This started as a diff from my 'gravitomagnetics +---' worksheet, which followed the Tajmar & De Matos paper.
It has additional ideas, but I don't want to muddle the last worksheet up too much.
Alright, I've made enough stupid sign errors that I'm just going to start from the top and make these two worksheets 1:1
Except that this worksheet is going to be signature -+++, just like all my others.
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:
$g_{ab} = \eta_{ab} + h_{ab}$ has units $[1]$
(Even though I'm using the same definition for either +--- or -+++ signatures, shouldn't one of the two be negative'd?
I suppose that doesn't matter if, instead, the $h_{ab}$ components of one signature are negative those of the other).
The coordinate basis is $\frac{\partial}{\partial x^\mu}
= (
\frac{\partial}{\partial x^0},
\frac{\partial}{\partial x^1},
\frac{\partial}{\partial x^2},
\frac{\partial}{\partial x^3}
) = (
\frac{1}{c} \frac{\partial}{\partial t},
\frac{\partial}{\partial x},
\frac{\partial}{\partial y},
\frac{\partial}{\partial z}
)$ has units $[\frac{1}{m}]$.
The dual basis is $dx^\mu
= (dx^0, dx^1, dx^2, dx^3)
= (c \cdot dt, dx, dy, dz)
$ has units $[m]$.
$g^{ab} \approx \eta^{ab}$ has units $[1]$,
So we will raise and lower with $\eta^{ab}$
$g_{ab,c} = h_{ab,c}$ has units $[\frac{1}{m}]$
$\Gamma_{abc} = \frac{1}{2} (h_{ab,c} + h_{ac,b} - h_{bc,a})$ has units $[\frac{1}{m}]$
${\Gamma^a}_{bc} = \frac{1}{2} ({h^a}_{b,c} + {h^a}_{c,b} - {h_{bc}}^{,a})$ has units $[\frac{1}{m}]$
${R^a}_{bcd} = 2 ({\Gamma^a}_{b[d,c]} + {\Gamma^a}_{e[c} {\Gamma^e}_{d]b})$ has units $[\frac{1}{m^2}]$
$= {h^a}_{b,[dc]} + {h^a}_{[d,c]b} - {{h_{b[d}}^{,a}}_{,c]}
+ \frac{1}{2} ({h^a}_{e,[c} + {h^a}_{[c|,e} - {h_{e[c}}^{,a})
({h^e}_{b,|d]} + {h^e}_{d],b} - {h_{d]b}}^{,e})$
$= {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}
)
$
$R_{ab} = {R^c}_{acb}$ has units $[\frac{1}{m^2}]$
$=
\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):
$(h_{ab} - \frac{1}{2} \eta_{ab} {h^c}_c)^{,b} = 0$ has units $[1]$
${h_{ab}}^{,b} = \frac{1}{2} {h_c}^{c,a}$
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:
$R = \eta^{ab} R_{ab} = -\frac{1}{2} {{{h^a}_a}^{,b}}_b$ has units $[\frac{1}{m^2}]$
$= -\frac{1}{2} \Box {h^a}_a$
$ = -\frac{1}{2} \Box h$
Einstein tensor:
$G_{ab} = R_{ab} - \frac{1}{2} g_{ab} R$ has units $[\frac{1}{m^2}]$
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:
$G_{ab} = 8 \pi \frac{G}{c^4} T_{ab}$ has units $[\frac{1}{m^2}]$
$T_{ab}$ has units $[\frac{kg}{m \cdot s^2}]$
so $8 \pi \frac{G}{c^4} T_{ab}$ has units $[\frac{1}{m^2}]$
$-\frac{1}{2} {\bar{h}_{ab,c}}^c = 8 \pi \frac{G}{c^4} T_{ab}$
${\bar{h}_{ab,c}}^c = -16 \pi \frac{G}{c^4} T_{ab}$
(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:
$\Delta \bar{h}_{ab} = -16 \pi \frac{G}{c^4} T_{ab}$
Let $\epsilon_g = \frac{1}{4 \pi G}$ has units $[\frac{kg \cdot s^2}{m^3}]$
so $G = \frac{1}{4\pi\epsilon_g}$
Let $\mu_g = \frac{4 \pi G}{c^2}$ has units $[\frac{m}{kg}]$
This gives the familiar formula: $c^2 = \frac{1}{\epsilon_g \mu_g}$ has units $[\frac{m^2}{s^2}]$
Poisson's equation:
$\Delta \phi = 4 \pi \rho$ has a solution $\phi = -\int_V \frac{\rho}{r} dV$
Combine with EFE to find:
$\Delta \bar{h}_{00} = -16 \pi \frac{G}{c^4} T_{00}$
$\Delta (\frac{c^2}{4} \bar{h}_{00}) = -4 \pi \frac{G}{c^2} T_{00}$
Let $\Phi^g = \frac{c^2}{4} \bar{h}_{00}$ has units $[\frac{m^2}{s^2}]$
$\Delta \Phi^g = -4 \pi \frac{G}{c^2} T_{00}$
$\Phi^g = \frac{G}{c^2} \int_V \frac{T_{00}}{r} dV$
$\Phi^g = \frac{1}{4\pi\epsilon_g} \int_V \frac{T_{00}}{c^2 r} dV$
Tada! Just like the electric field.
