Determining the Constants of The Mechanical Theory of Everything

My name is Joseph Milroy Brown (aka) Joe Brown. I was born on January 9, 1928 in the rural community of Gatewood near Fayetteville in southern West Virginia. I received two bachelor degrees in Mechanical Engineering from WVA University and Master and PhD degrees from Purdue with emphasis in Machine Design, Mechanics, and Mathematics. I worked in the Aerospace Industry for 20 years designing airplanes, missiles, space boosters, and satellites. I then taught mechanical engineering at Mississippi State University for 21 years. I am here to describe the unified theory of physical science presented in The Mechanical Theory of Everything.

I postulated that the universe is populated by an ether gas as required to transmit gravitational forces. The ether is a gas consisting of small elastic spheres moving at high velocities. The gas is very dense and very energetic. One cubic meter of the gas has enough energy to power the earth for millennia and the density is fifty trillion times as large as lead. How does nature obtain useful energy from this ether and how can we move so freely in such a dense ether?

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Synopsis and Further Developments to the Mechanical Theory of Everything

by Joseph M. Brown1

Quick Links: [ The Postulated Ether | Neutrinos | Matter | The Hydrogen Atom | Electrostatics and Gravitation | The Fine Structure | Accelerating an Electron | Accelerating Matter and the Demise of Einstein’s Relativity | The Schrödinger Equation | Kinematics of the Motion of a Translating Particle | Correcting the Schrödinger Equation | Model of a Moving Matter Particle | Derivation of the Corrected Schrödinger Equation | How We Get Our Energy | Analysis of Results | Appendix – Proof that the Energy of a Free Translating Matter Particle is $mv^2$ at High Velocity | Footnotes & References ]

The Postulated Ether

The brutino is the smallest thing in the universe. See Figure 1. Everything is made of brutinos—and nothing else.

The brutinos make up a rare gas which pervades the universe, and which extends indefinitely in all directions. Figure 2 shows the gas. Every cubic meter of the universe has $10^83$ brutinos in it, they move at a speed of $4 \times 10^9$ m/s (roughly ten times the speed of light), they have a mass of $10^{-66}$ kg. The energy in each cubic meter of space is

$\eta_0 m_b v_r^2 = 10^{83} \times 2.89 \times 10^{-66} \times (4 \times 10^9)^2 = 10^{37}$ joules/m3 [Eq. 1]

This is enough energy to supply the energy used on earth for millennia. The particles move and collide. All forces in the universe are the result of repeated collisions of the brutinos. The ether possesses a vast amount of energy. Nature, however, only lets us use a small amount of the energy. Getting useful energy out of the ether is akin to getting energy out of the atmosphere. But, the ether is different, which we will explain later, and show how all of our useful energy is derived from the ether.

The ether density is 50 trillion times that of lead. One would immediately wonder how we move in a sea of such a large density. When you learn what matter is it will become clear how things move without impediment from the dense ether.

Figure 1. The Brutino, the Ether Particle

$r_b = 4.052 \times 10^{-35}$ meters
$m_b = 2.89 \times 10^{-66}$ kilograms
$v_m = 3.510373 \times 10^9$ meters/second
$v_r = \sqrt{3\pi / 8} \times v_m = 3.813809 \times 10^9$ meters/second

The gas made up of brutinos is called the ether.

$\eta_0 = 1.46 \times 10^{83}$ brutinos per cubic meter of space
$\rho_0 = 4.23 \times 10^{17}$ kilograms per cubic meter of space
$ \ell = 2.35 \times 10^{-16}$ meters travel distance between impacts
$ s = \left( \frac{1}{1.46 \times 10^{83}} \right )^\frac{1}{3} = 1.899 \times 10^{-28} $, the average number of meters between brutinos.

Figure 2. The Brutino Gas of Ether Particles

The brutinos translate with a distribution of speeds, some very fast and some very slow. Figure 3 shows the distribution. The most probable speed, $v_{mp}$, is the highest ordinate shown in the figure. The average of the speeds $v_m$ also is shown, and is slightly larger than the most probable speed. The speed $v_r$ is the square root of the sum of the squares of the speeds of all particles divided by the number of particles. This speed

$v_r = \sqrt{\frac{v_1^2 + v_2^2 + v_3^2 + \cdots + v_n^2 }{n}}$ [Eq. 2]

is the RMS speed and is $ \sqrt{3 \pi / 8}= 1.085$ times the mean speed. The RMS speed is always greater then the mean speed, unless the speeds are the same.

Figure 3. Distribution of Ether Particles

Neutrinos

Complete condensations of these brutinos occur randomly and produce neutrinos. The neutrino is defined by its sonic sphere which is the surface at which brutinos flow into the neutrino at their speed of sound. The surface is almost spherical and its radius is approximately $10^{-15}$ m. The neutrino takes in mass at the rate of $10^{-2}$ kg/s. The neutrinos are formed with masses which vary by several orders of magnitude, with right- and left-handed twist. They expel the incoming mass out of two fine streams, one directed forward at velocity $v_r$ and one directed aft at velocity $v_m$. The neutrino develops a thrust of 1.43 meganewtons. It produces power $ 1.43 \times 10^6 \times 3 \times 10^8 = 4.29 \times 10^{14} $ watts. Figure 4 shows the neutrino. Neutrinos are made continually.

Figure 4. The Neutrino

The first secret of the neutrinos existence is our discovery of the microrocket pump. Just by aligning the background ether particles and squeezing them together produces a pump which produces a thrust of over a million Newtons and produces a vacuum to continually suck ether particles in, and then propels itself at the velocity of light through the dense ether. The second secret is that the pump had to be large enough so that the sonic sphere radius is approximately equal the mean free path so that incoming particles would not reach the other side of the sphere and, thus interrupt the inflow.

On the other hand, if the pump is too large and the sonic radius is larger than the mean free path then particles will bunch up as they enter the sonic sphere and greatly impede the input. For a given gas there is an optimum value of $r_c$ , and pump size, which will let the particles enter the sphere with no back pressure and thus be easily compressed as they flow inward. Thus, they can easily enter the small core. The gas also must leave a large enough MFP to particle radius so that the core is large enough to sustain itself.

For a given gas there is only one pump rate and $r_c$ which can produce solidification. This is why all neutrinos have the same angular momentum.

The ether gas requires a pump rate which gives an $r_c = 7.50 \times 10^{-16}$ m, which is approximately 3 times $ \ell (\ell = 2.35 \times 10^{-16} m)$.

Matter

On rare occasion a neutrino having the mass of a proton will begin orbiting in a circle with a radius of $10^{-16}$ m. The thrust of 1.43 meganewtons balances the centrifugal force and the angular momentum of $\hbar / 2$ requires that the mass be the proton mass. Such a process forms a proton. Figure 5 shows a proton.

Figure 5. The Proton

As the proton neutrino orbits, it sprays out brutinos at the velocity $v_r$ followed by a spray in the opposite direction at velocity $v_m$ . The overall effect of these outputs is to produce an in-and-out, expanding-contracting spherically symmetric flow similar to that of a breathing sphere. This breathing flow produces the Coulomb field. Polarity is produced by the right-hand twist of the proton neutrino.

The Coulomb field produces the attractive force between a proton and an electron making a hydrogen atom. The force is given by

$ F = \frac{e^2}{r_{ep}^2} $ [Eq. 3]

here $r_{ep}$ is the distance between electron and proton. Equating this to the centrifugal force we have

$ \frac{e^2}{r_{ep}^2} = \frac{mv^2}{r_{cg}} $ [Eq. 4]

here $r_{cg}$ is the distance from the electron to the center of gravity of the electron/proton system. Now

$ r_{ep} = r_{cg} \left ( 1 + \frac{m_e}{m_p} \right ) $ [Eq. 5]

Substituting from (5) into (4) gives

$ \frac{e^2}{r_{cg} \left ( 1 + \frac{m_e}{m_p} \right )^2} = m_e v^2 $ [Eq. 6]

From the dynamics of the photon impacting an electron we have the de Broglie wavelength for the resulting (translating) electron which travels around the proton in a wavelength

$ \lambda = \frac{h}{m_e v} = 2 \pi r_{cg}, \;\;\; v = \frac{h}{m_e r_{cg}} $ [Eq. 7]

Substituting this into (6) we have

$\frac{e^2}{r_{cg} \left ( 1 + \frac{m_e}{m_p} \right)^2} = \frac{\hbar}{r_{cg}} v, \;\;\; v = \frac{e^2}{\hbar \left ( 1 + \frac{m_e}{m_p} \right )^2} $ [Eq. 8]

Note here, if we ignore $m_e /m_p$ then (8) gives the fine structure constant $\alpha$.

