GEM field equation solutions

The task is to find a physically relevant solution to the GEM field equations, 5. The Poisson field equation of classical Newtonian gravity can be solved by an inverse distance potential, $1/R$. The potential has a point singularity where $R=0$. The GEM field equations are relativistic, so time must be incorporated. An inverse distance potential does not solve the field equations in four dimensions. The potential $A^{\mu}=(\frac{1}{\sigma^{2}},\overrightarrow{0})$ solves the field equations, where $\sigma^{2}$ is the Lorentz invariant distance squared, the negative of the Lorentz invariant interval squared, $-(c\tau)^{2}$. Distance is used instead of the interval because classical gravity depends on distance, not time. The potential has as a singularity that is the entire lightcone, where $\sigma^{2}=0$. The potential is not relevant in the classical domain since its derivative will not be an inverse square as required for a classical gravitational force.

Gravity is a weak effect. It is common in quantum mechanics to normalize a potential and study linear perturbations of weak fields, an approach that will be followed here. Assume spherical symmetry. Form a normalized potential with a linear perturbation:

$$A^{\mu}=(\frac{\sqrt{G}h}{c^{2}\sigma^{2}},\bi{0})\rightarro... ...R)^{2}-(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{2}}t)^{2}},\bi{0}).$$ (17)

Take the covariant derivative with respect to $t$ and $R$, keeping only terms to first order in the spring constant $k$:

$$\frac{\partial\phi}{\partial t} = \frac{c^{2}}{\sqrt{G}}\frac{k}{\sigma^{2}}+O(k^{2})$$ (18)

$$\bi{\nabla}\phi = -\frac{c^{2}}{\sqrt{G}}\frac{k}{\sigma^{2}}+O(k^{2}).$$

The change in the potential is a function of a spring constant $k$ over sigma squared. The classical Newtonian dependence on distance is an inverse square, so this is promising. A potential that applies exclusively to gravity is sought, yet the non-zero gradient of $\phi$ indicates an electric field. The sign of the spring constant $k$ does not effect the solution to the four dimensional wave field equations but does change the derivative of the potential. A potential that only has derivatives along the diagonal of the field strength tensor $\bigtriangledown_{\mu}A^{\nu}$ can be constructed from two potentials that differ by spring constants that either constructively interfere to create non-zero derivatives, or destructively interfere to eliminate derivatives. With this in mind, construct a potential that will have no electric field:

$$A^{\mu} = \frac{c}{\sqrt{G}}(\frac{1}{(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{... ...{2}}+\frac{k}{\sigma^{2}}z)^{2}-(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{2}}t)^{2}}$$

$$+ \frac{1}{(\frac{1}{\sqrt{2}}-\frac{k}{\sigma^{2}}x)^{2}+(\frac{1}... ...2}}-\frac{k}{\sigma^{2}}z)^{2}-(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{2}}t)^{2}},$$

$$\frac{1}{(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{2}}x)^{2}+(\frac{1}... ...{2}}+\frac{k}{\sigma^{2}}z)^{2}-(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{2}}t)^{2}}$$

$$+ \frac{1}{(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{2}}x)^{2}+(\frac{1}... ...2}}-\frac{k}{\sigma^{2}}z)^{2}-(\frac{1}{\sqrt{2}}-\frac{k}{\sigma^{2}}t)^{2}},$$

$$\frac{1}{(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{2}}x)^{2}+(\frac{1}... ...{2}}+\frac{k}{\sigma^{2}}z)^{2}-(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{2}}t)^{2}}$$ (19)

$$+ \frac{1}{(\frac{1}{\sqrt{2}}-\frac{k}{\sigma^{2}}x)^{2}+(\frac{1}... ...2}}-\frac{k}{\sigma^{2}}z)^{2}-(\frac{1}{\sqrt{2}}-\frac{k}{\sigma^{2}}t)^{2}},$$

$$\frac{1}{(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{2}}x)^{2}+(\frac{1}... ...{2}}+\frac{k}{\sigma^{2}}z)^{2}-(\frac{1}{\sqrt{2}}+\frac{k}{\sigma^{2}}t)^{2}}$$

$$+ \frac{1}{(\frac{1}{\sqrt{2}}-\frac{k}{\sigma^{2}}x)^{2}+(\frac{1}... ...}}+\frac{k}{\sigma^{2}}z)^{2}-(\frac{1}{\sqrt{2}}-\frac{k}{\sigma^{2}}t)^{2}}).$$

Take the covariant derivative of this potential, keeping only the terms to first order in the spring constant $k$.

$$\bigtriangledown_{\mu}A^{\nu}=\frac{c^{2}}{\sqrt{G}}\frac{k}... ... 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\end{array}\right]+O(k^{2}).$$ (20)

All this work was required to get a multiple of the Minkowski matrix!