from my "relativistic electromagnetic metric in medium" worksheet:
$T_{ab}$ is the stress-energy tensor, in units of $\frac{kg}{m \cdot s^2}$.
$\mathcal{S}_i$ is the Poynting field vector.
$\mathcal{E}_i = \frac{1}{\epsilon} \mathcal{D}_i$ is the electric field vector.
$\mathcal{H}_i = \frac{1}{\mu} \mathcal{B}_i$ is the magnetic field vector.
$T_{[uv]} = \pmatrix{
0 &
\frac{1}{2} \frac{1}{c} (n^2 - 1) (
\alpha \mathcal{S}_j
+ \frac{1}{\alpha} \epsilon_{jkl} \mathcal{H}^k \beta^l \mathcal{E}_m \beta^m
)
\\
-\frac{1}{2} \frac{1}{c} (n^2 - 1) (
\alpha \mathcal{S}_i
+ \frac{1}{\alpha} \epsilon_{ikl} \mathcal{H}^k \beta^l \mathcal{E}_m \beta^m
)
&
\frac{1}{c} (n^2 - 1) \frac{1}{\alpha} \mathcal{E}_{[i} \epsilon_{j]kl} \mathcal{H}^k \beta^l
}$
$T_{[0i]} = \frac{1}{2} \frac{1}{c} (n^2 - 1) (
\alpha \mathcal{S}_i
+ \frac{1}{\alpha} \epsilon_{jkl} \mathcal{H}^k \beta^l \mathcal{E}_m \beta^m
)$
$T_{[ij]} =
\frac{1}{c} (n^2 - 1) \frac{1}{\alpha} \mathcal{E}_{[i} \epsilon_{j]kl} \mathcal{H}^k \beta^l
$
antisymmetric Einstein field equations:
$R_{ab}$ is the Ricci curvature tensor.
$R_{[ab]} = 8 \pi \frac{G}{c^4} T_{[ab]}$
first Bianchi identity / antisymmetry of Ricci tensor with torsion:
${T^a}_{bc}$ is the torsion tensor
$R_{[ab]} = \frac{3}{2} (
\nabla_{[u} {T^u}_{ba]} + {T^u}_{v[u} {T^v}_{ba]}
)$
$R_{[ab]} = \frac{1}{2} (
\partial_u {T^u}_{ba}
+ \partial_b {T^u}_{au}
+ \partial_a {T^u}_{ub}
+ {\Gamma^u}_{uv} {T^v}_{ba}
+ {\Gamma^u}_{bv} {T^v}_{au}
+ {\Gamma^u}_{av} {T^v}_{ub}
- {\Gamma^v}_{ub} {T^u}_{va}
- {\Gamma^v}_{ba} {T^u}_{vu}
- {\Gamma^v}_{au} {T^u}_{vb}
- {\Gamma^v}_{ua} {T^u}_{bv}
- {\Gamma^v}_{ab} {T^u}_{uv}
- {\Gamma^v}_{bu} {T^u}_{av}
+ {T^u}_{vu} {T^v}_{ba}
+ {T^u}_{vb} {T^v}_{au}
+ {T^u}_{va} {T^v}_{ub}
)$
$R_{[ab]} = \frac{1}{2} (
\partial_u {T^u}_{ba}
+ \partial_b {T^u}_{au}
+ \partial_a {T^u}_{ub}
+ {\Gamma^v}_{vu} {T^u}_{ba}
+ {\Gamma^v}_{ua} {T^u}_{vb}
+ {\Gamma^v}_{ub} {T^u}_{av}
- {c_{ab}}^v {T^u}_{uv}
)$
geodesic component due to torsion:
$(\dot{u}^i)_{\Theta} = u_b {T^b}_{dc} u^c g^{di}$
$(\dot{u}^i)_{\Theta} = u_b T^{bic} u_c$
...low-velocity approximation ($u_0 \approx 1, u_i \approx \frac{v_i}{c}$)
$(\dot{u}^i)_{\Theta} = T^{0i0}$
Plane wave representation:
$q = \hat{q} exp(i \tilde{q}_u x^u)$
${T^a}_{bc} = {\hat{T}^a}_{bc} exp(i \tilde{T}_u x^u)$
${T^a}_{bc,d} = i \tilde{T}_d {\hat{T}^a}_{bc} exp(i \tilde{T}_u x^u) = i {T^a}_{bc} \tilde{T}_d$
$R_{[ab]} = \frac{1}{2} (
(\tilde{T}_u + {\Gamma^v}_{vu}) {T^u}_{ba}
+ (\delta^v_u \tilde{T}_b + {\Gamma^v}_{ub}) {T^u}_{av}
