4
$\begingroup$

According to this question , I want to calculate $DE(g_{ij})\widetilde g_{ij}$. If I treat $g^{ij}$ as irrelevant variable ,then I can get the same result as picture below. But I doubt that $g^{ij}$ has some connection with $g_{ij}$, so in the calculation should consider $g^{ij}$.

Consider the linearization of $\frac{\partial}{\partial x^i}\{g^{kl}\frac{\partial}{\partial x^j}g_{kl}\}$ , let $$ H=\frac{\partial}{\partial x^i}\{g^{kl}\frac{\partial}{\partial x^j}g_{kl}\} $$ Then $$ DH(g_{ij})\widetilde g_{ij}=\frac{d}{ds}|_{s=0}H(g_{ij}+s\widetilde g_{ij}) \\ =\frac{d}{ds}|_{s=0}\{\frac{\partial}{\partial x^i} ((g^{kl}+\frac{1}{s}\widetilde g^{kl})\frac{\partial}{\partial x^j}(g_{kl}+s\widetilde g_{kl}))\} \\ =\frac{d}{ds}|_{s=0}(\frac{1}{s}\widetilde g^{kl})\frac{\partial}{\partial x^i} \frac{\partial}{\partial x^j}(g_{kl}) + ... $$ In above calculation ,I am not sure it should be $\frac{1}{s}\widetilde g^{kl}$ or $s\widetilde g^{kl}$ in the term of $\widetilde g^{kl}$.But obviously, there are 2 order term different to the results in picture below.

enter image description here

$\endgroup$
  • $\begingroup$ Note that $(g + s\tilde{g})^{-1} \not = g^{-1} + s^{\pm 1} \tilde{g}^{-1}$ for both $\pm$. Therefore, your calculation is wrong. $\endgroup$ – mcd Oct 27 '16 at 21:56
  • $\begingroup$ @mcd I know there are some wrong , but it is important there will has 2 order term . $\endgroup$ – lanse7pty Oct 29 '16 at 2:28
2
$\begingroup$

You are right that one has to consider the derivatives of $g^{kl}$. The text only computes the second order terms. So if you have a derivative falling on some $g^{kl}$ term then the order is 1, because the Christoffel-symbols do not involve derivatives of $g^{kl}$.

By Jacobi's formula (cf. https://en.wikipedia.org/wiki/Jacobi%27s_formula ) we have that $\mathrm{tr} (A(x)^{-1} \partial_x A(x)) = \partial_x \log \det A(x)$. This implies that $H(g) = \partial_{ij} \log \det g$. Thus, $$\partial_s H(g + s\tilde{g})|_{s=0} = \partial_{ij} \mathrm{tr}(g^{-1}\tilde{g}).$$

You can also calculate $\partial_s (g + s\tilde{g})^{-1}|_{s=0} = -g^{-1}\tilde{g}g^{-1}$ and use this to calculate the terms that are not a trace.

$\endgroup$
  • $\begingroup$ Thanks ,I have edit my question , there are 2 order term different to the result of picture in my calculation. Could you give a detail process of derivative falling on $g^{kl}$ term ? $\endgroup$ – lanse7pty Oct 27 '16 at 2:49

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.