Problem 2.3b2

Prove that a metric tensor is symmetric.

Assume $\displaystyle{\eta_{\alpha\beta} \neq \eta_{\beta\alpha}}$ . Because it’s irrelevant what letter we use for our indices,

$\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}$ .

Then

$\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} (\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}$

So only the symmetric part of $\displaystyle{\eta_{\alpha\beta}}$ would survive the sum. As such we may as well take $\displaystyle{\eta_{\alpha\beta}}$ to be symmetric in its definition.

— edited Jun 15 ’15 at 22:48

— rob

— answered Jun 15 ’15 at 17:52

— FenderLesPaul

— Why is the metric tensor symmetric?

— Physics StackExchange

$\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}$

means that

$\displaystyle{\sum_{\alpha, \beta} \eta_{\alpha\beta}dx^{\alpha}dx^{\beta}=\sum_{\alpha, \beta}\eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}$

So in

$\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}$ ,

we cannot cancel out $\displaystyle{dx^{\alpha}dx^{\beta}}$ on both sides. In other words, we do NOT assume that $\displaystyle{\eta_{\alpha\beta} = \eta_{\beta\alpha}}$ in the first place.

means that

$\displaystyle{\sum_{\alpha, \beta}\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}\sum_{\alpha, \beta}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} \sum_{\alpha, \beta}(\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}$

3. “… only the symmetric part of $\displaystyle{\eta_{\alpha\beta}}$ would survive the sum” means that only the sum $\displaystyle{\left(\eta_{\alpha\beta} + \eta_{\beta\alpha}\right)}$ is physically meaningful.

— Me@2020-08-14 03:34:05 PM

Physics Town

continues

Problem 2.3b2

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