gEM: Unifying gravity, electricity, and magnetism by analogy

Douglas B Sweetser@alum.mit.edu


Contents

The goal

Create one mathematical structure for gravity and electromagnetism that can be quantized.

The problem

The difference between gravity and electromagnetism is the oldest core problem facing physics, going back to the first studies of electromagnetism in the seventeenth century. The best minds on the planet are trying to merge the Riemannian geometry of general relativity with the quantum field theory of the standard model. Three lines of attack are string theory in ten or eleven dimensions, loop quantum gravity, and non-commutative geometry. These approaches are technically challenging.

Some history

Gravity was the first inverse square law, discovered by Isaac Newton. After twenty years of effort, he was able to show that inside a hollow massive shell, the gravitational field would be zero. Ben Franklin, in his studies of electricity, demonstrated a similar property for an electrically charged hollow sphere. Joseph Priestly realized this meant that the electrostatic force was governed by an inverse square law just like gravity. Coulomb gets credit for the electrostatic force law modeled on Newton's law of gravity.

Hundreds of years later, Einstein started from the tensor formalism of electromagnetism on the road to general relativity. Instead of an antisymmetric field strength tensor, Einstein used a symmetric tensor. There is a precedence for transforming mathematical structures between gravity and electromagnetism.

The method

The process of transforming mathematical structures from electromagnetism to gravity will be continued. Specifically, the gravitational analog to the Lorentz force will be written, as well as the gravitational analog to the classic electromagnetic Lagrangian.

The results

A weak static gravitational field in a vacuum will be studied using standard methods: normalizing the potential and looking at perturbations. That field in the force equation leads to:

The Lagrangians

The classic electromagnetic Lagrangian

<alt>The electromagnetic Lagrangian equals minus m over gamma V the volume minus the electric charge q over c squared V times the 4-velocity U superscript mu over gamma (one over the square root of one minus the relativistic velocity squared) times the 4-potential A subscript mu minus one over four c squared times A super mu comma nu minus A nu comma mu times A sub mu comma nu minus A nu comma mu.</alt>

$\mathcal{L}_{EM}=-\frac{m}{\gamma V}-\frac{q}{c^{2}V}\frac{U^{\mu }}{\gamma }A_{\mu }-\frac{1}{4c^{2}}(A^{\mu ,\nu }-A^{\nu ,\mu })(A_{\mu ,\nu }-A_{\nu ,\mu })$

The structure

Lagrangian = mass - charge * velocity contracted with the potential - field strength tensor contracted with self.

The analogy

<alt>The electric charge q goes to the square root of G times the mass. </alt>

<alt>Minus in the tensor goes to plus. </alt>

<alt>The standard derivative represented by a comma goes to the covariant derivative semicolon.</alt>

The analogous Lagrangian

<alt>The analogous Lagrangian equals minus m over gamma V plus the square root of G over c squared times m times U super mu over gamma times A sub mu minus one over four c squared times A super mu semicolon nu plus A nu semicolon mu times A sub mu semicolon nu plus A nu semicolon mu.</alt>

$\mathcal{L}_{g}=-\frac{m}{\gamma V}+\frac{\sqrt{G}}{c^{2}}m\frac{U^{\mu }}{\gam... ...mu }-\frac{1}{4c^{2}}(A^{\mu ;\nu }+A^{\nu ;\mu })(A_{\mu ;\nu }+A_{\nu ;\mu })$

The total Lagrangian

<alt>The total Lagrangian L sub gEM equals minus m over gamma V plus the square root of G times m minus q over c squared V times U super mu over gamma times A sub mu minus one over two c squared times A super mu semicolon nu times A sub mu semicolon nu.</alt>

$\mathcal{L}_{gEM}=-\frac{m}{\gamma V}+\frac{(\sqrt{G}m-q)}{c^{2}V}\frac{U^{\mu }}{\gamma }A_{\mu }-\frac{1}{2c^{2}}A^{\mu ;\nu }A_{\mu ;\nu }$

