1
$\begingroup$

I know how to prove e.g. $$A^{ik}B_{lk}=A_{k}^iB^{k}_l.\tag{1}$$ (Raising and Lowering Indices Question). Today in a book, I find: $$g^{ik}\delta g_{lk}=-g_{kl}\delta g^{ki}.\tag{2}$$

$g^{ik}$ is the metric tensor. There is a negative sign. If I use the regular method to raise/lower the indices, I cannot get the negative sign. I guess this must be due to the variation $\delta$? Do you know how to prove this?

It then says because of the aforementioned equation, therefore:

$$T^{ik}\delta g_{ik}=-T_{ik}\delta g^{ik}.\tag{3}$$

$T^{ik}$ is the energy momentum tensor. Why they have such relationship?

$\endgroup$
  • $\begingroup$ The crucial observation here is that the object $\delta g^{ij}$ is not the 'raised index version of $\delta g_{ij}$', but rather the variation of the inverse metric $g^{ij}$. $\endgroup$ – gj255 Oct 30 '16 at 17:37
5
$\begingroup$

Remember that the metric tensor $g_{\rho\nu}$ and its inverse $g^{\rho\nu}$ fulfill the relation

$$ g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\nu.$$

In the above equation, if you take the variation on both sides you get

$$ \delta g^{\mu\rho}g_{\rho\nu}+g^{\mu\rho}\delta g_{\rho\nu}=0.$$

From this you get

$$g^{\mu\rho}\delta g_{\rho\nu}=-g_{\rho\nu}\delta g^{\mu\rho}.$$

Which is what you have in Eq. $(2)$. Equation $(3)$ follows from this.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thank you and sorry, I had typo in (3) before. I just corrected. And based on your proof, I can further prove: $$T_{ik}\delta g^{ik}=T^{nm}g_{ni}g_{mk}\delta g^{ik}=-T^{nm}g_{ni}g^{ik}\delta g_{mk}=-T^{nm}\delta^k_n\delta g_{km}=-T^{km}\delta g_{km}=-T^{ik}\delta g_{ik}$$ $\endgroup$ – HYW Oct 29 '16 at 16:36
  • $\begingroup$ That's it. Very good. $\endgroup$ – Apogee Oct 29 '16 at 16:51
  • $\begingroup$ Perfect answer. $\endgroup$ – Nogueira Oct 30 '16 at 17:44

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.