3
$\begingroup$

This is an equation (7.153) from Chapter-7 of Sean Carroll's An introduction to General Relativity: Spacetime and Geometry book. I think all of you who studied GR and went thorugh Carroll's book have already seen the equation. This is a perturbative expansion of Ricci tensor that is quadratic in h i.e. $R^{(2)}_{\mu\nu}[h]$. I wanted to derive the expression but no luck.For your convenience I have put the expression below that has to be derived from the expression of Ricci tensor.

$$R^{(2)}_{\mu\nu}[h]= \frac{1}{2} h^{\rho\sigma} \partial_\mu \partial_\nu h_{\rho\sigma}+ \frac{1}{4}(\partial_\mu h_{\rho\sigma}) \partial_\nu h^{\rho\sigma}+(\partial^\sigma h^\rho_\nu) \partial_{[\sigma}h_{\rho]\mu}-h^{\rho\sigma}\partial_\rho \partial_{(\mu}h_{\nu)\sigma}$$ $$+\frac{1}{2} \partial_\sigma(h^{\rho\sigma} \partial_\rho h_{\mu\nu})-\frac{1}{4} (\partial_\rho h_{\mu\nu})\partial^\rho h- (\partial_\sigma h^{\rho\sigma}-\frac{1}{2} \partial^\rho h) \partial_{(\mu}h_{\nu)\rho}$$

This is what I've done so far. I'm just giving the results that I've found after calculation.

Riemann Tensor: $$R^\rho_{\mu\tau\nu}=\partial_\tau \Gamma^\rho_{\mu\nu}+\Gamma^\sigma_{\mu\nu} \Gamma^\rho_{\sigma\tau}-\partial_\nu \Gamma^\rho_{\mu\tau}-\Gamma^\sigma_{\mu\tau} \Gamma^\rho_{\sigma\nu}$$ And Ricci Tensor: $$R^\rho_{\mu\rho\nu}=\partial_\rho \Gamma^\rho_{\mu\nu}+\Gamma^\sigma_{\mu\nu} \Gamma^\rho_{\sigma\rho}-\partial_\nu \Gamma^\rho_{\mu\rho}-\Gamma^\sigma_{\mu\rho} \Gamma^\rho_{\sigma\nu}$$

Now working out each term in the Ricci tensor: $$\Gamma^\sigma_{\mu\nu}=\frac{1}{2} (\partial_\nu h^\sigma_\mu + \partial_\mu h^\sigma_\nu - \partial^\sigma h_{\mu\nu})$$ $$\Gamma^\rho_{\sigma\rho}= \frac{1}{2} \partial_\sigma h$$

Therefore $$\Gamma^\sigma_{\mu\nu} \Gamma^\rho_{\sigma\rho}= \frac{1}{2} \partial^\rho h \partial_{(\mu} h_{\nu)\rho} - \frac{1}{4} (\partial_\rho h_{\mu\nu}) \partial^\rho h$$

Again $$\Gamma^\sigma_{\mu\rho}= \frac{1}{2} (\partial_\rho h^\sigma_\mu+\partial_\mu h^\sigma_\rho-\partial^\sigma h_{\mu\rho})$$

And $$\Gamma^\rho_{\sigma\nu}= \frac{1}{2} (\partial_\nu h^\rho_\sigma+\partial_\sigma h^\rho_\nu-\partial^\rho h_{\sigma\nu})$$

Therefore $$-\Gamma^\sigma_{\mu \rho} \Gamma^\rho_{\sigma \nu}= \partial^\sigma h^\rho_\nu \partial_{[\sigma} h_{\rho]\mu} - \frac{1}{4} (\partial_\mu h_{\rho\sigma}) \partial_\nu h^{\rho\sigma}$$

The derivative terms of Christoffel Symbols' didn't match with the expression of $R^{(2)}_{\mu\nu}[h]$. There's another mismatch of sign with $\frac{1}{4} (\partial_\mu h_{\rho\sigma}) \partial_\nu h^{\rho\sigma}$ in $-\Gamma^\sigma_{\mu \rho} \Gamma^\rho_{\sigma \nu}$ (meaning this term is positive in $R^{(2)}_{\mu\nu}[h]$)

Anyone who can resolve the issues will be very much appreciated. Also please check the terms in the Ricci tensor if they were accurately computed. Do correct me if I was mistaken.

$\endgroup$
  • $\begingroup$ Just to be clear, you should state you are perturbing around a Minkowski background, rather than a generic curved background, as then your formula would be missing terms. $\endgroup$ – JamalS Apr 4 at 20:29
1
$\begingroup$

I haven't checked your calculation, but if you've followed the right steps, the only way to make mistakes is algebra-related. I'll lay out the steps though. I've suppressed indices for clarity wherever possible.

$$R_{\mu \nu} = \partial \Gamma + \partial \Gamma + \Gamma \Gamma + \Gamma \Gamma$$

So $\mathcal{O}(h)$ perturbation to the Ricci tensor is given by

$$R_{\mu \nu}^{(1)} = \partial \Gamma^{(1)} + \partial \Gamma^{(1)} + \Gamma^{(1)} \Gamma + \Gamma \Gamma^{(1)} + \Gamma^{(1)} \Gamma + \Gamma \Gamma^{(1)}$$

and $\mathcal{O}(h^2)$ perturbation to the Ricci tensor is given by

$$R_{\mu \nu}^{(2)} = \partial \Gamma^{(2)} + \partial \Gamma^{(2)} + \Gamma^{(2)} \Gamma + \Gamma \Gamma^{(2)} + \Gamma^{(1)}\Gamma^{(1)} $$

Now, $(7.153)$ shows the part of $R_{\mu \nu}^{(2)}$ computed from $h_{\mu \nu}^{(1)}$ only. I noticed that you didn't write $\Gamma^{(2)}$ (coming from $h_{\mu \nu}^{(1)}$):

$$\Gamma^{(2)\sigma}_{\mu \nu} = -\frac{1}{2} h^{\sigma \lambda} (\nabla_\mu h_{\nu \lambda} + \nabla_\nu h_{\mu \lambda} - \nabla_\lambda h_{\mu \nu})$$

Maybe you missed this?

$\endgroup$
  • $\begingroup$ I have finally derived the expression for $R^{(2)}_{\mu\nu}[h]$ considering $\Gamma^{(2)\sigma}_{\mu\nu}$. It looks like I have really missed that. Thanks a lot. BTW, I need more insights on how one can get the expressions for Ricci tensor and Christoffel symbols in different orders i.e. $R^{(1)}_{\mu\nu}$ (in terms of $\Gamma ′s$) and $R^{(2)}_{\mu\nu}$ (in terms of $\Gamma ′s$) that you came up with in your answer. And different orders of $\Gamma ′s$ (in terms of physics metric $g_{\mu\nu}$) $\endgroup$ – SaidurRahman Apr 9 at 13:44
  • $\begingroup$ @SaidurRahman There's no physical insight, if that's what you're asking for. It's just brute-force computation. $\endgroup$ – Avantgarde Apr 9 at 15:04
  • $\begingroup$ Okay. Got that. One more thing I'd like to ask. In your answer, the $3^{rd}$ and $4^{th}$ terms in the expression of $R^{(2)}_{\mu\nu}$ contain $\Gamma$ times $\Gamma^{(2)}$. Is that $\Gamma$ with no superscript is zeroth order term, because it's already multiplied with $\Gamma^{(2)}$ ? $\endgroup$ – SaidurRahman Apr 9 at 17:03
  • $\begingroup$ @SaidurRahman Exactly. You got it. $\endgroup$ – Avantgarde Apr 9 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.