1 Fc = K´FLc = 3.2383 x 1013 x 1.000532 x 10—11 = 324 newtons.
.Now let see how are formed the gravity fields. As all the bodies are practically transparent to the microcorpuscles of gravity, we suppose that when gravity microcorpuscles move over an atom, the orbital electrons absorb some of them, and afterward they are emitted to the positive particles of their binary systems; because this, the atoms that receive such microcorpuscles, do this, in all the sphere corresponding to the seven shell. We will consider the action of gravity in accordance with the MKS system (meter, kilogram, second), for to simplify the explanations: L = 1 m . Here will work with the same particles m1, m2, employed before. When was study the theme: Microcorpuscles of Gravity was consider that in each active corpuscle there was also one active microcorpuscle. In the study of the Coulomb´s field, was consider that this was formed by: N o =
1.395 x 109 corpuscles / tj AMU; no = 10,285.5
When a particle or body (with radius r) produces a field with a given intensity; other particle at a distance L receives it with a reduced intensity, proportional to we call: unitary decrement (here: L = 1 . . .); tha is proportional to: u = (r / L)2 = (1.13763 x 10..10 / 1)2 = 1.2942 x 10—20
First we will consider the gravity interaction between two particles with mass: m1 = m2 =
1.673 x 10—27 Kg = 1 AMU; remembering that each active corpuscle has one active
microcorpuscle; so the mass of microcorpuscles emitted in each vertex of the binary system is:
Fo = no mg = 10,285.5 x 7.811 x 10—58 = 8.034 x 10—54 Kg. / tv7
mg = mass of a microcorpuscle.
Mass of the microcorpuscles received by the induced particle (m2) in a time of vertex: FL = u Fo = 1.2942 x 10—20 x 8.034 x 10—54 = 1.03976 x 10—73 Kg, / tv7
Force produced by FL moving at © velocity in the m2 particle: FL c = 1.103976 x 10—73 x 3 x 108 = 3.1928 10—65 Kg. m / tv7 = 3.1928 x 10—65 / 8.27 x 10–17 = 3.7718 x 10—49 Kg. m / Sec.; tv7 =8.27 x 10—17 Sec.
Applying Newton´s formula considering the mass of each particle: m1 = m2 = 1.673 x 10—27 Kg., we have: Fg = G m2 / l2 = 6.673 x 10—11 (1.673 x 10—27)2 / 12 = 1.8677 x 10—64 newton.
The induced force in form of gravity microcorpuscles field produced by particle m1 and received by particle m2 is: FL c; this force in form of gravity microcorpuscles field, is incremented by particle m2 , for to have the value Fg. In accordance with Newton´s formula G is the incremented term; the induced one is: m1 m2 / L2. In our formula the incrementer term is: G´ = Fg / FL c = 1.8677 x 10—64 / 3.7718 x 10—49 = 4.95175 x 10—16 ≈ 4.952 x 10--16
Numerical example: Important remark: In our model it is consider that all the atomic masses produce equal quantity of unitary decrement mass. Gravity attraction between two bodies in auteur space: m1 = 9 x 104 Kgs.; m2 = 7 x 105 Kgs: L = 9 m. With the Newton´s formula we have: Fg = G (ma mb) / L2 = 6.673 x 10—11 (9 x 104 x 7 x 105) / 92 = 0.0519011 newton.
The attraction of the bodies is proportional to: 1 / L2; because this we have the unitary decrement: u = (r7 / L)1 = (1.13763 x 10—10 / 9)2 = 1.59778 x 10—22
Quantity of AMU of the interacting bodies: n1 = m1 / m+ = 9 x 104 / 1.673 x 10—27 = 5.3795 x 1031AMU´s n2 = m2 / m+ = 7 x 105 / 1.673 x 10..27 = 4.1841 x 1032
Quantity of microcorpuscles (one active corpuscle by one active microcorpuscle) of the induced field produced by body m1: Noa = n1 mq / mc = 5.3796 x 1031 x 2.054 x 10—41 / 1.47236 x 10—50 = 7.50475 x 1040 microcorpuscles / tj.
Quantity of microcorpuscles that are emitted by body m1 in a time of vertex: no1 = No / 135,632 = 7.50475 x 1040 / 135,632 = 5.53317 x 1035 microcorpuscles / tv7
Mass of the microcorpuscles emitted by m1 in a time of vertex: Fo = no1 mg = 5.53317 x 1035 x 7.811 x 10—58 = 4.32196 x 10—22 Kg. / tv7
Mass of the microcorpuscles received by the induced body m2: FL = u Fo = 1.59778 x 10--22 x 4.32196 x 10—22 = 6.90554 x 19—44 Kg. / tv7
Force produced by FL moving at c velocity and acting in mb . n2 FL c = 4,1841 x 1032 x 6.90554 x 10—44 x 3 x 108 = 8.66804 x 10—3 Kg. m / tv7 = 8.66804 x 10—3 / 8.27 x 10—17 = 1.04813 x 1014 Kg. m / Sec. Fg = G´ n2 FL c = 4.95175 x 10—16 x 11,04813 x 1014 = 0.0519 newton
Monterrey, México, January 2004; January 2006: March, 2007 Manuel de Hoyos Robles
BINARY SYSTEMS SMALLER THAN THE SEVEN ONE.
I n the epochs of Galileo and Newton were solves the laws of the dynamics of the macroscopic bodies and with them the science of physics had a great progress. Till the end of the XIX century and the beginning of the XX one, was believe that were known all the laws of physics and that the progress of it will be due to detailed points of it. But in the XX century were find differences between the atomic and astronomic observations and the accepted theories; because this born modern physics, full of contradictory and ambiguous fundamentals. The year 2005 was named the year of physics, in order to commemorate the century of born of the mentioned physics; but this event resulted too dull, in contradiction with the importance they expected. The problem of this is that always, since born modern physics, have been many physicists that think that this physics can be improved to a degree that gets liberated of all its contradictions and ambiguities. Now that I am writing these lines, also there are many people that think so. Nevertheless I can confirm they are wrong, because the progress of a science is not as we wish, but in accordance with the observations and the good interpretation of them. A century of life of a deficient science is too long; specially if in this time many physicists live of this profession; always there are many problems that permit us to valuate this science as not definitive. In the decade 1970-80 were observed some quasars expanding at velocities faster than 9c. The great progress of modern physics was in the experimental way, not in the scientific one; and in the mentioned decades; neither in the experimental way.
There are many points I am interested in the investigation of physics; here will mention only one, that I consider is the root of this science, and that is the behavior of the smaller particles we know that exist, due to the effects they produce. In the same way that Galileo and Newton determined the dynamic of the macroscopic bodies, I am determining the dynamic of the corpuscles and of the microcorpuscles. I did not think of this, when I began this study, relating them with others phenomena; but in a direct way I thought so before 2003. I say all these, because the study of these particles have had no importance in modern physics; they considered the radiation of the electromagnetic and gravity fields of undulated character; so for them such particles do not exist. But my father in 1945, in determine the deflection of a ray of light passing near the Sun, and the precession of the perihelion of Mercury, considered the existence of such particles, and since then it began to know the behavior of them. Who understand this, can understand why Einstein failed, trying to find a unified field.
The born of a science, generally is not initiated investigating the roots of it; but the most evident effects produced by it. Physics science is consider that born with the Galileo and Newton investigations of the dynamic of the macroscopic bodies. Can not exist a universe without movement. When a science born, in a practically way it progresses, even employing wrong theories; but this progress is only in a material way, not in a scientific one.. Physical science is so ample, that always there are some facts could be investigated, and could be interpreted with a correct or a wrong theory; this last condition happened with modern physics; with the false fundamentals of this physics, it is not possible to study the roots of physics that are produced with the great velocity can have the corpuscles and the microcorpuscles. The great fail of modern physics was consider these phenomena were produced by waves; this, disregarding the root of physical science. Other great fail of modern physics was to suppose that light was a universal constant, and because this could not be aberrated There had been few interest to study the roots of physics. My father in considering the aberration of the energy had considerer the great velocity of the corpuscles and microcorpuscles. In all my investigation work I have done the same thing that my father; and because this, we are the creator of the dynamic of the corpuscles and microcorpuscles; although, practically our investigation is very much limited till now; we know these particles can vary their velocity from zero to light one, but we do not know how. Because this we are studying them in different conditions of pressure temperature, magnetism, gravity, etc. In this theme will be study such particles affected by great pressure.
In the theme: Unified Fields was explained in an objective way how are produced the attraction (and rejections) of the bodies or particles due to the field produced by the ultramicroscopic bodies. corpuscles and microcorpuscles; we consider them can move from zero to light or several light velocities. Till now we ignore why are produced such variations; so the most can pretend is to find the circumstances in which they act so, and in this theme the investigation will be so. We suppose that all the movements that exist in the universe are due to the two mentioned ultramicroscopic particles, that with their movement produce heat, gravity, electromagnetic effects, pressure, etc.. Here we meet with a vicious circle; for such particles to produce the fore mentioned effects they ought to be affected by one of such effects that they produce; so we are not able to define which existed first: if the particles that produced the effects, or the effects that only could be produced by the particles. Other problem we can not be able to understand is the concept of infinite; so our logic capacity is to accept the things are produced in the universe, as we appreciate them with our sensorial organs.
With our limited capacity, as was explained before, had been deduced and will be deduced all the problems we deal at. Here will consider the gravity action of particle ma over particle mb. ma produces an induced field that will affect to particle mb. In a given time particle ma emits an energy of corpuscles equal to: Fqn. When this energy reaches particle mb in form of induced field, has been reduced to: Fqn / L2.
We talked in a general way of the role of the corpuscles and microcorpuscles acting in the universe; but here will see them acting in the atoms that are the constituent of the universe. In a H atom, its binary system, in a stable state, is that of the seven shell, r7 = 1.13763 x 10!—10 m. When the atom is affected by a great pressure, as in a star, its diameter is reduced to: φn = 2 x 1.13763 x 10—10 n / 7. With all say before can get the idea that for to have a binary system with diameter smaller than φ7, it is required the mentioned pressure; nevertheless in most of the atom we have binary systems of the interior shells with diameter smaller than φ7 > φn. Next will be study the behavior of the corpuscles affected by a great pressure, and for this will consider the interaction of the two particles, positive and negative of the binary systems that have been compressed to the size of (n) shell (n < r1 =" r7" 7 =" 1.13763" 7 =" 1.6252" mq =" 2.054" tj7 =" 1,12" r1 =" r7" 7 =" 1.13763" 7 =" 1.6252" d1 =" 73" mq1 =" 73" mq7 =" 343" 41 =" 7.0452" u =" (r" r1 =" rn" 2 =" 8126" l1 =" 1.6252" u =" (r" r =" r7," r =" r7" l =" r1;" 73 =" 343" nq1 =" 73" mc =" 343" 50 =" 4.785" fq1 =" mc" nq1 =" 1.47236" 111 =" 7.04522" u =" (r1" 2 =" (8.126" 2 =" 0.25" fl1 =" u" fq1 =" 0.25" 39 =" 1.76131" nq1 =" 1.76131" 11 =" 2.52836" tj1 =" 2.52836" 11 =" 2.25746" tj1 =" 1.12" qa =" qb" 19 =" 5.488" fc =" 9" l2 =" 9.109" 2 =" 0.102626" qa =" qb" q1 =" 12.704" 19 =" 2.0326" nqa =" qa" mc =" 12.704" 50 =" 1.77226" fqa =" mc" nqa =" 1.47236" 1010 =" 2.6094" l =" 3" 11 =" 4.8756" u =" (r" 2 =" (8.126" 2 =" 0.02778" fl =" u" fqa =" 0.02778" 40 =" 7.2489" nqb =" 7.2489" 1010 =" 3.8541" tj7 =" 3.8541" 11 =" 3.4412" fc =" 9" l2 =" 9" 2 =" 1.5642" 6 =" 1.554" mc =" 1.47236" me =" 9.1091" 50 =" 6.1867" 50 =" 1.1363" 7 =" 4.833" 50 =" 3.2825" 7 =" 69,100" 6 =" 2" 5 =" 93,979" _5 =" 135,159" 4 =" 210,698" 7 =" 0.5" 72 =" 0.5" 7 =" me" 7 =" 9.1091" 2 =" 4.83269" m_l_6 =" 8.9391" 5 =" 1.849" 4 =" 4.4932" 7 =" m⊥" c =" 4.8327" 108 =" 1.4498" 6 =" 2.6817" 5 =" 5.547" 4 =" 1.348" nq7 =" 2" mc =" 2" 50 =" 2.7901" nq7 =" Nq7" tv7 =" 2.7901" 632 =" 20,586" u =" L7" 74 =" 2.27526" 2401 =" 1.18454" fq7 =" mg" nq7 =" 7.811" 586 =" 1.60797" fl7 =" u" fq7 =" 1.18454" 53 =" 1.9047" nq7 =" 1.9047" 586 =" 1.1763" tv7 =" 1.1763" 17 =" 1.42238" fg =" G”" nq7 =" 1.0134" 38 =" 1.4414" ma6 =" 2.12533" mb5 =" 3.6726" 5 =" 6.673" 2 =" 1.0061" a =" v7" w =" w2" r7 =" v72" 10 =" 8.37" fa =" me" a =" 9.1091" 1020 =" 7.6231" eo =" 2" fa =" 2" 10 =" 5.45"> = 0.5 me v⊥72 28 = 0.5 x 9.1091 x 10—31 x 69,0982 x 28 = 6.09 x 10—20 joule / to
The energy employed in a circular orbit is (n ) times bigger than that of the polygonal orbit:
n = Eo / E> = 5.45 x 10—19 / 6.09 x 10—20 = 8.95
The propeller particle of the seven shell has an impulse: m v⊥7, that can be determine as follow:
0.5 me v⊥72 = 0.5 x 9.1091 x 10—31 x 69,0982 = 2.174585 x 10—21 = 0.5 m⊥7 c2
0.5 m⊥7 (3 x 108)2 = 4.5 x 1016 m⊥7 m⊥7 = 2.174585 x 10—21 / 4.5 x 1016 = 4.8324 x 10—38 Kg. = 4.8324 x 10—38 / 1.47236 x 10—50 = 3.282 X 1012 corpuscles
The interaction between two particles (positive, negative) of a binary system is produced every time of vertex, in a continuous way, this is due that in both particles are produced some conditions that verify in the seven polygonal orbits model. In some space beyond the seven orbits, some of the fore mentioned conditions are verify, but other do not. If with some special circumstances would be possible that the continuous acting conditions, or at least part of them could act outside of the seven orbits, as in an eight, nine, etc. orbits; to this bigger system of orbits we will call: virtual orbits. These other orbits also will be polygonal, with straight line sides, Omitting a series of considerations, that will be understand better in the structure model of the virtual orbits, will be consider that the velocity (vn) of the orbital electron diminishes in a proportion equal to: (n) ; and also the length of the sides will grow.
n the seven shell the propeller particle has: 3.282 x 1012 active corpuscles: With this value can be determine the number of corpuscles there are in a propeller particle corresponding to the first shell, and also will find the number (n) of orbit (virtual ones) in which the propeller particle is minimum; as will be seen forward this is the biggest virtual orbit. In the polygonal orbits (the seven ones), if we multiply the quantity of corpuscles that has the propeller fluid (mp) of a vertex of an orbit by the number of orbit at the second power (n2), we have a constant value.
In order to apply the fore rule to the virtual orbits, will be done a compensative consideration in our mathematical operations, that will not affect our final result, but will simplify our operations. This will consist in consider the length of the sides of all the polygons as a constant value, equal to that of the system of the seven polygonal orbits; and the quantity of corpuscles of each mp varying in a uniform way from the first virtual orbit: n = 8 to mp = 1 corpuscle (maximum rn). As it is not possible to have a propeller particle with a fraction of quantum; then beyond the virtual orbits will be consider other orbits that we will call: Coulomb´s orbits . Practically, most of the virtual orbits are circular ones; same thing can be say of all the Coulomb´s orbits, so we can consider these last ones acting in a similar way than a star and its planets.
As was say before, the virtual and Coulomb´s orbits, are hypothetic ones, and this is because in they can not be produced all the effects that verify in the polygonal systems of the seven orbits. Nevertheless, some of the mentioned effects are produced, and because this we will be able to determine many properties, with their intensities and limitations, specially of the electrons. In the theme: The Masses of the Stars, was evidential that the propeller (interior) fluids that deflect the orbital electron in a time of vertex (more correct, in a time of deflection) work in all their magnitude in deflect the electron orbit; not so in deflect the orbits of the positive particles. In the Coulomb´s law, in which the effect of attraction is not limited to a time of vertex or of deflection. N n2 = constant = 3.282 x 1012 x 72 = 1.6082 x 1014 x 12 = 1 n2 The rule: N n2 = constant. N = quantity of active corpuscles in a vertex of the (n ) orbit = 1.6082 x 1014 corpuscles; when: n = 1. In the biggest virtual orbit: N = 1; n = (1.6082 x 1014)0.5. = 1.268 x 107
The radius of the n orbit (the biggest virtual one) is: rvn = n r7 / 7 = 1.268 x 107 x 1.13763 x 10—10 / 7 = 2.061 x 10—4 m. For determine the maximum virtual orbit was consider a propeller fluid formed only by one corpuscle. In the practice this is not possible, as will be seen forward. So, we can choose other del with other values for to determine the virtual orbits. With this will be reduced in a signifying way the number of orbits; and the sides of them will be bigger than that would correspond to orbits with N n2 = constant, in which N varies unitary values. With this last consideration, by advance can consider that the orbital electrons are affected by a longer time than an electric induction, and because this, the value of (vn) obtained first will be bigger. With a numerical example, will be evidential other facts that will be seen afterward. In the virtual orbit system will be consider the variation (reduction of the quantity of orbits) in a proportion of: n0.5) Velocity of the electron in its biggest virtual orbit: vn´ = v1 / n0.5 = 2,160,000 / (1.268 x 107)0.5 = 606.5707 m / Sec.
As was mentioned before, the definitive value of vn will be bigger that the value just obtained. α = 360º / 4 (1.268 x 107)0.5 = 0.o025275 = 2 x 0.o012637 v⊥n´/ Sec. = 2 vn Sin.(α / 2) = 2 x 606.5707 Sin.0o.012637 = 0.26757 m / Sec F⊥n . = me v⊥n´ = 9.1091 x 10—31 x 0.26757 = 2.4373 x 10—31 Kg m / Sec
In the biggest orbit there are: 4 ( 1.268 x 107)0.5 vertices the quantity of impuses produced in one rbit is: Σ F⊥n = 4 n F⊥n = 4 (1.268 x 107)0.5 x 2.4373 x 10—31 = 3.47159 x 10—27 Kg m / orbit This impulse was obtained considering the biggest virtual orbit with an electron moving at a velocity: vn = 606.5707 m / Sec.
