# Derivatives of distributions in general relativity

I am having some trouble when trying to reproduce some calculations involving the description of distributions (mostly used in spacetime junction conditions).

I am trying to reproduce the calculations in this paper. In the appendix, I can't derive this equality:

$$\nabla_\nu([R]n_\mu \, \underline{\delta}^\Sigma)= \underline{\delta}^\Sigma \left( [R]K^\Sigma_{\mu\nu} - [R]K^\rho_{\,\,\rho}n_\mu n_\nu +n_\mu \overline{\nabla}_\nu [R] \right) + \underline{\Delta}_{\mu\nu},$$ where an underline designates a distribution, $\overline{\nabla}_\nu$ is the covariant derivative on the hypersurface, $[R]$ is the scalar curvature jump across the hypersurface $[R]\equiv R^+|_\Sigma- R^-|_\Sigma$, the extrinsic curvature is $$K_{\mu\nu}=h^\rho_{\,\,\mu}h^\sigma_{\,\,\nu}\nabla_\rho n_\sigma,$$ where $n$ is the normal to the hypersurface and $h$ is the projector to the hypersurface $\Sigma$, $h_{\mu\nu}=g_{\mu\nu}-n_\mu n_\nu$. The distribution $\underline{\Delta}$ includes the $\delta'$ term. This is all in Eq.(37) of the paper, and the (unnumbered) equation right before it.

My problem is, instead of getting the above equality, I keep getting only $$\nabla_\nu([R]n_\mu \, \underline{\delta}^\Sigma) = \underline{\delta}^\Sigma \left( [R]K^\Sigma_{\mu\nu}+n_\mu \overline{\nabla}_\nu [R] \right) + \underline{\Delta}_{\mu\nu}.$$

If someone could point me out to what I might be doing wrong, I'd appreciate it.

The problem actually extends to eqn (38), mainly the $-2\alpha \delta$ term, which corresponds to the $f''(R)$ term in the eqn below eqn (31). I can't find where the $-K^\rho_{\,\,\rho} n_\mu n_\nu$ comes from, nor why there is no $-K^\rho_{\,\,\rho}g_{\mu \nu}$ term arising from $-g_{\alpha \beta}\nabla^\rho \nabla_\rho R$ of eqn (31).

• Your first equation is only the first term of the equation just before eqn (37). I guess you have computed the second term and found all the terms in eqn (37), and you are then left with your equation. Correct? If so, could you show all your computations? – user154997 Jul 8 '17 at 19:59
• Correct, the remaining term is decomposed as in eqn (35), directly onto eqn (37), which is why I removed that term and restricted the problem only to this second term. As for my computations, all I did was split the derivatives of the product, and since they are in the presence of a $\delta$, the derivatives are projected on the hypersurface ($\overline{\nabla}$). Hope I was clear enough – DisStudent Jul 9 '17 at 9:27
• Just to be sure we are on the same page: $\newcommand{\nablabar}{\overline{\nabla}}\newcommand{\s}{\phantom{m}}\nablabar_\mu v_\nu=h_\mu^{\s\rho}\nabla_\rho(h_\nu^{\s\sigma}v_\sigma)$, isn't it? – user154997 Jul 11 '17 at 11:51