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\begin{document}

\title{gEM: Unifying gravity, electricity, and magnetism by analogy}


\author{Douglas B Sweetser@alum.mit.edu}

\maketitle
\tableofcontents{}


\section{The goal}

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


\subsection*{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.


\subsection*{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.


\subsection*{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.


\subsection*{The results}

\begin{itemize}
\item The Lagrangian contains both terms with a connection and the Fermi
Lagrangian of electromagnetism.
\item The gravitational field equations are analogues to Gauss' and Ampere's
laws, so are not second rank tensor equations like Einstein's field
equations.
\item The Maxwell source equations result if all gravitational field terms
are zero.
\item The Maxwell homogeneous equations are unaffected.
\item The field equations have been quantized before, but new interpretations
will flow from the unification effort.
\item A link to the the Lagrangian of the standard model will be detailed.
\end{itemize}
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:

\begin{itemize}
\item A geodesic equation where the potential causes the curvature, something
which is missing from general relativity.
\item Newton's law of gravity if spacetime symmetry is broken.
\item A metric equation similar to the Schwarzschild metric if spacetime
symmetry is preserved, but the two metrics are different at second
order post-Newtonian accuracy.
\item A new class of solutions for the gravitational source where velocity
is constant, but the distribution of mass varies with distance. This
may provide new ways to look at problems with the rotation profiles
of disk galaxies and big bang cosmology.
\end{itemize}

\section{The Lagrangians}


\subsection*{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>

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


\subsection*{The structure}

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


\subsection*{The analogy}

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

\begin{itemize}
\item -q$\rightarrow +\sqrt{G}m$
\end{itemize}
<alt>Minus in the tensor goes to plus. </alt>

\begin{itemize}
\item -$\rightarrow +$
\end{itemize}
<alt>The standard derivative represented by a comma goes to the covariant
derivative semicolon.</alt>

\begin{itemize}
\item ,$\rightarrow $;
\end{itemize}

\subsection*{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>

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


\subsection*{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>

$=-\frac{m}{\gamma V}+\frac{(\sqrt{G}m-q)}{c^{2}V}(\phi -\overrightarrow{A}\cdot \overrightarrow{V})-\frac{1}{2c^{2}}(A^{\mu ,\nu }-\Gamma _{\omega }^{\: \mu \nu }A^{\omega })(A_{\mu ,\nu }-\Gamma _{\: \mu \nu }^{\omega }A_{\omega })$

<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>

$=-\frac{m}{\gamma V}+\frac{(\sqrt{G}m-q)}{c^{2}V}(\phi -\overrightarrow{A}\cdot \overrightarrow{V})-\frac{1}{2c^{2}}A^{\mu ,\nu }A_{\mu ,\nu }-\frac{1}{2c^{2}}\Gamma _{\: \mu \nu }^{\omega }(\Gamma _{\omega }^{\: \mu \nu }A^{\omega }A_{\omega }-2A^{\mu ,\nu }A_{\omega })$

\begin{itemize}
\item The kinetic energy term is for one particle experiencing both gravity
and electromagnetism, all other terms are added together.
\item The Fermi Lagrangian of electromagnetism is a subset.
\item Includes the connection squared and the connection contracted with
the field strength tensor and velocity.
\end{itemize}
<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>

$\mathcal{L}_{gEM}=-\frac{m}{V}\sqrt{1-(\frac{\partial \overrightarrow{R}}{\partial t})^{2}}+\frac{(\sqrt{G}m-q)}{c^{2}V}(\phi -\overrightarrow{A}\cdot \overrightarrow{V})-\frac{1}{2c^{2}}((\frac{\partial \phi }{\partial t})^{2}-(\nabla \phi )^{2}-(\frac{\partial \overrightarrow{A}}{\partial t})^{2}+(\nabla \overrightarrow{A})^{2})$


\section{The classical field equations}


\subsection*{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>

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


\subsection*{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>

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

<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>

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

<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>

\begin{itemize}
\item $g^{\mu \mu }$ is the diagonal of the field strength tensor.
\end{itemize}
<alt>g super u 0.</alt>

\begin{itemize}
\item $\overrightarrow{g^{u0}}$ added the electric field make up the first
row and column of the asymmetric field strength tensor.
\end{itemize}
<alt>g 3-vector.</alt>

\begin{itemize}
\item $\overrightarrow{g}$ added to the magnetic field makes up the rest
of the off-diagonal field strength tensor terms.
\end{itemize}

\subsection*{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>

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

\begin{itemize}
\item The first rank field equations for gravity are not equivalent to Einstein's
second rank field equations.
\item Gravitational field equations depend on a symmetric second rank tensor.
\end{itemize}
<alt>g equals zero</alt>

\begin{itemize}
\item With all derivatives of the g fields equal to zero, the Maxwell source
equations result.
\item The homogeneous Maxwell equations are unaffected.
\end{itemize}

\section{Connonical quantization}


\subsection*{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>

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


\subsection*{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>

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


\subsection*{Problems with quantization}

<alt>The Lagrangian of electromagnetism</alt>

$\mathcal{L}_{EM}$

\begin{itemize}
\item About EM only.
\item Must constrain modes to Lorenz gauge (Gupta/Bleuler).
\end{itemize}
<alt>The Lagrangian of gravity and electromagnetism</alt>

\noindent $\mathcal{L}_{gEM}$

\begin{itemize}
\item About EM and gravity.
\item Interpretation of quantization changes.
\end{itemize}
<alt>The contravariant derivative d super mu contracted with A sub
mu equals zero if and only if the mass M equals zero.</alt>

\begin{itemize}
\item $\partial ^{\mu }A_{\mu }=0\quad iff\: M=0$, get Maxwell field equations
only.
\item Mass breaks gauge symmetry. Get Maxwell field equations + 4-vector
gravitational field equations.
\end{itemize}
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.


