This is following Tajmar & Matos' paper, "Coupling of Electromagnetism and Gravitation in the Weak Field Approximation.
This paper uses the +,-,-,- signature.
This is starting out the same as "metric of flat space and EM potential" but with linearization.
Constants:
$c = 299792458 \frac{m}{s} = $ speed of light
$G = 6.67384 \cdot 10^{-11} \frac{m^3}{kg \cdot s^2} = $ gravitational constant
$[\frac{G}{c^4}] = [\frac{s^2}{kg \cdot m}]$ units
$\eta_{ab} = diag(1,-1,-1,-1)$
This much is distinct of the signature -- because it does not separate space+time:
Near field approximation:
$g_{ab} = \eta_{ab} + h_{ab}$
The coordinate basis is $\frac{\partial}{\partial x^\mu} = [
\frac{\partial}{c \partial t},
\frac{\partial}{\partial x},
\frac{\partial}{\partial y},
\frac{\partial}{\partial z}]$
$g^{ab} \approx \eta^{ab}$
Raising and lowering is with $\eta^{ab}$
$g_{ab,c} = h_{ab,c}$
$\Gamma_{abc} = \frac{1}{2} (h_{ab,c} + h_{ac,b} - h_{bc,a})$
${\Gamma^a}_{bc} = \frac{1}{2} ({h^a}_{b,c} + {h^a}_{c,b} - {h_{bc}}^{,a})$
${R^a}_{bcd} = 2 ({\Gamma^a}_{b[d,c]} + {\Gamma^a}_{e[c} {\Gamma^e}_{d]b})$
$= {h^a}_{b,[dc]} + {h^a}_{[d,c]b} - {{h_{b[d}}^{,a}}_{,c]}
+ \frac{1}{2} ({h^a}_{e,[c} + {h^a}_{[c|,e} - {h_{e[c}}^{,a})
({h^e}_{b,|d]} + {h^e}_{d],b} - {h_{d]b}}^{,e})$
$= {h^a}_{[d,c]b} - {h_{b[d,c]}}^{,a}
+ \frac{1}{2} (
+ {h^a}_{e,[c} {h^e}_{b,|d]}
+ {h^a}_{e,[c} {h^e}_{d],b}
- {h^a}_{e,[c} {h_{d]b}}^{,e}
+ {h^a}_{[c|,e} {h^e}_{b,|d]}
+ {h^a}_{[c|,e} {h^e}_{d],b}
- {h^a}_{[c|,e} {h_{d]b}}^{,e}
- {h_{e[c}}^{,a} {h^e}_{b,|d]}
- {h_{e[c}}^{,a} {h^e}_{d],b}
+ {h_{e[c}}^{,a} {h_{d]b}}^{,e}
)
$
$R_{ab} = {R^c}_{acb}$
$=
\frac{1}{2} (
{{h_{bc}}^{,c}}_a
- {h_{ab,c}}^{,c}
- {h^c}_{c,ab}
+ {{h_{ac}}^{,c}}_b
)
+ \frac{1}{4} (
{h^c}_{c,e} {h^e}_{a,b}
+ {h^c}_{c,e} {h^e}_{b,a}
- {h^c}_{c,e} {h_{ab}}^{,e}
- {h^c}_{e,b} {h^e}_{a,c}
- {h^c}_{e,b} {h^e}_{c,a}
+ {h^c}_{e,b} {h_{ac}}^{,e}
- 2 {{h_b}^c}_{,e} {{h_a}^e}_{,c}
+ 2 {{h_b}^c}_{,e} {h_{ac}}^{,e}
)
$
Choosing the gauge condition (de Donder gauge):
$(h_{ab} - \frac{1}{2} \eta_{ab} {h^c}_c)^{,b} = 0$
${h_{ab}}^{,b} = \frac{1}{2} {h_c}^{c,a}$
Let $h = {h^a}_a$
$\nabla \cdot \textbf{h} = \frac{1}{2} \nabla h$
$R_{ab} =
\frac{1}{2} (
\frac{1}{2} {{h_c}^c}_{,ba}
- {h_{ab,c}}^{,c}
- {h^c}_{c,ab}
\frac{1}{2} {{h_c}^c}_{,ba}
)
+ \frac{1}{4} (
{h^c}_{c,e} {h^e}_{a,b}
+ {h^c}_{c,e} {h^e}_{b,a}
- {h^c}_{c,e} {h_{ab}}^{,e}
- {h^c}_{e,b} {h^e}_{a,c}