$ \frac{v}{c} = \frac{e^2}{\hbar c} = \alpha $ [Eq. 9]

Returning to (6) and using $v$ from (8) gives

$m_e r_{cg} = \frac{\hbar^2 \left ( 1 + \frac{m_e}{m_p} \right )^2}{e^2} $ [Eq. 10]

All matter at rest consists of orbiting neutrinos moving in circular paths at the speed of light. In order for matter to move, a group of ether particles, making up a photon, impact matter and cause the orbiting neutrino to take a plane spiral path without changing its velocity. Thus, motion is easy to achieve in the super dense background.

The Hydrogen Atom

When the proton was formed the electron was formed at the same time. The electron was formed to balance the positive electrostatic field of the proton. However, what determines the mass of the electron? Using the Coulomb force and the resulting de Broglie wavelength all we can determine is the product of the mass and orbital radius (i.e., $m_e r_{cg}$).

For many years we have assumed that the neutrino can be formed having a large continuous spectrum of masses, certainly higher than the mass of the proton (or tauon) and lower than the mass of the electron. Recent research has shown an undisturbed neutrino can change masses from one mass (the tauon neutrino, muon neutrino, or electron neutrino) to another. This changing may not be a great surprise since the mass in each neutrino is completely changed some 20 orders of magnitude times per second.

Possibly there are only three classes of neutrinos: One which has a core which is nearly square (i.e., $\ell/d=1$) being the tauon/proton neutrino, one which possibly has the largest possible $\ell/d$ being the electron neutrino, and the other having an intermediate $\ell/d$ being the muon neutrino. Possibly there are only discrete masses and each one is due to a different resonant flow. In any case, however, the existence of the three classes of neutrinos could explain the existence of the three leptons (i.e., the electron, muon, and tauon). All three lepton structures would be similar to the electron structure with their inertial, electrostatic charge, and angular momentum loops.

Returning to formation of the hydrogen atom and the determination of the electron mass the electron may consist of the smallest possible neutrino and thus be the minimum energy solution to the hydrogen atom formation. Once the mass is determined then the orbital radius (the Bohr radius) is determined.

Electrostatics and Gravitation

A pair of breathing spheres immersed in water will attract each other when oscillating in-phase and will repulse each other when oscillating out-of-phase. The force varies inversely with the square of the separation distance. We have proven theoretically that orbiting neutrinos (making matter) produce the same effect as breathing spheres. Thus we have proven that the field produced by a proton models the electrostatic field, i.e., the Coulomb field. With this result we obtained the force $e^2 / r^2$ between two protons or any two electrostatic fields. Furthermore, we determined the (gravitational) field produced by two orbiting fields, such as that produced by the orbiting electron and proton in the hydrogen atom. These fields orbit at an amplitude, with respect to each other, equal to the ether particles radius. This amplitude is extremely small so the gravitational force is very much smaller than the electrostatic force.

Electrostatic and gravitational forces are due to breathing spheres. Einstein’s warped space is not required for gravitation.

The Fine Structure

In addition to the general breathing motion produced by the proton there are waves with crest-to-crest dimensions in the order of $10^{-16}$ m, the radius of the orbiting neutrino making the proton. The tangential motion in the field also produces $10^{-16}$ m sized waves. The resulting configuration consists of $10^{-16}$ m3 sized flows which we call wavespaces. The wavespaces are the fine structure of the electrostatic field. The wavespaces move radially from the charged particle at the velocity of light $c$ which is very slightly less than $v_r – v_m$. The wavespaces carry away the mass that is continually absorbed by the proton neutrino.

Figure 6 shows the fine structure of the proton field. Each of the wavespaces has a circulation flow which is right-handed for a positive charge and left-handed for a negative charge. Right-handedness is determined by the counterclockwise flow of the neutrino making the proton.

Figure 6. Fine Structure of the Photon Electrostatic Field

The fine structure constant $\alpha$ is the velocity at which an electron of hydrogen one (1H) travels in its lowest orbit, measured in speed of light units. Its value is

$ \alpha = \frac{e^2}{\hbar c} = \frac{1}{137.03599976} = 0.007297352536 $ [Eq. 11]

which is known to one part in 10 . We noted that the term

$ \left ( \frac{v_r – v_m}{v_m} \right )^2 = \left ( \sqrt{\frac{3 \pi}{8}} – 1 \right )^2 = \frac{1}{137.108733} = 0.007293481422 $ [Eq. 12]

was very close to the value of the fine structure constant. Current thinking is that the speed of light (i.e., the speed of the fine structure wavespaces) is slightly less than $ v_r – v_m$ since in the neutrino $v_r$ is generated at the same location as $v_m$ , while in the charged particle field $v_m$ follows $v_r$ by the amount of time to travel the length of the neutrino. As a result, the speed of light is slightly less than $v_r – v_m$ . If $c$ is $0.999720744 \times (v_r – v_m)$ then the square of the bracketed term below is $\alpha$ divided by $ \left ( 1 + \frac{m_e}{m_p} \right )^2 $.

$ \left [0.999720744 \times \frac{v_r – v_m}{v_m} \right ]^2 = \frac{\alpha}{\left ( 1 + \frac{m_e}{m_p} \right )^2} $ [Eq. 13]

from this

$ \alpha = \left ( 1 + \frac{m_e}{m_p} \right )^2 \left [ 0.999720744 \times \frac{v_r – v_m}{v_m} \right ]^2 $ [Eq. 14]

which is the formula for the fine structure constant.

Accelerating an Electron

Let us consider an atom which is just ready to emit a photon. The mass required to make the photon is believed to be stored in an elliptic string of wavespaces. For low energy interactions ($v/c \ll 1$) the ellipse is almost circular. Upon emission, the photon mass unwinds as a string unwinds from a spool. The amplitude of the resulting string of mass is the radius of the ellipse and the length is $2 \pi$ times the radius. The photon mass is equally distributed along the string—the same number of brutinos are in each wavespace. The photon then translates, without undulating, much as a metal wire bent in a harmonic shape would translate. The photon translates in the radial direction from the atom at the speed of the carriers of its mass (i.e., the wavespaces move at the speed of light.)

When a photon impacts an electron at rest a portion of its mass is transferred to an elliptic ring of the electron’s fine structure wavespaces, as shown in Figure 7.

Figure 7. Photon/Electron Interaction.

The approaching photon has the number of brutinos per wavespace equal to the sum of the captured brutinos plus the scattered brutinos per wavespace. The dots in the wavespaces are symbolic of many brutinos. Figure 7a shows the approaching photon. Figure 7b shows the electron at rest along with its wavespaces. Figure 7c shows the electron after impact and moving to the right at velocity $v$ where $v \ll c$. Figure 7d shows the scattered photon.

Figure 8. Geometry of the Accelerating Mass

Figure 8 shows the geometry of the photon/electron interaction. We show $A$ as a center for the photon mass which is collected in the wavespaces of the electron. We take $O$ as the center of mass of the electron/accelerating mass system and the electron rotates around this center but with an extremely small radius as compared to that of the captured mass. Note that the captured photon mass is still moving at the speed of light.

Accelerating Matter and the Demise of Einstein’s Relativity

When matter is accelerated mass impinges on the matter, some mass is scattered, some is captured by the field and the orbiting neutrino (which comprises the matter) takes a plane spiral path. Simply from writing and solving $F=ma$ produced the mass growth equation $m_v = m_0 / \sqrt{1-(v/c)^2}$. Furthermore, the spiral path takes longer for a cycle than the circular path so that the moving time for orbit is $\tau_v = \tau_0 / \sqrt{1-(v/c)^2}$. Finally, the orbits seen from a reference moving with the particle is an ellipse whose minor diameter is parallel to the velocity and is $ \sqrt{1-(v/c)^2}$ times the invariant major diameter. Thus, matter shortens as given by $\ell_v = \ell_0 \sqrt{1-(v/c)^2}$ where $\ell_0$ is the length at rest. These equations are identical to the Einstein space-time equations known as the theory of relativity. The newton and Einstein equations have been tested by immeasurable experiments during the past century—always verifying the simpler Newtonian equation and the Einstein equations. However, indications are that an electron accelerated in different directions at the same time and place on the earth would show the Einstein theory is incorrect.

