7

I was reading Schutz, A First Course in General Relativity. On page 9, he argued that the metric tensor is symmetric:

$$ ds^2~=~\sum_{\alpha,\beta}\eta_{\alpha\beta} ~dx^{\alpha}~dx^{\beta} $$ $\text{Note that we can suppose}$ $\eta_{\alpha\beta}=\eta_{\beta\alpha}$ $\text{for all}$ $\alpha$ $\text{and}$ $\beta$ $\text{since only the sum}$ $\eta_{\alpha\beta}+\eta_{\beta\alpha}$ $\text{ever appears in the above equation when}$ $\alpha\neq\beta$.

I don't understand his argument. If someone can explain why, I would really appreciate it.

Qmechanic
  • 201,751
snowball
  • 341
  • That question has some good answers already... – innisfree Jun 15 '15 at 18:02
  • 4
    I don't think this is a duplicate. The proposed dupe is asking if there are antisymmetric metrics; this question is asking how the quoted argument means that the metric is symmetric. – Kyle Kanos Jun 15 '15 at 18:37

2 Answers2

13

Assume $\eta_{\alpha\beta} \neq \eta_{\beta\alpha}$. Because it's irrelevant what letter we use for our indices, $$\eta_{\alpha\beta}dx^{\alpha}dx^{\beta}=\eta_{\beta\alpha}dx^{\beta}dx^{\alpha}.$$ Then $$\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2}(\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}$$ so only the symmetric part of $\eta_{\alpha\beta}$ would survive the sum. As such we may as well take $\eta_{\alpha\beta}$ to be symmetric in its definition.

rob
  • 89,569
FenderLesPaul
  • 1,086
  • 1
    what about the anti-symmetric contribution to e.g. $\eta_{ab} p^a q^b$ such that $p^a q^b \neq p^b q^a$ – innisfree Jun 15 '15 at 18:00
  • 1
    If you assume $\eta_{\alpha\beta}\ne\eta_{\beta\alpha}$ then you cannot claim the next relation in your answer. Because of the commutation between $dx^\alpha$ and $dx^\beta$, you have essentially said "$\eta_{\alpha\beta}$ is symmetric, therefore we can assume it is symmetric". Where have you established that the metric is equal to its symmetric components? – Jim Jun 15 '15 at 18:09
  • @Jims it says that only the symmetric part of $\eta_{ab}$ contributes in a contraction with a symmetric tensor $S^{ab}$ such as $S^{ab} = dx^a dx^b$. – innisfree Jun 15 '15 at 18:18
  • @innisfree But that should explicitly be stated. We aren't Math.SE. Identities like that can't be taken for granted – Jim Jun 15 '15 at 18:23
  • 1
    huh? i would have said, we aren't Math.SE so stuff like that can be taken for granted! – innisfree Jun 15 '15 at 18:28
  • There now both answers say exactly the same thing. Albeit with varying levels of technicality – Jim Jun 15 '15 at 18:40
  • 3
    What identity? I obviously just relabeled the indices. – FenderLesPaul Jun 15 '15 at 18:47
  • @FenderLesPaul Isn't that an identity? Ah who cares about terminology. Point is that making it clear made it better. You went up 4 points since I made that edit. Don't argue with good results – Jim Jun 15 '15 at 18:56
  • Sorry, I wasn't arguing, I just didn't know what exactly was being referred to. Regardless, thanks for pointing it out. – FenderLesPaul Jun 15 '15 at 18:57
  • In the last equality you swap ${\rm d}x^\beta$ and ${\rm d}x^\alpha$, I think @Jim is referring to this. – Alex Oct 04 '20 at 05:56
  • 1
    If your answer is correct, then any (0,2) tensor would be symmetric. – Alex Oct 04 '20 at 06:00
4

The metric tensor is created from the spacetime interval equation. On top of that, $[dx^\alpha,dx^\beta]=0$. Suppose we have a 1+1 dimensional spacetime, if you are given an interval equation resembling:

$$ds^2=-a\,dx_0^2+b\,dx_1^2+c\,dx_0dx_1$$

Obviously, $\eta_{0\,0}=-a$ and $\eta_{1\,1}=b$. Now, you can assume that, say, $\eta_{0\,1}=c/3$ and $\eta_{1\,0}=2c/3$ such that you still get the $c\,dx_0dx_1$ term, but why would you? You know $c=\eta_{1\,0}+\eta_{0\,1}$ and that's the only requirement you have. Since $dx_0$ and $dx_1$ commute, it is easier for everybody if you just say the tensor is symmetric and set $\eta_{1\,0}=\eta_{0\,1}=c/2$. Nothing changes except that you have saved yourself some trouble later on.

Jim
  • 24,458