Problem 2.3b2

Prove that a metric tensor is symmetric.

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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

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— Why is the metric tensor symmetric?

— Physics StackExchange

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1.

\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.

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2.

\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}}

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}}

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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

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2020.08.14 Friday (c) All rights reserved by ACHK