Indexes:
$A-Z$ = t,x,y,z in flat space
$i-n$ = x,y,z in flat space
$0123$ = specific coordinates in flat spacetime
$\alpha-\omega$ = t,x,y,z in curved-space
$a-h,o-z$ = x,y,z in curved space
$txyz$ = specific coordinates in curved spacetime

metric -> geodesic (gravitational force) -> Einstein tensor = stress-energy tensor

Popular spherical matter metrics like Schwarzschild, Kerr, etc. solutions take place at $r \rightarrow \infty$, therefore the matter is treated as a point source, therefore $T_{\mu\nu} = 0$. Likewise $G_{\mu\nu} = 0$.

Metrics which do consider matter terms include the FLRW metric, which uses $T_{\mu\nu} = diag(\rho, p, p, p) g_{\mu\nu}$. Is this relation always the case?

Also, MTW ch. 23 talks about embedding functions, and how metrics vary within the inside of a star.

Electromagnetic Stress-Energy

in flat space:

$E^i$ = electric field
$B^i$ = magnetic field
$S^i = {\epsilon^i}_{jk} E^j B^k$ = Poynting vector

Faraday tensor:
$F^{0j} = -F^{j0} = E^j$
$F^{jk} = {\epsilon^{jk}}_l B^l$
$F^{JK} = J(j) \downarrow \overset{ K(k) \rightarrow }{\left[\matrix{ 0 & E^k \\ -E^j & {\epsilon^{jk}}_l B^l }\right]}$

$F_{0j} = -F_{j0} = -E_j$
$F_{jk} = {\epsilon_{jk}}^l B^l$
$F_{JK} = J(j) \downarrow \overset{ K(k) \rightarrow }{\left[\matrix{ 0 & -E_k \\ E_j & {\epsilon_{jk}}^l B_l }\right]}$

$A^I$ = electromagnetic 4-potential
$A_I = [\phi, A_i]$
$\phi =$ electric potential.
$A_i =$ magnetic potential.
$F_{JK} = 2 A_{[K,J]}$
$F_{j0} = E_j = 2 A_{[t,j]}$
$F_{jk} = {\epsilon_{jk}}^l B_l = 2 A_{[k,j]}$