Circumference of the biggest virtual orbit:
Lo = 2 π rm = 2 π 2.061 x 10—4 = 1.294 x 10—3 m (see forward...)
At the same distance (rvn), with the Coulomb´s law are obtained the values will be given forward. By the condition of this problem, ought to be a numerical coincidence with the fore value expressed in distance, and the action of the Coulomb´s law, as will be proved with the values obtained forward, and those used in the theme: How Are Formed the Gravity and the Coulomb´s Fields.
Time of orbit: to´ = Lo / vo´ = 1.294 x 10—3 / 606.5707 = 2.1333 x 10—6 Sec. : The impulse produced by the Coulomb´s law by two unitary charges in a given instant, at a distance: (rn), is equal: Fc = K (q1 q2) / rn2 = 9 x 109 (1.6 x 10—19)2 / (2.061 x 10—4)2 = 5.4241 x 10—21 newton.
:The exterior impulse acting during a time of orbit (to) is: Σ Fc = Fc to = 5.4241 x 10—21 x 2.1333 x 10—6 = 1.15712 x 10—26 newton / orbit . . . .
We expected that the impulses of the Coulomb´s law would be equal to Σ F⊥n, but were produced bigger than that, for the reasons were said before in this theme. There are two different forces: one that produces a kinetic effect (exterior one) and the other that produces a deformation effect ( the interior one) Some paragraphs before was consider that the value of vn = 606.5707 m / Sec., is smaller than the real one. The Coulomb´s effects are produced in a longer time than the virtual ones: Σ Fc / Σ F⊥n = 1.15712 x 10--26 / 3.47159 x 10--27 = 3.3333.
The interpretation we give to the fore result is that in the biggest shell of the virtual orbit the velocity of the orbital electron has grown at a rate greater than in the polygonal orbits, that is: vn = 3.333330.5 x 606.5707 = 1,107.44 m / Sec.
In order the virtual orbits be in accordance with the polygonal and de Coulomb´s ones, the velocity of the orbital electron in the biggest virtual orbit ought to be equal to the velocity of the smaller Coulmb´s orbit; and the velocity of the orbital electron of the smaller virtual orbit ought to be equal to the velocity of the orbital electron of the biggest polygonal orbit.
vvn = 1,107.44 m / Sec. = vc1;
nmax = (1268 x 107)0.5 = 3561 orbits.
to = Lo / vn = 1.294 x 10—3 / 1107.44 = 1.168 x 10—6 Sec.
v⊥n = 2 vn Sin(α/2) = 2 x 1,107.44 Sin.0o.012637 = 0.4885 m / Sec.
F⊥n = me v⊥n = 9.1091 x 10--31 x 0.4885 = 4.45 x 10--31 Kg. m / Sec.
Σ F⊥n = n F⊥n = (4 x 1.268 x 107) 0.5 x 4.45 x 10--31 = 6.338 x 10--27 newton / orbit
From the Coulomb´s law we have: to = Lo / vn = 1.294 x 10--3 / 1,107.44 = 1.16846 x 10--6 Sec. Σ Fc = Fc to = 5.4241 x 10--21 x 1.16846 x 10--6 = 6.338 x 10--27 newton / orbit There is a coincidence between the impulse of the biggest virtual orbit and the smaller Coulomb´s one.
In order our model of the atomic orbits be correct, the biggest virtual orbit ought to be in accordance with the smaller Coulomb´s orbit; next will be given the data of the biggest virtual orbit. vn = 1,107.44 m / Sec.; n = 2,160,000 / 1,107.44 = 1,950.44 α = 360o / 4 x 1,950.44 = 0.o0461433 = 2 x 0.o0230717 v⊥n = 2 vn Sin (α/2) = 2 x 1107.44 Sin 0.o0230717 = 0.80269 m / Sec. F⊥n = v⊥n me = 0.80269 x 9.1091 x 10—31 = 7.3118 x10—31 newton Σ F⊥n = F⊥n 4 n = 7.3118 x 10—31 x 4 x 1950.44 = 5.7045 x 10—27 newton /to to = Lo / vn = 1.294 x 10—3 / 1107.44 = 1.16846 x 10—6 Sec.
The biggest virtual orbit has a radius: rn = 2.061 x 10—4 m, in the orbit number: n = 1950; a side has a length: ln = Lo / 4 n = 1.294 x 10—3 / 4 x 1950 =
1.66 x 10—7 m The space between the virtual shells is: d = 2.061 x 10—4 / 1950 = 1.057 x 10—7 m. ≈ rv1
Here we will consider that with the Coulomb´s law can continue applying the rule: N n2 = constant, but now admitting that N can be a quantity smaller than one; that is, considering that the Coulomb´s law works with microcorpuscles in applying the mentioned rule; in this way the biggest distance will be defined by one microcorpuscle = m⊥n. In order to continue with the same law of variation of mass of the propeller particles, will consider a fraction of corpuscle, that really is a fraction of 1.885 x 107 = mg / mm = 1.47236 x 10--50 / 7.811 x 10--58 negative microcorpuscles. For continue with the same process was done with the virtual orbits; for the Coulomb´s orbits will be established the proportion: N n2 = 3.282 x 1012 x 72 = 1.6082 x 1014 x 12 = 1 n2 = (1 x 1.885 x 107) r> . biggest Coulomb´s orbit.
n> = (1.6082 x 1014 x 1.885 x 107)0.5 5.506 x 1010 . The dimension (radius) of the maximum Coulomb´s orbit, that also is equal (?) to the return distance of the negative microcorpuscles is: r> = nr7 / 7 = 5.506 x 1010 x 1.13763 x 10—10 / 7 = 0.895 m.
Apparently there is a coincidence between the return distance and the intensity (of 1 h). Several, or many charges can produce a higher intensity than 1 h,, but not in the same line of action....
Monterrey, México, September 18, 2001 Manuel de Hoyos Robles
January 16, 2007
Con el tema: Sound Velocity (II) no nada mas se ha comprobado la velocidad del sonido, sino la validez de mi modelo de átomo con órbitas poligonales y el de la corriente eléctrica; con puros valores numéricos de macanofísica voy a comprobar esto: Una multimolécula se produce en un tiempo de tm = 28 x 10—5 Seg.; la órbita de un átomo tiene 28 lados; la influencia energética de un átomo afectado a los demás átomos de una mutimolécula es; no = (28 / 2.8 x 10—5)0.5 = 100, Un tiempo de deflexión es igual a t> = 4.47 x 10—20 Seg. Un átomo de multimolécula es afectado por un electrón de corriente eléctrica durante un tiempo de brinco tj = 1.12 x 10—11 Seg., o sea por el siguiente número de tiempos de deflexión: n1 = 1.12 x 10—11 / 4.47 x 10—20 = 2.5 x 108 efectos de deflexión. Estos efectos distribuidos en los átomos de una multimolécula nos dan: nm = n1 / no = 2.5 x 108 / 100 = 2.5 x 106 = número de átomos que tiene una multimolécula
Con el tema: Velocidades de Rechazo y de Admisión, complemento las leyes de gravitación universal de Newton. Con únicamente estos dos temas, entre otros muchos, demuestro de una manera irrebatible que la física moderna está equivocada. De cualquier manera le doy una honrosa despedida de una física óptimamente constituida,, por su buena voluntad, a Enstein y a todos los que contribuyeron en su estructuración a la física moderna; porque el progreso de una ciencia (y todos los progresos...) se han hecho a base de aciertos y fallas; desde luego tratando siempre de eliminar a estas últimas.
MECHANOPHYSICS Ib
THE COMPTON’S EFFECT (checked)
In a binary system in which the propeller particle m⊥ moves in the same direction than the axis of it, when moves along the diameter of the positive particle and of the orbital electron, this movement is appreciated as along practically straight lines, that are the axis of the binary systems ; this could be perceived in an easy way, because the position of all the interacting parts, but now suppose that a propeller particle different to that corresponding to the binary system is introduced in other direction, into the orbital electron, here will be some variation in the normal behavior of the binary system; but before we continue it is required to define what is a propeller particle. We can imagine it as a unitary particle moving along the diameters of the particles of the binary system, when the particle is into the binary system, this is all right, but if the orbital electron jumps to an interior shell; it is because the propeller particle acts as impulse fluid, this propeller particle is ejected out of the electron and afterward is spread open forming a ray. The fore means that the propeller particles could be represented by a unitary particle and by a spread open ray, that if introduced into the electron, gets integrated forming a unitary particle.
In the return distance effect there are many similitude between the interaction of the receding and the advancing corpuscles (that form the cosmic rays) , that intersect in the return distance effects, and the Compton’s effect; so we consider those phenomena as a variation of one, as is interpreted by mechanophysics. Using radiation as incident beam of monochromatic X ray to fall in a sample of scattering material (loose carbon particles), he examined the scattering radiation with a X ray spectrometer (an instrument which measure the corpuscular separation –wavelength- of X ray incident in it). In addition to the radiation at the same wavelength, as the incident X ray, he also found a scattered wavelength λ´ greater than that of the incident beam: wavelength λ’ greater than that of the incident beam: Δλ = λ´ -λ = h (1 – Cos Θ) / 0.5 m c2; m = mass of the incident ray (mass of all its corpuscles); we will talk of this forward...
In other themes have been give a astronomic model to explain how are integrated the atomic particles with the intersection of a ray of corpuscles, that after it reaches the return distance point, begins its receding trajectory and meet with another similar ray of corpuscles that is moving (advancing) toward the mentioned point. When both rays meet, they have the same velocity but in an opposite way. The corpuscles of the advancing ray have more remanent gravity than those of the receding ray; when an advancing corpuscle meets with a receding one, the first mentioned corpuscle will acts as if were a propeller fluid particle, because this yields its remanent gravity to the receding corpuscle, and with this remanent gravity the receding corpuscle will inverse the direction of the kinetic energy that had the advancing corpuscle, as if it were a reflection process, in such way that the receding corpuscle is not affected in its return velocity, after the intersection; this process is verify with all the corpuscles of both rays, in a similar way, so that the returning ray, beside does not lost its returning velocity, gets integrated in a single mass formed with all the corpuscles of both rays; really this is not a new process that was evidenced; this is the same mechanism that was given in our model to explain the movement of the propeller particles in the binary systems (see theme: The Return Distance in the Atomic Particles).
I said that the ideas mention in the fore paragraph are not new from the point of view of mechanophysics, also I am to dare or venture to say that neither to quantum physics, because in both physics there is the inquietude or restlessness to find a mechanism to explain the conservation of energy and mass into the atomic particles. But in mechanophysics there is not only try to find a different mechanism for different phenomenon, but also to have a connection and to harmonize them, one with each other. Here I will make a healthy critic of the quantum physics, because they have pretend to give elastic properties to all the atomic particles in a generalize way, but the elastic property is only one of the properties that have the particles in order could produce the interchange of energy without any lost of it. The idea got since Wolfgang Pauli and before him that the interchange of kinetic energy between one atomic particle and another one, are produced as if the particles were as tennis balls or billiard ones, can not have a fundamental value, because ignore in a more complete way the atomic structure. Between this idea of Pauli and the neutrino and that of Yukawa of interchange of matter or energy into the nucleus, there are many contradictions points, in some way here we have given them, so we will not have to get outside of this theme to explain them. Insisting: We can not work with the atomic particles as if they were tennis balls, although these balls are much more voluminous than the atomic particles, they do not have a mechanism to transmit their kinetic energy in form of inductive fluid to the other particles with which they bump. Because all mentioned this theme has a fundamental importance in mechanophysics and in all physics.
In one of our models of binary systems, the positive particle and the negative one (the orbital electron), there is an interchange of a propeller particle (mp). It is considered that this particle moves at light velocity between both binary particles; nevertheless, when it is absorbed by one of the particles, it is not produced any reflection effect that will try to reject one particle of the binary system to the other one. We suppose that this happens, because when the propeller particle (mp = m⊥) penetrates either in the positive particle or in the negative one, it is in the front side of them (parallel and collineal with the axis of the binary system). We imagine this diameter in which move the ppropeller partuicles as a hollow tube with equal length to the diameter of the particle (positive, or negative): In this tube is not produced any bump effect between the propeller particle and the binary system particle (+ or -); only is produced an interchange of inherent energy between both particles, in such way that when the particle mp has moved over one of the diameter of the binary system particle, this last particle is affected by a fluid whose energy is equal to: 0.5 mp c2, acting toward the other particle of the binary system. In normal conditions, the fore fluid is produced in an internal way that is not appreciated outside of the particle. In the case of an ionization (for any reason could be considered), the fluid is manifested in an exterior way, as a radiation. In our theory of the polygonal orbits the fore mentioned fluids act as have been explained in such theory.
When a radiation penetrates in a refringent medium, it continues moving along it with smaller velocity... When the radiation penetrates to an opaque medium only introduces a small space. Here all the corpuscles join, forming as a propeller particle that can penetrate to a binary particle, for instance, to an orbital electron: But the penetration not always is along the axis (or tube) of the particle. In the annex figure is indicated an orbital electron (in the 7 shell). Here can be disregarded the translation velocity (v7), compared with the radiation velocity (c ). The axis of the electron is indicated by the line: aOb. In the left side of the electron is indicated an approaching radiation with a frequency (ν): The force of the incident radiation is: h ν . This radiation in form of propeller fluid penetrates in the electron, and when meets with its axis having an angle Θ ; it will acts in two ways: 1) Will move the electron with an impulse: (h ν) Sin Θ, in the direction Od; the rest of the radiation: (h ν) Cos Θ will act as a propeller fluid that will be ejected by theorbital…….electron and at the same time will move the electron with the same (but opposite) imulse in the direction (Oa).
The vectorial sum of the two movements of the electron mentioned before will give an impulse equal to that of the incident radiation (h ν / c). Here we have the following values: (h ν / c) Cos Θ = h ν´/ c = to the impulse of the ejected fluid; h ν / c = to the impulse of the approaching radiation. The corpuscular separation of the incident ray is: λ = c / ν. ν = quantity of corpuscles / second, c = light velocity.
The corpuscular separation of the ejected ray is: λ´ = c / ν´ = c / (ν Cos Θ).
There is a proportion between the fore values:
λ : c / ν :: λ´ : (c / ν) Cos Θ; λ´-λ = (c / ν) (Cos Θ −1) . . . . (A)
In accordance with our model a part of the radiation: (ν h Sin.θ ) will act moving the electron in the direction O d. The other part ( ν h Cos. θ ) will act as a propeller particle (mp) that will be ejected by the electron in the direction: O b. We have::
0,5 mp c2 = ν h = c h / λ ; λ = h / 0.5 mp c ; λ´ = [ h / (0.5 mp c) ] Cos.θ
–
λ´ -λ = ( h / 0.5 mp c) [ (1 / Cos. θ)1] ≈ (h / mp c ) [ 1 – Cos θ ] ; when: Cos.θ≈ Sen θ
In the experiment of Compton with the X rays radiation with corpudcular separation λ. changed
after being scatered to λ´ . In accordance with the interpretation was given to this, was obtained the
following formula:
λ´ --λ = (h / m c) (1 – Cos θ) . . . . (B)
In this formula: λ´ = h / m c Cos.θ, and λ = h / m c
This is absurd, because: λ´ = (c ) / ν Cos.θ, and l = c / ν ´
.electron and at the same time will move the electron with the same (but opposite) imulse in the direction (Oa).
The vectorial sum of the two movements of the electron mentioned before will give an impulse equal to that of the incident radiation (h ν / c). Here we have the following values: (h ν / c) Cos Θ = h ν´/ c = to the impulse of the ejected fluid; h ν / c = to the impulse of the approaching radiation. The corpuscular separation of the incident ray is: λ = c / ν. ν = quantity of corpuscles / second, c = light velocity.
The corpuscular separation of the ejected ray is: λ´ = c / ν´ = c / (ν Cos Θ).
There is a proportion between the fore values:
λ : c / ν :: λ´ : (c / ν) Cos Θ; λ´-λ = (c / ν) (Cos Θ −1) . . . . (A)
In accordance with our model a part of the radiation: (ν h Sin.θ ) will act moving the electron in the direction O d. The other part ( ν h Cos. θ ) will act as a propeller particle (mp) that will be ejected by the electron in the direction: O b. We have::
0,5 mp c2 = ν h = c h / λ ; λ = h / 0.5 mp c ; λ´ = [ h / (0.5 mp c) ] Cos.θ
–
λ´ -λ = ( h / 0.5 mp c) [ (1 / Cos. θ)1] ≈ (h / mp c ) [ 1 – Cos θ ] ; when: Cos.θ≈ Sen θ
In the experiment of Compton with the X rays radiation with corpudcular separation λ. changed after being scatered to λ´ . In accordance with the interpretation was given to this, was obtained the following formula: λ´ --λ = (h / m c) (1 – Cos θ) . . . . (B)
In this formula: λ´ = h / m c Cos.θ, and λ = h / m This is absurd, because: λ´ = (c ) / ν Cos.θ, and l = c / ν ´ ...electron and at the same time will move the electron with the same (but opposite) imulse in the direction (Oa).
The vectorial sum of the two movements of the electron mentioned before will give an impulse equal to that of the incident radiation (h ν / c). Here we have the following values: (h ν / c) Cos Θ = h ν´/ c = to the impulse of the ejected fluid; h ν / c = to the impulse of the approaching radiation. The corpuscular separation of the incident ray is: λ = c / ν. ν = quantity of corpuscles / second, c = light velocity.
The corpuscular separation of the ejected ray is: λ´ = c / ν´ = c / (ν Cos Θ). There is a proportion between the fore values: λ : c / ν :: λ´ : (c / ν) Cos Θ; λ´-λ = (c / ν) (Cos Θ −1) . . . . (A).
