Are surface stress-energy tensors in a space-time junction conserved? Consider two space-times with respective metric $g^{\pm}_{\mu\nu}$ separated by a hypersurface junction. Assuming that the full metric given by
$$g_{\mu\nu} = \Theta(\ell) g^{+}_{\mu\nu}+\Theta(-\ell) g^{-}_{\mu\nu}\tag{3.48}$$
is continuous at the junction i.e. at $\ell = 0$, $$[g_{\mu\nu}] = g^{+}_{\mu\nu}-g^{-}_{\mu\nu} = 0.$$ Using the Einstein equations, one can compute the stress-energy tensor due to this metric which is given by (Eric Poisson's A relativists toolkit, Eq. (3.54))
$$T_{\mu\nu} = \Theta(\ell) T^{+}_{\mu\nu}+\Theta(-\ell) T^{-}_{\mu\nu}+ \delta(\ell)S_{\mu\nu},\tag{3.54}$$
where $S_{\mu\nu}$ is surface stress-energy tensor or a thin shell of matter. I do not see how the above stress-energy tensor satisfies $\nabla_{\mu} T^{\mu\nu} = 0$. We are bound to get derivatives of the $\delta$-function if one computes $\nabla_{\mu} T^{\mu\nu}$. Of course, $S_{\mu\nu} = 0$ is basically a junction condition that allows us to avoid this surface layer but in case $S_{\mu\nu}\neq 0$, its presence prevents the conservation of $T_{\mu\nu}$. So, how can we still interpret this as something physical (like it is done in Eric Poisson's A relativists toolkit)?
 A: *

*Yes, $$\nabla_{\mu} T^{\mu\nu}~\stackrel{m}{\approx}~0,$$ cf. e.g. my Phys.SE answer here.


*OP ponders if the covariant derivative $\nabla_{\mu} T^{\mu\nu}$ of the SEM tensor (3.54) could produce contributions proportional to the derivative of the Dirac delta distribution $\delta(\ell)$?


*Well, let's check. First of all, it is enough to consider the partial derivative $\partial_{\mu} T^{\mu\nu}$. We calculate
$$ \partial_{\mu}\delta(\ell)~\stackrel{(3.46)}{=}~\varepsilon n_{\mu}\delta^{\prime}(\ell). $$


*Next recall that the surface stress-energy tensor $$S^{\mu\nu}~=~S^{ab}e^{\mu}_ae^{\nu}_b \tag{3.55}$$ is tangent to the hypersurface $$S_{\mu\nu}n^{\nu}~=~0.\tag{above 3.55}$$


*Hence, the contribution proportional to $\delta^{\prime}(\ell)$ must vanish, cf.
$$ e^{\mu}_an_{\mu}~=~0.\tag{below 3.7}$$


*Alternatively, use that
$$ e^{\mu}_a\partial_{\mu}\delta(\ell)~\stackrel{(3.7)}{=}~\frac{\partial \delta(\ell)}{\partial y^a}~=~0, $$
which is zero because $(y^1,y^2,y^3,\ell)$ constitute independent coordinates of spacetime sufficiently close to the hypersurface $\Sigma$.
References:

*

*Eric Poisson, A Relativist's Toolkit, 2004; Subsections 3.4.2 + 3.7.4 + 3.7.5.