$J_I = [q, j_i]$ = 4-current
$q =$ charge density
$j_i =$ current density

Conservation of 4-current in flat space:
$J^I_{,I} = 0$

Conservation of 4-current in curved space:
charge vector conserved in curved space, using the covariant derivative:
${J^\mu}_{;\mu} = 0$
${J^\mu}_{,\mu} + {\Gamma^\mu}_{\nu\mu} J^\nu = 0$
using ${\Gamma^\mu}_{\nu\mu} = (log \sqrt{|g|})_{,\nu} = \frac{1}{\sqrt{|g|}} (\sqrt{|g|})_{,\nu}$
$\frac{\sqrt{|g|}}{\sqrt{|g|}} {J^\nu}_{,\nu} + \frac{1}{\sqrt{|g|}} (\sqrt{|g|})_{,\nu} J^\nu = 0$
$\frac{1}{\sqrt{|g|}} (\sqrt{|g|} J^\nu)_{,\nu} = 0$
...which is equivalent to charge, scaled by spatial volume, conserved in flat-space.
But using the curved-space basis.
In fact, if you convert the curved-space basis to flat-space basis...
$\frac{1}{\sqrt{|g|}} (\sqrt{|g|} J^I {e^\nu}_I )_{,J} {e_\nu}^J = 0$
$\frac{1}{\sqrt{|g|}} ( \sqrt{|g|}_{,J} {e_\nu}^J J^I {e^\nu}_I + \sqrt{|g|} {J^I}_{,J} {e_\nu}^J {e^\nu}_I + \sqrt{|g|} J^I {{e^\nu}_I}_{,J} {e_\nu}^J ) = 0$
$\frac{1}{\sqrt{|g|}} ( \sqrt{|g|}_{,J} \delta^J_I J^I + \sqrt{|g|} {J^I}_{,J} \delta^J_I + \sqrt{|g|} J^I {{e^\nu}_I}_{,J} {e_\nu}^J ) = 0$
$\frac{1}{\sqrt{|g|}} ( \sqrt{|g|}_{,I} J^I + \sqrt{|g|} {J^I}_{,I} + \sqrt{|g|} J^I {{e^\nu}_I}_{,J} {e_\nu}^J ) = 0$
$\frac{1}{\sqrt{|g|}} (\sqrt{|g|} J^I)_{,I} + J^I {{e^\nu}_I}_{,J} {e_\nu}^J = 0$
Note that the last, extra term is equal to the difference of the affine connection ${\Gamma^\rho}_{\mu\nu}$ and the vielbein spin-connection ${{\omega_\mu}^\rho}_\nu$,
defined with the relation ${{e_\nu}^I}_{,\mu} = {\Gamma^I}_{\mu\nu} - {{\omega_\mu}^I}_\nu$
So ${e^\nu}_{I,J} {e_\nu}^J = -{e^\nu}_I {{e_\nu}^J}_{,J} = -{e^\nu}_I ({\Gamma^J}_{J\nu} - {{\omega_J}^J}_\nu) = -{e^\nu}_I ({\Gamma^\rho}_{\rho\nu} - {{\omega_\rho}^\rho}_\nu) = -{e^\nu}_I ({\Gamma^\rho}_{\rho\nu} - {{\omega_\rho}^\rho}_\nu) = -{e^\nu}_I ({T^\rho}_{\rho\nu} + {\Gamma^\rho}_{\nu\rho} - {{\omega_\rho}^\rho}_\nu) = -{e^\nu}_I ({T^\rho}_{\rho\nu} + (ln \sqrt{|g|})_{,\nu} - {{\omega_\rho}^\rho}_\nu) $
So going from the curved basis to the flat basis adds us the extra term corresponding to the difference between the affine and the spin connections.
(speaking of which, now that ${\Gamma^\alpha}_{\beta\gamma}$ is no longer symmetric on $\beta\gamma$, is it ${\Gamma^\rho}_{\nu\rho} = (ln \sqrt{|g|})_{,\nu}$ or ${\Gamma^\rho}_{\rho\nu} = (ln \sqrt{|g|})_{,\nu}$ ?)

Relation between Faraday tensor and 4-current, in curved space:

Faraday tensor in curved space:
$F_{\mu\nu} = 2 A_{[\nu;\mu]} = 2 (A_{[\nu,\mu]} - {\Gamma^\rho}_{[\nu\mu]} A_\rho)$
Let ${T^\rho}_{\nu\mu} = 2 {\Gamma^\rho}_{[\nu\mu]} = $ the torsion tensor.
$F_{\mu\nu} = 2 A_{[\nu;\mu]} = 2 A_{[\nu,\mu]} + {T^\rho}_{\mu\nu} A_\rho$
In a torsion-free space, ${T^\rho}_{\mu\nu} = 0$ and we get
$F_{\mu\nu} = 2 A_{[\nu;\mu]} = 2 A_{[\nu,\mu]}$
So, aside from torsion, curvature of spacetime does not affect the Faraday metric.

Faraday and 4-current relation in flat space:
${F_{JK}}^{,K} = 4 \pi J_J$
$2 {A_{[K,J]}}^{,K} = 4 \pi J_J$
${A_{K,J}}^K - {A_{J,K}}^K = 4 \pi J_J$
${{A_K}^{,K}}_J - {A_{J,K}}^K = 4 \pi J_J$
Imposing the gauge ${A^K}_{,K} = 0$
$-{A_{J,K}}^K = 4 \pi J_J$

Curved-space Faraday and 4-current relation in curved space ... shouldn't match the conservation in curved space, because it doesn't consider the Riemann curvature tensor.
${F_{\mu\nu}}^{;\nu} = 4 \pi J_\mu$
Faraday and 4-current relation in curved space:

Electromagnetism stress-energy tensor:

$T^{\mu\nu} = \frac{1}{4\pi}(F^{\mu\alpha} {F^{\nu}}_\alpha - \frac{1}{4} g^{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} )$

