Same as "metric of curved space and EM potential" except that one was targetted with this crazy idea of reproducing the EM stress-energy tensor in curved space.
This is just about isolating higher order terms of a gravitomagnetic solution (i.e. neglecting all the DOF of gravitation except those that look analogous to EM).
I might pull in some influence from the Newtonian limit as well.
ADM decomposition:
$g_{ab} = \overset{a(i)\downarrow b(j)\rightarrow}{
\left[ \matrix{
-\alpha^2 + \beta_k \beta^k &
\beta_j \\
\beta_i &
\gamma_{ij}
} \right]}$
$g^{ab} = \overset{a(i)\downarrow b(j)\rightarrow}{
\left[ \matrix{
-1/\alpha^2 &
\beta^j /\alpha^2 \\
\beta^i /\alpha^2 &
\gamma^{ij} - \beta^i \beta^j /\alpha^2
} \right]}$
adding in the EM potential...
Let $\alpha^2 = \tilde{\alpha}^2 + 2 \Phi_\alpha + \gamma^{kl} A_k A_l$
Let $\beta_i = A_i$
So $\beta^i = \gamma^{ij} A_j$
Let $\gamma_{ij} = \tilde{\gamma}_{ij} - 2 \Phi_\gamma$
$g_{ab} = a(i)\downarrow \overset{b(j)\rightarrow}{
\left[ \matrix{
-\tilde{\alpha}^2 - 2 \Phi_\alpha & A_j \\
A_i & \tilde{\gamma}_{ij} - 2 \Phi_\gamma
} \right]}$
Now the metric looks like the Newtonian approximation of relativity (so long as $\Phi_\alpha = \Phi_\gamma$)
combined with the gravitomagnetic 4-potential $A_\mu$
Now the ADM connections:
$-{\Gamma^i}_{tt} =
\frac{1}{2} \alpha^{-2} \beta^i \beta^j \beta^k {\gamma_{jk}}_{,t}
- \frac{1}{2} \alpha^{-2} \beta^i \beta^j \beta^k \beta^l \gamma_{kl,j}
- \alpha^{-2} \beta^i \beta^j {\beta^k}_{,j} \beta^l \gamma_{kl}
+ \alpha^{-1} \alpha_{,j} \beta^i \beta^j
+ \alpha^{-1} \alpha_{,t} \beta^i
- \beta^k \gamma^{ij} \gamma_{jk,t}
+ \frac{1}{2} \beta^k \beta^l \gamma^{ij} \gamma_{kl,j}
+ {\beta^k}_{,j} \beta^l \gamma^{ij} \gamma_{kl}
- {\beta^i}_{,t}
- \alpha \alpha_{,j} \gamma^{ij}
$
$-{\Gamma^i}_{mt} = \frac{1}{2} (
\alpha^{-2} \beta^i \beta^j \gamma_{jm,t}
- \alpha^{-2} \beta^i \beta^j \beta^k \gamma_{km,j}
- \alpha^{-2} \beta^i \beta^j {\beta^k}_{,j} \gamma_{mk}
- \alpha^{-2} \beta^i \beta^j {\beta^k}_{,m} \gamma_{jk}
+ 2 \alpha^{-1} \alpha_{,m} \beta^i
- \gamma^{ij} \gamma_{jm,t}
- {\beta^i}_{,m}
+ {\beta^k}_{,j} \gamma^{ij} \gamma_{mk}
+ \beta^k \gamma^{ij} \gamma_{mk,j}
- \beta^k \gamma^{ij} \gamma_{jk,m}
)$
$-{\Gamma^i}_{mn} = \frac{1}{2} (
\alpha^{-2} \beta^i \gamma_{mn,t}
- \alpha^{-2} \beta^i \beta^j \gamma_{mn,j}
- \alpha^{-2} \beta^i {\beta^k}_{,n} \gamma_{mk}
- \alpha^{-2} \beta^i {\beta^k}_{,m} \gamma_{nk}
- \gamma^{ij} \gamma_{jm,n}
- \gamma^{ij} \gamma_{jn,m}
+ \gamma^{ij} \gamma_{mn,j}
)$
...and the ADM Riemann curvature tensor (will look ugly):
...and the ADM Ricci curvature tensor (will look ugly):
...and the ADM Gaussian curvature (will look ugly):
...and the ADM Gravitational curvature (will look ugly):
...what significance is the Gauss-Codazzi-Ricci relationship?
That will give the Riemann of the lower manifold in terms of the Riemann of the higher.
Contract to find relations between Riccis.
But can I say that the spatial components of the covariant Ricci of the lower equals the spatial components of the covariant of the Ricci of the higher?
Is $R_{ij}$ of 4D spacetime equal to $R_{ij}$ of 3D spacetime?
We have $R^\perp_{abcd} = {\gamma_a}^p {\gamma_b}^q {\gamma_c}^r {\gamma_d}^s R_{pqrs} - K_{ac} K_{bd} + K_{ad} K_{bd}$
What about $R^\perp_{ab} = \gamma^{cd} R^\perp_{cadb}$
Notice the $T_{ij}$ of EM stress-energy is $\gamma_{ij} (E^2 + B^2) - 2 (E_i E_j + B_i B_j)$
And notice using $P = \left[\vec{E} | \vec{B} | \vec{0} \right]$ gives us $tr (P^T P) = E^2 + B^2$ and $P P^T = E_i E_j + B_j B_j$
P's don't match K's because the $tr(P^T P)$ is off by that 2, and by the issue of K being symmetric, whereas P's are not.
Maybe K's can have imaginary, antisymmetric components?
$K_{ab} = -\perp \nabla_{(a} n_{b)}$ is extrinsic curvature
$F_{ab} = \nabla_{[a} A_{b]}$ is electromagnetism
And with gravitomagnetics the $A_a = g_{at}$ lines up close to $n_a$ ...