The Schrödinger Equation

When a photon impacts a free matter particle at rest, mass and momentum are imparted to the particle and it accelerates to velocity v. The center of mass of the imparted mass is captured at such a radius that the angular momentum imparted to the system is equal to Planck’s constant h. The imparted mass and the impacted particle mass remain at a fixed distance from each other and they begin rotating about their common center of mass as the system center of mass translates in a straight line. This motion is manifested as a matter particle undulating as it translates. The wavelength of this undulation is $h/(mv)$, where $m$ is the matter particle mass. The Schrödinger equation models the dynamics of the motion of the matter particle relative to a reference frame moving with the captured mass/matter particle system. Solution to the equation gives the velocity of the matter particle as a function of the location of the particle. We derive the Schrödinger equation by balancing the centrifugal forces against the centripetal forces. Thus, we show that the Schrödinger equation is a Newtonian equation.

The Schrödinger equation models the dynamics of matter particles when they translate. The equation was discovered in 1925 by the Austrian physicist Dr. Erwin Schrödinger. Discover is used since neither he nor any other scientist has been able to derive the equation from basic principles during the 90 plus years since its discovery. The equation, along with Einstein’s special theory of relativity, ushered in a whole new theory of physical science, namely Modern Physics consisting of quantum theory and relativistic theory. These theories both rejected classical, or Newtonian, theory. The equation has been used without fail by scientists throughout the world for almost a century. There can be no doubt about its capability for accurately predicting the behavior of nuclear particles.

The Schrödinger equation is a second order partial differential equation in two variables, space and time. The totalenergy of the translating particle is needed and physicists all over the world assume that the energy for a free translating particle is $mv^2$ , where m is the mass of the particle and v is its velocity. They conclude that this is the energy since the work done on the particle by the accelerating medium is $mv^2/2$, i.e., the kinetic energy. The difficulty with this value of energy shows up in solving the Schrödinger equation. For the free translating particle the Schrödinger equation is solved by separation of variables giving two ordinary differential equations—a time equation and a space equation. The wavelength for the particle is $h / (mv)$, the deBroglie wavelength. The wavelength from the time equation, using the total energy as $mv 2 / 2$ , is twice the deBroglie length, which is impossible. From this we conclude that the total energy is $mv^2$.

Next, we examine the space equation. Using the energy as $mv^2$ gives the result that the wavelength from the space equation is half the deBroglie length. By replacing the factor 2 in the denominator of one term of the Schrödinger equation by unity results in a space equation with the same wavelength as the deBroglie wavelength.

From the above two paragraphs we have proven that the Schrödinger equation has two errors: the particle total energy is half the actual energy and the factor 2 should not be in the original equation. The two errors used in the space equation balance each other so that predictions of the equation are accurate. The equation is useful, without fail.

With our model of a translating matter particle we show that the interpretation of the psi ($\psi$) function in the Schrödinger equation is not related to the probable location of the particle but is the velocity of the particle relative to a reference frame translating with the particle. We then show that the (corrected) Schrödinger equation results from summing forces, i.e., applying Newton’s equations.

Kinematics of the Motion of a Translating Particle

In the following discussion we consider the translation of an electron. Figure 9 shows a translating electron.

Figure 9. Geometry of the Translating Electron

The electron is labeled $e$ in the figure. It rotates in a circle of radius $r$ with velocity $v_{e/g}$ relative to the moving frame about the electron mass/captured mass center of gravity, g. The captured mass, $m_c$ , rotates about g with a radius R and a velocity $v_{c/g}$ relative to the moving frame. The angular velocity is $\omega$. The acceleration of the electron is $a_e$ and the centripetal force is F. The acceleration of the captured mass, $m_c$, is $a_c$ and its centripetal force also is F.

Figure 10 shows the absolute motion of the electron and captured mass as a function of distance traveled.

Figure 10. Paths of an Electron and Captured Mass

Figure 11 shows the path of the electron relative to the translating system. We note that the path is a circle.

Figure 11. Electron Path Relative to the Electron/Captured Mass System

While the electron travels one (deBroglie) wavelength the electron rotates one cycle, or a distance of $2 \pi r$. Thus, $2 \pi r$ is its path length and it travels this distance while the electron travels $\lambda$, the deBroglie wavelength. Now

$ 2 \pi r = \lambda $ [Eq 15]

Further, $v_{e/g}$ must be the same as v since the distances and elapsed times are the same.

The vertical and horizontal components of the velocity $v_{e/g}$ (= v) are harmonic functions, obviously. Figure 12 shows the horizontal component of the electron velocity relative to the translating reference system.

Figure 12. Horizontal Component of Electron Velocity vs. Location

This distribution is the same as obtained from the solution to the Schrödinger equation. However, quantum mechanicists incorrectly assume $\psi$ is related to the probable dwell time.

$p_{1-2} = A \int_{x_1}^{x_2} = \psi \psi^* dx $ [Eq. 16]

where $p_{1-2}$ is the probability that the electron is located between distances $x_1$ and $x_2$, A is a normalizing constant which makes the total probability unity, and $\psi^*$ is the complex conjugate of $\psi$.

The current quantum mechanics paradigm assumes that $\psi$ is related to the dwell time of the electron along its path. We now determine the dwell time distribution based on our model of the translating electron.

From the equation for velocity

$v_{e/{g_x}} = v_x = \frac{dx}{dt}$ [Eq. 17]

we have

$ dt = \frac{dx}{v_x} = \frac{dx}{v \sin \theta} $ [Eq. 18]

Then the time $\tau_{1-2}$ to move from $x_1$ to $x_2$ where the distances are measured relative to the moving frame is given by

$ \tau_{1-2} = \int_{x_1}^{x_2} \frac{dx}{v \sin \theta} = \int_{x_1}^{x_2} f_t dx $ [Eq. 19]

where $f_t$ is defined as the dwell-time frequency and is given by

$ f_t = \frac{1}{v \sin (x/ƛ)} = \frac{1}{v \sin (2 \pi x / \lambda)} $ [Eq. 20]

Now we can write

$ v \tau_{1-2} = \int_{x_1}^{x_2} \frac{dx}{sin (x/ƛ)}, \;\;\; v dt = \frac{dx}{sin(x/ƛ)}$ [Eq. 21]

A plot of $f_t v$ is given in Figure 13.

Figure 13. Electron Dwell Time Frequency Versus Distance Along Path

Comparing Figure 13 with Figure 12, the Schrödinger solution, we conclude that the psi ($ \psi $) function is actually the electron velocity. The real part and the complex part give the velocity components parallel to the x and y axes.

From the above we replace $ \psi $ by v in the Schrödinger equation. Now v is a complex number given by

$ v = v \sin \theta + iv \cos \theta $ [Eq. 22]

where $v \sin \theta $ is the horizontal component of the velocity and $ iv \cos \theta $ is the vertical component.

Correcting the Schrödinger Equation

The Schrödinger equation for the free translation of a particle of mass m is

$ \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + i \hbar \frac{\partial \psi}{\partial t} = 0 $ [Eq. 23]

where $\psi(x,t)$ is a function associated with the probable location of the particle. Assuming $\psi$ is a function of x and t separately the equation is solved by using

$\psi(x,t) = X(x) T(t)$ [Eq. 24]

where X is a function of x only and T is a function of t only. Substituting (24) into (23) gives the two ordinary differential equations

$\frac{i \hbar}{T} \frac{dT}{dt} = E$ [Eq. 25]

and

$ \frac{\hbar^2}{2m} \frac{1}{X} \frac{d^2 X}{dx^2} = -E$ [Eq. 26]

In these expressions E is the separation constant and is the total energy of the free particle.

