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Given the Hilbert action: $$ S_{H}=\int \sqrt{|g|}R d^{n}x $$ and the metric written in terms of Minkowski and perturbed metric: $$ g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}. $$

I am able to derived the suitable Lagrangian for Linearised theory is: $$ \mathcal{L}=\frac{1}{2}[ (\partial_{\mu}h^{\mu \nu})(\partial_{\nu}h)-(\partial_{\mu}h^{\rho \sigma})(\partial_{\rho}h_{\sigma}^{\mu})+\frac{1}{2}\eta^{\mu \nu}(\partial_{\mu}h^{\rho \sigma})(\partial_{\nu}h_{\rho \sigma})-\frac{1}{2}\eta^{\mu \nu} (\partial_{\mu}h)(\partial_{\nu}h) ] $$ which using the Euler-Lagrange equation should give rise to the linearised Einstein Tensor, nonetheless, I am not able to produce the result. So I have the following questions:

  1. The Lagrangian I obtained is it correct?

  2. If it's correct, can anyone show the remaining steps lead to linearised Einstein Tensor.

Qmechanic
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wong tom
  • 547

1 Answers1

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Below is a derivation of the linearized Einstein's equation from the given Lagrangian.

Since $h^{\mu\nu}$ does not appear in the Lagrangian, we have $\frac{\delta L}{\delta h^{\alpha\beta}} = 0 $. So one side of the Euler-Lagrangian equation is easy.

For the other side, we have

$$\frac{\delta L}{\delta \partial_\gamma h^{\alpha\beta}} = \frac{1}{2} [\delta^\gamma_{(\alpha} \partial_{\beta)} h + (\partial_\mu h^{\mu\gamma})( \eta_{\alpha\beta}) - 2 \partial_{(\alpha} {h^\gamma}_{\beta)} + \partial^\gamma h_{\alpha\beta} - \eta_{\alpha\beta}(\partial^\gamma h)], $$

and thus

$$\partial_\gamma \frac{\delta L}{\delta \partial_\gamma h^{\alpha\beta}} = \frac{1}{2} [\partial_\alpha\partial_\beta h + \eta_{\alpha\beta} (\partial_\mu\partial_\gamma h^{\mu\gamma}) - \partial_\alpha\partial_\gamma {h^\gamma}_\beta - \partial_\beta\partial_\gamma {h^\gamma}_\alpha + \partial_\gamma \partial^\gamma h_{\alpha\beta} - \eta_{\alpha\beta} (\partial_\gamma \partial^\gamma h)] $$

Setting $\frac{\delta L}{\delta h^{\alpha\beta}} = \partial_\gamma \frac{\delta L}{\delta \partial_\gamma h^{\alpha\beta}}$, we have $$\frac{1}{2} [\partial_\alpha\partial_\gamma {h^\gamma}_\beta + \partial_\beta\partial_\gamma {h^\gamma}_\alpha -\partial_\alpha\partial_\beta h - \partial_\gamma \partial^\gamma h_{\alpha\beta} - \eta_{\alpha\beta} (\partial_\mu\partial_\gamma h^{\mu\gamma}) + \eta_{\alpha\beta} (\partial_\gamma \partial^\gamma h) ] = 0.$$

Note, a tricky part is that $h$ is symmetric. This post explains how to handle it.

Petra Axolotl
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