Calculation of Einstein tensor for weak gravitational field

Question

I am studying A First Course in General Relativity (2nd Ed.) by Bernard Schutz. I have some difficulty in deriving Eq.(8.32) on P.193, the form of Einstein tensor for weak gravitational field, which is essential to derive the equation of gravitational wave in Lorentz gauge Eq.(9.1).

I noticed that in order to raise an index, one should use $g^{\mu\nu}$; the latter can be expanded using Eq.(8.12), $g_{\alpha\beta}=\eta_{\alpha\beta}+h_{\alpha\beta}$, with $h_{\alpha\beta}$ being small. Therefore by ignoring higher order terms in $h$ one uses $\eta^{\mu\nu}$ to raise the indices.

Using Eq.(8.31), $h^{\alpha\beta}=\bar{h}^{\alpha\beta}-\frac{1}{2}\eta^{\alpha\beta}\bar{h}$ and Eq.(8.29) $\bar{h}\equiv\bar{h}{^\lambda}_\lambda$. A typical term which will be used in the calculation of Riemann tensor has the form $$h_{\alpha\beta}=\bar{h}_{\alpha\beta}-\frac{1}{2}\eta_{\alpha\beta}{\bar{h}^\lambda}_\lambda$$ $$h_{\alpha\beta,\mu\nu}=\bar{h}_{\alpha\beta,\mu\nu}-\frac{1}{2}\eta_{\alpha\beta}{\bar{h}^\lambda}_{\lambda,\mu\nu}$$ By reordering the indices, I write down one term used in the calculation later $$h_{\mu\beta,\alpha\mu}=\bar{h}_{\mu\beta,\alpha\mu}-\frac{1}{2}\eta_{\mu\beta}{\bar{h}^\lambda}_{\lambda,\alpha\mu}=\bar{h}_{\mu\beta,\alpha\mu}-\frac{1}{2}{\bar{h}^\lambda}_{\lambda,\alpha\beta}$$

Now the Riemann tensor reads $$R{^{\alpha}}_{\beta\mu\nu}=\frac{1}{2}\left(h{^{\alpha}}_{\nu,\beta\mu}+h{_{\beta\mu,}}{^\alpha}_{\nu}-h{^{\alpha}}_{\mu,\beta\nu}-h{_{\beta\nu,}}{^\alpha}{_\mu}\right)$$ From which one obtains the Einstein tensor $$R_{\alpha\beta}=R{^{\mu}}_{\alpha\mu\beta}=\frac{1}{2}\left(h{^{\mu}}_{\beta,\alpha\mu}+h{_{\alpha\mu,}}{^\mu}_{\beta}-h{^{\mu}}_{\mu,\alpha\beta}-h{_{\alpha\beta,}}{^\mu}_{\mu}\right)$$ By substituting individual terms, one gets $$R_{\alpha\beta}=\frac{1}{2}\left(\bar{h}{^{\mu}}_{\beta,\alpha\mu}-\frac{1}{2}\bar{h}{^\lambda}_{\lambda,\alpha\beta}+\bar{h}{_{\alpha\mu,}}{^\mu}_{\beta}-\frac{1}{2}\bar{h}{^\lambda}_{\lambda,\alpha\beta}-\bar{h}{^{\mu}}_{\mu,\alpha\beta}+\frac{1}{2}4\bar{h}{^\lambda}_{\lambda,\alpha\beta}-\bar{h}{_{\alpha\beta,}}{^\mu}_{\mu}+\frac{1}{2}\eta_{\alpha\beta}\bar{h}{^\lambda}{_\lambda,}{^\mu}_\mu\right)$$ $$R_{\alpha\beta}=\frac{1}{2}\left(\bar{h}{^{\mu}}_{\beta,\alpha\mu}+\bar{h}{_{\alpha\mu,}}{^\mu}_{\beta}-\bar{h}{_{\alpha\beta,}}{^\mu}_{\mu}+\frac{1}{2}\eta_{\alpha\beta}\bar{h}{^\lambda}{_\lambda,}{^\mu}_\mu\right)$$

where one makes use of the fact $4=\eta{^\mu}_\mu$. But the resulting expression only matches three terms in Eq.(8.31), the last term is different and it does not vanish in Lorentz Gauge, I thought it over but just can't find the mistake. Many thanks!

Answer 1

$R_{\alpha\beta}=R{^{\mu}}_{\alpha\mu\beta}$ is not the Einstein tensor $G_{\alpha\beta}$, but the Ricci tensor. You get the Einstein tensor via $$ G_{\alpha\beta} = R_{\alpha\beta} - \frac{1}{2} g_{\alpha\beta} \mathcal{R} \textrm{,} $$ with $\mathcal{R} = R^{\alpha}_{\textrm{ }\alpha}$ being the Ricci scalar.

In your last equation for $R_{\alpha\beta}$, the term $\bar{h}{^{\mu}}_{\mu,\alpha\beta}$ should drop out in the calculations. Then calculating the Einstein tensor in the Lorentz gauge, you will get the correct wave equation for the trace reversed metric perturbation.

To show the next steps: We have the Ricci tensor as you wrote as $$ R_{\alpha\beta}=\frac{1}{2}\left(\bar{h}{^{\mu}}_{\beta,\alpha\mu}+\bar{h}{_{\alpha\mu,}}{^\mu}_{\beta}-\bar{h}{_{\alpha\beta,}}{^\mu}_{\mu}+\frac{1}{2}\eta_{\alpha\beta}\bar{h}{^\lambda}{_\lambda,}{^\mu}_\mu\right) $$ The Ricci scalar is $$ \mathcal{R} = \bar{h}_{\mu\nu,}{^{\mu\nu}}+\frac{1}{2}\bar{h}{^\lambda}{_\lambda,}{^\mu}_\mu $$ The Einstein tensor then evaluates to $$ G_{\alpha\beta}=\frac{1}{2}\left(\bar{h}{^{\mu}}_{\beta,\alpha\mu}+\bar{h}{_{\alpha\mu,}}{^\mu}_{\beta}-\bar{h}{_{\alpha\beta,}}{^\mu}_{\mu}-\eta_{\alpha\beta}\bar{h}{^\mu}{_\nu,}{^\nu}_\mu\right) $$ Imposing the Lorentz gauge $\bar{h}_{\mu\nu,}{^\mu} = 0$ we get $$ G_{\alpha\beta} = -\frac{1}{2}\bar{h}_{\alpha\beta,}{^\mu}_\mu $$ so the wave equation is $$ \bar{h}_{\alpha\beta,}{^\mu}_\mu = -\frac{16 \pi G}{c^4} T_{\alpha\beta} $$

Answer 2

you can find the detail in the manuscript