Solving (25) gives

$T(t) = \cos \frac{Et}{\hbar} – i \sin \frac{Et}{\hbar}$ [Eq. 27]

Quantum electrodynamacists use the total energy as $mv^2/2$. Substituting the energy into the arguments of cosine and sine in (27) and using $\hbar$ given by the deBroglie wavelength of the particle,

$ \hbar = mvr $ [Eq. 28]

gives

$\frac{Et}{\hbar} = \frac{\left ( mv^2 /2 \right) t}{mvr} = \frac{v}{2r} t$ [Eq. 29]

The period of this wave, $ \tau_t $, is obtained by setting $vt/(2r)$ equal to $2 \pi$. Thus

(30)

The period of the deBroglie wave, $\tau_d$, is

$ \tau_d = \frac{2 \pi r}{v} $ [Eq. 31]

From (30) and (31) the ratio of the period from the time equation to the deBroglie period is

$ \frac{\tau_t}{\tau_d} = \left ( 2 \frac{2 \pi r}{v} \right ) \div \left ( \frac{2 \pi r}{v} \right ) = 2$ [Eq. 32]

which, of course, is impossible. The ratio of the wavelengths is

$ \lambda_\tau / \lambda_d = 2 $ [Eq. 33]

since the time is doubled and the velocity is the same. If in (29) we use the energy as $mv^2$ then (29) becomes

$ \frac{Et}{\hbar} = \frac{\left ( mv^2 \right ) t}{mvr} = \frac{v}{r} t $ [Eq. 34]

The period $\tau_t$ now is given by

$\frac{v \tau_t}{r} = 2 \pi, \;\;\;\; \tau_t =\frac{2 \pi r}{v}$ [Eq. 35]

and

$ \frac{\tau_t}{\tau_d} = \left ( \frac{2 \pi r}{v} \right ) \div \left (\frac{2 \pi r}{v} \right )= 1$ [Eq. 36]

From this we conclude that the particle total energy is $mv^2$.

Let us return to the space equation (26). Letting

$ e^2 = \frac{2mE}{\hbar^2} $ [Eq. 37]

We solve (26) by writing

$ \frac{d^2 X}{dx^2} + e^2 X = 0 $ [Eq. 38]

A solution is

$ X = A \sin (ex) + B \cos (ex) $ [Eq. 39]

which is easily seen to be a solution by using (37) and substituting (39) into (38).

$ -Ae^2 \sin (ex) – B e^2 \cos (ex) + e^2 A \sin (ex) + e^2 B \cos (ex) = 0 $ [Eq. 40]

The wavelength, $\lambda_x$, for the wave represented by this equation is obtained by setting the sine and cosine arguments equal to $2 \pi$. Thus

$ e \lambda_x = 2 \pi = \sqrt{\frac{2mE}{\hbar^2}} \lambda_x $ [Eq. 41]

Using the incorrect energy $E = (mv^2)/2$ we have

$ \lambda_x = \frac{2 \pi \hbar}{\sqrt{2mmv^2 / 2}} = \frac{mv \lambda_d}{mv} = \lambda_d $ [Eq. 42]

Thus, the incorrect energy results in a space equation wavelength which agrees with the deBroglie equation. The reason for this agreement is that the Schrödinger equation is incorrect. Of course if in the term $\sqrt{2 mmv^2 / 2}$ of (42) the first 2 were replaced by 1 and the second 2 also were replaced by 1 then the space equation would agree with the deBroglie wavelength. Two wrongs make a right in this case.

The Schrödinger equation, i.e., (23), is corrected by removing the 2 in the denominator of the first term. In this case we let e have the value

$ e^2 = \frac{mE}{\hbar^2} $ [Eq. 43]

and E has the value

$ E = mv^2 $ [Eq. 44]

Now

$ e \lambda_x = 2 \pi = \frac{\sqrt{mE}}{\hbar} \lambda_x = \frac{\sqrt{m^2 v^2}}{\hbar} \lambda_x = \frac{mv}{mvr} = \frac{\lambda_x}{r}$ [Eq. 45]

Further

$ 2 \pi r = \lambda_x = \lambda_d $ [Eq. 46]

Using the corrected Schrödinger equation and the correct total energy results in a space equation wavelength the same as the deBroglie wavelength. The two must be the same.

Model of a Moving Matter Particle

To accelerate, a particle mass impacts the particle. Part of the mass scatters and part of the mass is captured by the field of the particle. The captured mass and the particle begin rotating about the combined mass center of the two masses as they translate. This motion is the undulation, or wave property, of translating particles. The center of mass of the captured mass is located at a position which results in the particle having an angular momentum of h. The resulting motion of the particle then has a flow component at velocity v (and an energy of $(1/2) mv^2$ ) plus a thermal component. The thermal component is the particle taking a circular path relative to the straight-line center of mass path. The circumference of this circular path is the same length as the wavelength so the velocity is v, same as the translational velocity. The thermal energy is, thus, also $(1/2) mv^2$. The total energy E then is

$E = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2$ [Eq. 47]

We, of course, had discovered that the total energy had to be $mv^2$, from solving the Schrödinger equation.

Making the above correction we obtain the equation for a free particle as

$ \frac{\hbar^2}{m} \frac{\partial^2 \psi}{\partial x^2} + i\hbar \frac{\partial \psi}{\partial t} = 0 $ [Eq. 48]

and the correct particle total energy is

$ E = mv^2$ [Eq. 49]

Now, recalling our discussion of the kinematics of the electronmotion we replace $ \psi $ by the particle relative velocity, v, so that the corrected Schrödinger equation is

$\frac{\hbar^2}{m} \frac{\partial^2 v}{\partial x^2} + i\hbar \frac{\partial v}{\partial t} = 0$ [Eq. 50]

Derivation of the Corrected Schrödinger Equation

An electron is accelerated by photons impacting it. The photons add momentum and mass to the particle. The mass added, $m_c$, is

$ m_c = \frac{m_0}{\sqrt{1 – (v/c)^2}} – m_0 $ [Eq. 51]

where $m_0$ is the matter particle mass when at rest, $v$ is the particle velocity, and $c$ is the speed of light.

The mass added, or captured mass, $m_c$, is assumed to continue moving at the velocity of the photon but in a closed loop around the electron. This is consistent with the energy of a moving matter particle having an energy of $ m_0 c^2 / \sqrt{1 – \beta^2} $. The electron of mass $m_0$ and the captured mass $m_c$ have a composite center of gravity g which translates in a straight path, of course.

The captured mass center c and the electron mass center e rotate around the center of gravity as the assembly translates. They are held together by electromagnetic forces. Figure 9 shows the two masses, $m_0$ and $m_c$ , their radii r and R, their velocities, $v_{e/g}$ and $ v_{c/g}$, the force pair, F and F holding them together, the accelerations, $a_c$ and $a_e$ , the center of gravity g of the electron/captured mass, and the angular velocity $\omega$. The electron (deBroglie) wavelength is $2 \pi r$. While point g translates one wavelength, point e travels a circle relative to point g, which circle has a circumference of $2 \pi r$. Thus, the velocity of e relative to g is v, the same as the absolute velocity of g.

The force balance for the electron, i.e., Newton’s equation for the electron is

$ F = \frac{m_0 v_{e/g}^2}{r} $ [Eq. 52]

where $v_{e/g}^2/r$ is the radial acceleration, which is inward in theradial direction. We can also write the force balance equation for the captured mass as

$ F = m_c \frac{\partial v_{c/g}}{\partial t} $ [Eq. 53]

where $ \partial v_{c/g} / \partial t $ is the radial acceleration of the captured mass.

We can represent the velocity v using complex numbers as

$ v = v \cos \theta + iv \sin \theta $ [Eq. 54]

This has the advantage of explicitly giving vertical and horizontal components of the parameters.

In the following analyses we need some partial differential relations. For the time equation

$ \frac{\partial v}{\partial t} = \frac{\partial v}{\partial \theta} \frac{\partial \theta}{\partial t} = \frac{\partial v}{\partial \theta} \omega = \frac{\partial v}{\partial \theta} \frac{v}{r} $ [Eq. 55]

For the space equation

$ \frac{\partial^2 v}{\partial \theta^2} = \frac{\partial}{\partial \theta} \left ( \frac{\partial v}{\partial \theta} \right ) $ [Eq. 56]

and

$ \frac{\partial v}{\partial \theta} = \frac{\partial v}{\partial x} \frac{\partial x}{\partial \theta} $ [Eq. 57]

Further

$ \frac{\partial x}{\partial \theta} = \frac{\partial x}{\partial t} \frac{\partial t}{\partial \theta} = \frac{v}{\omega} = \frac{vr}{v} = r $ [Eq. 58]

Now

$ \frac{\partial v}{\partial \theta} = \frac{\partial v}{\partial x} r = rf $ [Eq. 59]

where we have introduced

$ f = \partial v / \partial x$ [Eq. 60]

Returning to (56) and (59)

$ \frac{\partial^2 v}{\partial \theta^2} = \frac{\partial}{\partial \theta} \left( rf \right ) = r \frac{\partial f}{\partial \theta} $ [Eq. 61]

where we have placed r to the left of the differentiation since r is constant. We can write the derivative of f as

$ \frac{\partial f}{\partial \theta} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} = \frac{\partial f}{\partial x} r = \frac{\partial f}{\partial x} \frac{v}{\omega} = \frac{\partial f}{\partial x} \frac{v}{v} r = r \frac{\partial f}{\partial x} $ [Eq. 62]

Now, from (61), (62), and (60)

$ \frac{\partial^2 v}{\partial \theta^2} = r \left ( r \frac{\partial f}{\partial x} \right ) = r^2 \frac{\partial^2 v}{\partial x^2} $ [Eq. 63]

We will now show that (52) can be used to give the variation of v with the space coordinate x. We write $ m_0 v^2 / r $ as follows

$ F = \frac{m_0 v^2}{r} = \frac{m_0 v}{r} \left ( v \cos \theta + iv \sin \theta \right) $ [Eq. 64]
$ = – \frac{m_0 v}{r} \frac{\partial^2}{\partial \theta^2} \left( v \cos \theta + iv \sin \theta \right )$
$ = -\frac{m_0 v}{r} \frac{\partial^2}{\partial x^2} \left ( v \cos \theta + iv \sin \theta \right) r^2 $
$ = – \frac{m_0 v}{r} \frac{\partial^2 v}{\partial x^2} r^2 = -\hbar \frac{\partial^2 v}{\partial x^2} $

Let us now investigate the dependence on the time coordinate, t.