- {h^c}_{e,b} {h^e}_{c,a}
+ {h^c}_{e,b} {h_{ac}}^{,e}
- 2 {{h_b}^c}_{,e} {{h_a}^e}_{,c}
+ 2 {{h_b}^c}_{,e} {h_{ac}}^{,e}
)
$
$R_{ab} =
- \frac{1}{2} {h_{ab,c}}^{,c}
+ \frac{1}{4} (
{h^c}_{c,e} {h^e}_{a,b}
+ {h^c}_{c,e} {h^e}_{b,a}
- {h^c}_{c,e} {h_{ab}}^{,e}
- {h^c}_{e,b} {h^e}_{a,c}
- {h^c}_{e,b} {h^e}_{c,a}
+ {h^c}_{e,b} {h_{ac}}^{,e}
- 2 {{h_b}^c}_{,e} {{h_a}^e}_{,c}
+ 2 {{h_b}^c}_{,e} {h_{ac}}^{,e}
)
$
All the squared terms should go away because $h_{ab} << 1$, therefore $(h_{ab})^2 \approx 0$
$R_{ab} = -\frac{1}{2} {h_{ab,c}}^c$
$= -\frac{1}{2} \Box h_{ab}$
$= -\frac{1}{2} \Box \textbf{h}$
Next Gaussian curvature:
$R = \eta^{ab} R_{ab} = -\frac{1}{2} {{{h^a}_a}^{,b}}_b$
$= -\frac{1}{2} \Box {h^a}_a$
$ = -\frac{1}{2} \Box h$
Einstein tensor:
$G_{ab} = R_{ab} - \frac{1}{2} g_{ab} R$
Linerized approximation:
$G_{ab} = R_{ab} - \frac{1}{2} \eta_{ab} R$
$= -\frac{1}{2} {h_{ab,c}}^c + \frac{1}{4} \eta_{ab} {{{h^c}_c}^{,d}}_d$
Define gravitational potential:
$\bar{h}_{ab} = h_{ab} - \frac{1}{2} \eta_{ab} {h^c}_c$
${\bar{h}_{ab}}^{,b} = (h_{ab} - \frac{1}{2} \eta_{ab} {h^c}_c)^{,b} = 0$ by the de Donder gauge.
or $\nabla \cdot \bar{\textbf{h}} = 0$
${\bar{h}_{ab,c}}^c = {h_{ab,c}}^c - \frac{1}{2} \eta_{ab} {{{h^c}_c}^{,d}}_d$
(combined with units of $T_{ab}$ below, this means $\bar{h}_{ab,cd}$ has units $\frac{1}{m^2}$)
Substitute into Einstein tensor:
$G_{ab} = -\frac{1}{2} {h_{ab,c}}^c + \frac{1}{4} \eta_{ab} {{{h^c}_c}^{,d}}_d$
$= -\frac{1}{2} {\bar{h}_{ab,c}}^c$
$= -\frac{1}{2} \Box \bar{h}_{ab} = -\frac{1}{2} \Box \bar{\textbf{h}}$
....which makes perfect sense, because the Einstein tensor is the trace-reversal of the Ricci tensor,
and $\bar{\textbf{h}}$ is the trace-reversal of $\textbf{h}$
Einstein Field Equations:
$G_{ab} = 8 \pi \frac{G}{c^4} T_{ab}$
$-\frac{1}{2} {\bar{h}_{ab,c}}^c = 8 \pi \frac{G}{c^4} T_{ab}$
${\bar{h}_{ab,c}}^c = -16 \pi \frac{G}{c^4} T_{ab}$
(Ok this is the first part we separate space+time, so now metric signature matters)
Separating space and time:
$\frac{1}{c^2} \frac{\partial^2}{\partial t^2} \bar{h}_{ab} - \Delta \bar{h}_{ab} = -16 \pi \frac{G}{c^4} T_{ab}$
Assuming a steady state:
$\Delta \bar{h}_{ab} = 16 \pi \frac{G}{c^4} T_{ab}$
Stress energy of matter in curved space:
$T_{ab} = \rho u_a u_b + P (g_{ab} - u_a u_b) $
$T_{ab} = (\rho - P) u_a u_b + P g_{ab}$
(Looks like $u^a = u_a = (1,0,0,0)$ ... but does velocity sign matter?)