<alt>Expand the terms: equals minus m over gamma V plus the square root of G times m minus q over c squared V times phi minus A dot V minus one over two c squared times A super mu comma nu plus the Christoffel symbol sub omega super mu nu contracted with A super omega times A sub mu comma nu minus Christoffel super omega sub mu nu contracted with A sub omega.</alt>

$=-\frac{m}{\gamma V}+\frac{(\sqrt{G}m-q)}{c^{2}V}(\phi -\overrightarrow{A}\cdot... ...\mu \nu }A^{\omega })(A_{\mu ,\nu }-\Gamma _{\: \mu \nu }^{\omega }A_{\omega })$

<alt>Multiply out the terms with the Christoffel symbols: equals minus m over gamma V plus the square root of G times m minus q over c squared V times phi minus A dot the velocity v minus one over two c squared times A super mu comma nu A sub mu comma nu minus one over two c squared times Christoffel super omega sub mu nu times Christoffel sub omega super mu nu contracted with A super omega and a sub omega minus two A super mu comma nu A sub omega.</alt>

$=-\frac{m}{\gamma V}+\frac{(\sqrt{G}m-q)}{c^{2}V}(\phi -\overrightarrow{A}\cdot... ...Gamma _{\omega }^{\: \mu \nu }A^{\omega }A_{\omega }-2A^{\mu ,\nu }A_{\omega })$

<alt>The Christoffel symbol equals zero</alt>

For locally covariant coordinates, $\Gamma =0$, and

<alt>The Lagrangian L sub gEM equals minus m over V times the square root of one minus dR by dt squared plus the square root of G times m minus q over c squared V times phi minus A dot V minus one over two c squared times the time derivative of phi squared minus the gradient of phi squared minus the time derivative of A squared plus the vector gradient of A squared.</alt>

$\mathcal{L}_{gEM}=-\frac{m}{V}\sqrt{1-(\frac{\partial \overrightarrow{R}}{\part... ...{\partial \overrightarrow{A}}{\partial t})^{2}+(\nabla \overrightarrow{A})^{2})$

The classical field equations

Apply the Euler-Lagrange equation

<alt>The derivatives sub mu of the derivatives of the Lagrangian L with respect to the derivative of the derivatives sub mu of phi equals the derivative of L with respect to phi.</alt>

$\partial _{\mu }(\frac{\partial \mathcal{L}}{\partial (\partial _{\mu }\phi )})=\frac{\partial \mathcal{L}}{\partial \phi }$

<alt>Square root of G times mass times gamma over c V times the time derivative of R minus the scalar operator consisting of the second time derivative minus del squared acting on A sub mu equals minus the square root of G times m plus q over V times U sub mu over c gamma equals the mass current J sub m plus the electrical current J sub q if and only if the connection equals zero.</alt>

$\square ^{2}A_{\mu }=\frac{(\sqrt{G}m-q)}{V}\frac{U_{\mu }}{c\gamma }\qquad iff\: \Gamma =0$

The classical fields

<alt>The gravitational field g super mu equals the 4-vector the time derivative of g, minus c times the x derivative of A sub x, minus c times the y derivative of A sub y, minus c times the z derivative of A sub z.</alt>

$g^{\mu \mu }=(\frac{\partial \phi }{\partial t},-c\frac{\partial A_{x}}{\partia... ..._{y}}{\partial y},-c\frac{\partial A_{z}}{\partial z})=\partial ^{\mu }A^{\mu }$

<alt>The gravitational field super u zero equals the time derivative of A minus c times the gradient of phi.</alt>

$\overrightarrow{g^{u0}}=\frac{\partial \overrightarrow{A}}{\partial t}-c\overrightarrow{\nabla }\phi $

<alt>The gravitational field g equals c times the 3-vector operator the derivative with respect to y plus the derivative with respect to z, the derivative with respect to x plus the derivative with respect to z, the derivative with repsect ot x plus the derivative with respect to y acting on the 4-potential A equals del dot I times I minus del super u A super u.</alt>

$\overrightarrow{g}=c(\frac{\partial }{\partial y}+\frac{\partial }{\partial z},... ...{\partial }{\partial y})\overrightarrow{A}=((\nabla \cdot I)I-\nabla )^{u}A^{u}$

