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Summary:

Exercise statement: Prove that the stressenergy tensor is given by the functional derivative of the action with respect to ##\delta g^{\mu \nu}##
$$T_{\mu \nu} = \frac{2}{\sqrt{g}} \frac{\delta S}{\delta g^{\mu \nu}}$$
Main Question or Discussion Point
Tong proposes the following exercise in this lecture (around 25:30, section b)):
Exercise statement: Prove that the stressenergy tensor is given by the functional derivative of the action with respect to ##\delta g^{\mu \nu}##
$$T_{\mu \nu} = \frac{2}{\sqrt{g}} \frac{\delta S}{\delta g^{\mu \nu}}$$
I've been trying to solve it based on Tong's suggestion: using what he calls 'another cute trick' (I attached the PDF, owned by the University of Cambridge, UK).
After being stuck in it for a while I found a method in Carroll's book, which is as follows:
He takes the action for the KleinGordon scalar field ##\phi## (EQ 4.52 in Carroll's book)
$$S_{\phi} = \int \Big( \frac 1 2 g^{\mu \nu} (\nabla_{\mu} \phi)(\nabla_{\nu} \phi)V(\phi) \Big) \sqrt{g} d^n x$$
At this point, instead of varying the action with respect to ##\phi## we do so with respect to ##g^{\mu \nu}##. We get (the following EQs are 4.77 and 4.78 in Carroll's book):
$$\delta S_{\phi}= \int d^n x\Big[ \sqrt{g} \Big(  \frac 1 2 \delta g^{\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi \Big) + \delta \sqrt{g} \Big( \frac 1 2 g^{\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi V(\phi) \Big) \Big]$$
At this point we do some relabelling and we apply ##\delta \sqrt{g}=\frac 1 2 \sqrt{g} g_{\mu \nu} \delta g^{\mu \nu}## (please let me skip the proof of such a formula; it is explained in Carroll's book, page 163) to get
$$\delta S_{\phi}=\int d^n x \sqrt{g} \delta g^{\mu \nu} \Big[ \frac 1 2 \nabla_{\mu} \phi \nabla_{\nu} \phi + \Big(\frac 1 2 g_{\mu \nu} \Big) \Big( \frac 1 2 g^{\rho \sigma} \nabla_{\rho} \phi \nabla_{\sigma} \phi V(\phi) \Big) \Big]$$
From which we obtain the desired formula.
I have some questions:
1) Is this way of deriving the stressenergy tensor formula analogous to applying the 'another cute trick'?
2) It looks to me that everything boils down to seeing that the trick is varying the action with respect to the inverse metric instead of doing so with respect to ##\phi##. This doesn't look intuitive to me. Is there a reason one could come up with the idea or it is simply a matter of trying and trying till one gets it?
3) Is there any more illuminating method out there?
Any help is appreciated.
Thank you
Exercise statement: Prove that the stressenergy tensor is given by the functional derivative of the action with respect to ##\delta g^{\mu \nu}##
$$T_{\mu \nu} = \frac{2}{\sqrt{g}} \frac{\delta S}{\delta g^{\mu \nu}}$$
I've been trying to solve it based on Tong's suggestion: using what he calls 'another cute trick' (I attached the PDF, owned by the University of Cambridge, UK).
After being stuck in it for a while I found a method in Carroll's book, which is as follows:
He takes the action for the KleinGordon scalar field ##\phi## (EQ 4.52 in Carroll's book)
$$S_{\phi} = \int \Big( \frac 1 2 g^{\mu \nu} (\nabla_{\mu} \phi)(\nabla_{\nu} \phi)V(\phi) \Big) \sqrt{g} d^n x$$
At this point, instead of varying the action with respect to ##\phi## we do so with respect to ##g^{\mu \nu}##. We get (the following EQs are 4.77 and 4.78 in Carroll's book):
$$\delta S_{\phi}= \int d^n x\Big[ \sqrt{g} \Big(  \frac 1 2 \delta g^{\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi \Big) + \delta \sqrt{g} \Big( \frac 1 2 g^{\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi V(\phi) \Big) \Big]$$
At this point we do some relabelling and we apply ##\delta \sqrt{g}=\frac 1 2 \sqrt{g} g_{\mu \nu} \delta g^{\mu \nu}## (please let me skip the proof of such a formula; it is explained in Carroll's book, page 163) to get
$$\delta S_{\phi}=\int d^n x \sqrt{g} \delta g^{\mu \nu} \Big[ \frac 1 2 \nabla_{\mu} \phi \nabla_{\nu} \phi + \Big(\frac 1 2 g_{\mu \nu} \Big) \Big( \frac 1 2 g^{\rho \sigma} \nabla_{\rho} \phi \nabla_{\sigma} \phi V(\phi) \Big) \Big]$$
From which we obtain the desired formula.
I have some questions:
1) Is this way of deriving the stressenergy tensor formula analogous to applying the 'another cute trick'?
2) It looks to me that everything boils down to seeing that the trick is varying the action with respect to the inverse metric instead of doing so with respect to ##\phi##. This doesn't look intuitive to me. Is there a reason one could come up with the idea or it is simply a matter of trying and trying till one gets it?
3) Is there any more illuminating method out there?
Any help is appreciated.
Thank you
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