In accordance with our model a part of the radiation: (ν h Sin.θ ) will act moving the electron in the direction O d. The other part ( ν h Cos. θ ) will act as a propeller particle (mp) that will be ejected by the electron in the direction: O b. We have::
0,5 mp c2 = ν h = c h / λ ; λ = h / 0.5 mp c ; λ´ = [ h / (0.5 mp c) ] Cos.θ
–
λ´ -λ = ( h / 0.5 mp c) [ (1 / Cos. θ)1] => (h / mp c ) [ 1 – Cos θ ] ; when: Cos.θ≈ Sen θ
In the experiment of Compton with the X rays radiation with corpudcular separation λ. changed
after being scatered to λ´ . In accordance with the interpretation was given to this, was obtained the
following formula:
λ´ --λ = (h / m c) (1 – Cos θ) . . . . (B)
In this formula: λ´ = h / m c Cos.θ, and λ = h / m c
This is absurd, because: λ´ = (c ) / ν Cos.θ, and l = c / ν ´
SOME RETURN DISTANCE EFFECTS (checked)
Since I was a small boy I had singular conversations with my father about the nature of the structure of the Universe. He believed that the same laws that are applied to the macroscopic bodies, also are applied to the atomic particles ; for instance,, as he was supporter of the corpuscular theory of light, he considered that when a ray of light enters to a refringent medium its velocity does not change, but its trajectory became a zigzag one. He also considered that the radiation emitted by all the stars, including our Sun, in some way returned to the emitter body, this because after millions years had a reflection effect in the sidereal space, or because the Sun moving around our galaxy, or our galaxy around other ones, dragged such radiation, making their trajectory as a closed curve, etc. About the gravity and electric fields he considered similar conditions ; for him they were not fields of forces, but fields that induced forces into the atoms they affected, this because these ones were as rockets, that when affected by the mentioned fields, oriented theirs “jet motors” in accordance with the action of the mentioned fields. He was in accordance with the movement of the orbital electrons around the nuclei of the atoms, but not in a curved trajectory, because the great centrifugal forces, but in some polygonal orbits ; about this considered that in the atomic magnitudes were produced some changes in a rapid way, but not in a brusque and instantaneous one, as consider the quantum theory, but by small and rapid steps.
Maybe I was wrong, but all these and other ideas were assimilated by me in such way that I never believed in those of modern physics, as definitive ones ; either my father and me believed that if our ideas were not accepted by most physicists was not because the theories were wrong, but because most physicists have a conservative spirit ; since Galileo and Copernicus’ epochs had happened so. In my father’s epoch it was hard to divulge new ideas for a person with limited mediums or economic resources, because most of the physicists had a restrained mentality, as that of the physicist mentioned in the epilogue, but now with the Internet the situation has changed. When I was a young boy, in some way were an impulsive person, because this, I think I would not be a good student of physics. Now I think that a good student of the contemporary physics, could not be able to make a work as this one. I consider absurd and hurtful the idea some of them have for to defend their ideas as long as possible and at all intellectual cost (see commentaries with cursive letters in the theme : Super fluid Helium 3) ; nobody is owner of the knowledge of the Universe ; all us ought to be opened to the new ideas.
In the first part of this work my main interest was to divulge the idea of a model of a structured atom formed with binary systems, and when I began this task had not a clear idea of many important facts, as the ionization of atoms, the modulus of elasticity and other ones ; nevertheless, with my limited knowledge could advance till I got an idea how could be made an atom by uniting several binary systems ; but the medullar problem now is : How can be united such binary systems in a practical way ? If this were easy with our present knowledge, we could make precious metals as gold and platinum. We ignore many facts, Fermi knew how introduce low energy neutrons into the nuclei of the atoms, and because this, we think that in the future it will be easy to introduce the positive particles of the binary systems in the nuclei of the atoms, and with free electrons, form new binary systems ; maybe a neutron is formed by a proton and an electron fusion in one particle ( ?). But all these are problems corresponding to the first part of the protocol. Now we will try to make the first intent for to aboard the problem of the formation of electrons, positrons, protons and neutrons with corpuscles as those of light. When was obtained some information about the cosmic rays, with it we got aquatinted of the hypothesis of Millikan, than the particles of such rays are formed by corpuscles as those of light ; with this are confirmed some ideas we had before ; if you can make a house with bricks, an atomic particle could be made with corpuscles as those of light ; by the moment we will not pretend how is that one corpuscle unites to other one, a third to the two mentioned before, etc., but at least it will be tried to obtain a law can be applied in these processes.
Has been consider that the atomic particles could be formed because the intersection of the emitted radiation of our Sun, with the returning ones. The first ideas we get in this are: What velocity have the interacting particles? How much remnant gravity have? How interacts one particle with respect to the other one when they intersect ? How long time can live or remain a particle that is integrating, in accordance with the conditions in which is effectuated the integration ? etc. Some of the fore question can be answered with observation data ; unfortunately we have no access to them ; but if the behavior of the particles is governed by the same laws, even into the atomic particles than in the free spaces, we can obtain some important data from the atomic particles and apply them to the interstellar space movement of such particles. Those that were supporter of the theory of the expanding Universe and made astronomic observations, must know in which magnitude varied the velocity of the light corpuscles, due to the effect of the return distance ; some years ago I read in some place that the far celestial bodies were receding at a velocity manifested by a variation about 6 % of (c) ; unfortunately now that I need this value for my investigation work I was not able to find it ; but if we divide : c / 16 = 18,750 Kms./ Sec, we obtain a value of 6.25 % of c, that fix very well ; If it is considered that at : 5 x 1010 light years of distance = 1.58 x 1018 light seconds of distance (see theme : Return Distance) there are galaxies that recede at light velocity, in accordance with the theory of the expanding Universe, then the average velocity of light will be: c / 2 = 150,000 Kms./ Sec.
Lr = 1.58 x 1018 x 150,000 = 2.367 x 1023 Kms.; 2 Lr = 5.73 x 1023 Kms. (of receding and return distance)
With respect to our solar system it is not easy to determine the fore values, because there are intersection between the emitted and the returning corpuscles, in such way that with this intersections are formed the cosmic ray particles. Now our direct interest is to found how are formed such particles, but if when we were investigating in the first part of this work we had many doubts, now we have more than then and this is because the first part has many ignored points yet. There are two ways to get knowledge, one is by accepting them because are given by authorized investigators ; the other one is not only because the fore reason, but also because they can support an exhaustive logical analysis ; as the first one is the most easy way, most people accept it. Generally the progress of science has been attained because parting of known facts are obtained unknown or ignored ones ; in this theme we do not see that in a direct way can be obtained known facts that will help to attain unknown ones. Here will work in an opposite way, not in a direct one. We will make some intuitive suppositions, not scientific ones, and parting from them will try to arrive to true facts. It is logic to suppose that with the intersection of corpuscles the first atomic particles that are formed, are the smaller ones, as the electrons ; if the integration of corpuscles continue, are formed the positrons, next the protons and finally the neutrons.
Repeating : here will not explain how are formed the cosmic ray particles due to our ignorance in this field ; nevertheless, and in accordance with the return distance effects will be obtained some interesting conditions that eventually will help to attain positive results. We consider it is not necessary that the corpuscles travel distances of the magnitude of 5 x 1010 / 16 light years for to obtain some evidences of the return distance effects ; in our solar system can manifest such evidences. In this theme will be seen in a ratter semi empirical way a very much fundamental point. In all the theories have been see here of mechanophysics, always have been consider that either the gravity microcorpuscles, as the light corpuscles move at light velocity in their normal behavior and also have been seen that in some conditions, such particles could reduce that velocity, as in the return distance effects (astronomic and atomic ones), and do not doubt that such velocity could increase in other phenomena. Based in that gravity microcorpuscles and light corpuscles move at light velocity was consider the action of the propeller particles m⊥7, in the binary systems. Here in the Earth the fore consideration is satisfactory, but in the exterior planets of our solar system can not be sure that the gravity (of the Sun) has ( c) velocity, but c-Δc, and maybe a reduction of Δc could affect the movement of the propeller particles along the diameter of the binary particle in such way that are not produced the required deflections in the polygonal orbits in such way that all the molecules are affected and even destroyed. I consider that in the future will be require to write some themes about velocities slower than light velocity acting in the m⊥n particles.
Monterrey, México, January 4, 1999 Manuel de Hoyos Robles
PRESSURE AND TEMPERATURE (I) (corrected theme)
In this theme we will deal with great pressures enough to reduce the dimensions of the shells of the atoms, and these pressures only could be obtained at the great depth (diameter) of the celestial bodies as the stars and the planets. For to obtain the pressure over an atom that is near the center of the Earth, practically over it is acting all the matter corresponding to a pyramid, in this pyramid there are many variations data as density, liquid or solid states, the variation of gravity acting in its different caps, the interaction between the sides of the pyramids with the adjacent ones, etc.. Of course in a detailed and good study of this problem it would be necessary take in account all them; here will not pretend to do this, but only to have an idea of the problem, and this is enough, because some of the mentioned values vary with the time and other facts, in magnitudes greater than the approximations we can have. In the Sun it is considered that its temperature in its surface is of 6000º K, we can accept this value because is in accordance with the radiation it emits; in accordance with some theories or hypothesis (?) it is considered that in the nucleus its temperature is of: 12,000,000o K. If we accepted the first mentioned temperature for the given reasons,, we can not do the same with the last one mentioned. If we pretend to determine all the history of our Sun or of any star, it is necessary to define a principle and an end of them. In the infinite time of the existence of the Universe we can assure that all the celestial bodies are formed with the same matter, that millions of years the Earth was a part of our Sun, in this way we represent with anticipation the end will have our Sun when its great temperature gets exhausted; and also will represent the beginning of other solar system, if it is consider that a star born because two inert huge masses collide one with each other, or because any other reason the inert matter gets activated. In few words, the origin begins with inert matter and finish also with inert matter. But the fore affirmation is outside of our deduction capacity, because if it is considered infinite time and infinite distance, we are not able to define the beginning of the things neither the end of them; we can imagine an integration phase of our solar system due to some atomic or nuclear effect, our Sun can exploit and expel a great quantity of matter at the same temperature it had then; some of this matter, because their condition of kinetic energy and the attraction of the Sun, do not escape from it, neither return to it, but is transformed in satellites or planets; in millions of years the temperature of these planets diminish from its surface toward their nuclei.
All the fore circumstances were obtained parting from the condition we have in the present time of our solar system, and here it is pertinent to make a question: At the end of our solar system, will our Sun gets as an inert body with all its heat exhausted? If we accept in the Universe there is no lost of matter neither of energy and that with the return distance effect all these is recuperated after millions years, these incline us to consider we can not see an end of the radiation of the stars, including our Sun. The idea of an exhausted star, that does not emit any radiation, could not be probed in the practice, because if in the Universe there are stars in this condition, we are not able to appreciate them, because the only way we have now to do this, is by the radiation they emit. In this theme it will be seen that heat can be produced by pressure, and if this is correct, in the stars there is great pressure in their interior caps and because this, the accepted concepts of the evolution of the stars will have to change. With the ideas we obtain here it will be possible to define how our Sun got its great temperature, but for this we will have to part from the initial condition that had our Sun, of a celestial body in an inert state (low temperature), and not as is now; from the inert state it is possible to arrive to the present state by the increments of heat produced by the great pressure on its matter.
In other themes of this work it has been mentioned that the idea to consider that temperature is proportional to the second power of the velocity of the particles that are on the medium were the temperature is measured. But we pointed out that this is not true, because at zero K temperature, the orbital electrons into the medium move at their great velocity around the nuclei of their atoms.
From the fore explanation it is evidential a question: How it is possible that the orbital electrons move at their great velocity without producing heat? Since was made the model of the polygonal orbits of the atoms, first was consider that any orbital electron deflects in each vertex of its polygonal orbit due to the effect of a propeller particle m⊥7 ; this particle has a potential energy equal to: K = 0.5 m⊥7 v72 (see theme: The Return Distance in the Atomic Particles). In our model of the orbital electron it was considered that there was not any lost of energy, but only a transfer of it from a part of the particle to another one of it, and from this to the first mentioned, and so on, in this way it was consider there is a similitude between the electron and the Universe, because in this one we can consider no lost of kinetic energy but an interchange of it between the particles and bodies.
We can imagine a model in which the propeller particle m⊥n has a potential energy equal to:
0.5 m⊥7 c2, and the electron has other potential energy of the same magnitude. The propeller particle m⊥n transmits its energy to the orbital electron, as was explained on the mentioned theme. Either in classical as in modern physics has been consider that if a particle or a body that has a lineal impulse, can get a kinetic energy along a straight line for indefinite time, if the particle or body that has a lineal impulse, can get a kinetic energy along a straight lime for indefinite time if there are not resistance forces. With these examples and others can be given, we get a clear idea how a kinetic energy of a particle or of a body can remain for ever if there are not other energies that oppose its movement. In the Universe there is no lost of energy, but an interchange of it between the particles and bodies that are affected by the mentioned energy. Into an electron, a proton or any atomic particle, also we are forced to consider there is no lost of energy, but only an interchange of it, in order such particle work as have been observed; the way this interchange of energy is produced is understand in a very clear way if it is considered the return distance effects
With the fore explanations it is easy to understand that if the energy into the orbital electron is only manifested into it and does not affect other exterior particles, it will not be manifested any production of heat or any other exterior energy. In the surface of our Earth can be consider some exterior effects acting over the atoms, as the atmosphere pressure, but these effects can be disregarded if are compared with the kinetic energy can have the orbital electrons, so that never have been taken in account; these beside that there have not given a relation between pressure and temperature or heat. As is well known, the temperature have been obtained in function of the velocity of the particles in the medium is considered. If we analyze the problem from its most profound roots, and in accordance with the accepted theories, we can imagine the most perfect equilibrated Universe, in this there is no movement of any particle, except those into the atoms; maybe this will be the idea could be get of the origin of the Universe. In this perfect equilibrated Universe we can imagine that for an unknown reason is produced a minimum unstable equilibrium in a minimum particle; this unstable equilibrium will be transmitted to the adjacent particles, and so on, and after certain time we will have a Universe with many particles and bodies moving one with respect to each other, here also will have an equilibrated Universe, but not a static equilibrated one, but a dynamic equilibrated one.
In mechanophysics we prefer not to see the problems from their most profound roots, because we are not able to imagine an infinite distance, an infinite time, etc. We know that the orbital electrons move at great velocity around a point in the axis of a binary system, also the positive particles move around such points, all in accordance with the theory of the polygonal orbits, that is, that in each vertex of such orbits acts the propeller particles m⊥n, as we have say, interchanging kinetic energy with the orbital electron (or with the orbital positive particle). All right, this is very much correct; but now suppose that we can place the atoms in a medium in which the interchange of energy between the propeller particle m⊥n and the orbital electron is not produced. How can be produced this? This can be produced if in some way the propeller particle m⊥n does not interchanges its energy with the orbital electron, this will happen when in the medium there is a pressure equal or equivalent to the energy with which the propeller particle m⊥n is able to deflect the orbital electron in each vertex of its polygonal orbit; at the same time that is determined the mentioned pressure that we will call: propeller pressure we will determine the temperature that is produced by such propeller pressure, and that we will call: propeller temperature.
The fore problem will be aboard with a numerical example; we will consider a medium in which the atoms have such pressure that is equivalent to that of the propeller fluid m⊥n deflecting the orbital electron of the seven shell. The fore condition could be produced at certain depth of a planet, in which a cap of atoms with a height equal to the fore depth, are affected by the gravity of such planet; so that the weight of the cap produces a pressure equivalent to the effect produced by the propeller particle m⊥7 over the orbital electron. In our planet we can consider a density: w =
5.5 grs./ cm3 = 5,500 Kgs./ m3. An orbital electron is deflected in a vertex with an energy equal to: K⊥7 = 0.5 me v⊥72 = 0.5 x 9.1091 x 10--31 (6.91 x 104)2 = 2.1747 x 10--21 joule
If we consider that every time of vertex is produced the fore energy in one orbital electron of the atom, and that this has two orbital electrons in its seven shell, then the fore energy is equivalent to a force F acting in the surface of the atom, in such way that that force can oblige the orbital electrons to deflect in the vertices of their orbits (7 orbits). As the fore energy is produced every time of vertex (ty) in each electron, this would be equivalent to the mentioned force F acting in the surface of the atom, each second.
In some depth in the celestial bodies the pressure is so big that even the shells of their atoms cam be destroyed. Here will be seen some conditions to realize this. A celestial body in form of sphere could be divided in segments of pyramidal form, as show in the figure (1). All these segments coincide in the center of the sphere in their vertices (O), and with their triangular lateral sides with their adjacent ones. All the surfaces Ah if all the pyramids form the surface of the sphere. The height (h) is equal to the radius of the sphere. Parallel with surface Ah is the surface Ay at a distance (y) from the vertices (O) of the center of the sphere.
With the fore pyramid can be determined the different pressures can be produced at different depths of the spherical celestial body. For instance, in the surface Ay is acting all the volume of the pyramid corresponding to the volume there is between the surface Ay and the surface Ah: With a mathematical criterion, near vertex (O), (y) is a small quantity, also Ay; so the fatigue supporting by Ay tends to an infinite value. In this condition all the spherical celestial bodies in their nuclei can have a pressure enough to compress all the shells of the atoms. Based in the fore mentioned ideas will be determine the required pressure acting in one orbital electron. With the mathematical criteria can consider that with the same pressure it is possible to compress all the orbital electrons of each shell. But this is not so easy, because with the pressure is destroyed one shell, and all atoms of this shell get free with their great velocity and kinetic energy, opposing the action of the pressure. In this condition the solid particles get in a liquid state, and afterward in a gaseous one, as in the Sun. This by one side, by other side there is a limit in the mass of the celestial bodies to grow; but this is problem of other theme. Any way, here will be seen the problem in the most simple way, with some numerical examples. In Fig.2 of the theme: Hydrogenised Model of Atom, or: A Model of
Electric Current and its Application to Medicine, is represented with (z) a vertex of the seven shell;
the length of each side of the polygonal orbit is equal to: d7 = 2.55 x 10---11 m, its deflection angle;
δ = 360º / 28 = 12.o86, and the velocity of the orbital electron: v7 = 308,570 m / Sec., with all
these data is obtained the radial velocity: v⊥7 = 69,100 m / Sec., and the deflected distance: d⊥7 =
5.7102 x 10--12 m. (= 2 Lv Sin [360o / (28 x 2)]
Transforming the value: K⊥7 to Kg m by dividing it by (g) is obtained:
K⊥7 = 2.1747 x 10--21 / 9.81 = 2.21682 x 10--22 Kg m / one vertex
As was say before, if the atoms are in a medium in which there is a pressure enough big to nullify the
effect of m⊥7 over the orbital electron, then the force F that will produce this will be with an energy
in each vertex, as indicated in the following formula:
F7 d⊥7 = 5.7102 x 10--12 F7 = K⊥7 = 2.21682 x 10—22 Kg m2 vertex
F7 = 2.21682 x 10--22 / 5.7102 x 10--12 = 3.8822 x 10--11 Kg / one vertex
Next it is going to determine the force F over the surface of the atom; each atom occupies a surface equal: A7 = (2 r)2 = (2.27525 x 10--10)2 = 5.17676 x 10--20 m2 The resistance fatigue of the seven orbit :f7 = F7 / A7 = 3.8822 x 10--11 / 5.17676 x 10--20 = 7.5 x 108 Kgs./ m2 = 75,000 Kgs./ cm2 In the fore numerical example was given the fatigue required to nullify the effect of the radial velocity of the orbital electron of the seven shell; but if the deformation of the atom almost is reduce to the size of the six shell (without affecting this shell) v⊥6 = 9.398 x 104 m / Sec.