In flat space this simplifies to
$T^{00} = \frac{1}{8\pi} (E^2 + B^2)$
$T^{0j} = T^{j0} = \frac{1}{4\pi} {\epsilon_{ik}}^j E^i B^k$
$T^{jk} = \frac{1}{4\pi} ( -(E^j E^k + B^j B^k) + \frac{1}{2} (E^2 + B^2) \delta^{jk} )$
$T^{JK} = \frac{1}{8\pi} \cdot J(j) \downarrow \overset{ K(k) \rightarrow }{\left[\matrix{ E^2 + B^2 & 2 S^k \\ 2 S^j & -2 (E^j E^k + B^j B^k) + (E^2 + B^2) \delta^{jk} }\right]}$

ADM metric:
$\alpha$ = lapse
$\beta^i$ = shift
${(^3e)_u}^i$ = spatial triad

$g_{\mu\nu} = \mu(u)\downarrow \overset{\nu(v)\rightarrow}{\left[ \matrix{-\alpha^2 + \beta^a \beta_a& \beta_v \\ \beta_u & \gamma_{uv}} \right]}$
$g^{\mu\nu} = \mu(u)\downarrow \overset{\nu(v)\rightarrow}{\left[ \matrix{-1/\alpha^2 & \beta^v / \alpha^2 \\ \beta^u / \alpha^2 & \gamma^{uv} - \beta^u \beta^v / \alpha^2} \right]}$

ADM tetrad:
${e_\mu}^I = \mu(u) \downarrow \overset{ I(i) \rightarrow }{\left[\matrix{ \alpha & \beta^i \\ 0 & {(^3e)_u}^i }\right]}$
${e^\mu}_I = \mu(u) \downarrow \overset{ I(i) \rightarrow }{\left[\matrix{ 1/\alpha & 0 \\ -\beta^u / \alpha & {(^3e)^u}_i }\right]}$

Flat-space 3-vectors converted to curved-space 4-vectors:
Transforming a 3-vector by a triad gives $S^j {(^3e)^u}_j = S^u$
Transforming a 4-vector by a tetrad gives $S^\mu = S^J {e^\mu}_J = \left[\matrix{ S^0 / \alpha \\ -S^0 \beta^u / \alpha + S^u }\right]$
If $S^0 = 0$ (where $^0$ denotes the timelike component of the flat-space 4-vector) then this gives $ S^\mu = \left[\matrix{ 0 \\ S^u } \right]$
So I am going to assume the flat-space vectors $E^i, B^i, S^i$ have timelike component 0, and that only their 3-space components transform when transformed by the tetrad.

Curved-space Faraday tensor:
$F^{\mu\nu} = {e^\mu}_I F^{IJ} {e^\nu}_J$
$= \mu(u) \downarrow \overset{ I(i) \rightarrow }{\left[\matrix{ 1/\alpha & 0 \\ -\beta^u / \alpha & {(^3e)^u}_i }\right]} \cdot I(i) \downarrow \overset{ J(j) \rightarrow }{\left[\matrix{ 0 & E^j \\ -E^i & {\epsilon^{ij}}_k B^k }\right]} \cdot J(j) \downarrow \overset{ \nu(v) \rightarrow }{\left[\matrix{ 1/\alpha & -\beta^v / \alpha \\ 0 & {(^3e)^v}_j }\right]}$
$= \mu(u) \downarrow \overset{ J(j) \rightarrow }{ \left[\matrix{ 0 & E^j / \alpha \\ -E^u & -E^j \beta^u / \alpha + {\epsilon^{uj}}_k B^k }\right]} \cdot J(j) \downarrow \overset{ \nu(v) \rightarrow }{ \left[\matrix{ 1/\alpha & -\beta^v/\alpha \\ 0 & {(^3e)^v}_j }\right]}$
$= \mu(u) \downarrow \overset{ \nu(v) \rightarrow }{\left[\matrix{ 0 & E^v / \alpha \\ -E^u / \alpha & E^u \beta^v / \alpha - E^v \beta^u / \alpha + {\epsilon^{uv}}_k B^k }\right]}$

