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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!

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    $\begingroup$ maybe I misread your question, but to improve your chances of getting an answer, would you considering actully putting in the full form of the Schutz equations (8.31) etc, as you have them to hand. $\endgroup$ – user81619 Jun 19 '15 at 2:14
  • $\begingroup$ Done! Eq.(8.31) and Eq.(8.12) are both added. $\endgroup$ – gamebm Jun 19 '15 at 2:23
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    $\begingroup$ Me again, two things. did you read this: preposterousuniverse.com/grnotes/grnotes-six.pdf (or if not that particular page, moooch around Carroll's notes, it's there somewhere, look at the index or toc) and eq(9.1) is still, sorry about this, eq (9.1) :) $\endgroup$ – user81619 Jun 19 '15 at 2:58
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$R_{\alpha\beta}=R{^{\mu}}_{\alpha\mu\beta}$ 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 $$ \bar{h}_{\alpha\beta,}{^\mu}_\mu = -\frac{16 \pi G}{c^4} T_{\alpha\beta} $$

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  • $\begingroup$ Thanks a lot for pointing out the mistake. The extra term was a typo due to the copy-paste of latex equations, it was not there in my notes. I subtract the $\frac{1}{2}g_{\alpha\beta}R$, and I still am not getting the correct answer. $\endgroup$ – gamebm Jun 19 '15 at 12:12
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    $\begingroup$ The Ricci scalar evaluates to $$ \mathcal{R} = \bar{h}_{\mu\nu,}{^{\mu\nu}}+\frac{1}{2}\bar{h}{^\lambda}{_\lambda,}{^\mu}_\mu $$ Subtract $\frac{1}{2} g_{\alpha\beta} \mathcal{R}$ from your expression of the Ricci tensor and impose the Lorentz gauge and you should get the equation $$ \bar{h}_{\alpha\beta,}{^\mu}_\mu = -\frac{16 \pi G}{c^4} T_{\alpha\beta} $$ I don't exactly know what the result in your book is, but this is the wave equation you should get. $\endgroup$ – somebody Jun 19 '15 at 13:34
  • $\begingroup$ Thx! I got it... $\endgroup$ – gamebm Jun 21 '15 at 17:40
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enter image description here you can find the detail in the manuscript

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    $\begingroup$ Welcome to Physics! Note that this site has MathJax enabled, so you can write the equations using a Latex-like format. $\endgroup$ – Kyle Kanos Jun 19 '15 at 14:15
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    $\begingroup$ And you reference a manuscript - do you mean the book in the question, or some other reference? Please clarify. $\endgroup$ – Jon Custer Jun 19 '15 at 14:30
  • $\begingroup$ @JonCuster: I think Kai means the details are in the image $\endgroup$ – Kyle Kanos Jun 19 '15 at 15:37
  • $\begingroup$ @KyleKanos - That was another thought as well. Perhaps Kai will clarify. Thanks. $\endgroup$ – Jon Custer Jun 19 '15 at 15:40

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