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In space time metric in tensor form:

The distance is given by $$ds^2=c^2dt-dx^2-dy^2-dz^2$$

Which in tensor form is: $$ds^2=\sum_{\alpha \beta}g_{\alpha \beta}dx^\alpha dx^\beta$$

Using Einstein summation convention we have $$ds^2=g_{\alpha \beta}dx^\alpha dx^\beta$$

I was wondering what the explict form of this looks like:

I know that:

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I was wonder that $dx^\alpha$ and $dx^\beta$ look like?

All together does it look like this:

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(i know the above is wrong becuase it is invalid matrix)

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  • $\begingroup$ Please note that posting images of math is very strongly discouraged (even downvoted). Please use Mathjax for math. $\endgroup$ Nov 16, 2021 at 13:30

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If you use the fact that $g_{\alpha \beta} dx^{\alpha} = dx_{\beta}$ where $dx^{\alpha}$ is the column vector you have in your above expression, then $dx_{\beta}$ would be the row vector $(cdt,-dx,-dy,-dz)$. Then, the line element becomes $ds^{2} = dx_{\beta}dx^{\beta}$ which is a row vector multiplying a column vector, giving you a scalar and not a matrix.

Equivalently, you can view $g_{\mu \nu} dx^{\mu} dx^{\nu}$ as $\vec{dx}^{T} \cdot \textbf{g}\cdot \vec{dx}$ which gives a scalar and not a matrix.

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