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What is a lagrangian such that Euler-Lagrange equation (not sure if it's correct form for this case)

$$\frac{\partial \mathcal{L}}{\partial g_{\mu\nu}}=\partial_\lambda\frac{\partial \mathcal{L}}{\partial (\partial_\lambda g_{\mu\nu})}.$$

Gives us Einstein field equations?

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This is almost certainly answered elsewhere, but the Hilbert Action, from which Einstein's equation can be derived, is:

$$S = \int d^{4}x\;\left(\sqrt{|g|}\frac{1}{16\pi G}R + \mathcal{L}_{m}\right)$$

taking the variation is pretty complicated (there are second derivatives of the metric in the action, and you have to deal with gauge invariance) and best looked up in a textbook, though. But note, that by this definition, we define $T_{ab} = \frac{\delta \mathcal{L_m}}{\delta g^{ab}}$

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    $\begingroup$ The Wikipedia page does a pretty good job as an introduction of walking through the variation: en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action. Also, if it is not clear, the action is an integral over $d^4x$. $\endgroup$
    – Integral fan
    Jun 18, 2022 at 19:31
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    $\begingroup$ Perhaps you'd like to add a general matter Lagrangian to that as well, since the Einstein field equations are usually formulated with the presence of a stress-energy tensor. The sign difference between them and the general arbitration in the overall choice could also be emphasized. $\endgroup$
    – rhomaios
    Jun 18, 2022 at 19:34
  • $\begingroup$ So the lagrangian is just $\mathcal{L}=\frac{\sqrt{|g|}R}{16\pi G}$? $\endgroup$
    – JavaGamesJAR
    Jun 18, 2022 at 19:36
  • $\begingroup$ @JavaGamesJAR yes $\endgroup$
    – Zo the Relativist
    Jun 18, 2022 at 19:41
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    $\begingroup$ Really defining the volume integral means defining the theory of forms, which is usually beyond the scope of a first-year GR course. $\endgroup$
    – Zo the Relativist
    Jun 19, 2022 at 2:56
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With the cosmological constant $\Lambda$ included, the Hilbert action for empty space is $$S = \frac{c^4}{16 \pi G} \int (R-2 \Lambda) \sqrt{-g} \, \mathrm{d}^4 x. $$

Wikipedia calls it the Einstein-Hilbert action, but this is wrong. The action is due to Hilbert, not to Einstein.

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  • $\begingroup$ Notice that this action does not yield the stress-energy tensor on the right-hand side of Einstein's Equations $\endgroup$
    – Níckolas Alves
    Jun 19, 2022 at 11:50
  • $\begingroup$ Indeed, as stated, it is only for vacuum; matter can be added as above. $\endgroup$
    – KlausK
    Jun 19, 2022 at 11:51

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