<alt>The electric field E equals minus the time derivative of A minus c times the gradient of phi.</alt>

$\overrightarrow{E}=-\frac{\partial \overrightarrow{A}}{\partial t}-c\overrightarrow{\nabla }\phi $

<alt>The magnetic field B equals c times the curl of A.</alt>

$\overrightarrow{B}=c\overrightarrow{\nabla }\times \overrightarrow{A}$

<alt>g super mu mu.</alt>

<alt>g super u 0.</alt>

<alt>g 3-vector.</alt>

The Maxwell source equations + analogues for gravity

Source equations: Gauss' law and a dynamic g (or Newton's field equations under certain conditions).

<alt>The mass density rho sub m plus the charge density rho sub q equals the second time derivative of phi minus c squared times the Laplacian of phi: add two terms to this: the second time derivative of phi plus c times the time derivative of the divergence of A minus c times the divergence of the time derivative of A minus c squared times the Laplacian of A.</alt>

$-\rho _{m}+\rho _{q}=\frac{\partial ^{2}\phi }{\partial t^{2}}-c^{2}\nabla ^{2}\phi $

<alt>Equals the time derivative of g super zero zero plus a half of c times the divergence of g super u zero + c times the divergence of E.</alt>

$=\frac{\partial g^{00}}{\partial t}+\frac{c}{2}(\overrightarrow{\nabla }\cdot \overrightarrow{g^{u0}}+\overrightarrow{\nabla }\cdot \overrightarrow{E})$

One can recover Newton's field equation, $\rho _{m}=c^{2}\nabla ^{2}\phi $, in isolation if:

$\frac{\partial \overrightarrow{A}}{\partial t}=-c\overrightarrow{\nabla }\phi ,\quad \frac{\partial g^{tt}}{\partial t}=0$

Gauss' law, $\rho _{q}=\frac{\partial ^{2}\phi }{\partial t^{2}}-c^{2}\nabla ^{2}\phi $, can also be recover if:

$\frac{\partial \overrightarrow{A}}{\partial t}=c\overrightarrow{\nabla }\phi $

Ampere's law.

<alt>The mass current plus the electric current equals the square root of G times gamma times the mass over c V times the velocity plus the second time derivative of A minus c squared del squared A,</alt>

$-\overrightarrow{J}_{m}+\overrightarrow{J}_{q}=\frac{\partial ^{2}\overrightarrow{A}}{\partial t^{2}}-c^{2}\nabla ^{2}\overrightarrow{A}$

<alt>equals del g super u u plus one half the time derivative of g super u zero plus c times del dot I time I minus del acting on g 3-vector minus the time derivative of E plus c times the curl of B.</alt>

$=\nabla _{u}g^{uu}+\frac{1}{2}(\frac{\partial \overrightarrow{g^{u0}}}{\partial... ...rrightarrow{E}}{\partial t}+c\overrightarrow{\nabla }\times \overrightarrow{B})$

<alt>g equals zero</alt>

Connonical quantization

The classical electromagnetic Lagrangian fails

<alt>The generalized momentum pi super mu equals h the square root of G times the derivative of the Lagrangian with respect to the derivative of the time derivatives of the potential A super mu over c which equals minus F super mu zero.</alt>

$\pi ^{\mu }=h\sqrt{G}\frac{\partial \mathcal{L}}{\partial (\frac{\partial A^{\mu }}{c\partial t})}=-F^{\mu 0}$

<alt>F super zero zero equals zero.</alt>

$F^{00}=0$

<alt>The commutator of the position operator x super zero and the momentum operator pi zero (i.e. time and energy) equals zero.</alt>

$[x^{0},\pi ^{0}]=0$

The Lagrangian of gravity and electromagnetism works

<alt>Pi super mu equals h the square root of G times the time derivative of A super mu over c.</alt>

$\pi ^{\mu }=h\sqrt{G}\frac{\partial A^{\mu }}{c\partial t}$

<alt>The commutator of x super mu and pi super mu is not equal to zero.</alt>

$[x^{\mu },\pi ^{\mu }]\neq 0$

Problems with quantization

<alt>The Lagrangian of electromagnetism</alt>

$\mathcal{L}_{EM}$

<alt>The Lagrangian of gravity and electromagnetism</alt>

$\mathcal{L}_{gEM}$

<alt>The contravariant derivative d super mu contracted with A sub mu equals zero if and only if the mass M equals zero.</alt>

4 modes of transmission:

2 traverse (photons for EM),

1 scalar (gravitons for g),

1 longitudinal (gravitons for g).

