# Question about “different” equations of motion in dependence of indices

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?

• Could you explain what you did in the second way, to get (2)? – Siva Dec 25 '13 at 23:44
• @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 $\delta h^{\mu \sigma}$. – Andrew McAddams Dec 25 '13 at 23:53

If for any symmetric variation $\delta h^{\mu\nu}$, we have $$\delta h^{\mu\nu} A_{\mu\nu} = 0$$ 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 $A$, $$\delta h^{\mu\nu} (A_{\mu\nu} - A_{\nu\mu} ) = 0$$ The variation equation only makes the symmetric part of $A$ to be zero, $$A_{\mu\nu} +A_{\nu\mu} = 0$$
You choose the following variation to prove it $$\delta h_{\alpha \beta} = \begin{cases} \begin{array}{cc} \epsilon & \text{if } \alpha =\mu, \beta = \nu \text{ or } \alpha = \nu, \beta = \mu \\ 0 & \text{otherwise} \\ \end{array} \end{cases}$$