Curved-space electromagnetism stress-energy tensor:
$T^{\mu\nu} = {e^\mu}_J T^{JK} {e^\nu}_K$
$ = \mu(u) \downarrow \overset{ J(j) \rightarrow }{\left[\matrix{ 1/\alpha & 0 \\ -\beta^u / \alpha & {(^3e)^u}_j }\right]} \cdot \frac{1}{8\pi} \cdot J(j) \downarrow \overset{ K(k) \rightarrow }{\left[\matrix{ E^2 + B^2 & 2 S^k \\ 2 S^j & -2 (E^j E^k + B^j B^k) + (E^2 + B^2) \delta^{jk} }\right]} \cdot K(k) \downarrow \overset{ \nu(v) \rightarrow }{\left[\matrix{ 1/\alpha & -\beta^v / \alpha \\ 0 & {(^3e)^v}_k }\right]}$
$ = \frac{1}{8\pi} \cdot \mu(u) \downarrow \overset{ K(k) \rightarrow }{\left[\matrix{ (E^2 + B^2) / \alpha & 2 S^k / \alpha \\ -\beta^u (E^2 + B^2) / \alpha + 2 S^u & -2 S^k \beta^u / \alpha - 2 (E^u E^k + B^u B^k) + (E^2 + B^2) (^3e)^{uk}) }\right]} \cdot K(k) \downarrow \overset{ \nu(v) \rightarrow }{\left[\matrix{ 1/\alpha & -\beta^v / \alpha \\ 0 & {(^3e)^v}_k }\right]}$
$= \frac{1}{8\pi} \cdot \mu(u) \downarrow \overset{ \nu(v) \rightarrow }{\left[\matrix{ (E^2 + B^2) / \alpha^2 & -\beta^v (E^2 + B^2) / \alpha^2 + 2 S^v / \alpha \\ -\beta^u (E^2 + B^2) / \alpha^2 + 2 S^u / \alpha & \beta^u \beta^v (E^2 + B^2) / \alpha^2 - 2 S^u \beta^v / \alpha - 2 S^v \beta^u / \alpha - 2 (E^u E^v + B^u B^v) + (E^2 + B^2) \gamma^{uv} }\right]}$
Note that $(E^2 + B^2)$ denotes the flat-space vector norm.
Also note that, in absense of $\beta^u$, the stress-energy tensor takes on its more familiar form:
$T^{\mu\nu} = \frac{1}{8\pi} \cdot \mu(u) \downarrow \overset{ \nu(v) \rightarrow }{\left[\matrix{ (E^2 + B^2) / \alpha^2 & 2 S^v / \alpha \\ 2 S^u / \alpha & -2 (E^u E^v + B^u B^v) + (E^2 + B^2) \gamma^{uv} }\right]}$

If you wanted to build $T^{\mu\nu}$ from flat-space $E^i$ and $B^i$ then you would also need to know the tetrad components in order to compute $E^u$ and $B^u$
If you only had the curved-space components $E^u$ and $B^u$ then you could compute $E^2 = \gamma_{uv} E^u E^v$. The $\beta^u$ would already be provided by the metric.
The last necessary piece of information would be showing that $S^u = {\epsilon^u}_{vw} E^v B^w$.

Einstein field equations:
$G^{\mu\nu} = 8 \pi T^{\mu\nu}$

Einstein tensor:
$G^{\mu\nu} = R^{\mu\nu} - \frac{1}{2} g^{\mu\nu} R$

Einstein field equations for electromagnetism stress-energy:
$G^{\mu\nu} = 2 F^{\mu\alpha} {F^{\nu}}_\alpha - \frac{1}{2} g^{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} $
$R^{\mu\nu} - \frac{1}{2} g^{\mu\nu} R = 2 F^{\mu\alpha} {F^{\nu}}_\alpha - \frac{1}{2} g^{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} $
$R^{\mu\nu} - \frac{1}{2} g^{\mu\nu} R = 2 {e^\mu}_I F^{IK} {F^J}_K {e^\nu}_J - \frac{1}{2} g^{\mu\nu} F_{IJ} F^{IJ} $
$G^{00} = E^2 + B^2$
$G^{j0} = G^{0j} = 2 {\epsilon_{ik}}^j E^i B^k = 2 S^j$
$G^{jk} = -2 (E^j E^k + B^j B^k) + \gamma^{jk} (E^2 + B^2)$

The Ricci Tensor

Wikipedia says the Ricci curvature, applied to a vector (i.e. $ R(x,x) = x^a x^b R_{ab} $ ), gives a positive value when geodesics all traveling together near a point in direction $x^i$ converge, and a negative value when they diverge.

