Space-time metric in tensor form

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:

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

All together does it look like this:

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

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

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.