Let's have the action $$ S = \int (\partial_{\mu}h^{\mu \sigma}\partial^{\nu}h_{\nu \sigma} - \Lambda h^{\mu \nu}T_{\mu \nu}) d^{4}x. $$ For definiteness, $$ h_{\mu \nu} = h_{\nu \mu} , \quad T_{\mu \nu} = T_{\nu \mu} \neq f(h_{\mu \nu}), \quad h^{\mu \nu} = \eta^{\mu \alpha}\eta^{\nu \beta}h_{\alpha \beta}, \quad \eta^{\mu \nu} = diag (1, -1, -1, -1). $$ Let's take the variation of $S$ by $h_{\mu \nu}$ and set the surface terms to zero: $$ \delta S = \int \partial_{\mu} \delta h^{\mu \sigma} \partial^{\nu}h_{\nu \sigma} d^{4}x + \int \partial_{\mu}h^{\mu \sigma}\partial^{\nu}\delta h_{\nu \sigma} d^{4}x - \Lambda \int \delta h^{\mu \nu} T_{\mu \nu}d^{4}x = $$ $$ = -\int (\partial_{\mu}\partial^{\nu}h_{\nu \sigma}\delta h^{\mu \sigma} + \partial_{\mu}\partial^{\nu}h^{\mu \sigma} \delta h_{\nu \sigma} - \Lambda \delta h^{\mu \sigma} T_{\mu \sigma})d^{4}x. $$ Then it's time to my stupid question. The expressions under the integral are scalar form and we may rename the indices. But when we take out the variation $\delta h^{\mu \sigma}$, all is changed. I can do that by the two "ways": $$ \delta S = \int \delta h^{\mu \sigma}(\partial_{\mu}\partial^{\nu}h_{\nu \sigma} + \partial^{\alpha}\partial_{\mu}h_{\alpha \sigma} - \Lambda T_{\mu \sigma})d^{4}x = \int \delta h^{\mu \sigma}(2\partial_{\mu}\partial^{\nu}h_{\nu \sigma} - \Lambda T_{\mu \sigma})d^{4}x \qquad (1) $$ and $$ \delta S = \int \delta h^{\mu \sigma}(\partial_{\mu}\partial^{\nu}h_{\nu \sigma} + \partial^{\alpha}\partial_{\sigma}h_{\alpha \mu} - \Lambda T_{\mu \sigma})d^{4}x . \qquad (2) $$ By setting variation to zero I obtain different equations of motion $(1), (2)$. Where did I make the mistake?
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$\begingroup$ Could you explain what you did in the second way, to get (2)? $\endgroup$ – Siva Dec 25 '13 at 23:44
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$\begingroup$ @Siva : I wrote $$ \partial_{\mu}\partial^{\nu}h^{\mu \sigma}\delta h_{\nu \sigma} = \delta h^{\alpha \beta}\partial_{\mu}\partial_{\alpha}h^{\mu}_{\beta} = \delta h^{\alpha \beta}\partial^{\mu}\partial_{\alpha}h_{\mu \beta}. $$ After that I only rename $$ \mu \to \alpha , \alpha \to \mu , \beta \to \sigma $$ for $(1)$ and $$ \mu \to \alpha , \alpha \to \sigma , \beta \to \mu $$ for $(2)$. Then I take out $\delta h^{\mu \sigma}$. $\endgroup$ – Andrew McAddams Dec 25 '13 at 23:53
This question is related to quadratic form.
If for any symmetric variation $\delta h^{\mu\nu}$, we have \begin{equation} \delta h^{\mu\nu} A_{\mu\nu} = 0 \end{equation} can we conclude $A_{\mu\nu} = 0$? The answer is NO.
The symmetric variation has only $n(n+1)/2$ degrees of freedom, a full matrix $A$ has $n^2$. The rest piece is the antisymmetric part of $A$. Notice that for any $A$, \begin{equation} \delta h^{\mu\nu} (A_{\mu\nu} - A_{\nu\mu} ) = 0 \end{equation} The variation equation only makes the symmetric part of $A$ to be zero, \begin{equation} A_{\mu\nu} +A_{\nu\mu} = 0 \end{equation}
You choose the following variation to prove it $$\delta h_{\alpha \beta} = \begin{cases} \begin{array}{cc} \epsilon & \text{if } \alpha =\mu, \beta = \nu \text{ or } \alpha = \nu, \beta = \mu \\ 0 & \text{otherwise} \\ \end{array} \end{cases} $$
Symmetrizing the integrand makes your two equations of motions the same. So it always a good habit to make the integrand in the variational action symmetric, if you have a symmetric variation.
