# Ricci tensor manipulation of weak field metric in General Relativity

I'm having some problems trying to work through my university's GR notes, regarding the derivation of the weak field metric from the perturbation of the Minkowski metric. I'll provide the relevant material here.

Now in general we have a perturbed gravitational metric $$\mathbf{g} = \eta + \mathbf{h}$$ where $$\eta$$ is the Minkowski metric and $$\mathbf{h}$$ is some perturbation. Its component form is then just $$g_{\alpha\beta} = \eta_{\alpha\beta} + h_{\alpha\beta}$$ To solve the field equation $$\mathbf{G} = 8\pi \mathbf{T}$$, we have the Riemann tensor expressed in terms of the Christoffel symbols $$R^{\alpha}_{~~\beta\mu\gamma} = \Gamma^{\alpha}_{~~\beta\nu,\mu} - \Gamma^{\alpha}_{~~\beta\mu,\nu} +\Gamma^{\alpha}_{~~\sigma\mu}\Gamma^{\sigma}_{~~\beta\nu}-\Gamma^{\alpha}_{~~\sigma\nu}\Gamma^{\sigma}_{~~\beta\mu}$$ where the Christoffel symbols are $$\Gamma^{\alpha}_{~~\beta\gamma} = \frac{1}{2}g^{\alpha\sigma}(g_{\sigma\beta,\gamma}+g_{\sigma\gamma,\beta}-g_{\beta\gamma,\sigma})$$ Calculating the Christoffel symbol above gives $$\begin{split} \Gamma^{\alpha}_{~~\beta\gamma} &= \frac{1}{2}g^{\alpha\sigma}(g_{\sigma\beta,\gamma}+g_{\sigma\gamma,\beta}-g_{\beta\gamma,\sigma})\\ &= \frac{1}{2}(\eta^{\alpha\sigma}+h^{\alpha\sigma})((\eta_{\sigma\beta}+h_{\sigma\beta})_{,\gamma}+(\eta_{\sigma\gamma}+h_{\sigma\gamma})_{,\beta}-(\eta_{\beta\gamma}+h_{\beta\gamma})_{,\sigma})\\ &= \frac{1}{2}\eta^{\alpha\sigma}(h_{\alpha\beta,\gamma}+h_{\sigma\gamma,\beta}-h_{\beta\gamma,\sigma}) \end{split}$$ where the derivatives of Minkowski metric dissapears and we ignore the small term $$h^{\alpha\sigma}$$. So this becomes $$\begin{split} \Gamma^{\alpha}_{~~\beta\gamma} &= \frac{1}{2}(h^{\alpha}_{~~\beta,\gamma}+h^{\alpha}_{~~\gamma,\beta}+h_{\beta\gamma}^{~~~~~,\sigma}) \end{split}$$ The Riemann tensor is now $$\begin{split} R^{\alpha}_{~~\beta\mu\nu} &= \frac{1}{2}(h^{\alpha}_{~~\beta,\nu} + h^{\alpha}_{~~\nu,\beta} -h^{~~~~~,\alpha}_{\beta\nu})_{,\mu} - \frac{1}{2}(h^{\alpha}_{~~\beta,\mu} + h^{\alpha}_{~~\mu,\beta} -h^{~~~~~,\alpha}_{\beta\mu})_{,\nu}\\ &= \frac{1}{2}(h^{\alpha}_{~~\nu,\beta\mu}+ h_{\beta\mu~~~~,\nu}^{~~~~~,\alpha}-h^{\alpha}_{~~\mu,\beta\nu}-h_{\beta\nu~~~~,\mu}^{~~~~~,\alpha}) \end{split}$$ To calculate the Ricci tensor it is $$R_{\beta\nu} = R^{\alpha}_{~~\beta\alpha\nu}$$ So in the Riemann tensor all $$\mu\rightarrow\alpha$$ and becomes $$\begin{split} R^{\alpha}_{~~\beta\alpha\nu} &= \frac{1}{2}(h^{\alpha}_{~~\nu,\beta\alpha}+ h_{\beta\alpha~~~~,\nu}^{~~~~~,\alpha}-h^{\alpha}_{~~\alpha,\beta\nu}-h_{\beta\nu~~~~,\alpha}^{~~~~~,\alpha}) \end{split}$$ Here in my notes, it introduces a "trace reversed" perturbation $$\mathbf{h}$$ such that $$\bar{h}_{\alpha\beta} = h_{\alpha\beta}-\frac{1}{2}\eta_{\alpha\beta}h$$ where trace of $$\mathbf{h}$$ is $$h=h^{\alpha}_{~~\alpha}$$. The "trace reversed" is just $$\bar{h}=\bar{h}^{\alpha}_{~~\alpha} -\frac{1}{2}\eta^{\alpha\beta}\eta_{\alpha\beta}h = h - \frac{1}{2}4h = -h$$ Now so far it all made sense to me, it then went on to claim that this causes the Riemann tensor to become $$R_{\beta\nu} = \frac{1}{2}\left(\bar{h}^{\alpha}_{~~\nu,\beta\alpha} + \bar{h}_{\beta\alpha~~~,\nu}^{~~~~~,\alpha}-\bar{h}_{\beta\nu~~~,\alpha}^{~~~~~,\alpha}+\frac{1}{2}\eta_{\beta\nu}\bar{h}_{~~~,\alpha}^{,\alpha}\right)$$ and this is where I have my problem because I'm unable to get here.