The energy of the orbital electron is:
K⊥6 = 0.5 me v⊥62 = 0.5 x 9.1091 x 10--31 (9.398 x 10 4) 2 = 4.022688 x 10--21 joule = 4.022688 x
10—21/ 9.81 = 4.1006 x 10--22 Kg m
d⊥6 = 2 x 2.55 x 10--11 Sin 7.o5 = 6.6568 x 10—12 m A6´ = A7 x 62 / 72 = 5.17676 x 10—20 x 36 / 49 = 3.8033 x 10—20 m2 d⊥6 F⊥6 = 6.6568 x 10--12 F⊥6 = K⊥6 = 4.1006 x 10—22 Kg m F⊥6 = 4.1006 x 10--22 / 6.6568 x 10—12 = 6.1600 x 10—11 Kg./vertex f 6 = F6 / A6´ = 6.1600 x 10 --11/ 3.8033 x 10--20 Kgs./ m2. = 1.62 x 109 Kgs./ m2
As the deflection of the orbital electron in each vertex of its seven orbit can be effectuated due to the pressure of the Earth at the depth (h’), then the orbital electrons of the seven shell can move as free electrons; so in accordance with the Boltzmann’s formula have: K = 0.5 me v72 = 0.5 x 9.1091 x 10--31 (3.0857 x 105)2 = 4.3366 x 10--20 joule
T = 2 K / 3 k = 2 x 4.3366 x 10--20 / 3 x 1.38 x 10--23 = 2,095º K
In figure (1), we have: volume of pyramid with height (h): Vh = Ah h / 3; Ay = Ah y2 / h2;
volume between area Ah and Ay: Vh-y = Ah h / 3 – Ay y / 3 = (Ah / 3) (h – y3 / h2)
Δm = average density of the material:
Pressure in point (y) ≠ 0:
fy = Δm Vy / Ay = Δm (Ah / 3) (h – y3 / h2) / (Ah y2 / h2) = Δm / 3 (h3 / y2 –y) Kgs / m2 . . . (1)
For the Earth planet: r = h = 6.375 x 106 m,; Δm = 5,520 Kgs./ m3. In accordance with the theory, when the atoms receive a fatigue: fy = f7 = 7.5 x 108 Kgs./ m2, their volume is reduced: V6 = V7 (6 / 7)3 = 0.6297 V7. The Earth’s material; this is of lower density in the surface of the planet and grows as the depth grows. We are interested to know at which depth the
fatigue is: fy = f7 = 7.5 x 108 Kgs./ m2 . From the surface of the planet to a depth = h – y, the
pressure grows, and also the density; can be take as the average one, the fore mentioned.
fy y2 = f7 y2 = 7.5 x 108 y2 = (5,520 / 3) [ (6.375 x 106)3 - y3] . . . . (1)
407,608.7 y2 = [2.591 x 1020 – y3] y3 + 407,608.7 y2 = 2.591 x 1020; y = 6.2424 x 106 Depth in
whish is produced the mentioned fatigue:
h – y = 6.375 x 106 – 6.2424 x 106 = 132,600 m
If with a pressure f7 are destroyed the seven shells of the atoms, the electrons corresponding to such
shells get free and each one of them will produce an energy equal to:
K7 = 0.5 me v72 = 0.5 x 9.1091 x 10—31 x 308,5702 = 4.3366 x 10—20 joule In accordance with the
Boltzmann´s law, the fore energy is equivalent to a temperature equal to:
T7 = 2 K7 / (3 k) = 2 x 43366 x 10—20 / 3 x 1.38 x 10—23 = 2,095º K
Now we want to know the depth at which the fatigue is: f6 = 1.62 x 109 Kgs./ m2.; size of the
atoms equal to that of the sixth shell:
f6 y2 = 1.62 x 109 y2 = 5,520 / 3 [ (6.375 x 106)3 – y3]
880,434.8 = [2.591 x 1020 - y3]
y3 + 880,434.8 y2 = 2.591 x 1020; y = 6.094754 x 106
Depth of charge (y): = 6.375 x 106 – 6.094754 x 106 = 280,246 m.
K6 = 0.5 me v62 = 0.5 x 9.1091 x 10—31 x 360,0002 = 5.9027 x 10--20 joule
T6 = 2 x 5.9027 x 10--20 / 3 x 1.38 x 10—23 = 2,652o K
If we want to know at which depth the pressure of the Sun is equal to: f1, that is reduce the volume
of the atoms to that of the first shell, in accordance with the model given here, will obtain the
condition of the first shell (see Fig,2)
v⊥1 = 2,160 / Cos.45º = 3,054.7 Kms./ Sec.
r1 ≈ 1.13763 x 10—10 / 7 = 1.6251 x 10—11 m
A1 = (1.6251 x 10—11 x 2)2 = 1.05649 x 10—21 m2
K⊥1 = 0.5 x 9.1091 x 10—31 x (3,0547 x 106)2 = 4.25 x 10--18 joule = 4.3323 x 10—19 Kg. m
d⊥1 = L> / Cos.45º = 2.55 x 10—11 / 0.7071 = 3.6063 x 10—11 m
L1> = 2.55x 10—11 = length of a side of the polygonal orbit
F⊥1 = K⊥1 / d⊥1 = 4.3323 x 10—19 / 3.6063 x 10—11 = 1.2013 x 10—8 Kg.
fy = f1 = F⊥1 / A1 = 1.2013 x 10—8 / 1.0565 x 10—21 = 1.13706 x 1013 Kgs./ m2
?
Acceleration of gravity in the surface of the Sun: gs = G Ms / rs2 = 6.673 x 10—11 x 1.985 x 1030 / (7 x 108)2 = 270 m / Sec.2 Δs = 1,400 Kgs. / m3 Average weight of the Sun = Δm = 1,400 x 270 / 9.81 = 38,532 Kgs./ m3 fyy2 = f1y2 = 1.13706 x 1013 y2 = (38,532 / 3) [(7 x 108)3 - y3] ; y3 + 8.853 x 108 y2 = 3.43 x 1026 : y = 4.98 x 108 m
Depth from the surface of the Sun were the pressure is equal to: f1: r – y = 7 x 108 – 4.98 x 108 = 2.02 x 108 m In the determination of K and T have not been considering many facts, as the time required to produce them, if in some conditions they can be produced in a faster way, if the pressure produced by F is smaller than the required for eliminate the effect of m⊥7 acts in a given proportion, or does not act, etc. At temperature T can melt some metals, as iron; with the fore result we are able to understand many things and modify the interpretation of the behavior of them. At the present time the accepted theories for to explain how the Sun and the stars have high temperatures, are far from being satisfactory. If they want to imagine the origin of such celestial bodies, they also must imagine them as freezer celestial bodies, and from this condition, to a hot body as the Sun, there are many indeterminations. With the many models could be conceived, can be imagined two freeze celestial bodies colliding one with each other, the energy of the collapse was employed in produce the great heat of our Sun, and also ejected great quantity of matter that afterward was transformed in our planets.
Could be imagined that in the center of our galaxy, because a nuclear action, a huge celestial body exploited ejecting celestial bodies as our Sun and many stars and also many other smaller masses as the planets of our solar system; in this way were formed different solar systems. Could be given other models, but in all them the origin of the solar systems have been in a rather violent way. Whether have been the origin of the solar systems, have been consider they have a cycle of life they consider will end when the Sun or the stars will get in a freeze state; in accordance with the radiation that the stars emit, they consider the age of them. Maybe the fore mentioned theories (hypothesis) and other similar are correct, maybe not. I consider that with our present knowledge we are far to be able to give more definitive data. With what we have say in this theme we can give other model, we can imagine a Sun that can be formed in a soft way, a celestial body with a relative small mass with the time can grows attracting other celestial bodies that pass near it, some of them will increment the mass of it, other ones will move around it forming the planets; also we can imagine that the planets are formed after the original celestial body has grown in such magnitude that it exploits and ejects the bodies that afterward will be transformed in planets. With respect to the heat that gets the original body to be transformed in a Sun, this heat will be produced by pressure effects in a similar way as was explained in the numerical example. If we are conscious that with pressure it is possible to produce heat, this by one side, by other one, with the return distance effects we can imagine a star that does not gets exhausted in an infinite time; by one side does not lost matter, due to the returning distance effects; does not lost heat because its interior caps always have a pressure that produce heat. In the accepted theories, it has been considered that our Earth originally formed part of our Sun and was expelled from it, so that Earth originally was an incandescent body that began to cool from it surface toward its nucleus; they consider that if in the present time there are volcanic activities, is because the cooling effect has not affected the nucleus of our planet, but that in the long future all our planet will get freeze, in accordance with our model explained with the numerical example, this will never happens so, because the Earth. The Sun, the stars, etc. produce their own heat, due to pressure effects, and it is convenient to take in account this in the many facts could be explained with this new model.
LIBERATION VELOCITY (corrected theme)
With all have been say till here there are enough fundamentals for to accept mechanophysics, nevertheless there are many doubts about gravity force, for instance, it is not know if there is a return distance for gravity force, if a particle is saturated with gravity: Can be attracted by any other body? Can it attract to other particles, or not? The fore questions and others have some reasons to be made, because if the gravity can acts till infinite distances, as some ones suppose, in the infinite time of existence of the Universe, all the matter of it would be concentrated in a single place. It has been seen that there are some relations between the electromagnetic forces and the gravity ones; in some behavior of the electrons seem as if they are not affected by gravity, etc.; but is better do not continue pointing out the many doubts, they will be demonstrate as far as it is advanced in the investigation work.
Between the many facts can be seen of gravity is interpreted that a body is attracted at different velocity in different mediums, for instance in a medium in which the body weight more than in other, the velocity with which is attracted the body is faster in the more intense medium than in the less intense one. Also could be imagine that in all intensities the bodies are attracted at the same velocity (?), but with more energy in the more intensity medium (as if they have more weight), but this is not so, because (g) grows with the intensity. Can be remember the experiment of Galileo in the Piza’s tower, in which two bodies with different weight fall at the same velocity and time; all are affected with the same intensity. In order to see the problem in the most objective point of view of mechanophysics will deduce some formulas; in first place, will determine the formula that gives the velocity of escape or liberation that must have a body in order to escape from the return effect of the Earth, produced by gravity. The so called velocity of escape or liberation has a value equal to: ve ; its impulse is equal to: m vs; m = mass of the propelled body. The force of desacceleration produced by gravity on the body is: Ec = G m M / r 2 Kg. m / Sec2
In a brief way will be seen how is produced the liberation velocity in the surface of the Earth, in order to know what radial velocity must have a unitary body for to escape from the influence of the gravity of the Earth.
g = 9.81 m / Sec2 = desacceleration velocity in the surface of the Earth.
G = 6.673 x 10—11 m3 / Kg. Sec2. = constant of universal gravitation.
M = 5.975 x 1024 Kgs. = mass of the Earth.
r = 6.375,163 m = radius of the Earth.
g = G M / r2 = 9.81 m / Sec2 = Desacceleration of the unitary body in the surface of the Earth. After a time (t) the desacceleration of a body that is moving upward will be: ge ≈ 0; so the average desacceleration will be: g / 2 = 9.81 / 2.
In order the unitary body escapes from the influence of the gravity of the Earth (disregarding the
influence of the air), its liberation velocity must be: ve / 2 = g t; t = ve / (9.81 x 2).
In order the body start receding at velocity ve from the surface of the Earth, it would be able to
move a distance equal to R in a time (t); and from the surface of the Earth toward the outer space;
the velocity begins to decrement, as was specified; and with the same magnitude the value of (g).
R = 6.375,163 m =. ve t = ve (ve / 9.81 x 2) = ve2 / (9,81 x 2)
ve = (2 x 9.81 x 6,375,163)0.5 = 11,187 m / Sec
Now have been study the velocity of escape or of liberation, it would be interesting to study it in different cases, in order to begin to know its amplitude and its limitations, so it will be seen the case of a black hole; as we ignore many facts about this celestial bodies, here will imagine a body formed by a matter with equal density than a proton or a neutron, as with a neutron’s star.
Radius of proton: rp = 1.7143 x 10--14 m.
Volume of proton: Vp = (1.7143 x 10--14)3 4 π / 3 = 2.1103 x 10 --41 m3
Mass of proton: mp = 1.6725 x 10--27 Kg.
The density: Δp = 1.6725 x 10--27 / 2.11 x 10--41 = 7.92654 x 10 13 Kgs./m 3
The formula for a celestial body to work as a black hole is:
1 = G M / (2 rc c 2); M = 2 rc c 2 / G = 4π rc3 Δp / 3 .... (1)
Half of the liberated velocity is for to conteract the velocity (gt) = ve / 2, that is in the surface of the Earth; and the other half 9is for to nullify the action of attraction force of (gt) = ve / 2 toward Hearth, in free space The condition for the celestial body with uniform density... behave as black hole is (Eq. 1): rc = (3 c2 / 2 G π Δp)0.5 = [3 (3 x 108)2 / (2 x 6.673 x 10--11 π 7.92654 x 1013) ]0.5 = 2,850,293 m.
Mc = 2 rcc2 / G = 2 x 2,850,293 (3 x 108)2 / 6.673 x 10--11 = 7.688487 x 10 33 Kgs
2 c (c / rc) = 2 accelerations produced by black hole.
rc = radius of the blasck hole sphere.
D = average density of the black hole
In a black hole is consider that gravity force is so intense, that neither a corpuscle of light can escape from it. The fore conclusion was obtained because has not been observed any ray ejected from them. Here will be suppose that in such celestial body can be produced light radiation, in such way that each corpuscle in the instant is emitted has a kinetic impulse equal to: m c; if the corpuscle is emitted in a radial way, from the celestial body, from the beginning will be affected by the desacceleration effect produced by such body; m = mass of a light corpuscle = 1.47236 x 10--50 Kg.
The desacceleration impulse produced in a light corpuscle in a unitary time is: Fc = G Mc mc / rc2 = 6.673 x 10--11 x 7.688487 x 1033 x 1.47236 x 10--50 / 2,850,293 2 = 9.29816 x 10--40 Kg m./ Sec2
c
m c = 1.47236 x 10--50 (3 x 10 8) = 4.41708 x 10--42 Kg. m / Sec. Time (unitary) of braking:
tc = mc c / Fc = 4.41708 x 10--42 / 9.29816 x 10--40 = 4.7505 x 10--3 Sec. Space in which is produced the braking:
dc = 0.5 c tc = 0.5 x 3 x 108 x 4.7505 x 10--3 = 712,573 m Energy of braking: W = Fc dc = 9.29816 x 10--40 x 712,573 = 6.6256 x 10--34 joule =
1 quantum.
G Mc / 2 rc gc = c;
vc = G Mc / 2 rc c = 6.673 x 10--11 x 7.688487 x 10 33 / (2 x 2,850,293 x 3 x 108) = 3 x 10 8 m /
Sec = maximum velocity
..
Note: From a theoretical point of view the fore problem is all right, but can not exist a black hole as the fore one because its acceleration is bigger than c … gc = G Mc / rc2 = 6.673 x 10—11 x 7.688487 x 1033 / 2,850,2932 = 9.6896 x 1010 m / Sec2
Some speculations. In the fore model there are several indeterminations, beginning with the structure of the black hole; also is not defined if the gravity that acts over each light corpuscle is a remnant one, or is the one that meets with the corpuscle in each point of its receding trajectory; if both kind of gravity act; if the first law of the double fluid theory work here in an evident way, etc. .
With the fore result is proved that (g) is not a constant value and also that it is not possible there could be formed celestial bodies heavier than Mc, if in a distance bigger than dc there is matter that the body can attracts, this because it is not possible that (g) can produce velocities bigger than (c); with all these have been found some direct relation between gravity force and the electric one, but since long time ago have conceive a model of gravity force, in which the gravity is transmitted by so small particles that wl call: gravity’s microcorpuscle; this particles are saturated with gravity and of course, if they are emitted by a body with mass Mc they only can advance a distance dc, but it is premature to talk more of this by now
From all have say here, are get some doubts. Since was made the model of the return distance, was suspect that the electrons were not affected by gravity, being this correct or not, here is concluded that with so great: g = 3 x 108 m / Sec2, produced in the corpuscles of light, the electrons must be affected by gravity, if it is enough intense, maybe they are affected by the black hole and move around it as satellites, maybe they get exhausted by the great density of gravity and get disintegrated; it is possible that gravity can be transformed in electric negative charges; if the return distance of gravity is given by the distance dc, the capacity of attraction of a black hole is limited to the fore distance, but this does not mean that their mass could not grows in such way, that if the electrons move as was say before, around the black hole, in some aspects they behave as the orbital electrons in the binary systems; in this way it would be easy to determine the laws that govern the behavior of the propeller particles m⊥ acting in the orbital electrons and in the positive particles of the binary systems. With the fore doubts and others not mention, it is evident that the study of gravity is in a incipient state, there is much to be done about this .
Monterrey, México, August, 1999; July 11, 2002 Manuel de Hoyos Robles SPECIFIC HEAT (I) (corrected theme) Å-
The specific heat of a substance at a given temperature, is the quantity of calories required by a gram of the substance for to increment its temperature 1º C. In a complete study of physics can not be omitted that of specific heat. At the moment to write this theme, are ignored by us many facts related with this phenomenon, but, any way, have decided to write it, in spite of this, only because in mechanophysics the interpretation of this phenomenon is too much different than that given in modern physics. In few words, this theme is written here, not because we have many interesting data, but because at this phase of our investigations, feel obliged to write it.
