# Synopsis and Further Developments to the Mechanical Theory of Everything

by Joseph M. Brown1

#### 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 $mv^2$.

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