One vector:
$R_{ab} x^a x^b $ is the Ricci scalar curvature in the direction $x^a$.
If it is positive then geodesics travelling near the direction of $x^a$ tend to diverge. If it is positive then they converge. Also stated, if a volume travels along the geodesic then it changes by $-2 \delta V R_{ab} x^a x^b$

Two vectors:
$R_{ab} x^a y^b$ is the sectional curvature of the plane tangent to the metric at this point that passes through $x^a$ and $y^b$.
The sum of all sectional curvatures is the Gaussian curvature. Hence why $R = {R^a}_a$.

The Einstin tensor $G_{ab}$ is the trace-reversal of the Ricci tensor $R_{ab}$.
Defined by $G_{ab} = R_{ab} - \frac{2}{n} {R^c}_c g_{ab}$, for $n=4$ dimensions, gives $G_{ab} = R_{ab} - \frac{1}{2} {R^c}_c g_{ab}$
Because it's a trace-reversal, you can contract the equation and solve to find the opposite holds for restoring $R_{ab}$:
${G^a}_a = R - \frac{1}{2} R \delta^a_a = R - 2 R = -R$ for $R = {R^a}_a$ is the Gaussian curvature, is the sum of the sectional curvatures of individual ${R^i}_i$ (not summing i).

But in the timelike direction, because $g_tt$ is negative (courtesy of the signature), things get flipped back.
Let $T_{ab} = diag(\rho, P, P, P)$, $G_{ab} = 8 \pi T_{ab}$, $R_{ab} = G_{ab} - \frac{1}{2} {G^c}_c g_{ab}$
Then $R_{tt} = G_{tt} - \frac{1}{2} {G^c}_c g_{tt}$
For $g_{ab} \approx \eta_{ab}$ we then get...
$R_{tt} = 8 \pi \rho - \frac{1}{2} 8 \pi (-\rho + 3 P) (-1)$
$R_{tt} = 8 \pi \rho + 4 \pi (-\rho + 3 P)$
$R_{tt} = 4 \pi (\rho + 3 P)$
So positive mass and pressure means positive Ricci scalar curvature (for resing object) means tighter focusing of geodesics (into the center of the planet).

The connection coefficients determine the geodesics through a manifold. The Riemann tensor determines the deviation of geodesics.

If the Einstein tensor $ G_{ab} = R_{ab} - \frac{2}{n} R g_{ab} $ is the trace-reversal (subtracting $ \frac{1}{n} R g_{ab} $ from $ R_{ab} $ will produce the traceless version of the tensor) then, for a 2D example with diagonal $ {R^a}_b $, the Einstein tensor will be the negative of the Ricci tensor.

Look at some stress-energy tensors using $g_{ab} \approx \eta_{ab}$...
For $T_{ab} = diag(\rho, P, P, P)$ we get:
$R_{tt} = 4 \pi (\rho + 3 P)$
$R_{xx} = R_{yy} = R_{zz} = 4 \pi (\rho + P)$
so in matter fields, geodesics always converge.

For EM stress-energy tensor $T_{ab} = \left[\matrix{ \frac{1}{2}(E^2+B^2) & -S_j \\ -S_i & \frac{1}{2}(E^2+B^2) - E_i E_j - B_i B_j }\right]$
assume $E = \hat{x} E$, $B = \hat{y} B$, $S = E \times B = \hat{z} S$, we get:
$R_{tt} = R_{zz} = 4 \pi (E^2 + B^2)$
$R_{xx} = 4 \pi (B^2 - E^2)$
$R_{yy} = 4 \pi (E^2 - B^2)$
So it looks like geodesics along the timelike and Poynting vector converge.
But geodesics along the electric field converge when the electric field is strong than the magnetic field, and diverge when the magnetic is stronger than the electric.
And geodesics along the magnetic field likewise converges if the magnetic > electric, diverge if electric > magnetic.