Can not define when was observed with a rather scientific character the property of the substances to absorb heat: nevertheless in 1819, two young Frenchmen, Pierre Lous Dulong and Alexis Theres Petit made a unexpected discovery during the researches in the thermometry in which they have been engaged for some years with a dozen metals and sulfur (at room temperature), they found that what was name them as specific heat (c´ ) had practically the same value, approximately 6 cal mole. deg.: from this they inferred that the atoms of all the elements have the same heat capacity. Although they consider this as a general law that could be applied not only to solids (and liquids), but to gases: in 1830 this was not accepted in a satisfactory way, much remained to be learned about atomic weights. One of the first to find some discrepancies with the law of Dulong Petit, was Amadeo Avogadro that remarked some deviation in the case of carbon, but his measurements were not very precise. Matters got more serious in 1840, when two Swiss physicists: Auguste de la Rive and Francois Marcet, with diamond powder obtained c´ ≈ 1.4 (3º to 14º C). At almost the same time Henry Victor Reynault, with diamond obtained: c´ = 1.8 (9º to 98º C). During the next 20 years, he continue his studies of specific heat and find many other deviations from the general law, though none as large as for diamond.
In 1870, Henrich Friedrich Weber studied the specific heat considering big variation of temperatures, for instance between 0o to 200º C, he confirmed his conjecture for diamond (c´) vary by a factor of 3 over this range. The next time was hear from Weber is in 1875, when he presented his beautiful specific heat measurements for boron, silicon, graphite and diamond from - 100º to 1000º C. For the case of diamond © varied by a factor of 15 between these limits. Other investigator in 1872, James Dewar working at high temperatures as 2000º C found that carbon had c ≈ 6 at such temperatures. Dewar´s most important contribution to this subject deals with very low temperatures, for diamond he found c´ ≈ 0,05 in the interval between 20º to 85º K. Here have been given in a brief way the investigation work had been made in the practical field of the specific heat that was made mainly in the XIX century. About this have been given different interpretations, considering that the atoms of the samples are affected by heat due to vibrations in the three dimensions by velocity (kinetic energy) and rotation energy, also produced in the three dimensions.
Boltzmann was the first one to give theoretical interpretation to the Dulong Petit law. In 1871 he showed that the average kinetic energy equals the average potential energy for a system of particles, each one of whish was suppose oscillates under the influence of external harmonic forces. In 1876 he applied these results to a three dimensional lattice. This gave him an average energy 3RT ≈ 6 cal / mol. As Boltzmann himself out it, his result was in good agreement with experiment for all sample solids with the exception of carbon, boron, and silicon went on to speculate that this anomalies might be a consequence of a lost of degree of freedom due to a sticking together” at low temperatures of atoms at neighboring lattice points. As the equipartition theorem of classical statistic mechanics used by Boltzmann did not fixed well in diamond, gases molecules, the theory of Boltzmann was consider incomplete or not definitive, they considered there were other unknown vibrations. For instance, Maxwell found that: cp / cv = 5 / 3, for mercury vapor.
Other considerations were made with gases combined with the ether, in which there was not attained equilibrium. Kelvin considered that the classical equipartition theorem was wrong. Einstein study the problem from the point of view of quantum physics, his first formula fails at low temperatures, as was proved by Nernst. The correct temperature dependence at low temperatures was first obtained by Peter Debye, for nonmetallic substances. Einstein ended this active research of specific heat of solids. Debye, Max Born and Theodore von Karman were working for more exact data of specific heat. In 1913 Einstein returned to the investigation of specific heat in gases, this due to the discovery made by Arnold Euken; it was known then that cv = 5.0 for molecular hydrogen at room temperature, he showed that at T = 60º K; cv ≈ 3, that is consider to the following of two rotational degree of freedom.
Heramm Walther Nernst, in 1905 proposed a new hypothesis for the thermal behavior of liquid and solids at absolute zero, for example, graphite and diamond, in the case of carbon, their specific heat tends to zero as T ≈ 0. Therefore it does not exclude a nonzero specific heat at zero temperature. In fact in 1906 (without concluyent tests) Nernst assumed that all specific heat tends to
1.5 cal / deg at T = 0o. In 1910 he and after many experimental the zero specific heat at low temperature was imposed.
In a brief way (taking data from the book: The Science and the Life of Einstein, by Abraham Pais) has been given the history of the specific heat since the XIX and XX centuries; the data obtained here will be very much useful for to obtain a model from the point of view of mechanophysics. All that dedicate to the physical investigation feel a great admiration and respect for the persons that in some way contributed to the progress and knowledge of a branch of science (physics). If it is consider by us that they took a wrong way, will not be criticized by this; since Galileo and Newton always have been respected the experimental data and if the theoretical interpretation is wrong, this is not a simple decision.
Talking in a general way, in an atom there are two kind of orbital electrons; the interior shell and the exterior shell ones. This last can form the molecules uniting one atom with other one; to this electrons we will call: molecable electrons, and in most atoms they correspond to the seven shell. In carbon and silicon there are also two molecable electrons in the sixth shell, than can move to the seven one. In our model of specific heat will consider that all the molecable electrons try to absorb the caloric fluid when the atom is in a medium of higher temperature; and this fluid is distributed in all the orbital electrons of the atom.
Also can eliminate heat when is in a medium with lower temperature. As the carbon atom has four molecable electrons can absorb more heat than the atoms that have two molecable electrons; so its specify heat is smaller But not all the caloric fluid is employed in produce the specifics heat, but also in change the substances from a solid state to a liquid one, or to a gaseous one. At very low temperature, the atoms have few calorific corpuscles that oppose the absorption of corpuscles that can increment their temperature in an easy way; an opposite effects happens with a high temperature.
But the binary systems with molecable electrons not only contribute to change the temperature of the bodies affected by heat, but also to change their physical conditions; for instance, water is a solid substance at a temperature lower than 0o C, and transform in vapor at 100º C. When water is in an ice or in a vapor state its specific heat is reduced to half. The quantity of calories required to change a substance from solid state to liquid one is called: latent heat of fusion. To transform ice at 0o C to water at the same temperature are required 80 calories / gr. In accordance with our model of interchange effect; this means that ice has the fore capacity of cohesion. When water gets 100º C, it is required 540 calories / gr. to transform it in vapor (in gas). Next are given some latent heat of fusion: cf and of vaporization cv:
ºº cf cv cal. /gr.
Water 80 540
Alcohol 25.8 204
Cu 49.4 1150
Fe 59.1 1500
Hg 2.7 71
Au 15.0 411
Äg 25.6 556
Pb 5.6 207
Äs we have seen, a more complete study of specific heat will not be complete without explaining the latent heat, maybe in the future will have to do this. By now will limit to give some simple numerical examples for to explain our elementary model of specific heat, so, next will be given some technical concepts.
An atomic mass unit (amu) is approximate the mass = 1.672 x 10—27 Kgs. One mole is the quantity of matter whose mass is numerically equal to its atomic mass expressed in grams. One kilo mole of a substance equals the quantity of it, whose mass in kilograms is numerically equal to its atomic (or molecular) mass. The fore means that one kilogram of hydrogen is practically equivalent to one kilo mole. The Avogadro´s number (6.02252 x 1026 molecules / k mole) multiply by one (amu) is equal to one kilogram:
1.672 x 10—27 x 6.02252 x 1026 = 1.0 Kg. / k mole.
If it is known the dimension of an atom and its (amu), could be determine the Avogadro´s number, I suppose there are several methods to determine the fore values; for instance, if the X rays have a corpuscular separation similar to the diameter of an atom: d ≈ 10—8 cm, by some interference effect could be determine the diameter of the atoms in a given body. In most of the problems given in mechanophysics have consider as diameter of the atoms the following value: 2 r = d = 2.27 x 10—8 cm.; with this value can obtain the Avogadro´s number (N) in an approximate way for all the elements, also can obtain the specific heat / amu. From the book: Chemistry by Michell J. Sienko and Robert Plane, were obtained the following values:
ce amu ce / k mole 2r Al 0.216 26.98 5.8 2.50 x 10—8 cm.
Cu 0.0922 63.54 5.9 2.34 ´´
¨Fe 0.108 55.85 6.0 2.34 ´´
¨Ni 0.105 58.71 6.2 2.30 ´´
Äu 0.0306 197.0 6.0 2.68 ´´
Äg 0.0565 107.88 6.1 2.68 ´´
The specific heat of water is: ce = 1 calory / cm3 = 4.2 joule / cm3. In accordance with the Avogadro number (6.02252 x 1023 atoms / cm3) there are the following quantity of water molecules, that are formed by two H atoms and one O atom = 3 atoms: n = 6.02252 x 1023 / 3 = 2.0075 x 1023 molecules.
The energy for to increment the temperature of one water molecule is: K1 = 4.2 / 2.0075 x 1023 = 2.09215 x 10—23 joule / molecule
With the Boltzmann´s formula we have. T = 2 K / 3 k = 2 x 209215 x 10—23 / 3 x 1.38 x 10—23 = 1o.0 C.
Considering an isometric distribution of the atoms will be determine the volume occupied by one copper atom: VCu = (2rCu)3 = (2.34 x 10—8)3 = 1.28 x 10—23 cm3
Number of atoms there are in 1 cm3: n1 = 1 / 1.28 x 10—23 = 7.8 x 1022 atoms / cm3 = 7.8 x 1022 / 8.9 = 8.764 x 10—21 atoms / grm
The density of copper is: ΔCu = 8.9 gr. / cm3 N = 7.8 x 1022 x 63.54 / 8.9 = 5.572 x 1023 atoms / mol
63.54 = (amu) of Cu.
As the fore value of N is smaller than the obtained: N = 6.02 x 10--23,… the first idea we get is that the copper atoms are not distributed in an isometric way in the sample, but in a more compact one, could be other reasons that will be seen after our model will be better integrated
Next will be given the models that explain the specific heat of different atoms and molecules, seen from the most simple way, so avoiding the details that complicate the explanations and clarity of the problem and that only contribute in a minimum part to the exactness of the problem. As most of the atoms have 2 molecable electrons, here will be consider most of them have practically the same orbital velocity and practically and proportional incremented the orbital velocity when the temperature grows 1º K.
When the temperature in a sample is incremented 1º C, this increment is not made in an instantaneous way, but in a uniform one, during a lapse of time (t); at the beginning the increment is practically 0o K, and at the end of the mentioned lapse the increment is 1º K, so the average increment is; T = 0.5 º K, so the increment of the energy in the mentioned lapse in each atom or molecule of the sample is; Kl´ = 1.5 k T = 1.5 x 1.38 x 10--23 x 0.5 = 1.035 x 10 –23 joule. Number of molecules in a mole: N = 6.02252 x 1023 = Avogadro number.
In the molecules of a mole of sample have: Kl = 6.02252 x 1023 x 1.035 x 10 –23 = 6.23 = specific heat /atomic mass. When the temperature of the medium in which is the sample, is incremented 1º K, the atoms of the sample begin to absorb the calorific corpuscles corresponding to the incremented temperature; and afterward they are distributed in all the orbital electrons, as will seen afterwards. till the sample gets its temperature equilibrated with that of the medium. Generally most of the atoms have two molecable electrons in the seven shell; but silicon and carbon atoms have the fore mentioned electrons in the seven shell plus two molecable electrons in the six one. The capacity to interchange heat (change of the specific heat) can vary with the number of molecable electrons in each atom; the interchange of heat is produced in the vertices of the polygonal orbits in a magnitude proportional to the energy of deflection.
The atoms absorb the corpuscles that produce heat, by conduct of the molecable electrons; most of the atoms have two molecable electrons in the seven shell, with them the atoms can absorb and emit corpuscles producer of heat. Few atoms as C, Si, also have two molecable electrons in the sixth shell, so they can absorb and emit corpuscles producer of heat more faster than than the atoms of two molecable electrons; we suppose they do this operation proportional to their deflection energy. For instance, the atoms with two molecable electrons require a certain time to reach an incremented temperature and maintain its specific heat of their atomic mass at a value of ± 6.2. This due to their capacity to absorb and emit corpuscles. The atoms with four molecable electrons obtain the fore process in less time (t/2…?).
If the fore process is produced in a medium with very low temperature, the molecable electrons can absorb more rapid the corpuscles producer of heat and the values of the specific heat of the atomic masses try to diminish. The opposite happens when the temperature grows.: In molecules formed with several atoms; the interior atoms with all their molecable electrons linking in both sides the adjacent atoms; and only in the two extreme atoms with their molecable electrons linking in one side with an atom and in the opposite side free; these last electrons are the only can absorb corpuscles when the temperature grows and emit heat corpuscles when the temperature diminishes. These and other problems could be appreciate with numerical problems. Let see an example of a sample of Cu atoms when their temperature is incremented 1o C
Specific heat: ce = 0.0922 Cal. / gr 1oC; amu = 63.54 ce = 0.0922 Cal. = 0.0922 x 4.2 = 0.38724 joule mCu = 63.54 amu = 63.54 x 1.6752 x 10—24 = 1.0644 x 10—222 gr.
Energy required for increment its temperature 1oC
K = ce mCu = 0.38724 x 1.0644 x 10—22 = 4.12136 x 10—-23 joule
Increment of temperature of the structure of the atoms + the vacuum space covering such structure:
T = 2 K / 3 k = 2 x 4.12136 x 10—23 / 3 x 1.38 x 10—23 = 2oC (= 1o + 1o)
The energy of deflection of two molecable electrons in the seven shell is:
E7 = me v⊥72 = 9.1091 x 10—31 x 69,1002 = 4.3494 x 10--21 joule
The energy of deflection of two molecable electrons in the six shell is:
E6 = me v⊥62 = 9.1091 x 10—31 x 93,9802 = 8.0454 x 10 --21 joule.
The specific heat for a molecule with two molecable electrons in the seven shell is:
ce2 = 6.23 E7 / E7 = 6.23 x 4.3494 x 10--21 / 4.3494 x 10--21 = 6.23
The specific heat for a molecule with two molecable electrons in the seven shell, plus two in the six shell is: ce4 = 6.23 E7 / (E7 + E6) = 6.23 x 4.3494 x 10--21 / (4.3494 x 10--21 + 8.0454 x 10--21) = 2.19
In our original theory was consider that the molecules would be formed by orbital electrons that move in the seven shell, and also by electrons that move in the sixth one; as happens with molecules formed by carbon atoms. There is other alternative that consider that the molecable electrons of the six shell can move to the seven one, for to form the molecules: If this is so, the molecable electrons of the seven shell diminish their capacity to absorb heat (corpuscles of heat), so we have:
6.23 x 4.3494 x 10--21 / 2 x 8.0454 x 10—21 = 1.68
For to have a more complete theory of specific heat it would be necessary to include the molecules;
my knowledge in chemistry are practically null, so I can not explain with fundamental reasons the
behavior of them from the point of view of specific heat. Here it is necessary to clear out the
interpretation of Avogadro´s number: in the book: Introduction to Modern Physics (second edition)
by: C. H. Blanchard, C. R. Brunet, R. G. Stoner and R. L. Weber was copied the following paragraph:
Let NA represent the number of atoms in 12 Kgs. of carbon. The mass of NA helium atoms will be 4 / 12 as great, or 4 Kgs. In general, the mass of NA atoms of a subsistence of atomic mass A· will be A· Kgs. This is just one kilo mole of the substance. A kilo mole of any element (compound), then, contain the same number of atoms. This number NA, is called Avogadro¨s number. Notice that we do not at this point know anything about the size of Avogadro´s number, except that it must be very large. As a matter of fact, Avogadro´s number is not easy to measure. We shall see later that its current best value is NA = (6.02252
0.0038) x 1026 molecules / kmole.
In the fore explanation they do not give the quantity of molecules that there are in a unitary volume (one liter); if they refer to atoms, the number of these in the unitary volume is given by the Avogadro´s number: if they refer to molecules, the Avogadro´s number gives the quantity of atoms that there are in a unitary volume and also the quantity of molecules divided by the quantity of atoms that has each molecule, that there are in a unitary volume, the fore considering that most of the atoms have more or less the same volume, it is easy to conceive from a mathematical point of view, the validity of the fore explanation, but here will be given other interpretation from the point of view of the specific heat, for instance, considering the water molecule, it has: c = 1.0, but the ice and vapor have: c = 0.5
The water molecule in both extremes haas 2 H atoms, in one side of them are connected to the O atom, and in the opposite they are free; in accordance with the Pierre Lous Dulong and Alexis Theres Petit rule, in the free sides are absorbrd 3 calories x 2 = 6; in accordance with the experiment are absorbed 2.1 x 2 = 4.2
Monterrey, México, November 24, 1999, March 20,2001; Manuel de Hoyos Robles
THE MASSES OF THE STARS (corrected theme) .
Maybe the investigators that have had most difficulty to be accepted their theories, were my father and me. My father considered that the so called fields of forces, were not so, only were fields of induce ultramicroscopic particles moving at light velocity that affected the atoms of the celestial bodies in a way that the atoms emit inducers fluids in the opposite side were they received the mentioned induce ultramicroscopic particles, so that between two celestial bodies there are attraction forces. He considered that in the infinite time of existence of the Universe the fluid of the ultramicroscopic particles did not exhaust, this was because if a celestial body emits its fluid, by other side this celestial body receives the same quantity of such fluid from all the bodies of the Universe. Although from a mathematics point of view, the idea of my father was a satisfactory one, from a strict point of view has some unsatisfactory conditions, so, was conceived other idea, that is that of the return distance of all the emitted particles of the fluids. Here will not talk more of this, in order do not get outside of this theme, only will mention again, that about 1945 my father proved that the theory of relativity was founded in wrong concepts; and till now the work of my father has not been recognized in an official way. A similar thing can say of my model of polygonal orbits of the atoms and of the binary systems. All right, they give the solution of a part of physics, but not of all, and in this way our theories do not solve all the present problems. But with our work do not pretend to displace the modern physics, only to give an alternative, in order all physicists have an ampler field for to choose better ideas, that in the future will define which is the most correct physical science.
In the fore theme was defined the maximum temperature could exist in the Universe, and from the point of view of the atoms, the fore definition seems too logic, because an atom can not have but limited quantity of caloric fluid particles. If the atoms are forced to yield more quantity of caloric fluid, the only thing could be attained is to reduce their masses. In an acceleration machine, if a proton or an electron are accelerated at light velocity, the only thing is attained, is to disintegrate their structures.
