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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)?

Qmechanic
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Dr. user44690
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1 Answers1

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

  2. 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 $\delta(\ell)$?

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

  4. 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 $$S_{\mu\nu}n^{\nu}~=~0.\tag{above 3.55}$$

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

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

  7. For completeness let us mention that the terms in (\$) proportional to the Dirac delta distribution $\delta(\ell)$ read $$ \varepsilon n_{\mu}(T^{\mu\nu}_+-T^{\mu\nu}_-)+\nabla_{\mu} S^{\mu\nu}~=~0. $$ This should be compared with corresponding curvature terms in the Bianchi identity (4.25).

References:

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