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I have found confusing definitions in various places regarding the stress-energy tensor, in particular when used to derive Einstein GR equations from the principle of stationary action. Some of these various definitions are $${T}_{\mu\nu}=-\frac{2}{\sqrt{-g}} \frac{\delta{\mathcal{L}}_M}{\delta{g}^{\mu\nu}}, \tag{1}$$ $${T}_{\mu\nu}=-\frac{2}{\sqrt{-g}} \frac{\delta{\mathcal{S}}_M}{\delta{g}^{\mu\nu}}, \tag{2}$$ or even $${T}_{\mu\nu}=-\frac{2}{\sqrt{-g}} \frac{\delta(\mathcal{L}_M\sqrt{-g})}{\delta{g}^{\mu\nu}}. \tag{3}$$

I have been able to follow the derivation leading to the G.R. equations using the definition $(1)$, which I have also seen in these questions. But then I found the other definitions here which really confused me. Is $(1)$ the correct one? Otherwise, which one is correct?

[Here I'm using the Minkowski sign convention $(-,+,+,+)$.]

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  • $\begingroup$ Where have you seen the first equation being used? $\endgroup$ Commented Feb 17, 2021 at 10:22
  • $\begingroup$ Corrected the missing parentheses. Thanks. I've seen the first definition for instance here: physics.stackexchange.com/q/119838 but in other places as well $\endgroup$
    – Carla
    Commented Feb 17, 2021 at 10:24
  • $\begingroup$ or here: physics.stackexchange.com/q/54856 $\endgroup$
    – Carla
    Commented Feb 17, 2021 at 10:36
  • $\begingroup$ Note that the third equation you wrote is not identical to the one in the Wikipedia link you gave. One has variation and one has partial derivatives. $\endgroup$
    – Rd Basha
    Commented Feb 17, 2021 at 10:50
  • $\begingroup$ @RdBasha I didn't give that link (someone edited my question...) But you can see an example here : en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action $\endgroup$
    – Carla
    Commented Feb 17, 2021 at 15:04

1 Answer 1

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All definitions (1)-(3) are in principle the same. However, the various notations$^1$ may warrant some explanation:

  • Eq. (2) uses the standard/traditional definition of a functional/variational derivative (FD) of the action functional $S=\int\!d^dx ~{\cal L}$ in $d$ spacetime dimensions.

  • Eq. (1) uses a 'same-spacetime' FD $$ \frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}~:=~ \frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - d_{\mu} \left(\frac{\partial{\cal L}(x) }{\partial\partial_{\mu}\phi^{\alpha} (x)} \right)+\ldots, $$ which obscures/betrays its variational origin, but is often used for notational convenience. See e.g. this, this, & this Phys.SE posts and links therein.

  • Eq. (3) is the same as eq. (1), except the Lagrangian density ${\cal L}=\sqrt{|g|}L$ is written$^1$ as a product of a density $\sqrt{|g|}$ and a scalar function $L$.

--

$^1$ As always, be aware that that different authors use different notation.

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  • $\begingroup$ I have corrected the last equation. Thank you for the answer, but before I validate it, I would like to understand how the variation of the action means the same thing as the variation of the Lagrangian, and why multiplying by the square root of the metric determinant does not change the result. $\endgroup$
    – Carla
    Commented Feb 17, 2021 at 12:33
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    $\begingroup$ @Carla It is not that the result doesnt change, it is whether one includes the metric factor in the definition of the Lagrangian or not. $\endgroup$
    – NDewolf
    Commented Feb 17, 2021 at 12:41
  • $\begingroup$ @Carla you seemed to have rolled back the edit to the question, but note that the third equation in the question should not be $\mathcal{L}_M\sqrt{-g}$, but rather $L_M\sqrt{-g} \equiv \mathcal{L}_M$ (as Qmechanic had edited your question to reflect) $\endgroup$ Commented Feb 17, 2021 at 12:52
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    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Feb 17, 2021 at 13:26
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    $\begingroup$ Well, I was not sure which dimension that you work in. $\endgroup$
    – Qmechanic
    Commented Feb 17, 2021 at 16:10

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