Returning to (53) we write

$ F = m_c \frac{\partial v_{c/g}}{\partial t} = m_c \frac{\partial v_{c/g}}{\partial \theta} \frac{v}{r} $ [Eq. 65]
$ = m_c \frac{\partial}{\partial \theta} \left ( v_{c/g} \cos \theta + iv_{c/g} \sin \theta \right) \frac{v}{r} $
$ = m_c \left ( -v_{c/g} \sin \theta + iv_{c/g} \cos \theta \right ) \frac{v}{r} $
$ im_c \left ( v_{c/g} \cos \theta + iv_{c/g} \sin \theta \right ) \frac{v}{r} $
$ im_c v_{c/g} v / r $
$ im_0 \frac{r}{R} v \frac{R}{r} \frac{v}{r} = im_0 v^2 /r $

From the last and second terms in (64) we can express the force on the captured mass as

$ \frac{m_0 v^2}{r} = \frac{1}{i} m_c \frac{\partial v_{c/g}}{\partial t} = -im_0 \frac{r}{R} \frac{\partial v}{\partial t} \frac{R}{r} = -im_0 \frac{\partial v}{\partial t} $ [Eq. 66]

where we have represented the velocity by a complex number. The sum of the force on the electron added to the force on the captured mass must be zero. Thus from (64) and (66)

$ – \hbar \frac{\partial^2 v}{\partial x^2} – im_0 \frac{\partial v}{\partial t} = 0 $ [Eq. 67]

Or, changing signs, multiplying by $\hbar$, and dividing by $m_0$ gives

$ \frac{\hbar^2}{m_0} \frac{\partial^2 v}{\partial x^2} + i \frac{\hbar \partial v}{\partial t} = 0 $ [Eq. 68]

Equation (68) is very similar to the Schrödinger equation for the free translation of a charged particle. The differences are that the Schrödinger equation has a factor 2 in the denominator of the first term and $\psi$, presumably, is not a velocity. We have shown that the factor 2 is incorrect and that $\psi$ should be replacedby the particle (complex) velocity.

Einstein’s theory of relativity and Schrödinger’s equation of quantum mechanics gave birth to modern physics. The results of both have just been shown to be derived from Newtonian mechanic’s. Thus, The Mechanical Theory of Everything could signal the death of modern physics.

How We Get Our Energy

Every cubic meter of space has enough energy to power everything on the earth for millennia. How do we get the meager supply we use? There are 10 steps in the process.

  1. Nature developed a micro-rocket pump which sucks out ether particles, makes a vacuum, and thus produces neutrinos.
  2. One mass of neutrinos makes protons (and simultaneously electrons and simultaneously makes hydrogen atoms) continually.
  3. Hydrogen atoms have gravitational fields, Thus hydrogen atoms congregate and make hydrogen stars.
  4. Hydrogen stars grow and make larger atom stars.
  5. Larger atom stars continue to grow and their gravitational fields get strong enough to break down the atomic structure and then make neutron stars.
  6. Neutrino stars grow to lightyear size and break down nuclear structure—then the big-big-big bang!
  7. Big-big-big bang makes dark mass, smaller neutrino stars, hydrogen stars (suns), planets (earth), asteroids, comets, space dust, and many neutrinos.
  8. Earth evolves, makes anaerobic bacteria, makes atmosphere, makes aerobic bacteria, makes fungi, makes plants, and makes animals.
  9. Organic material makes oil, gas, and coal. Also the sun continually dumps energy on the earth.
  10. Humans mine coal, pump the oil and gas, and use solar cells and wind mills to meet our energy needs.

Analysis of Results

We show that there are two errors in the Schrödinger equation. In solving the second degree partial differential equation which is a function of space, x, and time, t, the separation of variables technique is used. The result is two ordinary differential equations, the time equation and the space equation. The Schrödinger equation models the dynamics of a translating matter particle. The particle undulates and its wavelength is given by $h/(mv)$, i.e., the deBroglie wavelength. The time equation involves the total energy of the particle and conventional practice is to use the energy as $(1/2) mv^2$. With this value the wavelength from the time equation is double the particle wavelength, which is impossible. We present a dynamic model of a translating particle. With this model the particle total energy is $mv^2$. The second error in the Schrödinger equation is that the factor 2 in the equation must be replaced by unity.

With the above corrections to the Schrödinger equation along with considering the psi complex function as the complex velocity of the particle it is shown that the (correct) Schrödinger equation is a classical mechanics equation—the Schrödinger equation is an example of Newton’s equation F = ma.

Using our dynamic model of the translating particle, simply from the kinematics of the motion we show that the particle horizontal velocity is a periodic oscillating function same as the harmonic function resulting from the corrected Schrödinger solution using psi as the velocity.

Appendix: Proof that the Energy of a Free Translating Matter Particle is $mv^2$ at High Velocity

We have shown that a free matter particle moving at velocity v has a flow energy of $mv^2/2$ and a thermal energy of $mv^2$ for a total energy of mv 2 . The work required to accelerate a matter particle to velocity v is $mv^2 / 2$. We call this work the kinetic energy. In addition to the work added additional energy must be added to produce the thermal energy. That energy is $mv^2 / 2$ , of course.

We now consider the energy of a free matter particle moving at a velocity approaching the speed of light. As a matter particle approaches the speed of light the captured mass which, incidentally becomes much more massive than the particle itself, and the particle both approaches the speed of light. Thus, the captured mass and the particle are at near their maximum velocity. Thus, their thermal velocity approaches zero.

The ratio of the thermal energy to the flow energy approaches unity at low-velocity and approaches zero at high velocity. Figure A-1 illustrates these ratios.

Figure A-1. Thermal Energy Divided by Flow Energy Versus Velocity

From the above arguments, we conclude that the total energy at all velocities of a free translating particle is always $mv^2$, independent of a particle’s velocity. We make this conclusion notwithstanding the fact that we haven’t determined the shape of the variation of $E_{th} / E_{fe}$ from $v/c =0$ to $v/c = 1.0$


Footnotes:

  1. Retired professor of mechanical engineering at Mississippi State University and President of Basic Research Press. This paper is based upon the analyses in The Mechanical Theory of Everything by the author. See the reference.

Bibliography

Brown, Joseph M. The Mechanical Theory of Everything. ISBN 978-0-9712944-9-3. Basic Research Press. Starkville, MS. 2015.

 
 

The Rise and Fall of Modern Physics

My name is Joseph Milroy Brown, aka Joe Brown. I was born on January 9, 1928 in the rural community of Gatewood near Fayetteville, which is a small town in southern West Virginia. I am here to tell you about my book, The Mechanical Theory of Everything. I received two bachelor degrees in engineering from West Virginia University. Also, I received master and PhD degrees from Purdue University with majors in Machine Design, Engineering Mechanics, and Mathematics.

I worked in the aerospace industry for 20 years designing airplanes, missiles, space boosters, and satellites. I taught engineering part-time at WVU, USC, and UCLA while an undergraduate, in graduate school, and while in industry. I taught machine design, mechanics, and thermodynamics at Mississippi State University for 21 years, ending in 1991. I then managed a chain of bookstores from 1991 until now which my wife and I own and operate.