+ (\delta^v_u \tilde{T}_a + {\Gamma^v}_{ua}) {T^u}_{vb}
- {c_{ab}}^v {T^u}_{uv}
)$
$R_{[ab]} = 8 \pi \frac{G}{c^4} T_{[ab]}$
$\frac{1}{2} (
(\tilde{T}_u + {\Gamma^v}_{vu}) {T^u}_{ba}
+ (\delta^v_u \tilde{T}_b + {\Gamma^v}_{ub}) {T^u}_{av}
+ (\delta^v_u \tilde{T}_a + {\Gamma^v}_{ua}) {T^u}_{vb}
- {c_{ab}}^v {T^u}_{uv}
) = 8 \pi \frac{G}{c^4} T_{[ab]}$
separate into space/time
$\frac{1}{2} (
(\tilde{T}_u + {\Gamma^v}_{vu}) {T^u}_{i0}
+ (\delta^v_u \tilde{T}_i + {\Gamma^v}_{ui}) {T^u}_{0v}
+ (\delta^v_u \tilde{T}_0 + {\Gamma^v}_{u0}) {T^u}_{vi}
- {c_{0i}}^v {T^u}_{uv}
) = 8 \pi \frac{G}{c^4} T_{[0i]}
= 4 \pi \frac{G}{c^5} (n^2 - 1) (
\alpha \mathcal{S}_i
+ \frac{1}{\alpha} \epsilon_{jkl} \mathcal{H}^k \beta^l \mathcal{E}_m \beta^m
)$
$\frac{1}{2} (
(\tilde{T}_u + {\Gamma^v}_{vu}) {T^u}_{ji}
+ (\delta^v_u \tilde{T}_j + {\Gamma^v}_{uj}) {T^u}_{iv}
+ (\delta^v_u \tilde{T}_i + {\Gamma^v}_{ui}) {T^u}_{vj}
- {c_{ij}}^v {T^u}_{uv}
) = 8 \pi \frac{G}{c^4} T_{[ij]}
= 8 \pi \frac{G}{c^5} (n^2 - 1) \frac{1}{\alpha} \mathcal{E}_{[i} \epsilon_{j]kl} \mathcal{H}^k \beta^l$
Hmm, maybe I can isolate the torsion before splitting into space/time?
$(
\delta^r_a \delta^q_b \tilde{T}_p
+ \delta^q_a \delta^r_p \tilde{T}_b
+ \delta^r_b \delta^q_p \tilde{T}_a
+ \delta^r_a \delta^q_b {\Gamma^v}_{vp}
+ \delta^q_a {\Gamma^r}_{pb}
+ \delta^r_b {\Gamma^q}_{pa}
- \delta^q_p {c_{ab}}^r
) {T^p}_{qr}
= 16 \pi \frac{G}{c^4} T_{[ab]}$
...and how do I invert this, to solve for ${T^a}_{bc}$?
One thing to note, the torsion geodesic component is only influenced by the symmetric components of the 1st and 3rd indexes of the torsion tensor... maybe I can isolate that?
That term does uniquely show up in the relation with stress energy as...
$\frac{1}{2} {c_{ab}}^v {T^u}_{vu} + \tilde{T}_{[b} {T^u}_{a]u} + ... = 8 \pi \frac{G}{c^4} T_{[ab]}$
${T^u}_{iu} \approx {T^0}_{i0} \approx 16 \pi \frac{G}{c^4} \frac{1}{\tilde{T}_{0}} T_{i0} = -8 \pi \frac{G}{c^5} \frac{1}{\tilde{T}_{0}} (n^2 - 1) (\alpha \mathcal{S}_i + \frac{1}{\alpha} \epsilon_{jkl} \mathcal{H}^k \beta^l \mathcal{E}_m \beta^m)$
...for $g_{ab} \approx \eta_{ab}$, so ${T^0}_{i0} \approx -T_{0i0}$
$T_{0i0} \approx 8 \pi \frac{G}{c^5} \frac{1}{\tilde{T}_{0}} (n^2 - 1) (\alpha \mathcal{S}_i + \frac{1}{\alpha} \epsilon_{jkl} \mathcal{H}^k \beta^l \mathcal{E}_m \beta^m)$
$(\dot{u}_i)_{\Theta} \approx 8 \pi \frac{G}{c^5} \frac{1}{\tilde{T}_{0}} (n^2 - 1) (\alpha \mathcal{S}_i + \frac{1}{\alpha} \epsilon_{jkl} \mathcal{H}^k \beta^l \mathcal{E}_m \beta^m)$
...for $g_{ab} \approx \eta_{ab}$ ...
