Picked up a book on General Relativity for Mathematicians, but I'm a bit unclear on some of the tensor notation. For example, the Minkowski Metric

$$\eta_{\mu \nu} (\Delta x^\mu)(\Delta x^\nu)$$

Where $\eta_{\mu \nu}$ is the standard $4 \times 4$ identity matrix with an $-1$ in the (1,1) place.

I'm a bit confused as to how this notation works. Are $x^\mu$ and $x^\nu$ $4 \times 1$ matrices? (i.e. $\in \mathbb R^4$) If so, how does the product make sense? How is it then summed over to get

$$\eta_{\mu \nu} (\Delta x^\mu)(\Delta x^\nu) = -(\Delta x^0)^2 + \sum_{i=1}^{3} (\Delta x^i)^2?$$

Moreover, given a linear transformation $\Lambda$, how are

$$\Lambda^{\mu'}_\nu x^\nu$$ and $$x' = \Lambda x$$ the same?

Reference: http://arxiv.org/pdf/gr-qc/9712019.pdf

  • $\begingroup$ The notation $\Lambda_{\nu}^{\mu} x^{\nu}$ says that you get a vector out of it (since you have one non-contracted index) and that the $mu$th element is given by $\Lambda_{\nu}^{\mu} x^{\nu}$. This is just the usual matrix-vector product. $\endgroup$ – Cameron Williams Mar 13 '16 at 3:21
  • $\begingroup$ @CameronWilliams Wonderful. So then how does the metric notation work? $\endgroup$ – Anthony Peter Mar 13 '16 at 3:23

The book you are using employs the Einstein summation convention. This means that something of the form $f_{\mu}g^{\mu}$ is actually the same as $\sum\limits_{\mu=0}^{3}f_{\mu}g^{\mu}$. So, in $\eta_{\mu \nu} (\Delta x^\mu)(\Delta x^\nu)$, everything is being written as a scalar. In the Minkowski metric, this yields the summation you have written above.

For the second part, writing $x'=\Lambda x$ (regular matrix times vector multiplication) out component-wise yields $$x'=\Lambda x=\sum_{\nu} \Lambda^{\mu'}_{\nu}x^{\nu}=\Lambda^{\mu'}_\nu x^\nu$$

  • $\begingroup$ Ah! This was not clear from the text. It defined $\eta_{\mu \nu}$ equal to the matrix, not saying that $\eta_{ \mu \nu }$ meant the $(\mu, \nu)$ entry $\endgroup$ – Anthony Peter Mar 13 '16 at 3:28
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    $\begingroup$ @AnthonyPeter It's an abuse of notation. Differential geometry and special/general relativity are awful for this. $\endgroup$ – Cameron Williams Mar 13 '16 at 3:30
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    $\begingroup$ @CameronWilliams Really, physics in general. $\endgroup$ – Ben Sheller Mar 13 '16 at 4:37
  • $\begingroup$ @BenS. This is very true. $\endgroup$ – Cameron Williams Mar 13 '16 at 5:56

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