Stress-Energy Equivalence of Matter and EM

Stress energy of a perfect fluid at rest: $T_{ab} = diag(\rho, P, P, P)$

If you use the Schwarzschild metric and compute gravity around a celestial body, I can plug in the numbers for Earth or any other planet in the solar system and accurately compute the relativistic gravitational forces and it comes out to $9.8 \frac{m}{s}$.
If you wanted to create an equivalent stress-energy of electromagnetism, my first thought was to compute equivalent values within the electromagnetism stress-energy tensor, especially the $T_{tt}$ component. In the matter stress-energy tensor $T_{tt} = \rho$.

Density of Earth: $\rho = 5.5147098661212 \frac{g}{cm^3} = 5514.7098661212 \frac{kg}{m^3}$
Using $c = 299792458 m/s = \tilde{c} \cdot m/s$, so $1 s = \tilde{c} m$
and $G = 6.67384 \cdot 10^{-11} \frac{m^3}{kg \cdot s^2} = \tilde{G} \cdot \frac{m^3}{kg \cdot s^2}$, so $1 kg = \frac{\tilde{G}}{\tilde{c}^2} m$.
$\rho = 5514.7098661212 \frac{kg}{m^3} \cdot \frac{ \frac{\tilde{G}}{\tilde{c}^2} m}{1 kg} $
$\rho = 4.0950296770075 \cdot 10^{-24} \frac{1}{m^2}$

The Schwarzschild metric, and equations of structure associated with its calculation of the interior of a spherical body, involve integrating across the enclosed volume of the radius at the coordinate evaluated (to go from $\frac{1}{m^2}$ to $m$).
Earth radius $= 6.37101 \cdot 10^6 m$
Earth mass $ = \frac{4}{3} \pi r^3 \rho = 0.0044357853581474 m$, half a centimeter, which is half the Schwarzschild radius (how big the event horizon would be if the Earth collapsed into a black hole).

In the electromagetism stress-energy tensor $T_{tt} = \frac{1}{2} (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) = \frac{1}{8\pi} (\frac{E^2}{c^2} + B^2)$.
For .45 Gauss at the earth's surface (in absence of any electric field) we get $B = .45 \cdot \frac{1}{\tilde{c}} \sqrt{ \frac{1}{10} \tilde{G} } \cdot \frac{1}{m} = 3.8777491037247 \cdot 10^{-15} \frac{1}{m}$.
This gives $T_{tt} = \frac{1}{8 \pi} B^2 = 5.9830075735057 \cdot 10^{-31} \frac{1}{m^2}$.
My first thought on canceling Earth's gravity is to find an equivalent stress-energy $T_{tt}$ of magnetic field strength equal to the $T_{tt}$ of matter.
In other words, a magnetic field of $4.1 \cdot 10^{-24} \frac{1}{m^2}$, or of 1177.28406095 Gauss.
This really isn't that strong of a field, but this is only the equivalent stress-energy of a $m^3$ of the Earth's density.
That's a strange concept: a small neodymium magnet (2000 Gauss) has as much stress-energy as a $m^3$ of the average Earth density of $5.51 \frac{g}{cm^3}$ (iron is $7.87 \frac{g}{cm^3}$ and silicon is $2.33 \frac{g}{cm^3}$, so the Earth is somewhere between).

For entertainment, assuming constant magnetic field throughout a volume, how big to get the equivalent stress-energy of the Earth's matter?
$\frac{4}{3} \pi r^3 \cdot 5.9830075735057 \cdot 10^{-31} \frac{1}{m^2} = 0.0044357853581474 m$, solve for $r$ to get $r = 1.769956 \cdot 10^27 m$. Andromeda is closer than this distance.
How strong of a field is needed for a specific radius?
$\frac{4}{3} \pi r^3 \cdot \frac{1}{8 \pi} (B)^2 = 0.0044357853581474 m$
$ B = \sqrt{\frac{1}{r^3} 6 \cdot 0.0044357853581474} \sqrt{m} = 0.16314016105449 \sqrt{\frac{1}{r^3}} \sqrt{m} = 18931877878330.0 / \sqrt{r^3} Gauss = 1893187787.833 / \sqrt{r^3} Teslas $
Now I know this is wrong because magnetic fields aren't evenly distributed. I have to do a real integral over the field's function, which tends to have inverse square falloff and is nothing near constant.
But, at this field, how big would it have to be in order to lower the power to 8.5 Teslas, the CERN recent maximum level?
$r = 367437.24750191 m$, so about $\frac{1}{20}$th the radius of the earth.
How about, for a radius of $4000 m$, how big of a field? Note this is only considering the magnetic field energy, not the electric or matter.
$B = 7483.4818099598 Teslas$. We still have a ways to go there at least.