Phrase ``scalar photon'' is non-sense because photons transform like vectors.

When gravitational waves are detected, this proposal predicts they will have scalar (timelike) polarity or longitudinal polarity, not transverse polarity as predicted by general relativity.

Integration with the standard model

The standard model does not obviously deal with curved spacetime. The unitary aspects of the symmetries U(1), SU(2), and SU(3) will be condensed with the 4-vectors and metric. Start with the standard model Lagrangian:

<alt>The Lagrangian L equals psi bar gamma (Dirac matrix) super mu contracted with the covariant derivative of the standard model psi.</alt>

$\mathcal{L}=\bar{\psi }\gamma ^{\mu }D_{\mu }\psi $

where

<alt>The covariant derivative D sub mu equals the standard derivative d sub mu minus i the coupling constant for electromagnetism times Y the U(1) symmetry times the potential A sub mu minus the coupling constant for the weak force times tau super a which runs from zero to two over two times the potential W super a sub mu minus the coupling constant for the strong force times lambda super b which runs from zero to seven over two times the potential G super b sub mu.</alt>

$D_{\mu }=\partial _{\mu }-ig_{EM}YA_{\mu }-ig_{weak}\frac{\tau ^{a}}{2}W_{\mu }^{a}-ig_{strong}\frac{\lambda ^{b}}{2}\textrm{G}_{\mu }^{b}$

The electromagnetic potential <alt>A sub mu</alt>

$A_{\mu }$ is a complex-valued 4-vector. The only way to form a scalar with a 4-vector is to use a metric. Since it is complex-valued, use the conjugate like so:

<alt>A super mu A super nu conjugated contracted with the metric g sub mu nu equals the absolute value of A super zero squared minus the absolute value of A super one squared minus the absolute value of A super two squared minus the absolute value of A super three squared.</alt>

$A^{\mu }A^{\nu *}g_{\mu \nu }=\mid A^{0}\mid ^{2}-\mid A^{1}\mid ^{2}-\mid A^{2}\mid ^{2}-\mid A^{3}\mid ^{2}$

The parity operator flips the sign of the spatial part of a 4-vector.

<alt>A super mu A super nu conjugated operated on by the parity operator contracted with g sub mu nu equals the absolute value of A super zero squared plus the absolute value of A super one squared plus the absolute value of A super two squared plus the absolute value of A super three squared.</alt>

$A^{\mu }A^{\nu *p}g_{\mu \nu }=\mid A^{0}\mid ^{2}+\mid A^{1}\mid ^{2}+\mid A^{2}\mid ^{2}+\mid A^{3}\mid ^{2}$

Normalize the potential.

<alt>A super mu over the absolute value of A times A super nu conjugated and acted on by the parity operator over the absolute value of A contracted with g sub mu nu equals one.</alt>

$\frac{A}{\mid A\mid }^{\mu }\frac{A}{\mid A\mid }^{\nu *p}g_{\mu \nu }=1$

From this, it can be concluded that the normalized 4-vector is an element of the symmetry group U(1) if the multiplication operator is the metric combined with the parity and conjugate operators. The same logic applies to the 4-vector potentials for the weak and the strong forces which happen to have internal symmetries.

In curved spacetime,

<alt>A super mu over the absolute value of A times A super nu conjugated and acted on by the parity operator over the absolute value of A contracted with g sub mu nu does not equal one.</alt>

$\frac{A}{\mid A\mid }^{\mu }\frac{A}{\mid A\mid }^{\nu *p}g_{\mu \nu }\neq 1$

<alt>A super mu over the absolute value of A times A super nu conjugated and acted on by the parity operator over the absolute value of A contracted with g sub mu nu.</alt>

Mass breaks U(1), SU(2), and SU(3) symmetry, but does so in a precise way (meaning you can calculate what $\frac{A}{\mid A\mid }^{\mu }\frac{A}{\mid A\mid }^{\nu *p}g_{\mu \nu }$ should equal). There is no need for the Higgs mechanism to give particles mass while preserving U(1)xSU(2)xSU(3) symmetry, so this proposal predicts no Higgs particle will be found.

