Now, taking from my "relativistic electromagnetic metric in medium" worksheet...
from the section "special relativity electromagnetic stress-energy in medium":

$T^{(uv)} = \downarrow u(i) \overset{\rightarrow v(j)}{\left[\begin{matrix} \frac{1}{2} (\frac{1}{\epsilon} D^2 + \frac{1}{\mu} B^2) & \frac{1}{2} \frac{1}{c} (n^2 + 1) S_j \\ \frac{1}{2} \frac{1}{c} (n^2 + 1) S_i & \frac{1}{2} \eta^{ij} (\frac{1}{\epsilon} D^2 + \frac{1}{\mu} B^2) - \frac{1}{\epsilon} D^i D^j - \frac{1}{\mu} B^i B^j \end{matrix}\right]}$
$T^{uv}$ has units $\frac{kg}{m \cdot s^2}$

For electromagnetism, $T_{00}$ and $T_{0i}$ are given by:
$T_{00} = \frac{1}{2} ((\epsilon)^{-1} D^2 + (\mu)^{-1} B^2)$
$T_{0i} = -c^2 (\frac{v_p}{c})^2 \epsilon_{jkl} D_k B_l$
$T_{i0} = -c^2 \epsilon_{ikl} D_k B_l$
What do we do about the asymmetry? That brings on a whole other level of complexity.
In fact, if we are using the de Donder gauge then we are stuck with
$-\frac{1}{2} \Box \bar{h}_{ab} = G_{ab}$
aka
$-\frac{1}{2} \Box (h^{TR})_{ab} = (R^{TR})_{ab}$
With 'TR' meaning 'trace-reversal'.
This means that any asymmetry in the Ricci curvature will also have to show up in the metric (right? or will the $\Box$ get rid of it? I can't see how...).
It does make me think I do, in fact, need to use a symmetrizing mechanism. Options?
There's the simple $T'_{ab} := T_{(ab)}$.
There's the Belinfante and Rosenfeld symmetrization: $T'^{ab} = T^{ab} + \frac{1}{2} (s^{bac} + s^{bca} + s^{acb})_{,c}$
But this means also introducing our torsion term $s_{abc}$ which should be coupled to our particle spins... $2 T^{[ab]} = {s^{abc}}_{,c}$

if we simply symmetrize we get:
$T_{0i} = -\frac{1}{2} \frac{1}{c} (n^2 + 1) S_i$
For, $\vec{D} \times \vec{B} = \epsilon \vec{E} \times \mu \vec{H} = \frac{1}{(v_p)^2} \vec{S} = \frac{1}{c^2} n^2 \vec{S}$, for $\vec{S}$ the Poynting vector.