In the Earth a body moving in a horizontal direction at a velocity of 7,908 m / Sec. (with no resistance of air....), does not fall toward the Earth; this is because the fore velocity would be equal to that of a satellite moving around our planet. In accordance with the theory of the binary systems, seventh orbit, a proton moves around its nucleus at a velocity: v+7 = 672.268 m / Sec.; the electron moves at a velocity: v7 = 308,570 m / Sec. v+7 = 4 x 308,570 / 1836 = 672.268 m / Sec. = velocity of the positive particle of the binary system, seven orbit. 1836 me = 1836 x 9.1091 x 10--31 = 1.6724 x 10--27 Kg. = mass of proton
With all say in this paragraph, we think that if there is a star that has a: g7 = 672.3 m / Sec2, in this star could not exist binary systems corresponding to the seven shell. g7 = v+7 / Sec.; this is because the acceleration capacity of the star can nullify the translation velocity of the positive particle of the binary system of the seven shell. Here is pertinent to mention that the electrons are not affected in the same way by gravity as the nuclei of the atoms (see theme: The Return Distance
- No Growing of Entropy-) In this star can not be formed any hydrogen atom, or any binary system corresponding to the seven orbit or shell.
In the surface of the Earth its gravity force can accelerate the falling bodies the following value: g = 9.81 m / Sec2. In the surface of the Sun the acceleration effect of its gravity is bigger, and can be determine by the Newton’s law:
F = m gs = Gm Ms / rs2
m = mass of the attracted body by the Sun.
gs = acceleration of the body by gravity of Sun in its surface.
Ms = mass of the Sun = 1.985 x 1030 Kgs.
rs = radius of the Sun = 7 x 108 m
G = universal gravitation coefficient = 6.673 x 10--11 m3 / Kgs. Sec2
gs = G Ms / rs2 = 6.67 3 x 10--11 x 1.985 x 1030 / (7 x 108)2 = 270 m / Sec2
Our first problem in this theme is to define all the characteristics of a star that has a: g that can nullify in its surface, the velocity: v+7 = 672.268 m / Sec. of the positive particle (a proton) corresponding to the seventh orbit. In this way can not be formed binary systems corresponding to the seven shell in such planet. In this problem and in all will be seen here intervene several phenomena acting simultaneously, in some way are not well known by us, and because this, the solutions will be given here will be rather qualitative, than quantitative. Our main objective will be to fix the limits in which the phenomena verify and avoid some ideas that do not consider such limits, and because this, some times give very much wrong results. For instance, if it is consider that the density of a celestial body grows with its mass, the matter of the Sun (Δ = 1.4) would be denser than that of the Earth (Δ = 5.5). In our fore theme was given a model in which it is required high temperature to produce the binary systems of the different shells. In that model the temperature at which is formed a binary system is defined by the kinetic energy of the negative particle (the electron). The electron moves from one vertex to the adjacent one (time of vertex) without the propeller particle into it, that in some cases acts as calorific particles.
This propeller particles only remain into the electron during the time is produced the deflection (time of deflection); after this such particles are emitted toward the positive particle or proton. In this proton the propeller particles remain into it during the time it moves from a vertex to the adjacent one (time of vertex). The fore explanation is very much important, because, as was explained in our fore theme, the calorific energy (and any other one....) of an atomic particle is given not only by its kinetic energy, but by the calorific corpuscles the particle has into it. As it is convenient that the reader have a clear idea of this, will be obtained the calorific energies of both particles, in a binary system corresponding to the first shell.
mp / me = 1.6725 x 10--27 / 9.1091 x 10--31 = 1,836.
Velocity of the positive particle of a binary system corresponding to the first shell:
v+1 = 4 v1 / 1,836 = 4 x 2,160,000 / 1,836 = 4,706 m / Sec.
Kinetic energy of the orbital electron (negative particle): E = Et = 0.5 x 9.1091 x 10--31 (2,160,000)2 = 2.125 x 10--18 joule = calorific energy of the electron.
Calorific energy of the positive particle: E+ = 0.5 x 1.6725 x 10--27 (4,706)2 = 7.4467 x 10--21 joule Et = energy that has the positive particle due to the calorific corpuscles it has into it....
In our Sun almost all atoms are hydrogen ones, maybe there are many binary systems corresponding to the interior shells, with out heavier atoms. Here will not be defined this, neither other complex problems. For to obtain the star with: g = 672.268 m / Sec2, it will be consider that it has all its other characteristics similar to that of our Sun , and this because we believe there are not enough perturbing differences. With the fore criteria and using the Newton’s law will be obtained the required characteristics of the star without binary systems corresponding to the seven shell.
672.268 / 270 = 2.49
Its radius is: r = 2.49 rs = 2.49 x 7 x 108 = 1.7429 x 109 m.
Its mass: M = 2.493 Ms = 15.44 x 1.985 x 1030 = 3.0645 x 1031 Kgs.
g = G M / r2 = 6.67 x 10--11 x 3.0645 x 1031 / (1.7429 x 109)2 = 672.9 m / Sec.
The idea of not giving limit to some processes or to some phenomena have consider the existence of the neutron stars. These stars could be obtained if it is not take in account a limit to the action of gravity. In the fore theme was define the maximum temperature could exist, and was consider that with this temperature could be formed a binary system corresponding to the first orbit or shell; in other words, at this temperature could not exist binary systems corresponding to bigger orbits than the first one, and because this, we suppose that if this maximum temperature “increments” its rhythm of action, neither could have binary systems corresponding to the first shell. If with gravity force was eliminate the possibility to obtain binary systems of the seven orbit, as was seen in the fore numerical example, it is possible to eliminate the binary systems corresponding to the first shell. With this we have two ways to obtain the same thing and find a relation between temperature and gravity, and many other things.
It is well known that in the interior of our Sun there are zones in which can not exist binary systems corresponding to the exterior shells (7, 6, ...). In the examples will be made here will be consider that all the atoms of the stars deform in a uniform way, it does not matter they are exterior or interior ones; this will be made because in this way are simplified the solutions without having evident errors. In order to simplify our problem, will be consider that the Sun is formed practically of binary systems.... that is near 100% hydrogen atoms, some of them affected by great compressions. Ws = 1.985 x 1030 Kgs. = weight of Sun. Wh = 1.6725 x 10--27 Kgs. = weight of a proton.
N = Ws / Wh = 1.985 x 1030 / 1.67252 x 10--27 = 1.18683 x 1057 = quantity of H atoms there are in .the Sun (in the supposition all atoms are of H). 2rh = dh = 2 x 1.1376 x 10--10 = 2.2752 x 10--10 m = diameter of an atom.
If it is suppose that the atoms are distributed in an isometric way, we have that the volume that occupy an atom is (a cube): Vh = (2.2752 x 10--10)3 = 1.18 x 10--29 m3 2 rs = ds = 1.4 x 109 m = diameter of the Sun.
Volume that occupy the Sun: Vs = (1.4 x 109)3 π / 6 = 1.4368 x 1027 m3
Volume that occupy Vn atoms: V n = N Vh = 1.18683 x 1057 x 1.18 x 10--29 = 1.4 x 1028 m3 Vn / Vs = 1.4 x 1028 / 1.4368 x 1027 = 9.744 = times that reduces its volume the Sun due to its gravity. For a star to reduce the volume of its atoms from the seven shell to the first shell, there will be a reduction of volume of: 73 = 343 times. For the Sun to reduces 9.7 times, some atoms will have a diameter equal to that of the 3 shell; 33 = 27; 343 / 27 = 12.7 > 9.744.
With a process of try and error, and parting from the data of the Sun, will find a star without any binary system that is, here is seen the problem from a mathematical point of view, with polygonal orbits, because the density of the caps of the star grows from the surface to its center...: M1’ = 73 x 1.985 x 1030 = 6.81 x 1032 Kgs.
r1’ = 7 x 7 x108 = 4.9 x 109 m.
g1’ = G M1 / r12 = 6.67 x 10--11 x 6.81 x 1032 / (4.9 x 109)2 = 1,891.82 m / Sec2
g1 > 2,160,000 x 4 / 1836 = 4706
Second try:
4706 / 1,891.82 = 2.49
M1 = 2.493 x 6.81 x 1032 = 1.05 x 1034 Kgs.
r1 = 2.49 x 4.9 x 109 = 1.22 x 1010 m
g1 = 6.67 x 10--11 x 1.05 x 1034 / (1.22 x 1010)2 = 4,705.4 m / Sec2.
In the theme: Liberation Velocity is given the escape velocity: ve = (2 g r)0.5 , for any celestial body. In order the star works as a black hole: ve > c = 3 x 108 m / Sec. The fore star with mass M1 has a escape velocity: ve1 = (2 g1 r1)0.5 = (2 x 4,705.4 x 1.22 x 1010)0.5 = 10,715,025 m / Sec. c / ve1 = 300,000,000 / 10,715,025 = 28. With our model of atom of the seven shell, we have the liberty to conceive many models of seven nushells that harmonize in a geometrical way with that of the seven shells, as will be seen forward. The first one model conceived was that in which there is a value of the shell with respect to the nushell equal to: 1,836 / 4 = 459. In this model only was take in account geometrical values. But for a more definitive model will be necessary to take in account the time of action of all the elements that intervene here; and also there ought to be a complete harmony between the action of the atoms and the action of the celestial bodies. In the fore mentioned star was consider that in each space equal to the volume of the first shell there is a proton (can be two....). In our theory of the polygonal orbits, it is consider that all the sides of all the orbits have practically the same length (L).
r7 = radius of an atom = 1.1376 x 10--10 m
L = 2 π r7 / 28 = 2 π x 1.1376 x 10--10 / 28 = 2.55 x 10--11 m = length of a side of the 7 orbit.
The sides of all orbits are equal.
Length of a side of the seven nushell = 4 x 2.55 x 10—11 / 1,836 = 5.56 x 10—14 m .
In the fore length (L) moves an orbital electron corresponding to a binary system. The orbital electron or negative particle of a binary system, moves around a shell; the proton or positive particle moves around a nushell (see theme: Hydrogenized Model of Atom). A binary system corresponding to the first shell has a radius of the magnitude: r1 = r7 / 7 = 1.1376 x 10--10 / 7 = 1.625 x 10--11 m.
The radius of the first nushell has a magnitude: r1’ = 4 r1 / 1,836 = 4 x 1.625 x 10--11 / 1,836 = 3.54 x 10--14 m.
In a coarse way in each side of a nushell or nuorbit, corresponding to a binary system of the first shell, there is space for 3 protons. Also in a coarse way, in the volume of a nushell (first orbit) there is space for the following number (n) of protons: n = 33 = 27 protons (27 binary systems corresponding to the first shell) ≈ 28.....
With the fore geometrical consideration it is possible to have a star of the same diameter:
d1 = 2 r1 = 2 x 1.22 x 1010 m. = do, as the fore one obtained, but with a mass equal to: Mo = n1M1. Omitting some physical variations, here will be respected all the geometrical considerations mentioned before; in this way the star with mass Mo (see forward) could not be consider a neutron star, in which all the neutrons are in material contact one with each other.
Without changing the dimension: r1 = ro’ = 1.22 x 1010 m ; and without affecting the geometrical conditions mentioned here, can be incremented the mass of a star:
Mo’ = 28 M1 = 28 x 1.05 x 1034 = 2.94 x 1035 Kgs.
go’ = 6.67 x 10--11 x 2.94 x 1035 / (1.22 x 1010)2 = 131,751 m / Sec2.
veo’ = (2 x 131,751 x 1.22 x 1010)0.5 = 56,698,512 m / Sec.
300,000,000 / 56,698,512 = 5.29
Second try
Mo = 5.293 x 2.94 x 1035 = 4.355 x 1037 Kgs.
ro = 5.29 x 1.22 x 1010 = 6.455 x 1010 m.
go = 6.67 x 10--11 x 4.355 x 1037 / (6.455 x 1010)2 = 697,142 m / Sec2.
veo = (2 x 697,142 x 6.455 x 1010)0.5 = 3 x 108 m / Sec = c = liberation velocity.
In the fore problems the volume of the stars grew, when was reduce the volume of their atoms; this was correct till the deformation reached to the first shell of the atoms. In order to simplify the problems, in all cases was consider that all the atoms of the stars have the same deformation from the core of them to their surfaces. In accordance with the deformation of the atoms could be determined their temperature. When the temperature grows. the radius of each star grows.
If it is consider that with the growing of the mass of a star, never there is a diminish of its volume, so the fore explanations are satisfactory, nevertheless, if the volume of the star could diminishes, because into the first shell of its atoms there is space for many protons, that can be in a material contact, one from each other, this star would have the same density than one neutron’s star,
The volume that occupies an atom is: Va = (2.2752 x 10--10)3 π / 6 = 1.1778 x 10--29 m3 Volume of the first shell: V1 = Va / 73 = 1.1778 x 10--29 / 343 = 3.4338 x 10--32 m3 Diameter of a proton: dp = 2 rp = 2 x 1.7143 x 10--14 = 3.4286 x 10--14 m
Considering the volume occupied by a proton (a cube): Vp = dp3 = (3.4286 x 10--14)3 = 4.03 x 10--41 m3. If it is consider that the protons do not get deformed when they occupy all the volume of the first shell, although are in material contact one with each other, are required the following quantity of them: N1 = V1 / Vp = 3.4338 x 10--32 / 4.03 x 10--41 = 8.52 x 108 Taking in account all the fore information, it can be obtained a neutron star, in which obviously the escape velocity ve ought to be higher than: ve = 3.0 x 108 m / Sec. If the star with mass Mo could be compressed to a volume in which its density would be equal to that of a neutron’s star, its radius would be reduced to the following value : rn’ = ro / N11/3 = 6.455 x 1010 / (8.52 x 108)1/3 = 6.81 x 107 m gn’ = G Mo / rn’ = 6.67 x 10--11 x 4.355 x 1037 / (6.81 x 107))2 = 6.2654 x 1011 m / Sec2 > c / Sec.
ven’ = (2 G rn´ )0.5 = (2 x 6.2654 x 1011 x 6.81 x 107)0.5 = 9.2377 x 109 m / Sec. vm´ also has a value bigger than (c )... c / vn´ = 300,000,000 / 9.2377 x 109 = 0.032476 Mn = c Mo vm´ rn’ = 0.0324763 x 4.355 x 1037 = 1.4916 x 1033 Kgs. rn = c rn’ / vn´ = 0.032476 x 6.81 x 107 = 2.21 x 106 m gn = G Mn / rn2 = 6.67 x 10--11 x 1.4916 x 1033 / (2.21 x 106)2 = 2.037 x 1010 m / Sec2 ven = (2 gn rn)0.5 = (2 x 2.037 x 1010 x 2.21 x 106)0.5 = 3 x 108 m / Sec = c.
As: gn > c / Sec, we can not accept the theory of the neutron’s stars. From an astronomy book were take the following values: the Betelgence and the Antares stars, by far are more voluminous than the Sun. Exist the idea that a huge celestial body has more capacity to attract other bodies; but if it is take in account that in a big celestial body its escape velocity does not permit, neither the light radiation to escape; this make us to think that also the gravity fluid could not escape. Also we do notwhich is the return distance of the gravity in a huge celestial body.
Monterrey, México, August 15, 2001; January / 2007 Manuel de Hoyos Robles.
MICROCORPUSCLES OF GRAVITY (corrected theme)
In our investigation work, the smaller particles have been study are the corpuscles as those of light and the calorific ones. Nevertheless are other particles very much smaller than those mentioned, and those are the gravity microcorpuscles. From them there is few information, but as have been say in other themes, all the particles and bodies of the Universe, are formed from the smallest and most elementary particles ones, and as far as we know, this particles ought to be those that produce the gravity force. These particles are so small and move at light velocity, in such way that all the particles and bodies of the Universe, known by us are transparent to such microparticles.
For all us it is not difficult to understand that if light and gravity are formed by material particles, and these particles move at light velocity, they were emitted at the fore velocity by the atoms of any body. Who read this, immediately get some doubts; for instance. Why the corpuscles of light and the microcorpuscles of gravity are emitted at the light velocity and not at any other one? The fore question will not be answered immediately, first will be seen how are structured some atomic particles, as the protons, electrons, etc.
From the smallest particles to the biggest bodies of the Universe, it is easy to conceive they are formed from the most small particles could exist in the Universe. Here we get inclined to consider that the smaller particles, all them have the same characteristics. This does not happens in the macroscopic world, in which can be structured different bodies with different elements (more than one hundred of them), Omitting a series of considerations, explanations, even the corpuscles as those of light, as the microcorpuscles as those of gravity are emitted at light velocity, because both have some similar characteristics and conditions. Heavier particles formed by corpuscles as those of light and microcorpuscles as those of gravity: Why do not this other particles are moving in the Universe at light velocity?
For Einstein the light velocity was the maximum could exist in the Universe, he considered it as a limited one and a reference for all the processes of the Universe. But all these are very much feeble arguments for an objective physics, because if a ray of light is emitted from a body that is moving at a velocity v, this ray will be moving at a velocity: c
v. Since I conceived the theory of the polygonal orbits, also conceived many things, one of them was that into an atomic particle, as a proton or an electron, there are particles formed by corpuscles, as those of light. For to explain the action and behavior of them I was forced to consider that into any atomic particle working in normal conditions, there is neither a win, nor a lost of energy, and that if into such particles there is a movement of other smaller ones, this could be so, because there is an interchange of energy between the particles, and this interchange is manifested by the kinetic energy of the particles. This could be understand better if it is consider that any particle of the Universe, by itself has an inherent energy equal to that given by its mass moving at light velocity.
For to explain the structure of an atom formed by several orbital electrons, it was consider that each orbital electron corresponds to a binary system, whose the negative particle is the orbital electron, and the positive one is a proton alone, or integrated with one or two neutrons. The positive and the negative particles work together, as if they were a unitary particle, because between them is shared a propeller particle that acts in the negative particle (deflect the orbit of the electron) when it is in a vertex of its orbit; after that the propeller particle is ejected by the orbital electron toward the positive particle, were it works in a similar way with such particle. The propeller particle works into the positive or the negative particles, because it has an energy equivalent to a kinetic one, equal to its mass moving at light velocity. -With the propeller particle m⊥n , for instance, in the negative particle, the fore axial kinetic energy is manifested by the velocity of the orbital electron moving with an energy equal to: 0.5 me v⊥n2 (= 0.5 m⊥n c2); v⊥n = axial velocity of the orbital electron; c = velocity of light;; me = mass of the orbital electron; m⊥n = mass of the propeller particle. For instance, when the orbital electron is in a vertex of its orbit, the propeller particle and the electron have the interchanged energies given before. . . In this process there is neither a lost nor a win of interchange energy, only an interchange of it. In the macroscopic world, it is observe how the planets move around the Sun by indefinite time without any lost of energy. . .