The Forces

The Lorentz Force

<alt>The electromagnetic force F super mu equals q times U sub nu over c times A super mu comma nu minus A nu comma mu.</alt>

$F_{EM}^{\mu }=q\frac{U_{\nu }}{c}(A^{\mu .\nu }-A^{\nu ,\mu })$

The structure:

Force = charge * velocity * field strength tensor.

The analogy

<alt>minus q goes to the positive square root of G times m,</alt>

<alt>minus goes to plus,</alt>

<alt>comma goes to semicolon.</alt>

The analogous force

<alt>The proposed gravitational force F super mu equals minus the square root of G times m times U sub nu over c times A super mu semicolon nu plus A nu semicolon mu.</alt>

$F_{g}^{\mu }=-\sqrt{G}m\frac{U_{\nu }}{c}(A^{\mu ;\nu }+A^{\nu ;\mu })$

Note: the gravitational force and the electromagnetic force behave differently under charge inversion, if mass goes to negative m or electric charge goes to negative q.

The total force

<alt>The total force F super mu equals minus the square root of G times m plus q times U sub nu over c times A super mu semicolon nu minus the square root of G times m plus q times U sub nu over c times A nu semicolon mu.</alt>

fbox

$F_{gEM}^{\mu }=\frac{\partial mU^{\mu }}{\partial \tau }=(-\sqrt{G}m+q)\frac{U_{\nu }}{c}A^{\mu ;\nu }-(\sqrt{G}m+q)\frac{U_{\nu }}{c}A^{\nu ;\mu }$

If
F is zero, this has the form of a Killing's equation.

Geodesics and their sources

<alt>The force F super mu equals the derivative of the momentum m with respect to tau times U super mu equals m times the derivative of U super mu with respect to tau plus U super mu times the derivative of m with respect to tau,</alt>

$F^{\mu }=\frac{\partial mU^{\mu }}{\partial \tau }=m\frac{\partial U^{\mu }}{\partial \tau }+U^{\mu }\frac{\partial m}{\partial \tau }=0$

<alt>equals m times U super nu times U super mu sub semicolon nu equals m times U super nu times U super mu sub comma nu plus m times the Christoffel symbol super mu sub nu omega times U super mu U super omega equals m times the second derivative of x super mu with respect to tau plus m times the Christoffel symbol super mu sub nu omega times U super nu U super omega,</alt>

$=mU^{\nu }U_{\: ;\nu }^{\mu }=mU^{\nu }U_{\: ,\nu }^{\mu }+m\Gamma _{\: \nu \om... ...{\mu }}{\partial \tau ^{2}}+m\Gamma _{\: \nu \omega }^{\mu }U^{\nu }U^{\omega }$

<alt>equals minus the square root of G times m plus q times one over c times the derivative of A super mu comma nu minus the square root of G times m plus q times the derivative of x sub nu with respect to tau times A super nu comma mu minus two over c times m times the Christoffel symbol sub omega super mu nu times U sub nu U super omega.</alt>

$=(-\sqrt{G}m+q)\frac{\partial x_{\nu }}{c\partial \tau }A^{\mu ,\nu }-(\sqrt{G}... ...u }A^{\nu ,\mu }-\frac{2}{c}m\Gamma _{\omega }^{\: \mu \nu }U_{\nu }U^{\omega }$

Therefore, general relativity's description of geodesics is incomplete.

Note: A geodesic equation applies to electromagnetism!

