This question is in reference to Exercise 30.4.2 in Thomas Moore's A General Relativity Workbook, which asks you to show that a gauge transformation of the trace-reversed metric perturbation $H_{\mu\nu}$ transforms as follows (his notation):

\begin{align} H'_{\mu\nu} &= H_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu + \eta_{\mu\nu}\partial _\alpha \xi^\alpha \end{align}

He says to begin the proof by substituting the results of the previous exercise,

\begin{align} h'_{\mu\nu} &= h_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu \end{align}


\begin{align} H_{\mu\nu} &= h_{\mu\nu} - \frac{1}{2} \eta_{\mu\nu} h \end{align}

When I do this I get the following:

\begin{align} H'_{\mu\nu} &= \frac{\partial x^\alpha}{\partial x^{'\mu}} \frac{\partial x^\beta}{\partial x^{'\nu}} H_{\alpha\beta} \\ &= (\delta^\alpha_\mu - \frac{\partial \xi^\alpha}{\partial x^{'\mu}}) (\delta^\beta_\nu - \frac{\partial \xi^\beta}{\partial x^{'\nu}}) (h'_{\alpha\beta} + \partial _\alpha \xi_\beta + \partial _\beta \xi_\alpha - \frac{1}{2} \eta_{\alpha\beta} h)\\ &= (\delta^\alpha_\mu \delta^\beta_\nu - \delta^\beta_\nu \partial _{\mu} \xi^\alpha - \delta^\alpha_\mu \partial _{\nu} \xi^\beta + \partial _{\mu} \xi^\alpha \partial _{\nu} \xi^\beta) (h'_{\alpha\beta} + \partial _\alpha \xi_\beta + \partial _\beta \xi_\alpha - \frac{1}{2} \eta_{\alpha\beta} h)\\ &= h'_{\mu\nu} + \partial _\mu \xi_\nu + \partial _\nu \xi_\mu - \frac{1}{2} \eta_{\mu\nu} h - h'_{\alpha\nu} \partial _\mu \xi^\alpha - h'_{\beta\mu} \partial _\nu \xi^\beta + \frac{1}{2} h \eta_{\alpha\nu} \partial _\mu \xi^\alpha + \frac{1}{2} h \eta_{\mu\beta} \partial _\nu \xi^\beta \end{align}

after dropping terms of order $\mathcal{O}(|\partial _\mu \xi^\nu|^2)$. Simplifying,

\begin{align} H'_{\mu\nu} &= H_{\mu\nu} - h'_{\alpha\nu} \partial _\mu \xi^\alpha - h'_{\beta\mu} \partial _\nu \xi^\beta + \frac{1}{2} h (\eta_{\alpha\nu} \partial _\mu \xi^\alpha + \eta_{\mu\beta} \partial _\nu \xi^\beta)\\&= H_{\mu\nu} - h'_{\alpha\nu} \partial _\mu \xi^\alpha - h'_{\beta\mu} \partial _\nu \xi^\beta + \frac{1}{2} h (\partial _\mu \xi_\nu + \partial _\nu \xi_\mu) \end{align}

But I don't see how to get from here (assuming it's correct to this point) to the desired result

\begin{align} H'_{\mu\nu} &= H_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu + \eta_{\mu\nu}\partial _\alpha \xi^\alpha \end{align}

I've tried different combinations of raising/lowering indices with $\eta_{\mu\nu}$, renaming dummy indices, substituting for h and h', etc., but I just don't see it. Any help would be appreciated.

Following the approach suggested by Drake Marquis in his comment:

\begin{align} H'_{\mu\nu} &= h'_{\mu\nu} - \frac{1}{2} \eta_{\mu\nu} h'\\ &= h_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu - \frac{1}{2} \eta_{\mu\nu} h'\\ &= H_{\mu\nu} + \frac{1}{2} \eta_{\mu\nu} h - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu - \frac{1}{2} \eta_{\mu\nu} h'\\ &= H_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu + \frac{1}{2} \eta_{\mu\nu} (h - h')\\ &= H_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu + \frac{1}{2} \eta_{\mu\nu} (\eta^{\alpha\beta} h_{\alpha\beta} - \eta^{\alpha\beta} h'_{\alpha\beta})\\ &= H_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu + \frac{1}{2} \eta_{\mu\nu} (\eta^{\alpha\beta} (h_{\alpha\beta} - h'_{\alpha\beta}))\\ &= H_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu + \frac{1}{2} \eta_{\mu\nu} (\eta^{\alpha\beta} (\partial _\alpha \xi_\beta + \partial _\beta \xi_\alpha))\\ &= H_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu + \frac{1}{2} \eta_{\mu\nu} (\partial _\alpha \xi^\alpha + \partial _\beta \xi^\beta)\\ &= H_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu + \frac{1}{2} \eta_{\mu\nu} (2 \partial _\alpha \xi^\alpha )\\ &= H_{\mu\nu} - \partial _\mu \xi_\nu - \partial _\nu \xi_\mu + \eta_{\mu\nu}\partial _\alpha \xi^\alpha \end{align} the desired result.

  • $\begingroup$ You know, by definition, $H'_{\mu\nu}=h'_{\mu\nu}-h'\eta_{\mu\nu}/2$, then substitute the transformation of $h_{\mu\nu}$ to give the final result. This is the simplest method. $\endgroup$ – Drake Marquis Dec 21 '17 at 1:02
  • $\begingroup$ Thanks very much for the suggestion, it's a more straightforward approach, that I hope I documented correctly in the main post. I still don't know why what I was originally doing led me into the weeds, but I'm satisfied with the proof. $\endgroup$ – P. Gallez Dec 22 '17 at 0:13

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