Low-velocity approximation: $u_t \approx 1, u_i \approx v_i$
$T_{00} = (\rho - P) (u_0)^2 + P g_{00} \approx \rho$
$T_{0i} = \rho u_0 u_i + P (g_{0i} - u_0 u_i) \approx -\rho v_i$
$T_{ij} = \rho v_i v_j + P \delta_{ij}$
$T_{ab} = \left[\matrix{
c^2 \rho_m & -\rho_m c v_i \\
-\rho_m c v_i & \rho_m v_i v_j + P \delta_{ij}
}\right]$
Combine with EFE to find:
$\Delta \bar{h}_{tt} = 16 \pi \frac{G}{c^4} T_{tt}$
$\Delta (\frac{c^2}{4} \bar{h})_{tt} = 4 \pi G \rho_m$
Let $\Phi^g = \frac{c^2}{4} \bar{h}_{tt}$
$\Delta \Phi^g = 4 \pi G \rho_m$
Poisson's equation:
$\Delta \phi = 4 \pi \rho$ has a solution $\phi = -\int_V \frac{\rho}{r} dV$
$\Phi^g = \frac{c^2}{4} \bar{h}_{tt} = -G \int_V \frac{\rho_m}{r} dV$
Let $G = \frac{1}{4\pi\epsilon_g}$, so $\epsilon_g = \frac{1}{4 \pi G}$
$\Phi^g = -\frac{1}{4\pi\epsilon_g} \int_V \frac{\rho_m}{r} dV$
Tada! Just like the electric field.
Next do the same trick with $T_{ti} = -\rho_m c^2 \frac{v_i}{c}$
Let $p_i = \rho_m v_i$ be the fluid momentum variable.
$T_{ti} = -c^2 \frac{p_i}{c}$
$\Delta \bar{h}_{ti} = 16 \pi \frac{G}{c^4} T_{ti}$
$\Delta (-\frac{c}{4}\bar{h}_{ti}) = 4 \pi \frac{G}{c^2} p_i$
Let $(A^g)_i = -\frac{c}{4} \bar{h}_{ti}$ be the gravitomagnetic potential vector.
Use Poisson's equation:
$(A^g)_i = -\frac{c}{4} \bar{h}_{ti} = -\frac{G}{c^2} \int_V \frac{p_i}{r} dV$
Let $\mu_g = \frac{4 \pi G}{c^2}$.
$(A^g)_i = -\frac{\mu_g}{4 \pi} \int_V \frac{p_i}{r} dV$
Just like the magnetic field.
This gives the familiar formula: $c^2 = \frac{1}{\epsilon_g \mu_g}$
Notice that $T_{ij}$ doesn't play into account at all.
If you stop here, take what we know of $\bar{h}_{ab}$, and substitute it back into $h_{ab}$, you find something interesting.
$\bar{h}_{tt} = \frac{4}{c^2} \Phi^g$
$\bar{h}_{ti} = \bar{h}_{it} = -\frac{4}{c} (A^g)_i$
Let's assume the rest is zero for now: $\bar{h}_{ij} = 0$
Now let's calculate $h_{ab}$ from $\bar{h}_{ab}$
$\bar{h}$ definition:
$\bar{h}_{ab} = h_{ab} - \frac{1}{2} \eta_{ab} {h^c}_c$
contract with $\eta^{ab}$:
${\bar{h}^a}_a = {h^a}_a - \frac{1}{2} \eta^{ab} \eta_{ab} {h^c}_c$
${\bar{h}^a}_a = {h^a}_a - \frac{4}{2} {h^a}_a$
${\bar{h}^a}_a = -{h^a}_a$
substitute back into definition:
$\bar{h}_{ab} = h_{ab} + \frac{1}{2} \eta_{ab} {\bar{h}^c}_c$
solve for $h_{ab}$
$h_{ab} = \bar{h}_{ab} - \frac{1}{2} \eta_{ab} {\bar{h}^c}_c$
${\bar{h}^a}_a = \bar{h}_{tt} - \Sigma_i \bar{h}_{ii} = \bar{h}_{tt}$
Now solve for $h_{ba}$
$h_{tt} = \bar{h}_{tt} - \frac{1}{2} \eta_{tt} {\bar{h}^a}_a = \frac{1}{2} \bar{h}_{tt}$