Have you ever heard of the equation $E=mc^2$ ? How about the mass of a particle getting larger because it begins moving? How about a clock slowing down as it moves? And, how about material shrinking as it moves? All of these occur and the work of two different scientists explain how such things occur. One scientists was Albert Einstein (1879 – 1955) and the other was Isaac Newton (1643 – 1727). Einstein developed the equations explaining the phenomena using a weird four-dimensional system where time is the fourth dimension. Using Newton’s work, the exact same equations are developed describing the phenomena, i.e., strictly using classical mechanics. Actually, Newton did not develop the equations since he didn’t know the structure of matter. Knowing the structure of matter, which we do, and using Newtonian mechanics, gives straightforward answers to the mass growth, time variation, and matter shrinking questions. We now show you how this occurs. We also show some other interesting phenomena using classical mechanics.

We begin with an ether theory. We have an ether consisting of a gas of perfectly elastic, smooth, spherical, identical particles which are moving at high speed and colliding with each other. The ether is very energetic. One cubic meter of the ether gas has enough energy to power the earth for millennia. The ether is very dense.  Its density is 50 trillion times as great as lead. With such a great density, you may wonder how we can move. We’ll explain how we move in this talk. Not only is space filled with ether gas particles, but all matter, photons, everything in the universe is made of these gas particles. Our ground rules for the development of our theory are the laws of classical, or Newtonian, mechanics. From the postulated ether it is easy to derive the conservation laws:

  1. Mass is conserved;
  2. Linear Momentum is conserved (in three directions);
  3. Angular Momentum is conserved (in three directions);
  4. Energy is conserved.

Furthermore, defining force as the time rate of momentum imparted by collisions, it is easy to derive Newton’s famous law $F = ma$.

First, we describe the structure of matter. Matter at rest consists of a mass m taking a circular path with a velocity equal the speed of light c. Thus, the energy E of matter is given by

$E = mc^2$  [Eq. 1]

which is the formula postulated by Einstein.

In order for matter to move, photons impinge on the matter, deposit mass to the matter, and scatter off at lower energy (and a longer wavelength). The momentum imparted to the matter is equal to the mass captured times c, plus the momentum imparted by the scattered photon. The average scatter angle is 90° so that the average momentum imparted by the scattered photon is its mass times its speed, c. Thus, the average total momentum imparted is the total momentum of the impacting photon. Assuming many photons are used, the Newtonian analysis of the acceleration gives the equation for the mass at velocity, $m_v$ , as

$m_v = m_0 / \sqrt{1-(v/c)^2}$  [Eq. 2]

where $m_0$ is the mass at rest, $v$ is the particle velocity, and $c$ is the speed of light. The captured mass $m_c$ is

$m_c = m_v – m_0 = m_0 \left ( \frac{1}{\sqrt{1-(v/c)^2}} – 1 \right )$  [Eq. 3]

The circular path of the mass making matter changes to a plane spiral path as shown here.

Figure 1.  Rest and Moving Particle

Motion is accomplished simply by changing the direction of the moving mass – not by changing its speed. Impact by matter with a single ether particle will change the direction slightly. Thus, it is easy to move in the dense ether. The orbital path for one cycle of a moving particle is longer than that for a rest particle and the mass still moves at  velocity c. Thus, it takes longer to execute a cycle. The time for a cycle when moving, $T_v$ , is

$T_v = T_0 / \sqrt{1-(v/c)^2}$  [Eq. 4]

where $T_0$ is the time for a circular orbit.

The orbit seen from a reference frame moving with the particle is an ellipse with its minor diameter parallel to the velocity shortened by the factor

$d_v = d_0 \sqrt{1-(v/c)^2}$ [Eq. 5]

where $d_v$ is the minor diameter and $d_0$ is the diameter of the path at rest.

The five Newtonian equations are identical to the Einstein equations. However, the Einstein interpretations differ radically from the Newtonian interpretations. According to the Einstein theory:

  1. In equation (1) the particle does not have mass moving at velocity c. It is simply energy.
  2. Using equation (2) if you see a particle moving past you at velocity v, the Einstein theory says its mass is larger than $m_0$. If you run and catch up with the particle, its mass decreases to $m_0$.
  3. In equation (4) if the particle moves at velocity v relative to you, the time for an orbit is increased. If you catch up with the particle the orbit time goes back to the rest orbit time and the orbit shape goes back to circular.
  4. In equation (5), again, the length goes back to $l_0$ if you catch up with that particle.
  5. The Einstein theory postulates that photons always move at velocity c with respect to all frames, no matter what their velocity is. This is not true. The measurement of the velocity of a photon, using a clock and length measure on the given frame, always gives the same result since, for instance, the clock on a moving frame runs slower and the measuring rod shortens so that the ratio of length divided time is the same for all reference systems.
  6. A really absurd result of the Einstein theory is illustrated by two space ships approaching each other. An observer on the earth sees each one moving at 0.8 times the speed of light. The observer on one space ship sees the other space ship coming toward him at only 0.8 times the speed of light, according to the Einstein theory. Another interesting aspect of accelerating matter is what happens to the mass captured by the accelerated particle. Our Newtonian analysis results in the center of gravity of the mass being captured at a fixed distance from the particle. The two masses then rotate as they translate with their joint center of mass taking a straight path, of course, as shown below.

Another interesting aspect of accelerating matter is what happens to the mass captured by the accelerated particle. Our Newtonian analysis results in the center of gravity of the mass being captured at a fixed distance from the particle. The two masses then rotate as they translate with their joint center of mass taking a straight path, of course, as shown below.

Fig. 2.  Center of Gravity of Captured Mass/Particle

Thus, the particle undulates as it translates – giving it a wave property. The equation of this motion is obtained by balancing the centrifugal force on each mass by a centripetal force holding the mass in the curved path, i.e., F = ma. Writing F = ma for both masses gives the famous Schrödinger equation which spawned quantum theory. but the Schrödinger equation is simply a Newtonian (classical) mechanics equation.

Earlier we showed you that the Einstein relativity equations are derivable from classical (Newtonian) theory. We have just indicated that the foundation of quantum mechanics, i.e., the Schrödinger equation is derived from classical mechanics. Physicists almost universally believe this is impossible. They say the Schrödinger equation can not be derived from more fundamental principles. But, actually, the Schrödinger equation is a Newtonian system. Now modern physics was ushered in by Einstein’s special relativity and quantum mechanics. However, both theories are Newtonian theory. We believe Newton should usher out modern physics and usher back in classical physics.

 
 

What’s New & What’s Old With Joe Brown’s Theory?

What’s New & What’s Old With Joe Brown’s Theory?

 

Joe Brown uses an ether gas, which is old as the hills. He also uses Newton’s mechanics—the space and time are separate; mass, momentum, angular momentum, and energy are conserved—also F=ma.

Let us now talk about the new.