$$\underline{\textbf{My attempt:}}$$

Starting with $$\begin{split} R_{\beta\nu} = R^{\alpha}_{~~\beta\alpha\nu}&= \frac{1}{2}(h^{\alpha}_{~~\nu,\beta\alpha}+ h_{\beta\alpha~~~~,\nu}^{~~~~~,\alpha}-h^{\alpha}_{~~\alpha,\beta\nu}-h_{\beta\nu~~~~,\alpha}^{~~~~~,\alpha}) \end{split}$$ Now I replace all the $$h$$ by their $$\bar{h}$$ counterpart where $$\begin{split} \bar{h}^{\alpha}_{~~\nu,\beta\alpha} &= h^{\alpha}_{~~\nu,\beta\alpha} - \frac{1}{2}\eta^{\alpha}_{~~~\nu}h_{,\beta\alpha}\\ \bar{h}^{~~~~~,\alpha}_{\beta\alpha~~~,\nu} &= h^{~~~~~,\alpha}_{\beta\alpha~~~,\nu} - \frac{1}{2}\eta_{\beta\alpha}h^{,\alpha}_{~~~,\nu}\\ -\bar{h}^{\alpha}_{~~\alpha,\beta\nu} &= -h^{\alpha}_{~~\alpha,\beta\nu} + \frac{1}{2}\eta^{\alpha}_{~~~\alpha}h_{,\beta\nu}\\ -\bar{h}^{~~~~~,\alpha}_{\beta\nu~~~,\alpha} &= -h^{~~~~~,\alpha}_{\beta\nu~~~,\alpha} + \frac{1}{2}\eta_{\beta\nu}h^{,\alpha}_{~~~,\alpha} \end{split}$$ Rearranging the Riemann tensor gives $$\begin{split} R_{\beta\nu} = R^{\alpha}_{~~\beta\alpha\nu}&= \frac{1}{2}\left(\bar{h}^{\alpha}_{~~\nu,\beta\alpha} + \bar{h}_{\beta\alpha~~~,\nu}^{~~~~~,\alpha}-\bar{h}_{\beta\nu~~~,\alpha}^{~~~~~,\alpha}\underbrace{-\bar{h}^{\alpha}_{~~\alpha,\beta\nu}}_{-(-h)_{,\beta\nu}}-\frac{1}{2}\eta_{\beta\nu}\bar{h}_{~~~,\alpha}^{,\alpha} \underbrace{- \frac{1}{2}\eta^{\alpha}_{~~~\alpha}h_{,\beta\nu}}_{-\frac{1}{2}2h_{\beta\nu}} + \frac{1}{2}\eta^{\alpha}_{~~~\nu}h_{,\beta\alpha} +\frac{1}{2}\eta_{\beta\alpha}h^{,\alpha}_{~~~,\nu} \right) \\ &= \frac{1}{2}\left(\bar{h}^{\alpha}_{~~\nu,\beta\alpha} + \bar{h}_{\beta\alpha~~~,\nu}^{~~~~~,\alpha}-\bar{h}_{\beta\nu~~~,\alpha}^{~~~~~,\alpha}-\frac{1}{2}\eta_{\beta\nu}h_{~~~,\alpha}^{,\alpha} + \frac{1}{2}\eta^{\alpha}_{~~~\nu}h_{,\beta\alpha} +\frac{1}{2}\eta_{\beta\alpha}h^{,\alpha}_{~~~,\nu} \right) \end{split}$$ where both the underbraced terms cancel with each other(I believe). This is as far as I can get, how can I show that $$-\frac{1}{2}\eta_{\beta\nu}h_{~~~,\alpha}^{,\alpha} + \frac{1}{2}\eta^{\alpha}_{~~~\nu}h_{,\beta\alpha} +\frac{1}{2}\eta_{\beta\alpha}h^{,\alpha}_{~~~,\nu} = \frac{1}{2}\eta_{\beta\nu}\bar{h}^{,\alpha}_{~~~\alpha}$$