$h_{ti} = \bar{h}_{ti} - \frac{1}{2} \eta_{ti} {\bar{h}^a}_a = \bar{h}_{ti}$
$h_{ij} = \bar{h}_{ij} - \frac{1}{2} \eta_{ij} {\bar{h}^a}_a = \frac{1}{2} \delta_{ij} \bar{h}_{tt}$
We get
$h_{ab} = \left[\matrix{
\frac{2}{c^2} \Phi^g & -\frac{4}{c} A_x & -\frac{4}{c} A_y & -\frac{4}{c} A_z \\
-\frac{4}{c} A_x & \frac{2}{c^2} \Phi^g & 0 & 0 \\
-\frac{4}{c} A_y & 0 & \frac{2}{c^2} \Phi^g & 0 \\
-\frac{4}{c} A_z & 0 & 0 & \frac{2}{c^2} \Phi^g
}\right]$
Add this to the flat-space metric $\eta_{ab}$ to find: ($g_{ab} = \eta_{ab} + h_{ab}$)
$g_{ab} = \left[\matrix{
1 + \frac{2}{c^2} \Phi^g & -\frac{4}{c} A_x & -\frac{4}{c} A_y & -\frac{4}{c} A_z \\
-\frac{4}{c} A_x & -1 + \frac{2}{c^2} \Phi^g & 0 & 0 \\
-\frac{4}{c} A_y & 0 & -1 + \frac{2}{c^2} \Phi^g & 0 \\
-\frac{4}{c} A_z & 0 & 0 & -1 + \frac{2}{c^2} \Phi^g
}\right]$
...and this looks surprisingly similar to a combination of the metric representing the Newtonian limit of relativity ($g_{ab} = \eta_{ab} + 2 \delta_{ab} \Phi$),
as well as the metric (similar to Kaluza-Klein) whose geodesic looks similar to the Lorentz force ($g_{ab} = \eta_{ab} + 2 \delta_{t(a} A_{b)})$
Coincidentally those are the forces which we are recreating using this model.
Faraday tensor:
$F_{ab} = \downarrow a \overset{\rightarrow b}{\left[\matrix{
0 & -\frac{1}{c} E_j \\
\frac{1}{c} E_i & {\epsilon_{ij}}^k B_k
}\right]}$
$F^{ab} = \left[\matrix{
0 & \frac{1}{c} E^j \\
-\frac{1}{c} E^i & \epsilon^{ijk} B_k
}\right]$
Funny how raising/lowering twice on antisymmetric tensors looks the same whether you're using +--- or -+++
Maxwell's laws look like
${F^{ab}}_{,b} = \mu_0 J^a$
for $J^a = [ c \rho, J^i ]$ gives
$\frac{1}{c} {E^i}_{,i} = c \mu_0 \rho$ becomes ${E^i}_{,i} = \frac{\rho}{\epsilon_0}$
$-\frac{1}{c^2} {E^i}_{,t} + \epsilon^{ijk} B_k = \mu_0 J^i$
And ${\star F^{ab}}_{,b} = 0$
$\star F_{ab} = \frac{1}{2} \epsilon_{abuv} F^{uv}$
for $\epsilon_{abcd} = [0,1,2,3]$ (notice Rovelli, Vidotto use $\epsilon_{abcd} = -[0,1,2,3]$?)
$\star F_{ab} = \downarrow a \overset{\rightarrow b}{\left[\matrix{
0 & -B_j \\
B_i & -\frac{1}{c} {\epsilon_{ijk}} E^k
}\right]}$
$-\frac{1}{c} {B^i}_{,i} = 0$
$\frac{1}{c} {B^i}_{,t} + \frac{1}{c} \epsilon^{ijk} E_k = 0$
Now define the gravitoelectric and gravitomagnetic fields:
(Should this be upper or lower? Tajmar & Matos' paper uses vector notation, which implies upper, which implies a sign change using this +--- metric.)
$E^g_i = -\Phi^g_{,i} - {A^g}_{i,t}$
$B^g_i = {\epsilon_i}^{jk} {A^g}_{k,j}$
...and now we get Maxwell's laws for gravitation...