\section{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>

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

\noindent 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>

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

\noindent 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>

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

\noindent 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.


\section{The Forces}


\subsection*{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>

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


\subsection*{The structure:}

Force = charge {*} velocity {*} field strength tensor.


\subsection*{The analogy}

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

\begin{itemize}
\item $-q\rightarrow +\sqrt{G}m$
\end{itemize}
<alt>minus goes to plus,</alt>

\begin{itemize}
\item -$\rightarrow +$
\end{itemize}
<alt>comma goes to semicolon.</alt>

\begin{itemize}
\item ,$\rightarrow $;
\end{itemize}

\subsection*{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.


\subsection*{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

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

\begin{description}
\item [If]F is zero, this has the form of a Killing's equation.
\end{description}

\subsection*{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>

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

<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>

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

\begin{itemize}
\item With no force, spacetime curvature causes acceleration (in general
relativity).
\item With no force, spacetime curvature is caused by the potential (not
in general relativity).
\end{itemize}
Therefore, general relativity's description of geodesics is incomplete.

Note: A geodesic equation applies to electromagnetism!


\section{A weak field gravitational force}


\subsection*{Newton's Law}

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

\begin{itemize}
\item Solution: $\phi =\frac{1}{R}$ in 3D.
\end{itemize}
<alt>R equals zero</alt>

\begin{itemize}
\item Singularity: point where $R=0$.
\end{itemize}

\subsection*{Solution to gEM field equations}

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

\begin{itemize}
\item Want gravity only, no EM, therefore $A_{\mu }=(A_{0},\overrightarrow{0})$.
\end{itemize}
<alt>A sub mu equals the four-vector one over tau squared, the three-vector
zero</alt>

\begin{itemize}
\item Solution: $A_{\mu }=(\frac{1}{\sigma ^{2}},\overrightarrow{0})$ in
4D, where $\sigma ^{2}=R^{2}-t^{2}$.
\end{itemize}
<alt>tau squared equals zero</alt>

\begin{itemize}
\item Singularity: the lightcone where $\tau ^{2}=0$, and the Maxwell equations
take over.
\end{itemize}

\subsection*{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>

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

\noindent 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>

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


\subsection*{Gravitational force for a weak field}

\noindent 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>

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

<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

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


\section{Newton's law of gravity}

Must break spacetime symmetry.


\subsection*{Three assumptions}

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

\begin{itemize}
\item Gravitational source spring constant: $k=GM$.
\end{itemize}
<alt>sigma squared equals R squared minus t squared is approximately
equal to R prime squared.</alt>

\begin{itemize}
\item Static field approximation: $\sigma ^{2}=R^{2}-t^{2}\cong R'^{2}$.
\end{itemize}
<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>

\begin{itemize}
\item Break spacetime symmetry: $(\frac{\partial t}{\partial \tau },\frac{\partial \overrightarrow{R}}{\partial \tau })\rightarrow (\frac{\partial t}{\partial \mid R\mid },\frac{\partial \overrightarrow{R}}{\partial \mid R\mid })=(0,\hat{R})$.
\end{itemize}
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>

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

\noindent 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>

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


\subsection*{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>

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


\section{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>

\noindent 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>

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

\noindent 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>

\noindent 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>

\noindent 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>

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


\subsection*{Assumptions}

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

\begin{itemize}
\item Gravitational source spring constant: $k=GM/c^{2}$.
\end{itemize}
<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>

\begin{itemize}
\item Static field approximation: $\sigma ^{2}=R^{2}-t^{2}\cong R'^{2}$.
\item Both sigma and tau must have the same magnitude, R. To make the metric
extressed in terms of tau real, assume sigma is i R, so tau is R.
\end{itemize}
<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>

\noindent 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

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

\begin{itemize}
\item Will pass all classical test to first degree of post Newtonian accuracy. 
\item Will differ in predictions about second degree tests.
\end{itemize}

\section{{}``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>

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

\noindent Assume:

\begin{itemize}
\item The weak field approximation for the 1/interval squared potential.
\end{itemize}
<alt>the derivative of U super mu with respect to tau equals zero</alt>

\begin{itemize}
\item No change in velocity, $\frac{\partial U^{\mu }}{\partial \tau }=0$.
\end{itemize}
<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>

\begin{itemize}
\item Break spacetime symmetry: $(\frac{\partial t}{\partial \tau },\frac{\partial \overrightarrow{R}}{\partial \tau })\rightarrow (\frac{\partial t}{\partial \mid R\mid },\frac{\partial \overrightarrow{R}}{\partial \mid R\mid })=(0,\hat{R})$.
\end{itemize}
<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>

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

\noindent 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>

\begin{itemize}
\item Gravitational source spring constant: $k=GM/c^{2}$.
\end{itemize}
<alt>sigma squared equals R squared minus t squared is approximately
equal to R squared</alt>

\begin{itemize}
\item \begin{flushleft}Static field approximation: $\sigma ^{2}=R^{2}-t^{2}\cong R'^{2}$. \end{flushleft}
\item \begin{flushleft}Both sigma and tau must have the same magnitude,
R. To make the mass distribution extressed in terms of tau real, assume
sigma is i R. Tau could either be positive or negative R. Choose negative
so the mass decays with distance.\end{flushleft}
\end{itemize}
<alt>The initial mass m equals c times e to the G source mass M over
c squared R.</alt>

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

\noindent 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.


\section{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.
\end{document}