A weak field gravitational force

Newton's Law

<alt>phi equals one over R</alt>

<alt>R equals zero</alt>

Solution to gEM field equations

<alt>therefore A sub mu equals the four-vector A sub zero, the three-vector zero</alt>

<alt>A sub mu equals the four-vector one over tau squared, the three-vector zero</alt>

<alt>tau squared equals zero</alt>

The weak field approximation

Normalize and study perturbations:

<alt>The four-vector one over sigma squared comma zero 3-vector for a weak field goes to the four-vector one over R initial plus a spring constant k over two R initial times R that sum squared minus T initial plus k over two T initial times t that sum squared all over one over R initial squared minus T initial squared, a three-vector zero.</alt>

$(\frac{1}{\sigma ^{2}},\overrightarrow{0})\rightarrow (\frac{\frac{1}{(R_{0}+\f... ...{0}+\frac{k}{2T_{0}}t)^{2}}}{\frac{1}{R_{0}^{2}-T_{0}^{2}}},\overrightarrow{0})$

Take the derivative with respect to $t$ and $R$.

<alt>The derivative of A super zero with respect to t equals k over sigma squared plus terms of order k squared.</alt>

$\frac{\partial A^{0}}{\partial t}=\frac{k}{\sigma ^{2}}+O(k^{2})$

<alt>The derivative of A super zero with respect to R equals minus k over sigma squared plus terms of order k squared.</alt>

$\frac{\partial A^{0}}{\partial R}=-\frac{k}{\sigma ^{2}}+O(k^{2})$

Gravitational force for a weak field

Plug into the force of gravity proposal, assuming the Christoffel symbol is zero which will be the case for a weak field in Euclidean coordinates. Note the contravariant derivative flips a sign.

<alt>F for gravity super mu equals minus the square root of G times m times U sub nu over c times A super mu comma nu plus A super nu comma mu equals m times the four-vector the derivative of t with respect to tau, minus the derivative of R with respect to tau contracted with k over sigma squared, k over sigma squared.</alt>

$F_{g}^{\mu }=-\sqrt{G}m\frac{U_{\nu }}{c}(A^{u,\nu }+A^{\nu ,\mu })=m(\frac{\pa... ...\frac{\partial R}{\partial \tau })(\frac{k}{\sigma ^{2}},\frac{k}{\sigma ^{2}})$

<alt>For gravity in a weak field, F super mu equals m times the four-vector k over sigma squared times the derivative of t with respect to tau, minus k over sigma squared times the derivative of R with respect to tau.</alt>

fbox

$F_{g}^{\mu }=m(\frac{k}{\sigma ^{2}}\frac{\partial t}{\partial \tau },-\frac{k}{\sigma ^{2}}\frac{\partial R}{\partial \tau })$

Newton's law of gravity

Must break spacetime symmetry.

Three assumptions

<alt>The spring constant k equals Newton's gravitational constant G times the source mass M.</alt>

<alt>sigma squared equals R squared minus t squared is approximately equal to R prime squared.</alt>

<alt>The four-vector the derivative of t with respect to tau, the derivative of R with respect to tau goes to the derivative of t with respect to the absolute value of R, the derivative of R with respect to the absolute value of R equals the four-vector zero, the unit three-vector R hat.</alt>

Lightlike events in Minkowski spacetime.

<alt>Two lightlike events on lightcone of Minkowski spacetime are on the forty-five degree of the light cone.</alt>

\includegraphics{minkowski.eps}

<alt>The difference element of tau squared or d tau squared equals d t squared minus d R squared equals zero.</alt>

$d\tau ^{2}=dt^{2}-d\overrightarrow{R}\cdot d\overrightarrow{R}=0$

<alt>The 4-vector the derivative of t with respect to tau, the derivative of R with respect to tau equals the 4-vector gamma, gamma beta</alt>

$(\frac{\partial t}{\partial \tau },\frac{\partial \overrightarrow{R}}{\partial \tau })=(\gamma ,\gamma \overrightarrow{\beta })$

Lightlike events in Newtonian spacetime.

