2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Could you explain what you did in the second way, to get (2)? $\endgroup$ – Siva Dec 25 '13 at 23:44
  • $\begingroup$ @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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.