$(\dot{u}_i)_{\Theta} \approx 8 \pi \frac{G}{c^5} \frac{1}{\tilde{T}_{0}} (n^2 - 1) \mathcal{S}_i$
Now for the influence of Noether current and spin tensor onto torsion.
Going off of the Einstein-Cartan-Sciama-Kibble Wikipedia page.
(I'm moving the last index to the first, so it matches with my 'Differential Geometry' worksheets...)
TODO more background and derivation of these relations.
${T^c}_{ab} + {T^d}_{da} \delta_b^c - {T^d}_{db} \delta_a^c = \frac{8 \pi G}{c^4} {S^c}_{ab}$
(But wait, isn't torsion tensor antisymmetric wrt the two spinning indexes, which in this paper I think is the 1st and 2nd? Doesn't that mean the 1st and 2nd are always zero when contracted with a symmetric tensor? So where did this relation come from anyways?)
So ${T^a}_{bc}$ is in units of $\frac{1}{m}$
And ${S^a}_{bc}$ is in units of $\frac{kg}{s^2}$
From spin tensor Wikipedia page:
$e_v(T^{uv}) = 0$
four-momentum: $P^u = \int T^{u0}(x,t) d^4 x$
$M_y^{abu}(x) = M_0^{abu}(x) + y^a T^{bu}(x) - y^b T^{au}(x)$
$M_y^{abu}(x) = (y^a - x^a) T^{bu}(x) - (y^b - x^b) T^{au}(x)$
angular momentum tensor $M^{uv} = \int M_0^{uv0} (x,t) d^4 x$
$S^{cab}(x) = M_x^{abu}(x) = M_0^{abu}(x) + x^a T^{bu} (x) - x^b T^{au}(x)$
and $\partial_u M_0^{abu}(x) = 0$
means $e_u (S^{uab}) = T^{ba} - T^{ab}$
Misner, Thorne, Wheeler "Gravitation". (Notice MTW operates on the assumption that $T^{[ab]} = 0$.)
$J^{abc}(\vec{x}, \vec{y}) = (x^a - y^a) T^{bc}(\vec{x}) - (x^b - y^b) T^{ac}(\vec{x})$ eqn. 5.46, where $\vec{x}$ is the 'field point' parameter of $J^{abc}$ and $\vec{y}$ is the 'origin' of the angular momentum field.
$\frac{\partial}{\partial x^c}( J^{abc} ) = \delta^a_c T^{bc} + (x^a - y^a) \frac{\partial}{\partial x^c}(T^{bc}) - \delta^b_c T^{ac} - (x^b - y^b) \frac{\partial}{\partial x^c}(T^{ac})$ eqn. 5.46
(how to distinguish between $\frac{\partial}{\partial x^c}$ and $e_c(\cdot)$?)
$\frac{\partial}{\partial x^c}( J^{abc} ) = T^{ba} - T^{ab} + (x^a - y^a) \frac{\partial}{\partial x^c}(T^{bc}) - (x^b - y^b) \frac{\partial}{\partial x^c}(T^{ac})$ eqn. 5.46
While MTW has $T^{[ab]} = 0$ and torsion metrics don't, both MTW and Wiki say $\frac{\partial}{\partial x^v}( T^{uv} ) = 0$, though Wiki says $\frac{\partial}{\partial x^u}( T^{uv} )$ is not necessarily zero.
$\frac{\partial}{\partial x^c}( J^{abc} ) = T^{ba} - T^{ab}$ eqn 5.47
$e_c( J^{abc} ) = T^{ba} - T^{ab}$
Wiki says $P^u = \int T^{u0}(\vec{x},t) d^4 x = \int T^{uv} \Sigma_{vabc} dx^a dx^b dx^c$ ... ? Did I do that right?
So Wikipedia's M is the same as MTW's J, and S is just the M tensor at $x=0$
But where does the relation between torsion T and spin S come from?
${T^c}_{ab} + {T^d}_{da} \delta_b^c - {T^d}_{db} \delta_a^c = \frac{8 \pi G}{c^4} {S^c}_{ab}$
${T^a}_{ab} + {T^d}_{db} - n {T^d}_{db} = \frac{8 \pi G}{c^4} {S^a}_{ab}$
${T^a}_{ab} = \frac{8 \pi G}{c^4 (2 - n)} {S^a}_{ab}$
But wait, aren't these all antisymmetric on the 1st and 2nd indexes?
Then ${T^a}_{ab} = {S^a}_{ab} = 0$ anyways, and we just get
${T^c}_{ab} = \frac{8 \pi G}{c^4} {S^c}_{ab}$
$[S^{abc}] = [M^{abc}] = [J^{abc}] = [T^{ab}] \cdot m = \frac{kg}{s^2}$
$[T^{abc}] = \frac{1}{m}$