<alt>Two simultaneous lightlike events in Newtonian spacetime are simultaneous.</alt>

\includegraphics{newton.eps}

<alt>d R squared is not equal to zero and is not a function of t</alt>

$dR^{2}\neq 0\neq f(t)$

<alt>The four-vector the derivative of t with respect to the absolute value of R, the derivative of R with respect to the absolute value of R equals the four-vector zero, the unit vector R hat.</alt>

$(\frac{\partial t}{\partial \mid R\mid },\frac{\partial \overrightarrow{R}}{\partial \mid R\mid })=(0,\hat{R})$

Plug into force equation

<alt>The gravitational force F super mu equals the four-vector zero, minus G source M test m over R squared times the unit vector R hat.</alt>

$F_{g}^{\mu }=(0,-\frac{GMm}{R^{2}}\hat{R})$

A metric equation

The force equation is two second order differential equations (assuming radial symmetry) [sigma squared equals minus tau squared].

<alt>The second derivative of t with respect to tau minus k over tau squared times the derivative of t with respect to tau equals zero.</alt>

$\frac{\partial ^{2}t}{\partial \tau ^{2}}+\frac{k}{\tau ^{2}}\frac{\partial t}{\partial \tau }=0$

<alt>The second derivative of R with respect to tau plus k over tau squared times the derivative of R with respect to tau equals zero.</alt>

$\frac{\partial ^{2}\overrightarrow{R}}{\partial \tau ^{2}}-\frac{k}{\tau ^{2}}\frac{\partial \overrightarrow{R}}{\partial \tau }=0$

<alt>The four-vector U super zero, the three-vector U equals the four-vector the derivative of t with respect to tau, the derivative of R with respect to tau.</alt>

Substitute $(U^{0},\overrightarrow{U})=(\frac{\partial t}{\partial \tau },\frac{\partial \overrightarrow{R}}{\partial \tau })$, to create first order equations.

<alt>The tau derivative of U super zero plus k over tau squared U super zero equals zero.</alt>

$\frac{\partial U^{0}}{\partial \tau }+\frac{k}{\tau ^{2}}U^{0}=0$

<alt>The tau derivative of the three-vector U minus k over tau squared U equals zero.</alt>

$\frac{\partial \overrightarrow{U}}{\partial \tau }-\frac{k}{\tau ^{2}}\overrightarrow{U}=0$

Solve.

<alt>U super mu equals the four-vector a constant z times e to the k over tau, a constant three-vector Z times e to minus k over tau.</alt>

$U^{\mu }=(ze^{\frac{k}{\tau }},\overrightarrow{Z}e^{-\frac{k}{\tau }})$

<alt>U super mu equals the four-vector z, Z.</alt>

For flat spacetime, $U^{\mu }=(z,\overrightarrow{Z})$. The constraint on relativistic velocities in flat spacetime is:

<alt>U super mu contracted with U sub mu equals d t squared minus d R squared over d tau squared equals one equals z squared minus three-vector Z squared.</alt>

$U^{\mu }U_{\mu }=\frac{dt^{2}-dR^{2}}{d\tau ^{2}}=1=z^{2}-\overrightarrow{Z}\cdot \overrightarrow{Z}$

<alt>d tau squared</alt>

Solve for the constants, and plug back into the constraint, multiplying through by $d\tau ^{2}$.

<alt>d tau squared equals e to the minus two k over tau times d t squared minus e to the 2 k over tau times d R squared.</alt>

$d\tau ^{2}=e^{-\frac{2k}{\tau }}dt^{2}-e^{\frac{2k}{\tau }}dR^{2}$

Assumptions

<alt>k equals G M over the speed of light c squared.</alt>

<alt>tau squared equals R squared minus t squared is approximately equal to R squared, so tau is approximately equal to tau plus or minus R.</alt>

<alt>d tau squared equals e to the two G M over c squared R times d R squared minus e to the minus 2 G M over c squared R times d t squared.</alt>

$d\tau ^{2}=e^{-\frac{2GM}{c^{2}R}}dt^{2}-e^{\frac{2GM}{c^{2}R}}dR^{2}$

<alt>G M over c squared R</alt>

Taylor series expansion to second order in $\frac{GM}{c^{2}R}$.