In order to compare the electromagnetic stress-energy tensor to the matter stress-energy tensor, I should first find a metric that generates a curvature tensor which equates to a nonzero metric -- specifically the matter stress-energy tensor -- which is something the Schwarzschild metric doesn't do. Once I have that I can then start thinking about how the two equate or cancel one another. I am trying to numerical solve for such a metric in the presence of evenly distributed matter, using the Earth as a test case, but my numerical finite difference approximations of gravity turn out to be anywhere up to a factor of 2 times more powerful than the analytical computations.

So that is what I'm wondering right now -- if there is, for certain predefined EM fields, any computed metric tensors that satisfy the EFE equations? I suppose I'd have to pick out a particular EM field formula/shape and work it backwards to find the metric, and then forward to find the connections and the subsequent gravitational force. That's what I'm trying to do numerically at the moment. It's a computationally costly problem, and not necessarily all that accurate.

Gravity


The goal is to find some configuration of EM fields that can cancel out the Earth's gravity, and hopefully allow the creation of a flying machine =P
At the moment I'm solving for steady states of metrics (i.e. $\partial_t g_{ab} = 0$), but I wonder if removing this constraint and solving for the time derivatives as well -- which influences the connections (since gravity is $\ddot{x}^i = {\Gamma^i}_{tt}$) -- could make things easier. What if a good solution was a constantly-rotating spacetime? One with an energy source off-axis? The energy source could produce the well in the lapse function. The rotation would give it a time derivative.



Geodesics trace from outer space through the Earth's surface down to the center of the Earth, courtesy of local time slowing down when measured by reference frame time (so that less coordinate time fills each tick of a local clock), in the center of the Earth, and local space length stretching out to filling a greater coordinate volume (so that more coordinate space is squished into the same region of local space).

Gravitational geodesics accelerate stationary objects traveling solely through time according to the equation $\ddot{x}^i = -{\Gamma^i}_{tt}$, which in absence of the metric shift vector $0 = g_{it} = \beta_i = \beta^j \gamma_{ij}$, and subsequently $0 = g^{it} = \beta^i / \alpha^2 $ (as the Schwarzschild metric is) reduces to $ \ddot{x}^i = \frac{1}{2} {g_{tt}}^i $. This means, of all the Schwarzschild metric, that acceleration (outward) is derived by the gradient of the timelike component of the metric. So if the $ g_{tt} $ function has a maxima at $ r = 0 $ then the acceleration will always be inwards towards the center of the planet.

But in the Schwarzschild metric, outside the planet, $R_{ab} = 0$, so the rate at which geodesics are converging onto the planet shouldn't be contracting together. Only inside the planet do we get $R_{ab} = G_{ab} + \frac{1}{2} R g_{ab} = ...$

as said before, in absence of g_it, as is in Schwarzschild metrics, gravity equals $ -{\Gamma^i}_{tt} = -g^{it} \Gamma_{ttt} - g^{ij} \Gamma_{jtt} $ ... but $g^{it} = 0$ because $g_{it} = 0$, so gravity is only $-g^{ij} \Gamma_{jtt}$, (the inverse spatial metric transformation of $-\Gamma_{itt}$), and $\Gamma_{itt} = g_{jt,t} - \frac{1}{2} g_{tt,j}$. But as we said before, $g_{jt} = 0$, so we just have gravity = inverse spatial metric times $\frac{1}{2} g_{tt,j}$ ... $g_{tt}$ is the timelike component of the metric tensor. It is how time coordinates slow when you are completely at rest (courtesy of $s = \int ds = \int \sqrt{ds^2} = \int_{\lambda=0}^{\lambda=s} \sqrt{g_{ab} \frac{dx^a}{d\lambda} \frac{dx^b}{d\lambda}} d\lambda$ ... if you are at rest then $\frac{dx^a}{dt} = [1,0,0,0]$ (only time component) ... and $s = \int ds = \int_{\lambda=0}^{\lambda=s} \sqrt{g_{tt}} d\lambda$

so in Schwarzschild metric, gravity is completley based on the gradient of the gravitational time dilation.