${(E^g)^i}_{,i} = -\frac{\rho_m}{\epsilon_g}$
${(B^g)^i}_{,i} = 0$ (by symmetric of 2nd derivatives and antisymmetry of Levi-Civita tensor)
${\epsilon_i}^{jk} (E^g)_{k,j} = -(B^g)_{i,t}$
${\epsilon_i}^{jk} (B^g)_{k,j} = -\mu_g \rho_m v_i + \frac{1}{c^2} (E^g)_{,t}$
Now comparing the ratios between matter and charge...
$\rho_m = \frac{m}{e} \rho$
$m = $ particle mass, $e = $ particle charge.
2014 CODATA electron e/m is $1.75882002411 \cdot 10^{11} \frac{C}{kg}$
$\Phi^g = -\frac{m}{e} \frac{\epsilon_0}{\epsilon_g} \Phi$
$(A^g)_i = -\frac{m}{e} \frac{\mu_g}{\mu_0} A_i$
Next note $\epsilon_0 \mu_0 = c^{-2}$ and $\epsilon_g \mu_g = c^{-2}$
$\rightarrow \frac{\epsilon_0 \mu_0}{\epsilon_g \mu_g} = 1$
$\rightarrow \frac{\epsilon_0}{\epsilon_g} = \frac{\mu_g}{\mu_0}$
Let $\kappa = -\frac{m}{e} \frac{\mu_g}{\mu_0} = -\frac{m}{e} \frac{\epsilon_0}{\epsilon_g} = -7.41\cdot 10^{-21} \frac{m}{e}$
Substitute to find ...
$\phi^g = \kappa \cdot \phi$
$(A^g)_i = \kappa \cdot A_i$
$(E^g)_i = \kappa \cdot E_i$
$(B^g)_i = \kappa \cdot B_i$
Now you have gravitation defined in terms of electromagnetism
Both sources are created by the particle -- the electromagnetic and the "gravitic-gravitomagnetic" fields.
Coupling of particles in a vacuum:
electron: $\kappa = 4.22 \cdot 10^{-32} \frac{1}{T s}$
proton: $\kappa = -7.6 \cdot 10^{-29} \frac{1}{T s}$
lead ions: $\kappa = -1.6 \cdot 10^{-26} \frac{1}{T s}$
Rewriting gravitation in terms of its electromagnetic-analogous components:
Gravitation $\rightarrow$ Electromagnetism:
${\epsilon_i}^{jk} E_{k,j} = -\frac{1}{\kappa} (B^g)_{i,t}$
${\epsilon_i}^{jk} B_{k,j} = \frac{e}{m} \mu_0 \rho_m v_i + \frac{1}{\kappa} \frac{1}{c^2} (E^g)_{i,t}$
Electromagnetism $\rightarrow$ Gravitation :
${\epsilon_i}^{jk} (E^g)_{k,j} = -\kappa B_{i,t}$
${\epsilon_i}^{jk} (B^g)_{k,j} = -\frac{m}{e} \mu_g \rho v_i - \kappa \frac{1}{c^2} E_{i,t}$
In hyperbolic conservation law form:
Electromagnetism $\rightarrow$ Gravitation :
$B_{i,t} + \frac{1}{\kappa} {\epsilon_i}^{jk} (E^g)_{k,j} = 0$
$\epsilon_0 E_{i,t} + \frac{1}{\kappa \mu_0} {\epsilon_i}^{jk} (B^g)_{k,j} = \rho v_i$
Gravitation $\rightarrow$ Electromagnetism:
$(B^g)_{i,t} + \kappa \frac{1}{\epsilon_0} {\epsilon_i}^{jk} \epsilon_0 E_{k,j} = 0$
$(E^g)_{i,t} - \kappa c^2 {\epsilon_i}^{jk} B_{k,j} = \frac{1}{\epsilon_g} \rho_m v_i$
...concludes magnetic fields of 10 T and electron currents at frequency 1 GHz only create a gravitational field of $(E^g) = 2.65 \cdot 10^{-21} \frac{m}{s^2}$
Solutions of increasing $(E^g)$...
1) increase $\frac{m}{e}$...
Electron is given above
Proton is 1836 times heavier than an electron
Lead is 207.2 times heavier than a proton
Mercury is also 200.592 times heavier than a proton ... and has a high magnetic susceptibility...
2) gravitomagnetic analogue to magnetism ($B^g$ vs $B$)
This could create a gravitomagnetic relative permeability...