Of course if it is required to obtain a complete model for to understand how is produced the interchange of the inherent energy, it is required to know how this is produced. In the macroscopic world, when an elastic body shocks with other one, part of the kinetic energy of the first body, that we will call: active body, is transmitted to the second body, that we will call: passive body, this in accordance with the laws of impulse and momentum. The time both bodies shock, we will call: time of elasticity. Into the atomic particles (proton, electron), there are active particles (as corpuscles) that can yield its inherent energy, for instance, in an electron, the action of interchange energy would require a longer time than the elastic time; this time of interchange can be produced in a continuous way, or by a series of pulsation, in accordance with one model.
If the interchange of inherent energy is produced by simple elastic effects, can be used the formula: K2 / K = 4& / (1 + &)2 = energy yield by the active particle / inherent energy of the same particle; being done this in an elastic pulsation; & = mp / ma; mp = mass of the passive particle; ma = mass of the active particle. Each pulsation is equivalent to the time of elasticity. It is possible to make others models. By now we consider it is not essential to know this mechanism, so here will be omitted; maybe in the future will be require to know if the gravity microcorpuscles can be reflected; what is positive and negative charge, etc. When we have talked of the binary systems, it has been consider that the positive and the negative particles are united by the action of a particle that moves between them at light velocity, as have been explained in other themes. To this particle we have not given a definitive name, some times we have named: charged particle, other propeller particle, etc.; of the two mentioned names we consider by now, more correct to name as propeller particle.
One of the main objectives of this theme, is to determine the mass of a gravity microcorpuscle, and this could be obtained if it is know, at least in part, how is the interaction of the atomic particles. As we use to do, with a numerical example can be better explained the theory. Here will be consider the orbital electron of a binary system corresponding to the seven orbit or shell.
The interaction between the propeller particle and the orbital electron has been explained in other themes, nevertheless, in order to make more accessible the explanations, will be repeated some ideas obtained in other themes. The radius of the seven orbit is: r7 = 1.13763 x 10—10 m. The seven orbit has 28 equal sides of length: L = 2 r7 Sin.(180º / 28) = 2.54748 x 10—11 m. The velocity of the orbital electron is: v7 = 2,160 / 7 = 308.57 Kms./ Sec. When the orbital electron arrives to a vertex of its polygonal orbit, its trajectory is deflected an angle: α7 = 360º / 28. In order to produce such deflection it is required an axial velocity equal to: v⊥7 = 2 v7 Sin.(180º / 28) = 69,098 m / Sec.
The kinetic energy required to deflect the orbital electron an angle α7 is equal to: E⊥7 = 0.5 me v⊥72 = 0.5 x 9.1091 x 10—31 x 69,0982 = 2.174585 x 10—21 joule. The fore energy E⊥7 is obtained because (in accordance with our model) the propeller particle yields the inherent energy to the orbital electrons; this energy is equal to:
0.5 mp c2 = 0.5 (3 x 108)2 mp = E┴7
mp = 2.174585 x 10—21 / 4.5 x 1016 = 4.8324 x 10—38 Kg. (mp = m⊥7 )
The fore mass mp is formed by a big quantity of corpuscles, as those of light. A corpuscle has an inherent energy equal to one quantum = h, so: h = 6.6256 x 10—34 joule = 0.5 mc c2 mc = 6.6256 x 10—34 / 4.5 x 1016 = 1.47236 x 10—50 Kg. = mass of a corpuscle.
So a propeller particle is formed with n corpuscles: n = mp / mc = 4.8324 x 10—38 / 1.47236 x 10—50 = 3.282 x 1012 corpuscles. Although we ignore many facts, are able to anticipate some results by logic deductions. For instance, for an active particle to yields its inherent energy to other passive particle or body, it is required that both particles do not have any movement, or both have similar movements. In other words, ought to have a satisfactory contact between them in a polygonal orbit. If the orbital electron moves with a velocity v⊥7, the propeller particle ought to move parallel at v⊥7, also. Other condition for the interchange of inherent energy, before is produced the interaction between both particles; is the time in which it is produced; if the active particle is one corpuscle, it is required one time t; if the active particle is formed by n corpuscles, by this reason its energy grows n times in the same instant; the velocity of interaction (time of vertex) grows in a proportion of n times and the kinetic energy grows in a proportion of n2; so the total energy grows in a proportion of n3.. In order the propeller particle moves at the same velocity v⊥7 of the orbital electron, it is required that the mentioned particle be affected by the following energy :0.5 mp v⊥72 = 0.5 x 4.8324 x 10—38 x 69,0982 = 1.15362 x 10—28 joule.
For the propeller particle to get the fore energy, it is required that into the corpuscles there be a quantity of active gravity microcorpuscles, forming a kind of propeller particle that we will call: nugravity. In this example the nugravity of the propeller particle, is formed by n = 3.282 x 1012 gravity microcorpuscles. If we denominate by mg the mass of a microcorpuscle of gravity, then:
0.5 n mg c2 = 0.5 mp v⊥72 = 0.5 x 3.282 x 1012 (3 x 108)2 mg = 1.15362 x 10—28 joule mg = 1.15362 x 10—28 / (4.5 x 1016 x 3.282 x 1012) = 7.811 x 10—58 Kg. = mass of a microcorpuscle of gravity.
The mass of the nugravity of the propeller particle: n mg = 3.282 x 1012 x 7.811 x 10—58 = 2.56357 x 10—45 Kg.
It is suppose that each corpuscle of light, when is emitted has a given quantity of microcorpuscles that with their inherent energy makes each corpuscle and its microcorpuscles to have equal movements in order they continue united.. We consider that one active microcorpuscle of gravity has each corpuscle, many times acts an interchange of inherent energy during the return distance time. Knowing the mass of a gravity microcorpuscle, have a datum for determine the time of the mentioned particle to yield its inherent energy to the corpuscle that contain it. This will be done in an indirect way, that is, during the return distance effects.
We are interested in which time (2 t>) is produced the deflection of the trajectory of the orbital electron in a vertex of the seven orbit. With the knowledge we have till here, will obtain this value. With the solution of other problem of theme: Velocity of Sound (II) will confirm this. Our first idea was consider the movement of the propeller particle m⊥n = m⊥7, doing this along the diameter of the electron (2 φ), similar to the movement of a corpuscle along the return distance: 2 Lr. In both cases the particle start with a proportional velocity (v⊥7 and c / 2…); when each reaches the point of return, the fore velocities are reduced to zero, and when they arrive to the starting point, they recuperate their original velocity. In accordance with this we have:, for the electron: re = 1.4 x 10—15 m = radius of the electron; then: 2 T> = 4 re / v⊥n = 4 x 1.4 x 10—15 /( 69100 x 0.5) Sec = 4.05 x 10—20 ≈ 4.47 x 10—20 / 1.1. = half time of return distance into the diameter of the electron (with out considering the time of interchange of inherent energy. . . 1.1 = abarralogic effect.. . .
Here we have that 2 T> is proportional to 2 Tr; but we have not take in account that in the trajectory equal to: 4 re the particle m⊥7 delay several times of interchange of inherent energy, between the orbital electron and the propeller particle. This time is very much big compared with 2 .T>, but practically null compared with 2 Tr.
In the theme: Polygonal, Virtual and Coulomb´s Orbits, was obtained the constant value N nx2; in which N = quantity of active corpuscles that are in a vertex of (n ) orbit; nx = number of orbit, that value is proportional to the radius of the orbit, and to other values will be seen forward.. It is possible to find other constant with N, considering the volume in which it acts; and also the time Tx, required for a corpuscle (of light) to move over the diameter of the mentioned spheres; first into the diameter of the ejecter electron, and after in the outer space diameter in which is produced the return distance effect. In our investigation with the polygonal orbits we are familiarized with some characteristic of them, so we supposed there are other constant values using the quantity of corpuscles N; and for to find this constant will consider the variation of (nx) will be inverse proportional to the volume in which it acts.. The volume will be the sphere obtained with a radius equal to the return distance. . . The half return distance: rr = 0.5 c Tr. In a vertex of the seven orbit of the electron: N = m⊥n7 = 3.282 x 1012 corpuscles. The volume of a sphere with radius of an electron: re = 1.4 x 10—15 m, the value of: N1 = 3.282 x 1012 corpuscles
We consider that an orbital electron of the seven shell has a propeller fluid of: m⊥7 = 3.282 x 1012 corpuscles; that are proportional to the volumen of the electron, or to itas radius: r7. Here will make a relation between the movement of the propeller fluid along the diameter of the electron, and the movemente of the ejected fluid (light radiation) along the the eturn distance Lr.
It was say berore that that the variation of time; and also of distance, is proportional to the distance L3. For these will consider the minimun effect of the fore action;, that produce in a time of deflectio: t> ; that can be produced when the propeller fluid moves from one side to the opposite one ( 2 d at c/2 velocity = 4 r at c velocity) in a time t>. This could be consider is produced by one corpuscle: T> x 12. The maximun would be: T> (3,2682 x 1012)3 = Tr = 1.58 x 1018 = return time.
Tr = T> nx3 = T> (3.282 x 1012)3 = 1.58 x 1018 T> = 1.58 x 1018 / (3.282 x 1012)3 = 4.47 x 10—20 Sec. = timee of vertex
Here was made a relation between an astronomic return distance and the movement of the propeller particle m⊥n along the diameter of the electron The time of vertex is: Tlv7 = dv7 / v7 = 2.55 x 10—11 / 308,570 = 8.27 x 10—17 Sec.; Tv7 / T> = 8.27 x 10—17 / 4.47 x 10—20 ≈ m+ / me . . . For the deflection of the orbital electron there is one microcorpuscle acting over one corpuscle.
Monterrey, Néxico, August 30, 2001; January / 2007 Manuel de Hoyos Robles
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. 21 cm. RADIATION OF INTERSTELLAR HYDROGEN (checked)..
In 1949 was discovered the 21 cm. radiation produced in clouds of interstellar hydrogen. I doubt that with modern physics could be given a satisfactory explanation of these phenomenon. In the year 2000 I approached this problem, because I had some ideas how could be produced, and I made a theme with the same name than this one (April 18 / 2000). Now I decided to eliminate that theme, because is more complex than illustrative, in opposition to this one. In the theme: The Gravity Force, was obtained the liberation velocity for a body to escape from the attraction of gravity by a celestial body: vL = (2 g r)0.5 . . . . (A). Here: r is the radius of the planet and g the acceleration produced by it. This same formula (A) can be applied to atomic particles, for instance, to an electron and a particle emitted by it.
In the case of a hydrogen atom that is in a interstellar cloud, it is supposed that can absorb some corpuscles from the other atoms. In this condition its propeller particle m⊥7 = mp gets over saturated; the excess of matter will be ejected by the atom, and here we need to know how much is this excess, and other data. We know that all the atoms can get their propeller particles with more mass than that they have at normal temperature (21ºC), if they are warmed. The excess of heat is emitted in form of caloric fluid. In the atomic particles there are gravity microcorpuscles. In other theme have been say that the orbital electrons are not affected in a direct way by gravity. In some way, here will prove this again, because the propeller particles that move between the positive particles and the orbital electrons have a quantity of gravity microcorpuscles, and if their masses are incremented, the quantity of microcorpuscles will be incremented. Omitting a series of considerations, that could be understand with the numerical example, here will be consider that the interaction between the oversaturated propeller particle and the radiation of: λL = 21 cm Half of the energy of rejection will be employed for produce the λ = 21 cm. radiation, vL = 2 (gL re)0.5 . Here: re = radius of the electron = 1.4 x 10—15 m; v⊥7 = g⊥7 = 69,100 m / Sec.; m⊥7 = 4.8324 x 10—38 Kgs. vL = 2 (69,100 x 1.4 x 10—15)0.5 = 1.967 x 10—5 m / Sec.;
A radiation with a corpuscular separation equal to: λ1 = 1 m has an energy equal to: h (c). Here © has no physical dimension, only a numerical one, equal to 3 x 108. The energy that has the propeller particle of the binary system corresponding to the seven orbital electron is equal to:
0.5 m⊥7 c2; this by one side; by other side we have that the terms: 0.5 m⊥7c2 and h (c) / vh ; corresponding to the energy of propeller fluids, have similar variation of energy...so we can give the following proportion: 0.5 m⊥7c2 : h (c) x / vh. Because the fore characteristic, it is possible to give to (x) any value we which, with out to affect the fore proportion; if we choose: x = 1 we obtain the following equation:
0.5 m⊥7c2 = hν = h (c) / vh vh = h (c) / 0.5 m⊥7 c2 = 6.6256 x 10—34 / 0.5 x 4.8324 x 10—38 x 3 x 108 = 9.14 x 10—5 m / Sec. But if we know that the liberation velocity is equal to: vL Also there is a lineal proportion between vh and vL; so, we have: vh : 1 m :: vL: λL
λL = vL / vh = 1.967 x 10—5 / 9.14 x 10—5 = 0.215 m = 21.5 cm. Impulse of the liberation radiation: mL = mc (c) / (λL) = 1.47236 x 10—50 x 3 x 108 / 0.215 =
2.054 x 10—41 Kgs. mc = mass of a corpuscle.- - - - -
Monterrey, México, March 25, 2002; Manuel de Hoyos Robles
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THE MAXIMUM TEMPERATURE COULD EXIST (CONTROVERTIAL THEME….)
To explain any structure in the nature we ought begin from the most elementary parts. But in the Universe can be imagined from the infinite small toward the infinite big. In this way we are not able to define what are the smaller particles could have an atom. In accordance with mechanophysics (and all physics....) every atom emits fluids of particles as those of light, gravity, electric and magnetic ones. But those particles do not are the structure, but only make to work such structure, we part from them considering they make the structure of the atomic particles, supposing they give not only the form, but their behavior The most elementary atoms are those of hydrogen, and with them is formed most of the matter of the stars, as our Sun.
A hydrogen atom is formed by a proton and an electron, as have been explained in the theme of binary systems. There is not a material contact between both particles, that spin around a point that we can consider as the center of the nucleus of the atom. To understand better this it is required to study the theory of the double fluid, the theory of the polygonal orbits, and of course that of the binary systems, and others ones of mechanophysics. The interaction between the proton or electric positive particle and the electron, the negative one, is due to another particle (a propeller one) , that when is into the electron transmits all its kinetic energy to it, in such a way that this kinetic energy and those due to its translation velocity makes the electron to move around the center of the nucleus, as was explained before. Similar thing can be say with respect to the proton or positive particle of the binary system.
A neutron is formed by a proton and an electron, united in a material way, so that between a neutron and a proton or an electron, there is no rejection, neither attraction forces. Considering this, or the binary systems, it is easy to imagine that uniting two or more binary systems and / or neutrons, it is easy to form more complex atoms, as helium, lithium, etc. This idea, as many great ideas, is easy to understand. As a neutron has not a manifested electric charge, it could be introduced in an easy way into an atom with atomic mass m, and obtain other atom with mass (m + 1), but this is not so simple, because it is required some temperature and pressure, as those are in the stars and some celestial bodies, as will be seen forward with some numerical examples.. In nature there is abundant material to obtain any element it is required; so this first step is no problem. In a second step it is required that each binary system, forming an atom, works in a determined way, different to the other ones.
Before to aboard a second step, it is necessary to understand and interpret other problems in a different way than those accepted by modern physics. In modern physics has been consider that heat is produced by the movement of the atoms and molecules, and with this consideration have been obtained many complexities, as entropy, statistical physics, etc., that we reject. For instance, in a stable gas, all its atoms or molecules have the same temperature, of course in an average way, because if two atomic particles bump, are suppose there is a momentary interchange of inherent energy in accordance with our model; it does not matter with which velocity they move; what determine their temperature is their propeller fluid (caloric fluid) they have into them, and not the kinetic, or any other movement of the particles (atoms or molecules) that move into the volume of the gas, at different velocities. And with this it is visualized that can not be a temperature, higher than a certain limit. In modern physics it is talk of temperatures equivalent to millions degrees Kelvin, produced in some atomic process, and these because it is easy to imagine, for instance, a proton moving at light velocity, in accordance with the actual accepted theory of heat. But in accordance with our theories and models, if the electron that is moving at light velocity has few excess of this propeller fluid into it, its temperature could be near zero Kelvin.
It is not easy to measure the temperature of a free proton, or a free electron, or an atom, moving into a gas, only considering their kinetic energy. It is required to determine such temperature by the quantity of propeller particles there are into each atomic particle of the gas, that could be obtained under certain conditions, that we suppose are in the binary systems, in an easy way, and this is due they are in an equilibrated state. The most simple binary system is that that form the hydrogen atom, because this is not affected by other binary system, as happen in more complex atoms, in which are produced the ce and the ci effects, In the theory will be study here will not be consider these, neither other effects could complicate our explanations. In a binary system free of the ce and the ci effects can consider that the translation velocity of the orbital electron around the nucleus, gives the kinetic energy of it and is equivalent to a calorific temperature. In our theory of the polygonal orbits, it is well known for all have study it, that the most exterior orbit and less energetic one are equivalent to the seven shell, and the most interior one and more energetic is the first one.. In our theory of the polygonal orbits in each shell (n orbit) there is a velocity of the orbital electron equal to: vn = 2,160 / n Kms./ Sec. With this velocity and in accordance with the Boltzmann´s formula, can determine the equivalent temperature Tn of such electron moving in its corresponding orbit.
If it is consider that the velocity of an orbital electron gives a kinetic energy equal to its temperature. Then seen the problem in an opposite way, in a medium in which there are free electrons and protons, in order they unite each other forming binary systems; can be obtained them if the medium is at a temperature Tn , equivalent to the velocity of the electron: vn . The free electron in the medium with temperature Tn , in a fast way gets the velocity vn and after this, the electron continue receiving calorific fluid of the medium. In order the electron gets more caloric fluid than that of equilibrium, it begins to emits the excess of fluid at a rhythm equal to a time of vertex and a quantity proportional to a propeller fluid of orbit. Similar conditions could happens in the free proton into the medium, in such way that when an electron and a proton meet, must form a binary system, It is suppose that also is required some pressure. After is formed in the mentioned way a binary system, corresponding to an atom, this new atom with its new binary system has to move to other zone with less heat, in order the new atom gets in a stable condition. In a metal atom this condition is obtained at a temperature lower than that of the melting point.. If n = 7, is formed a hydrogen atom. If n = 6: v6 = 2,160 / 6 = 360 Kms./ Sec.; K6 = 0.5 m6 v62 = 0.5 x 9.1091 x 10--31 x 360,0002 = 5.9027 x 10—20 joule
In accordance with Boltzmann´s law, the temperature for to form a binary system corresponding to the sixth orbit is: T6 = K6 / (3/2) k = 5.9027 x 10--20 / 1.5 x 1.38 x 10--23 = 2,851o.5 K.