How do we mess with gravity? Look at $\Gamma_{itt}$. It equals $g_{jt,t} - \frac{1}{2} g_{tt,j}$. The gravitational time dilation gradient is $g_{tt,j}$. What's left? $g_{jt,t}$. The time derivative (i.e. it has to be changing wrt time) of $g_{jt}$. This was zero for Schwarzschild. It is the inner product of time and space frame basis vectors. It is not zero for a Kerr metric -- a rotating object. Probably not for a charged rotating object as well (Kerr-Newman metric)
Note that this remaining term is a time derivative, so it only gets value when the metric is changing. So maybe if we put an energy source on something that spins in a circle...

Kerr-Newman metric of charged rotating bodies, Kerr-Schild representation (pseudo-Cartesian):
$g_{uv} = \eta_{uv} + 2 H l_u l_v$
$g^{uv} = \eta^{uv} - 2 H l^u_* l^v_*$
$l^u_* = \eta^{uv} l_v$
$\eta_{uv} = diag(-1,1,1,1)$
$H = \frac{r M - Q^2 / 2}{r^2 + a^2 z^2 / r^2}$
$l_u = (1, \frac{rx+ay}{r^2+a^2}, \frac{ry-ax}{r^2+a^2}, \frac{z}{r})$
$r$ is defined by $\frac{x^2+y^2}{r^2+a^2} + \frac{z^2}{r^2} = 1$
$a = J / M$ is the angular momentum in terms of total mass
$Q$ = total enclosed charge

Geodesics: $\ddot{x}^a = -{\Gamma^a}_{bc} \dot{x}^b \dot{x}^c$
Gravity. Spatial geodesics components of resting objects: $\ddot{x}^i = -{\Gamma^i}_{tt} \dot{t}^a \dot{t}^b = -{\Gamma^i}_{tt}$
...in terms of 1st kind Christoffels: $\ddot{x}^i = -g^{it} \Gamma_{ttt} - g^{ij} \Gamma_{jtt}$
From there, just looking at the 1st kind Christoffel:
$\Gamma_{itt} = g_{it,t} - \frac{1}{2} g_{tt,i}$
$= \beta_{i,t} - \frac{1}{2} (-\alpha^2 + \beta^2)_{,i}$
$= \beta_{i,t} + \alpha \alpha_{,i} -\beta^j \beta_{j,i}$

So if you want to cancel the typical gravity contribution of $\alpha \alpha_{,i}$, you either need a negative counteracting $\beta_{i,t}$ (decrease in $\beta$ over time), or you need a large enough $\beta^j \beta_{j,i}$ (change in $\beta$ along $\beta$).

If you look at the Kerr metric - which is a metric for rotating spherical bodies - you see that the g_jt components correspond with rotation. The "catalogue of spacetimes" also shows a rotating Schwarzschild metric which had dt dphi terms added to it.

Coincidentally, when White developed off of the Alcubierre drive, he first criticized that a space ship of a few meters would take a mass of the size of Jupiter to produce the necessary warp in spacetime ... but then revised his statement that, if the object were rotating sufficiently fast enough, that the massive source object would.

Never mind White didn't say rotate, he said oscillate, in fact a rotating metric, where beta is rotating, will have grad beta of beta be zero ... I think ... because at least for a rotating v, grad v of v is zero, right? So it has to be oscillating ... ?

SO how to manipulate ${\beta^i}_{,t}$? Well, $\beta^i$ is the rotation / frame-dragging component of the metric.
So if that needs a time derivative, you would have to oscillate the rotation.

But for the record, White's "oscillations" are, in fact, science-fiction, as you can read about here. He is calling for oscillations into higher-dimensional planes of existence (distinct of the 3+1 spacetime dimensions).