# Perturbed Ricci tensor due to metric perturbation i.e. $R^{(2)}_{\mu\nu}[h]$ in Linearized theory of Einstein field equation

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.

• 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. Apr 4 '19 at 20:29

## 1 Answer

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?

• 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 $g_{\mu\nu}$) Apr 9 '19 at 13:44
• @SaidurRahman There's no physical insight, if that's what you're asking for. It's just brute-force computation. Apr 9 '19 at 15:04
• 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 $\Gamma^{(2)}$ ? Apr 9 '19 at 17:03
• @SaidurRahman Exactly. You got it. Apr 9 '19 at 17:51