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

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

For electromagnetism, $T_{00}$ and $T_{0i}$ are given by:
$T_{00}=\frac{1}{2}((ϵ{)}^{-1}{D}^2+(\mu {)}^{-1}{B}^2)$
$T_{0i}=-c^2(\frac{v_p}{c}{)}^2{ϵ}_{jkl}{D}_k{B}_l$
$T_{i0}=-c^2{ϵ}_{ikl}{D}_k{B}_l$
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
$-\frac{1}{2}◻{\overline{h}}_{ab}=G_{ab}$
aka
$-\frac{1}{2}◻(h^{TR}{)}_{ab}=(R^{TR}{)}_{ab}$
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 $◻$ 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 $T_{ab}^{\prime }:=T_{(ab)}$.
There's the Belinfante and Rosenfeld symmetrization: $T^{\prime ab}=T^{ab}+\frac{1}{2}(s^{bac}+s^{bca}+s^{acb}{)}_{,c}$
But this means also introducing our torsion term $s_{abc}$ which should be coupled to our particle spins... $2T^{[ab]}={s^{abc}}_{,c}$

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