  1. The ether is made up of a rare gas consisting of perfectly elastic, smooth, hard, identical, fast-moving spheres. The ether has a long mean free path—l/d =1019.
  2. No one before Brown bothered to define a physically realizable ether. Complete condensations of the ether occur. The condensations are of the same scale as the mean free path. The long mean free path permits condensations on the scale of the mean free path since particles coming in don’t meet particles coming out. But a pump is required to bring the ingoing particles out of the condensation in very small streams. Brown discovered the pump. It is simple, particles being aligned without changing their speeds then being squeezed together without changing their energy flow 8% faster than before being squeezed. This phenomenon produces the pump for removing particles from the core of the condensation and provides a thrust of 1.4 meganewtons to propel the condensations at the speed of light. Any mechanics person worth his salt can understand this process if he/she will just think. These condensations are the neutrinos. The condensations and the pump discoveries are new.
  3. What controls the mass of the proton? All neutrinos have the same angular momentum and all masses of neutrinos have the same thrust, 1.4 meganewtons. A neutrino with the mass of a proton can orbit with its thrust balancing the centrifugal force. There is only one mass, which can produce the correct angular momentum and balance the centrifugal force. That mass is the mass of the proton, and that orbiting mass is the proton. This is new and it is not complicated like the 8-fold way.
  4. When translating, the neutrino emits two fine streams of ether particles, one forward at the RMS velocity, vr, of the background gas, and the other at the mean speed, vm, of the background gas. The velocity vr is 8% larger than vm so the neutrino translates at velocity vr-vm, which is slightly greater than the speed of light. The mass flow out these two streams balances the mass rate flowing into the neutrino.
  5. When the neutrino is in orbit, as in the proton, it establishes a gross in-out flow similar to that produced by an expanding-contracting sphere in the atmosphere. This is the Coulomb field. Polarity is produced by the right-hand spin of the neutrino making the proton. This field interacting with a similar field produces inverse square forces with attraction between fields produced by unlike handedness and repulsion by like handedness. We know of no other theory of the force produced by interacting charges. Further, our theory is of the utmost simplicity.
  6. The proton neutrino orbits with a radius of 10-16 The outflow of the forward stream at velocity vr followed by an opposite stream at velocity vm produces a radial expanding wave with an amplitude of 10-16m and a radial flow advancing at the velocity of light. The velocity is slightly less than vr-vm. The transverse flow produced by the neutrino translating at vr-vm also produces waves with a dimension of 10-16m. The radial and transverse flows produce wavespaces with the dimension 10-16m by 10-16m by 10-16m. These wavespaces carry the mass away from the proton at the speed of light. These wavespaces are the fine structure of the electrostatic field. Finally (2016) we have discovered what the fine structure is. Quantum electrodynamicists have discovered its magnitude to one part in 1011, but they do not know what the fine structure is.
  7. When a matter particle is moving at velocity v there is a closed string of wavespaces in the shape of an ellipse, which encircles the particle. The minor axis of the ellipse is parallel to the velocity and the eccentricity of the ellipse is v/c, where c is the velocity of light. The mass mc is the mass captured by the matter particle when it was accelerated from rest. This mass is captured off-center and it rotates about the particle as it translates. This produces the wave property of matter and the magnetic field of the charged particle. This is new.
  8. As a charged particle translates the captured mass and the particle attract each other (by an electromagnetic force) so that the distance between them is stable. Applying Newton’s law (F=ma) to the motion, an equation almost exactly the same as the Schrödinger equation results. This is new with Joe Brown’s theory, and it is earth shaking. Most physicists believe that quantum mechanics cannot be derived from Newtonian (classical) mechanics.
  9. The Schrödinger equation is a second order partial differential equation in two variables, space x and time t. The equation for the free translation is always separable into two ordinary differential equations. The separation constant is the total energy of the particle. We have proved that the energy transferred to a particle of mass m to accelerate it to velocity v is mv2, even though the work done on the particle is ½ mv2. Thus, the kinetic energy is only half of the total energy of the particle and of the energy transmitted by the accelerating medium. Using the total energy as mv2, rather than the (incorrect) conventionally used total energy of ½ mv2 results in a period from the Schrödinger time equation, which is equal to the period of the electron. On the other hand if the incorrect energy $mv^2/2$ in the time equation is used the period is double the period of the translating electron, which is impossible. This gives further evidence that the particle total energy is mv2 for the separation constant. Using mv2 in the space equation shows that the factor 2 in the Schrödinger equation should be unity. With this correction to the Schrödinger equation we prove that the Schrödinger equation is simply an application of Newton’s equation F=ma. These results of course, are new (and earth shaking).
  10. In order to accelerate a matter particle, which like all matter particles consists of an orbiting neutrino, it is necessary to change the path of the neutrino from circular to a plane spiral. Since matter particles produce electrostatic fields, part of the accelerating photon can be captured by the field and re-direct the path of the circularly orbiting matter particles. This captured mass adds momentum to the particle and thus causes it to translate. The resulting mass mv of the particle moving at velocity c is given by $m_v=m_o/\sqrt{1-(v/c)^2}$ is the rest mass. The mass growth with velocity was observed experimentally and Einstein postulated (rather than derived) the above expression in his development of the special theory of relativity. This classical derivation of the mass growth formula was obtained by Dr. Darrell B. Harmon, Jr. in 1969 – but it still is not recognized by the physics community. It is the first step in proving the space-time assumptions of special relativity are erroneous.
  11. The plane spiral path of a neutrino making a translating matter particle has the shape of an ellipse when viewed from a frame moving with the particle. The minor axis of this ellipse is parallel to the particle velocity and the ellipse eccentricity is v/c. As a result the particle, and all matter, shortens with motion by the factor $\sqrt{1-(v/c)^2}$. This is just another nail in the special theory of relativity coffin – discovered by the developers of the kinetic particle theory of physics.
  12. The time required for one orbit of the neutrino making a matter particle is longer when the particle is moving. The relation is $T_v=T_o/\sqrt{1-(v/c)^2}$ where Tv is the cycle time when moving and To is the cycle time when at rest. This means that nucleii decay times will increase when moving. Surely this is the last coffin nail, which will be required to eliminate the special theory of relativity from physical science.
  13. We prove that the orbiting neutrino, such as that making the proton, produces an oscillation in the background gas, which is the Coulomb field. Further, we show that two particles of opposite charge in orbit with each other produces the gravitational field. To produce the electrostatic field the half amplitude of the oscillation is the orbital radius of the proton. To produce the gravitational field the half amplitude of the oscillation of the electron relative to the proton is the radius of the basic ether particle, 10-35 Thus, we have derived Einstein’s unified field theory. This is new, scientists should not spend much time studying a complex space warp in the presence of matter for producing gravity – while gravitation is just caused by a mechanical breathing sphere phenomenon. This is a new with Joe Brown’s theory. Nonetheless Einstein’s theories have been useful and no reasonable alternatives have been available until now. Einstein was an unparalleled scientist.
  14. The universe is not expanding! In1924 Edwin Hubble discovered that photons emitted from distant stars were shifted toward the red more than those from close stars. The physics community (eventually) interpreted the shift as a Doppler effect and, thus that more distant stars were moving away from the earth faster than close stars. They said, the universe is expanding. Actually the universe is not expanding. The angular momentum of the emitting atom/photon remains constant as the photon translates. The photon loses one ether particle for each wavelength of travel. Photons leaving a star 1010 light years from the earth will use the last of the ether particles when they reach the earth. Light wears out. This mechanism is a new development of the kinetic particle theory of physics.

 

Joseph M. Brown

 
 

What is the Difference Between Energy & Kinetic Energy?

The energy of matter is $ mc^2 $ where $ m $ is the mass of the matter and $ c $ is the velocity of light. We consider an example of a Newtonian system accelerating a mass $m $ from zero to velocity $ v $, where $ v \ll c $. The energy given up by the Newtonian system is $ mv^2 $. We also show that the work done by the accelerating system is half the energy given up by the accelerating system.

We know that the energy of matter is $ mc^2 $, i.e., its mass times the square of velocity. We say that a ball of mass $ m $ translating at velocity $ v $ has a kinetic energy of $ \frac{1}{2} {mv^2} $. But, isn’t its energy $ mv^2 $? We think so. How do we reconcile the fact that the work, i.e. energy expended, which is $ \int F dx $ required to bring the energy of the ball up to $ mv^2 $, when only half that amount of work is required?

Read the entire article text here:

What is the Difference Between Energy and Kinetic Energy

 
 

What’s New with Brown’s Theory of Everything?

blog_header4

  • Everything is made up of elastic balls, including ether, neutrinos, matter, and radiation.
  • Postulates immediately and rigorously give classical mechanics.
  • Neutrinos are nanoscale particles.
    • The size of a neutrino is the mean free path.
    • Neutrinos result from complete condensations of the ether gas.
    • Neutrinos violate the Second Law of Thermodynamics.
  • Neutrinos propel themselves.
    • Neutrinos initially align ether particles without changing their speeds.
    • Ether particles simply aligned and then squeezed together without changing their energy flow at speed $ v_r $ – the background RMS speed.
    • Since $ v_r $ > $ v_m $ ($ v_r = 1.08v_m $) neutrino flows at speed $ v_r – v_m $.
    • Thus, the speed of a neutrino is $ c = v_r – v_m $ where c is the speed of light.
  • Neutrino mechanism provides means for moving in an ether gas which is a trillion times as dense as lead.
  • Matter is made up of mass orbiting at the speed of light:  $ E = mc^2 $ (Famous formula derived by Brown’s group.)
  • To accelerate matter, mass is added at high speed.

3-8-16-fig1

  • Nuclear processes take longer when moving  (Famous formula derived by Einstein and by Brown’s group.)
  • Matter shortens when it moves.