<alt>d tau squared equals one minus two times G M over c squared R plus two times the square of G M over c squared R times d t squared minus one plus two G M over c squared R plus two times the square of G M over c squared R times d t squared.</alt>

fbox

$d\tau ^{2}=(1-2\frac{GM}{c^{2}R}+2(\frac{GM}{c^{2}R})^{2})dt^{2}-(1+2\frac{GM}{c^{2}R}+2(\frac{GM}{c^{2}R})^{2})dR^{2}$

``Newtonian'' Constant velocity solutions

Apply the chain rule to the force equation.

<alt>F super mu equals the derivative with respect to tau of the inertial mass m times U super mu equals inertial m times the derivative of U super mu with respect to tau plus U super mu times the derivative of inertial m with respect to tau.</alt>

$F^{\mu }=\frac{\partial mU^{\mu }}{\partial \tau }=m\frac{\partial U^{\mu }}{\partial \tau }+U^{\mu }\frac{\partial m}{\partial \tau }$

Assume:

<alt>the derivative of U super mu with respect to tau equals zero</alt>

<alt>The 4-vector the derivative of t with respect to tau, the derivative of R with respect to tau goes to the derivative of t with respect to the absolute value of R, the derivative of R with respect to the absolute value of R equals the 4-vector 0, the unit vector R hat.</alt>

<alt>The four-vector zero, the derivative of the inertial mass m with respect to tau times the three-vector V equals the four-vector zero, minus k times the inertial over tau squared times the unit vector R hat.</alt>

$(0,\frac{\partial m}{\partial \tau }\overrightarrow{V})=(0,-\frac{km}{\sigma ^{2}}\hat{R})=(0,\frac{km}{\tau ^{2}}\hat{R})$

Both the three-vectors are constants, creating a first order differential equation.

<alt>The derivative of inertial m with respect to tau plus k times the inertial m over tau squared equals zero.</alt>

$\frac{\partial m}{\partial \tau }-\frac{km}{\tau ^{2}}=0$

Solve.

<alt>m equals a constant c time e to the k over tau.</alt>

$m=ce^{-\frac{k}{\tau }}$

<alt>k equals G the source mass M over the speed of light c squared</alt>

<alt>sigma squared equals R squared minus t squared is approximately equal to R squared</alt>

<alt>The initial mass m equals c times e to the G source mass M over c squared R.</alt>

$m=ce^{\frac{GM}{c^{2}R}}$

For thin disk galaxies

<alt>The velocity and mass profile of a galaxy with respect to distance. The X axis is the radius of the galaxy, on the Y is velocity and mass. The velocity curve is a straight horizontal line. The mass decays exponentially.</alt>

\includegraphics{constant_v.eps}

<alt>V maximum</alt>

After reaching $V_{max}$, the rotation profile of thin disk galaxies has a constant velocity while the mass decays exponentially. It might match this equation, but needs to be looked at in detail.

For the early Universe:

<alt>The velocity and mass profile of the early Universe. X axis is the radius of the Universe, on the Y is velocity and mass. The velocity curve is a straight horizontal line, indicating all parts of the Universe had the same velocity. The mass gets exponentially more dense closer to the big bang. </alt>

\includegraphics{bang.eps}

The very high density of the early Universe combined with its uniform velocity distribution would be consistent with the constant velocity solution of the gravitational force equation dominating the dynamics. The solution is stable.

Conclusions

Using a nineteenth century approach, two pillars of twentieth century physics have been fused, general relativity and the standard model. Both required technical modifications. The description of geodesics by general relativity is not complete because it does not explicitly show how the potential source causes curvature. A dynamic metric equation is found but it uses a simpler set of field equations (a rank 1 tensor instead of 2). In the standard model as elsewhere, combining two 4-vectors requires a metric. By normalizing the 4-vectors, the unitary aspect of the standard model can be self-evident. Although not mentioned before, the normalization process is essential to getting the dimensional analysis right.

This theory makes three testable predictions, two subtle, one not. First, the polarity of gravitational waves will be scalar or longitudinal, not transverse as predicted by general relativity. Second, if gravitation effects are measured to secondary post Newtonian accuracy, the coefficients for the metric derived here are different from the Schwarzschild metric in isotropic coordinates. Such an experiment will be quite difficult to do. The third test is to see if the complete relativistic force equation matches all the data for a thin spiral galaxy. It is this test which should be investigated first.