In modern physics has not been defined a limit to the maximum temperature could exist, and some times they talk of temperatures of millions degrees Kelvin, and we consider this is absurd. In accordance with what has been say, the maximum temperature could exist is: T1.
K1 = 0.5 x 9.1091 x 10--31 x 2,160,0002 = 2.125 x 10--18 joule T1 = 2.125 x 10--18 / 1.5 x 1.38 x 10--23 = 102,656o K.
Some atoms have several orbital electrons in a given shell; the shells with more orbital electrons are the fourth and the third one. The Md atom has 32 orbital electrons in the 4 shell, and 30 in the 3 shell. In accordance with our model, the 32 binary systems of the four shell are formed and united to structure the Md atom at a temperature T4 , and practically in a simultaneously way; of course the binary systems more affected by the ce effect require a higher temperature, and the more affected by the ci effect require a lower one. About all these there are many things to be seen, because there are many atoms with the same quantity of orbital electrons in a given shell and ought to be some differences. With all say here can get some ideas: Why the atoms only have seven shells? Why can not exist temperatures higher than T1 ? For instance in the seven shell there is much free space for the two orbital electrons to move. In the first shell the temperature T1 is high, but the space is too small to admit more than 2 orbital electrons . Here have not talk of pressure, chemical properties, etc.
Monterrey, México, July 30, 2001 Manuel de Hoyos Robles
THE MAXIMUM TEMPERATURE COULD EXIST (II)
In all books of physics they consider there could be temperatures of several millions of Kelvin degree. I consider this ides very much unsatisfactory for many reasons, specially, because with this will be required many properties of the matter. Now that I have study more problems in mechanophysics have consider that the velocity of particles or bodies is not the cause that produce the heat, but an effect produced by it. And that for produce the heat it is required the action of the atom; there are many things to talk about this, and this require other themes.
When Boltzmman deduced the formula: T = K / 1.5 k, this was good for temperatures smaller than: T <> 2,160,000 m / Seg., with respect to the Earth. If it shock with a body that is fixed with respect to the Earth, will penetrate in the shell of the atoms of the body, and reduce its velocity. In the distant stars the electromagnetic radiation are received in the Earth at lower velocity (red shift). It has been observed galaxies expanding at velocity bigger than 9 c, and the light we receive of them has a velocity of 1 c…. not 9 c… . . . . . ,
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VELOCITY OF SOUND (I) (corrected theme) Å---- pendiente
For to study the sound conductor we will imagine an isometric distribution of atoms, and imagine the volume of the sample is going to be study (could be a cube of nm3 atoms) is divided by transversal planes separated one from each other a distance equal to the diameter of an atom (2r), so that between two adjacent planes there is a slice of conductor in such way that if the sound is produced in one slice in a given instant, in the next instant the next slice is affected by such sound, and so on. In accordance with the fore explanation will be made our model of sound conductor. Can be considered that only a small part of the atoms of the cross section work as there will be specified, and that the other atoms do not interfere in the production of the sound effect.
nm = length of a multimolecule = 2,497,996 atoms = L
A = nm2 = area of a plane (multiplane)
w = density of the material of the conductor (gr./ cm3)
M = mass of the conductor = A L w = (gr)
m = mass of a slice of conductor = 2r A w
V = volume of sound conductor = L A (cm3 )
dV = volume of a slice of conductor = 2r A
f = fatigue produced in a transversal section of the conductor due to the vibration of sound (gr./ cm2)
E = modulus of elasticity (Kgs./ cm2)
g = acceleration produced by the gravity (cm./ Sec2 )
vs = velocity of sound in a medium whose gravity has a value of: g = 981 cm./ Sec2 (.... ?....
This is only a proportion of a fatigue that is proportional to the intensity of the sound; for a given intensity the value (f) could be equal to the real one.
A f g = force (newton) produced by sound in a plane of the sample in a medium with a gravity that
produces an acceleration equal to (g).
Due to vibration of sound each slice of conductor gets a kinetic energy: 0.5 m vs2 (gr cm2 / Sec2 )
When the actions get regulated in a time of multimolecule, we have a fatigue f (gr./ cm2) acting in a
time t2r in each one of the slices of a unitary space of conductor with length: L = 1 multimolecule.
nm = 2,497,996 atoms; this is mathematically equivalent to a fatigue (f) acting in one slice of the
mentioned conductor, in the mentioned time.
m vs2 = f nm g dV = E g dV (fundamental deduction).
As the mass of the conductor is: M = m nm , then, multiplying the fore equation by: nm
nm m vs2 = M vs2 = E g nm dV = E g V ; here: nm dV = V; f nm = E;
The equation that gives the velocity of sound is: vs = (E g V / M)0.5 = (E g / w)0.5 . . . . (A) Density of material: w = M / V
In order to have a better idea how works our model in accordance with the polygonal orbits will be made a numerical example with iron atoms.
Some times it is necessary to deal with some unsolved problems for to be able to structure a new model that works in accordance with our requirements. Have been commented that each element has its own modulus of elasticity different to the modulus of the other ones elements; till here has not been seen why are produced such differences and in this problem will not be done this, because it would get outside the present theme; here will be limited to consider such modulus of elasticity in accordance with the experimental data, without pretend to make any technical or scientific analysis. In my investigation work I have had some negative experiences trying to explain something before I have study some other correlated problems, in such way that have appeared some contradictions and wrong concepts in some of the theories, because the fore reason, so, I prefer to omit some ideas I have in the present time, and this is because I am preparing this work for the internet.
In a sample of a material, the atoms of the sample are in contact one from each other, in their 7 shell zone, in order the modulus of elasticity E works in the atoms of the sample it is required to compress it applying some forces; because this the shell of their seven orbits will try to penetrate in the other one, because this there will be an opposition between the adjacent atoms, and this is due to the modulus of elasticity. In Fig.(1), with a, b, c, d is represented a piece of the trajectory of the orbital electron of an atom that coincide with an adjacent one, whose trajectory of its corresponding orbital electron is: e, b, c, f. If it is applied a compression force to the sample, the trajectory of the orbital electron of the first atom will change in accordance with the pointed line: a’, b’, c’, d’, in this way the side bc of the first atom penetrate in the shell of the second atom, in accordance with the side b´c´, because this, the modulus of elasticity of the sample gets working opposing such penetration as is well known from experimental data, the fore opposition of the modulus of elasticity is not only produced by the action of mechanical forces of compression, but also by tension, and by vibrations, change of temperature, etc. The tension resistance could be explained considering the link there is between the atoms of each multimolecule, by the orbital electrons.
In our problem of sound, when this begin to work, the atoms of the sample get some kinetic energy in such way that the velocity of the orbital electrons is incremented, of them our immediate interest are those of the seven orbit. It is supposed that the sample is at 0o K when the orbits of two adjacent atoms are as indicated in the two pieces of orbits by the continuous line a, b, c, d and e, b, c,
f. In this condition, the orbits of the two orbital electrons of the two adjacent atoms move freely at their fundamental velocities. Next, because it is produced a sound in the sample, the orbit of the first atom moves from b to b’, then from c to c’ and from d to d’ etc. In this way the first atom will transmit its sound effect to the second one; but this transmission is not produced by one step, or one time of orbit, but by several ones; this is because the transmission is not complete while the orbital electron of the second atom does not get an equal increment than that has the orbital electron of the first atom. For the orbital electron to advance from one side to the opposite one of its orbit, it moves at an average velocity equal to: 308,570 / π = 98,221 m / Sec. For an iron sample the velocity of sound advance a diameter of atom after the following steps (orbits): nFe = 98,221 / 5,080 = 19 .3 orbits. For water: nw = 98,221 / 1,450 = 68 orbits. With the interchange effect can be obtained a very much satisfactory model of sound conductor, because when the atoms vibrate produce tension and compression.
In the structure of our model of sound we see some problems that in some way the practical observations have helped us in a great deal to solve them, and because this, maybe this structuration will not be made in the most logical way, that is, deducing the data with which we work, and with them make all the steps of our structure, some times we will depart from semistructured data. Let aboard our problems; here will be considered all the problems limited exclusively to the effect of sound, avoiding to mix them with others due to other observations, as heat, electricity, magnetism, etc., if the orbital electrons have some change in their velocities, here will be attributed exclusively to an effect of sound; if the atoms of the sample, also have some variation in their movements, also will be attributed exclusively to an effect of sound, and because this will be able to relate both kind of movements one with each other.
Sound, as heat is a kind of energy could be consider, in some way as a particular manifestation of heat, in this way the atoms get some movements and these movements are produced by the interior particles of them, that is, the orbital electrons, or viceversa, it does not matter how it is. One important fact is that if all the orbital electrons of the atoms, change their movements due to sound, only with considering the effect produced by the orbital electrons of the seven shell we will be able to determine the exterior movements of the atoms of the sample, and because this, the structure of our model will simplify in a great deal. The orbital electrons of the seven shell are those that are in contact with their adjacent atoms, and because this, they are responsible of the exterior movements of the atoms of the sample; in these electrons are manifested in a direct way some properties of the atoms, as their modulus of elasticity, in which is based our model; about this modulus of elasticity, that gives resistance of compression and tension to the atoms, as was say, here will omit give details have not been study yet; here we will be limited to apply its properties to the model. In the integration of our models, by preference will be considered binary system formed with a positive particle of one proton. When the sample is not affected by sound, would be considered the orbital electrons (7 shell) have a velocity = ve7. When the sound acts, this velocity is incremented in a quantity equal to: Δ ve7 It was seen before, that for sound advance the diameter of an atom, the seven orbit electron of the iron atom makes nFe = 19.3 orbits:
nFe = v7 / π vs = 308,570 /3.1416 x 5080 = 19.3
In all these have been consider that all the atoms of the sample are affected by sound in a given time(“mathematical” time). In the He atoms at lambda temperature, only 9180.5 (see theme; Superfluid Helium 3) orbital electrons from 2,497,996 atoms in a cross section of the sample are affected, in this way the modulus of elasticity E could not be apply in the same way as in the fore example, because the fatigue produced in one atom will be distributed in: 2,497,996 / 9180.5 atoms.
For He4: ma = 4 x 1.6725 x 10- 27 = 6.69 x 10—27 Kg.
In a weekly popular science magazine: “Ciencia Ilustrada”, edited in Sao Pablo, Brazil some years ago I read an article about the Russian investigator Lev Davidovish Landau, with respect to an investigation work with liquid helium at very low temperatures, in 1941, he foresaw the existence of a kind of vibrator propagation in the liquid helium, different to the ordinary sound, this because the superfluity of helium. He named it sound second and show that near absolute zero those waves would have a velocity of: cs / 30.5 , were: cs = velocity of ordinary sound. In 1944 the sound second was confirmed in the practice. About ten years later he studied a new type of wave in helium 3 that he named: sound zero = 30.5 cs all these results were obtained by very much complex deductions. In our theme: Superfluid Helium 3 was say that at the same time it is tried to obtain 0o K also it is tried to produce an ionization effect; although here will not be solved this problem, it is logic to consider that the effects of sound zero and sound second, are related with this.
With our model of sound conductor the fore results can be obtained in a simple way and at the same time are confirmed very important facts of mechanophysics, as the theories of the polygonal orbits; of the multimolecules; of the binary systems; of the caloric fluids that can vary from zero to light velocity; etc. As we have not access to data practically exclusive of specialized investigators, as some sound velocities at different temperatures, pressure, modulus of elasticity, etc., the explanations will be given without these data; maybe in the future will be able to deduce them theoretically.
Water density: ww = 1 gr./ cm3 Helium 3 “ : w3 = 1 x 3 / 18 = 0.1667 gr./ cm3 Helium 4 “ w4 = 1 x 4 / 18 = 0.222 “
If we confine a liquid into a closed recipient and reduce in a mechanical way its volume, the liquid substance will oppose to this. Knowing the energy required to reduce a given volume of the liquid, can be determined its modulus of elasticity. The fore explanations are well known facts for any physicist, but seen from the point of view of mechanophysics they have a special importance as will be seen in the determination of sound zero. Because the helium is at lambda temperature: T = 2o.2 K; applying Boltzmann’s formula, can be obtained the increment of energy of the orbital electron due to the lambda temperature: K2 = K He = 1.5 k T = 1.5 x 1.38 x 10-- 23 x 2o.2 = 4.554 x 10-- 23 joule / atom.
In classic physics it is considered that sound is propagated by waves, in accordance with this is a frequency and a wave length. For an equal wave length one system of waves has more amplitude while more energy has. Considering a similar medium in which sound propagates, the velocity of sound in this medium depends of the properties of such medium. The length [λ] and the frequency [ν] of a wave of sound vary with the intensity of it, but as: λν = constant, also the velocity of sound is constant in such medium, it does not matter that are higher or lower tones of sound.
At very low temperatures, as the lambda, it can not be consider that He is affected by sound in a similar way that if He is at higher temperature, because at lambda temperature the exterior energy: K = 4.554 x 10-- 23 joule / atom, can not correspond to an interior energy characteristic of an ordinary sound; from this point of view can not be considered that He at 2o.2 K is a similar medium than He at 14o.85 K, or higher. At very low temperatures the fatigue (f) is not constant but grows in the same proportion than the temperature.
In the theme: Superfluid Helium 3, was considered that from 2,497,996 multimolecules, only 9180.5 ≈ 30 multimolecules were working for produce sound second from the helium in a liquid state (lambda temperature = 2o2 K). If it is applied formula (A), can not be considered that the acting multimolecules (30), can produce the normal fatigue (f) in the transversal section (multiplane) of the sample of the sound conductor, so that: Σ f2 = n f2 < n =" 2,497,996" k2 =" 0.5" vs22 =" 4.554" vs22 =" 4.554" 27 =" 13,614" vs2 =" 116.7" khe3 =" 0.5" 12 =" 3.075" k2 =" 3.075" 23 =" 6.75;" to =" 6.75" t2 =" 6.75" 2 =" 14o.85" mo =" 3" 4 =" 0.75" fo =" 0.75" f2 =" 6.75" fo =" n" vs0 =" 3" vs2 =" 3" 7 =" 350.1" vs =" 202" tj =" 1.12" tm =" n" tj =" 2,497,996" 11 =" 2.798" orbit =" 2.3084" n12 =" 1.9317" 2 =" 1" n =" 2,497,996" atoms =" number" r =" 2.275" cm =" diameter" n2 =" 2,497,9962" vs =" (E" e =" w" vs =" sound" ma =" mass" vs =" sound" e =" modulus" g =" gravity" 3 =" volume" vs2 =" 0.5" e =" 2.1" vs =" [E" 3 =" (2" 3 =" 1.1778" m3 =" effective" atom =" effective" ma =" 56" 27 =" 9.366" 1=" ma" 3 =" 9.366" 29 =" 7952" m3 =" density" vs =" (E" 5 =" (2.1" 5 =" 5090" vs2 =" 0.5" 50902 =" 1.213276" 3 =" 0.5" 29 =" 1.2132" fe =" ma" vs =" 9.366" 5090 =" 4.7673" fi =" E" 2 =" 2.1" 2 =" 1.06647" rfe =" 1.13763" vs =" sound" atoms =" n" tj =" 1.12"> = 4.47 x 10—20 Sec. (see theme: The Deflection Time). But as the force produced by E in the so called interior effect is manifested in the atomic magnitude, as a liquid substance. A solid body does not distribute the tension or compression fatigues in a uniform way in all directions of its atoms, as does a liquid substance. But this does not matter, because the fatigue produced by sound is distributed in a uniform way in all the atoms of the solid body.
The effect of E is not only manifested in a vertex (the direct contacted one) of the seven shell, but in all them, that is in the 28 vertices. The orbital electron continually is moving. In the seven shell of almost all atoms (except H), there are two orbital electrons, and this means that both are affected by sound at the same time. If here is only consider the action of one acting in half of the system, will give the same final result than two acting in all the system.. Note: the advance in each one of the two of the 7 obits of them. is only in half of them…
The external fatigue (fe) will acts in the extreme atom of a multimolecule, but will affect all the atoms of it, in similar way that a force acting in the extremes of a fixed rod, or column will affect all the atoms of the rod. The external force is acting in one atom; this atom is the extreme one of the rod or chain of (n) atoms, that also are affected, as was say.. Fe tj n = 4.7673 x 10—22 x 1.12 x 10—11 n = 5.3394 x 10—33 n Kg. m 28 Fi t> = 14 x 2 x 1.06647 x 10—8 x 4.47 x 10—20 = 1.3348 x 10—26 Kg. m n = 28 Fi t> / Fe tj = 1.3348 x 10—26 / 5.3394 x 10—33 = 2.5 x 106 = quantity of atoms that has a multimolecule
As (n) resulted equal to the number of atoms that has a multimolecule, this gives a preponderant value to all mechanophysics. With this we are able to determine the energy required to produce sound in iron. Such energy is produced in a time of multimolecule: tm = 1.12 x 10—11 x 2.5 x 106 = 2.8 x 10—5 Sec. . Energy of sound: Ks = tm Ki = 2.8 x 10—5 x 1.2132 x 10—18 = 3.4 x 10—23 joule / Sec.. We do not know the details of the problems that had Landau and other physicist in studding the helium at very low temperatures. With respect to the property of helium to climbs up in the recipient, we deduced there is a good cohesion between the helium and the wall of such recipient; being the cohesion stronger than its capacity of evaporation due to its light weight and if it climbs up by the wall of the recipient, it is because the helium is lighter than the air. If the recipient and the helium were in a vacuum space, the helium will not try to climbs. With respect to sound second, we have at lambda temperature: t2 = 2.o2 K.
Applying Boltzmann´s formula: K2 = 1.5 k t2 = 1.5 x 1.38 x 10—23 x 2.2 = 4.554 x 10—23 joule / atom K2 = 0.5 ma vs2 = 0.5 x 4 x 1.6725 x 10—27 vs22 = 3.345 x 10—27 vs22 vs2 = (4.554 x 10—23 / 3.345 x 10—27)0.5 = 116.7 m / Sec. vs = 30.5 x 116.7 = 202 m / Sec .
With respect to sound zero, in He 3:
vs0 = 3 x 116,7 = 350 m / Sec.
Ko = (3 / 4) x 4.554 x 10—23 x 32 = 3.074 x 10—22 joule / atom.
to = (3 / 4) 2.2 x 32 = 14o.85 K
Monterrey, México, February 19, 2002 Manuel de Hoyos Robles
jueves, 12 de noviembre de 2009
MECHANO PHYSICS
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