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I've been trying to learn how to multiply two tensors in order to go from $$g_{\mu\nu} dr^\mu dr^\nu$$ to $$c^{2}\,dt^2-dx^2-dy^3-dz^2$$ But I can't figure it out. $g_{\mu\nu}$ is a $4\times4$ matrix, and after the tensor multiplication you get a scalar.

Can someone explain all the steps to me?

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  • $\begingroup$ This link helps einsteinrelativelyeasy.com/index.php/special-relativity/… $\endgroup$
    – Sancol.
    Feb 23 at 18:37
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    $\begingroup$ That’s not really tensor multiplication; it’s (double) contraction. When you multiply tensors (without contracting) you end up with more indices (a higher-rank tensor). Here you end up with fewer: a scalar. $\endgroup$
    – Ghoster
    Feb 23 at 18:40
  • $\begingroup$ If you try to think in terms of matrices and row or column vectors, what will you do when you encounter a tensor with three or more indices? $\endgroup$
    – Ghoster
    Feb 23 at 19:04
  • $\begingroup$ Keep in mind $g_{\mu\nu}, g^\mu_\nu, $ and $g^{\mu\nu}$ have different matrix representations. So you could get the wrong results if you use a matrix to do it and choose the wrong one. $\endgroup$
    – R. Romero
    Feb 24 at 9:20
  • $\begingroup$ Do you know the difference between one forms and vectors? It's very important to get the matrix representations of the contractions right. Often you have to actually use a transpose or the inverse of a transpose to get the matrix multiplication approach to work right. $\endgroup$
    – R. Romero
    Feb 25 at 6:27

2 Answers 2

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Perhaps you are unaware of the use of Einstein notation for brevity. In this case, we wish to perform a scalar contraction: $$\sum_{\alpha, \beta}g_{\alpha \beta} \ dx^\alpha dx^\beta = \sum_{\alpha} \sum_{\beta}g_{\alpha \beta} \ dx^\alpha dx^\beta $$

Based on the result you want, the metric signature is $(+, -, -, -)$ ; so the corresponding metric tensor is: $$g = \begin{pmatrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\0 & 0 & 0 & -1\end{pmatrix}$$

The off diagonal terms are all zero, so $g_{ij} = 0 \ \forall \ i \ne j$ $$\sum_{\alpha} \sum_{\beta}g_{\alpha \beta} \ dx^\alpha dx^\beta = g_{00} \ dx^0 dx^0 + g_{01}(...) + g_{02}(...) + g_{03}(...) \\+g_{10}(...) + g_{11} \ dx^1 dx^ 1 + g_{12}(...) + g_{13}(...) \ + \ [...]$$

It is clear that this reduces to: $$\sum_{\alpha} \sum_{\beta}g_{\alpha \beta} \ dx^\alpha dx^\beta = g_{00} \ dx^0 dx^0 + g_{11} \ dx^1 dx^1 + g_{22} \ dx^2 dx^2 + g_{33} \ dx^3 dx^3$$

Substitute $g_{00} = 1,\ g_{11} = -1,\ g_{22} = -1,\ g_{33} = -1$ and with $dx^0 = cdt,\ dx^1 = dx,\ dx^2 = dy,\ dx^3 = dz$, you get your desired result: $$\boxed{\sum_{\alpha} \sum_{\beta}g_{\alpha \beta} \ dx^\alpha dx^\beta = c^2dt^2 - dr^2}$$

Hope this helps.

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    $\begingroup$ That was exactly what I was looking for. Thank you. $\endgroup$
    – lee pappas
    Feb 23 at 19:29
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The expression $g_{\mu\nu} dr^\mu dr^\nu$ is shorthand for the following summation: $$g_{\mu\nu} dr^\mu dr^\nu=\sum_{i=0}^3\sum_{j=0}^3\ g_{ij}\ dr_i \ dr_j$$ where the expression on the right is written using more standard matrix indexing notation.

With $$g_{\mu\nu}=\left( \begin{array}{cccc} -c^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$ and $$dr^\mu = \left(dt,dx,dy,dz \right)$$ we simply apply the above summation formula to obtain $$g_{\mu\nu} dr^\mu dr^\nu=-c^2 \ dt \ dt + 0 \ dt \ dx + ... + 1 \ dx \ dx \ + ... + 1 \ dy \ dy + ... + 0 \ dy \ dz + 1 \ dz \ dz$$$$g_{\mu\nu} dr^\mu dr^\nu=-c^2 \ dt^2 + dx^2 + dy^2 + dz^2$$

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