Next do the same trick with $T_{0i}$
$\Delta \bar{h}_{0i} = -16 \pi \frac{G}{c^4} T_{0i}$
$\Delta (-\frac{c}{4} \bar{h}_{0i}) = 4 \pi \frac{G}{c^3} T_{0i}$
Let $(A^g)_i = -\frac{c}{4} \bar{h}_{0i}$ be the gravitomagnetic potential vector, with units $[\frac{m}{s}]$
$\Delta (A^g)_i = 4 \pi \frac{G}{c^3} T_{0i}$
$(A^g)_i = -\frac{G}{c^3} \int_V \frac{T_{0i}}{r} dV$
$(A^g)_i = -\frac{\mu_g}{4 \pi} \int_V \frac{T_{0i}}{c r} dV$
Just like the magnetic field.
Notice that we aren't considering $T_{ij}$ 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.
$\bar{h}_{00} = \frac{4}{c^2} \Phi^g$
$\bar{h}_{0i} = \bar{h}_{i0} = -\frac{4}{c} (A^g)_i$
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 $h_{ab}$
$h_{ab} = \bar{h}_{ab} - \frac{1}{2} \eta_{ab} {\bar{h}^c}_c$
${\bar{h}^a}_a = -\bar{h}_{00} + \Sigma_i \bar{h}_{ii} = -\bar{h}_{00}$
Now solve for $h_{ba}$
$h_{00} = \bar{h}_{00} - \frac{1}{2} \eta_{00} {\bar{h}^a}_a = \frac{1}{2} \bar{h}_{00} = \frac{2}{c^2} \Phi^g$
$h_{0i} = \bar{h}_{0i} - \frac{1}{2} \eta_{0i} {\bar{h}^a}_a = \bar{h}_{0i} = -\frac{4}{c} (A^g)_i$
$h_{ij} = \bar{h}_{ij} - \frac{1}{2} \eta_{ij} {\bar{h}^a}_a = \frac{1}{2} \delta_{ij} \bar{h}_{00} = \frac{2}{c^2} \Phi^g \delta_{ij}$
We get
$h_{ab} = \left[\begin{matrix}
\frac{2}{c^2} \Phi^g & -\frac{4}{c} A_x & -\frac{4}{c} A_y & -\frac{4}{c} A_z \\
-\frac{4}{c} A_x & \frac{2}{c^2} \Phi^g & 0 & 0 \\
-\frac{4}{c} A_y & 0 & \frac{2}{c^2} \Phi^g & 0 \\
-\frac{4}{c} A_z & 0 & 0 & \frac{2}{c^2} \Phi^g
\end{matrix}\right]$
Add this to the flat-space metric $\eta_{ab}$ to find: ($g_{ab} = \eta_{ab} + h_{ab}$)
$g_{ab} = \left[\begin{matrix}
-1 + \frac{2}{c^2} \Phi^g & -\frac{4}{c} A_x & -\frac{4}{c} A_y & -\frac{4}{c} A_z \\
-\frac{4}{c} A_x & 1 + \frac{2}{c^2} \Phi^g & 0 & 0 \\
-\frac{4}{c} A_y & 0 & 1 + \frac{2}{c^2} \Phi^g & 0 \\
-\frac{4}{c} A_z & 0 & 0 & 1 + \frac{2}{c^2} \Phi^g
\end{matrix}\right]$
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.