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Let's have interaction between gravitational and scalar real fields. For an action of gravitational field in vacuum I add term $S_{m} = \int d^{4}x\sqrt{-g}L_{m}$, where $$ L_{m} = \frac{1}{2}g^{\mu \nu}\partial_{\mu}\varphi \partial_{\nu} \varphi - V(\varphi ) $$ So the variation $\delta S_{m}$ must give $\frac{1}{2}\int d^{4}x \sqrt{-g}\delta (g^{\mu \nu})T_{\mu \nu}$, where $T_{\mu \nu}$ refers to the stress-energy tensor of scalar real field. I tried to do it, but this leads me to the answer $$ \delta S_{m} = \int d^{4}x\delta (\sqrt{-g})L_{m} + $$ $$ +\int d^{4}x \sqrt{-g}\left(\frac{1}{2}\delta (g^{\mu \nu})\partial_{\mu} \varphi \partial_{\nu} \varphi - g^{\mu \nu} \partial_{\mu}\varphi \partial_{\nu} \delta \varphi - \frac{\partial V( \varphi )}{\partial \varphi }\delta \varphi\right)= $$ $$ =\frac{1}{2}\int d^{4}x\sqrt{-g}\delta (g^{\mu \nu})T_{\mu \nu} + \int d^{4}x\sqrt{-g}\left( g^{\mu \nu} \partial_{\mu}\varphi \partial_{\nu} \delta \varphi - \frac{\partial V( \varphi )}{\partial \varphi }\delta \varphi\right). $$ What to do with the second integral? Does it equal to zero according to Euler-Lagrange equation, or I can't say so?

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Just a quibble, but you should be using covariant derivatives in your action. It doesn't matter for first integrals, but if you want to integrate by parts and get the EOM for your klein-gordon field, you will have a much easier time of it if everything is covariant. –  Jerry Schirmer Dec 2 '13 at 22:02
    
@JerrySchirmer : yes, I'm agree. I used non-covariant derivative only because I worked with scalar function. –  Andrew McAddams Dec 3 '13 at 4:41

2 Answers 2

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The expression for $\delta S_m$ that you're expecting holds provided the variation you are performing is the variation with respect to the inverse metric only; there should be no $\delta\varphi$ terms. In other words; set $\delta\varphi = 0$, and you obtain the desired expression.

See, for example, Carroll Spacetime and Geometry p.164, he does the same computation and explicitly remarks

"Now vary this action with respect, not to $\phi$, but to the inverse metric..."

Generally speaking, in fact, the stress tensor is defined to be proportional to the functional derivative of the action with respect to the inverse metric; \begin{align} T_{\mu\nu} = -2\frac{1}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu\nu}} \end{align}

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This is the correct answer. –  Trimok Dec 2 '13 at 21:16

Variations of the action must be performed with respect the field of which you want to get the equations of motion. You have varied with respect the metric field and the $\phi$ field at the same time. What have you done is not strictly incorrect, since the variations are independent, but led you in confusion. In fact, you are not sure how to behave. However, you can apply Euler-Lagrange equation for the field $\phi$ and then reach the conclusion (as well as put $\delta \phi = 0$ since you are not varying with respect to $\phi$).

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Thanks for the answer. But I think that your second equality has wrong sign: $$ g_{\mu \nu}g^{\mu \nu} = 4 \Rightarrow \delta(g_{\mu \nu}g^{\mu \nu}) = 0 \Rightarrow g_{\mu \nu}\delta g^{\mu \nu} = -g^{\mu \nu}\delta g_{\mu \nu}. $$ –  Andrew McAddams Dec 2 '13 at 20:44
    
You're right, sorry. I have fixed it. –  Federico Dec 2 '13 at 21:12

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