3-8-16-fig2

  • How much mass is added?
    • Mass added depends upon velocity.  The more mass is added, the higher the velocity.
    • Mass added $ m_{added} = m_v – m_0 = \frac{m_0}{\sqrt{1-(\frac{v}{c})^2}} – m_0 $
    • Famous formula postulated by Einstein and derived by Brown’s group.
  • Where is the added mass stored?
    3-8-16-fig3
  • The captured mass and electron undulate as they translate.
    3-8-16-fig4
  • The wavelength of the electron
    $ \lambda = \frac{h}{m_{e}v} $
  • $ h $ is Planck’s Constant, $ v $ is the electron’s velocity, $ m_e $ is the electron mass.
  • This equation is the wavelength postulated by de Broglie
  • We show the course of the oscillation.
  • All matter undulates as it translates.
  • Einstein’s theory is invalid.  Newton’s theory is correct.
    • Mass growth is a Newtonian phenomenon.
    • Matter shortening is a Newtonian phenomenon.
    • Time dilation is a Newtonian phenomenon.

Experimental Example

What Einstein Sees What Newton Sees
Start with a Bar at Rest mass: 100 kg
length: 10 m
undulation: no
mass: 100 kg
length: 10 m
undulation: no
Accelerate bar to 0.9c, Einstein and Newton at rest mass: 229 kg
length 4.36 m
undulation: yes
mass: 229 kg
length 4.36 m
undulation: yes
Accelerate Einstein and Newton to 0.9c mass: 100 kg
length 10 m
undulation: no
mass: 229 kg
length 4.36 m
undulation: yes

What Einstein would see is ridiculous.  Einstein’s theory absolutely must be incorrect.

 
 

Where Did We Come From?

blog_header3

  • In the beginning there was space filled with a gas of Brutinos
  • Neutrinos are made continually by condensing brutino gas
  • Orbiting Neutrinos continually make hydrogen atoms
  • Hydrogen atoms congregate to make hydrogen gas
  • Hydrogen gas congregates to make hydrogen stars
  • Hydrogen stars continually grow to make heavy atom stars
  • Heavy atom stars continually grow to make neutron stars
  • Neutron stars continually grow until nuclear structure collapses
    • Star has light year diameter
  • Super-sized neutron star explodes and produces
    • Dark mass
    • Small neutron stars
    • Large atom stars
    • Small hydrogen stars
    • Planets – including Earth
    • Asteroids and space junk
  • Earth becomes habitable
 
 

Mystery of Supernova ASASSN-15lh

I wonder if Supernova ASASSN-15lh is the explosion of a light-year sized neutron star exploding at a far distance from the Earth – a distance much larger than 1011 light years. In Chapter 9 Section D, the Human Cosmos, of my book The Mechanical Theory of Everything it is speculated that the universe we see was formed from the explosion of a giant (light year sized) neutron star – a star that was large enough so that gravity collapsed the nuclear structure. It is speculated that the large amount of dark mass in our observable part of the universe consists of large neutron stars, which are the debris of our giant neutron star.

Joseph M. Brown

 
 
 
 

The Fine Structure Constant

tfsc_blog_header

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The fine structure constant, alpha (α), has a value of 1/137…. It first appeared as the velocity in speed of light units of the orbiting electron in its lowest energy state in a hydrogen atom. The analysis of the orbital velocity gives the equation

tfsc_equation_1

where e is the charge of the electron ħ is Planck’s constant, and c is the speed of light.

The fine structure constant is the ratio of electromagnetic forces to nuclear forces. Furthermore, alpha appears throughout quantum electrodynamic (QED) theory. QED physicists use the constant throughout their analyses of the interactions of electron and photons. However, they have not the least idea of its origin. Many physicists reading a book, when they come to page 137, will pause and wonder what the origin of α is.

During the years 1967 to 1970, the McDonnell Douglas Company funded the Advanced Propulsion Research Group to develop a new physical theory which, hopefully, would lead to extremely advanced propulsion. The group consisted of Joseph M. Brown (PhD Purdue – Mechanical Engineering), Darell B. Harmon, Jr. (PhD UCLA – Physics), Leon A. Steinert (PhD Colorado – Physics), and Robert M. Wood (PhD Cornell – Physics).

The research centered on developing a theory of physics based on an absolute space – separate absolute time universe filled with an ether of extremely small, smooth, elastic spheres. We dubbed this the kinetic particle theory of physics. Two parameters characterizing such a gas are the particle mean speed vm and the particle RMS speed vr. Toward the end of the research effort, we discovered the relation

tfsc_equation_2

The value 1/137….. results since vr/vm =√(3π/8). This expression, modified by the orbital analysis of the atom using the atom center of mass system, gave the factor 1/137……. This quantity agreed with the value of the fine structure constant within one part in 70,000, see the paper by Brown, Harmon, and Wood [1].

Based upon the precise agreement of this arrangement of the kinetic particle gas parameters and the fine structure constant, our group felt greatly encouraged that the kinetic particle theory of physics should be developed.

We assumed that nuclear particles consisted of condensations of the ether gas and that a condensation would act like a (fluid mechanic) doublet. Further, matter had to consist of mass orbiting at the speed of light. This was required to give matter energy as E=mc2. The computation of the interaction of one nucleon with another consisted of the analysis of two doublets. From this it was deduced that the strength of the strong nuclear force was proportional to the square of the mean velocity and, of course, proportional to the ether mass density. Thus, nuclear forces are proportional to ρvm2. This analysis was reported by Brown and Harmon in reference [2].

We soon concluded that these condensations moving at the speed of light in a circle making matter had been neutrinos translating in a straight line (at the speed of light, of course). For a condensation to be stable, it had to suck in gas particles, align them, and then expel them in two extremely fine streams. This required a pumping mechanism and streams that were so fine that this outflow would not interfere with the inflow. This fine stream requirement necessitated an extremely large mean free path to particle diameter ℓ/d ratio. The ℓ/d for the ether gas is 1018, see page 51 of [3].

It was noted that when particles were taken from a Maxwell-Boltzmann gas, aligned to be parallel and directed with the same sense, then squeezed together so that they formed a solid stream without changing their energy, then the translational velocity will jump from vm to vr, an 8.5% increase in velocity. The neutrino does what is specified above. This means that the neutrino will translate at the velocity vr-vm. Therefore, the speed of light is c=vr-vm. This discovery was reported in reference [4] and [5].

Knowing that the speed of light is vr-vm, we know that electromagnetic forces are the background density ρ times (vr-vm)2. With this information and that the strong nuclear force is proportional to ρvm2 we now know that

tfsc_equation_3

is the ratio of the electromagnetic force to the strong nuclear force.

The neutrino has two fine streams of ether particles exiting in opposite directions from the spherical neutrino. The stream directed forward is a solid stream translating at the velocity vr. The stream directed aft is at the velocity vm. When a neutrino is in a circular orbit, being a nuclear particle, such as the proton, then it produces a spherical wave with a velocity amplitude vr followed by a velocity amplitude vm. Thus, one of the primary characteristics of the electrostatic field is wave spaces with dimensions of 10-16m (the orbital radius of the proton, see page 600 of [3]). The waves advance spherically symmetric from the proton at the velocity vr-vm, or c.

The interactions of photons and electrostatic charge are dominated by these wave spaces. A photon consists of a string of the ether gas particles strung out uniformly over a harmonic wave. For high energy photons each wave space has many particles and for very low energy photons many wave spaces along the harmonic curve may have no particles. The wave spaces always travel radially from the charged particle at the speed of light. Thus, the speed of light is always the same, which is vr-vm. The wave space encapsulate the photon particles making up the photon.

From the above discussion and analyses it is clear that we know much about the fundamental mechanism of the fine structure constant. However, we still do not know why the electron in its lowest state in the hydrogen atom has the velocity [(vr-vm)/vm]2.

References
1. Brown, J.M., Harmon Jr., D.B., and Wood, R.M., “A Note on the Fine Structure Constant,” McDonnell Douglas Astronautics Company Paper MDAC WD 1372 Huntington Beach, CA, June 1970.
2. Brown, J.M., Harmon, Jr. D.B., “A Kinetic Par- ticle Theory of Physics”, J. Mississippi Academy of Sciences, VXVIII, Pages 1-26, 1972. Avail- able from Basic Research Press.
3. Brown, J.M., The Mechanical Theory of Everything, ISBN: 978-0-9712944-9-3, Basic Research Press, Starkville, MS, 2015.
4. Brown, J.M., “A Counter Example to the Second Law of Thermodynamics”, Abstract p.98, Jour- nal of the Mississippi Academy of Sciences, Vol. XXVI Supplement, 1981. Available from Basic Research Press.
5. Brown, J.M., “Force Production from Interacting Gas Flows for BMD Applications”, Final Report on U. S. Army Contract DAS6-80-C-0034 Administered by U. S. Army Ballistic Missile Defense Agency, Box 1500, Huntsville, Al. 35807, October 1, 1981.

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