First, the two underbraced terms do not cancel. $$\eta^\alpha_{\;\;\alpha}=4$$, not $$2$$. So, you have an extra term of $$\partial_\beta\partial_\nu \bar{h}$$.
First note these three things: $$\bar{h}=-h$$ (as you have already calculated), $$\eta_{\beta\alpha}\partial^\alpha=\partial_\beta$$, and finally $$\eta^\alpha_{\;\;\nu}\partial_\alpha=\partial_\nu$$. Then, we have \begin{align} -\frac{1}{2}\partial^\alpha\partial_\alpha\eta_{\beta\nu}h+\frac{1}{2}\partial_{\beta}\partial_\alpha\eta^\alpha_{\;\;\nu}h+\frac{1}{2}\partial^\alpha\partial_\nu\eta_{\beta\alpha}h+\partial_\beta\partial_\nu \bar{h} & =\frac{1}{2}\partial^\alpha\partial_\alpha\eta_{\beta\nu}\bar{h}-\frac{1}{2}\partial_\beta\partial_\nu \bar{h}-\frac{1}{2}\partial_\beta\partial_\nu \bar{h}+\partial_\beta\partial_\nu \bar{h} \\ &=\frac{1}{2}\partial^\alpha\partial_\alpha\eta_{\beta\nu}\bar{h} \end{align} which is the required simplification. I took the liberty of pulling out the Minkowski metric from the partial derivatives, as their derivative is zero.
• Is $\eta^{\alpha}_{~~\alpha} = 4$ the same way $\eta^{\alpha\beta}_{~~~~\alpha\beta} = 4$? Also another question, is $\bar{h}^{\alpha\beta}_{~~~~~,\alpha\beta} = \bar{h}^{~~~~~,\alpha\beta}_{\alpha\beta}$? how do you change between them? May 18, 2020 at 19:59
• Would you do $\eta^{\alpha\alpha'}\eta^{\beta\beta'}\eta_{\alpha\alpha'}\eta_{\beta\beta'}\bar{h}^{~~~~~,\alpha\beta}_{\alpha\beta} = \bar{h}^{\alpha'\beta'}_{~~~~~,\alpha'\beta'} = \bar{h}^{\alpha\beta}_{~~~~~,\alpha\beta}$? May 18, 2020 at 20:02
• $\eta^{\alpha\beta}\eta_{\alpha\beta}=\eta^\alpha_{\;\;\alpha}=\delta^\alpha_{\alpha}=4$; the trace of the metric is the number of dimensions. Yes, that is how you would do that, as the derivative of the Minkowski metric is 0. May 18, 2020 at 20:18