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could anyone please explain or show some simple steps how using matter action: $S = \int d^4x \sqrt{-g} L(X, \phi)$, where $X = \frac{1}{2} g^{\mu \nu} \nabla_\mu \phi \nabla_\nu \phi$

We can derive energy-momentum tensor: $T_{\mu \nu} = \frac{2}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu \nu}}$

I understand that we should take variation of action $\delta S$ with respect to $g^{\mu \nu}$ , but I don't understand how to fit X into $L(X, \phi)$ Can I just put instead of $L(X, \phi)$, $X = \frac{1}{2} g^{\mu \nu} \nabla_\mu \phi \nabla_\nu \phi$, into action and take variation or there is anything more in the process ? Am I wrong about my variant of just putting X into action ?

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  • $\begingroup$ Yes, you can't just put in $X$ instead. Hint: $\delta L(X) = \delta X L'(X)$ $\endgroup$
    – Eletie
    Commented Jun 9, 2022 at 11:00
  • $\begingroup$ So if we talk about $ L(X, \phi)$ it will be: $\delta L(X, \phi) = \delta X L(X, \phi) + \delta \phi L(X, \phi)$? $\endgroup$
    – John Wayne
    Commented Jun 9, 2022 at 14:52
  • $\begingroup$ You're missing the derivatives of $L$ (that's what the prime denotes). I'll write this explicitly in an answer $\endgroup$
    – Eletie
    Commented Jun 9, 2022 at 22:39

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When making the variations you need to use the following relation $$\delta L (X, \Phi) = \delta X \frac{\partial L(X,\Phi)}{\partial X} + \delta \Phi \frac{\partial L(X,\Phi)}{\partial \Phi} \ ,$$ which is just a general fact for derivatives/variations of functions.

The variations with respect to the metric will only include the first term on the RHS - the other term will contribute to the $\Phi$ equations of motion.

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  • $\begingroup$ Eletie, thank you very much for answering my question $\endgroup$
    – John Wayne
    Commented Jun 10, 2022 at 7:07

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