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In almost all the books and lectures people compute $(ct)^2-(x)^2$ and show that it is a Lorentz invariant without any motivation to do. Or equivalently they state that Minkowski Metric is

$$\begin{pmatrix}1 & 0\\\ 0 & -1 \end{pmatrix}$$

But where does this come from?

GedankenExperimentalist
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  • In a comment to this answer to a similar question John Rennie wrote "Sadly, it was arbitrary and just worked. Well, no, it wasn't arbitrary. When Einstein wrote his original paper on special relativity he had to write time with an opposite sign to to space, but he didn't write the metric in the way we do these days. He was more driven by the physics. When Minkowski reformulated SR we got the metric with the signature we use today." – garyp Oct 12 '23 at 17:49
  • Possibly useful: My answer (Bondi motivates using "radar measurements") https://physics.stackexchange.com/a/508251/148184 to Minkowski Metric Signature. – robphy Oct 12 '23 at 17:50
  • @garyp I have written an answer below which tries to show that its not as arbitrary as it may seem – GedankenExperimentalist Oct 12 '23 at 17:53
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    Related and/or dupes of: https://physics.stackexchange.com/q/779000/25301, https://physics.stackexchange.com/q/112038/25301, https://physics.stackexchange.com/q/229864/25301 & probably others – Kyle Kanos Oct 12 '23 at 18:59
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    I've got a few books that cover SR left over from undergrad/grad school days and they all seem to motivate the metric from the constancy of the speed of light and a change between frames. Can you identify some of the "almost all books" you've looked at? – Kyle Kanos Oct 12 '23 at 19:01
  • @GedankenExperimentalist Really, the wording of the OP's question leaves room for interpretation. When I first read it, my take was "why is the universe the way it is?" That's what I had in mind by copying the quote. But I might have missed the mark. It might have been "what observations lead to the M. metric?" You answer that. Both bases covered. – garyp Oct 13 '23 at 11:23

1 Answers1

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A nice approach to derive this is to work in four-vector formalism from the start as Eigenchris does. Now, we want, by the orthogonality of time and space coordinates that $$\hat{e}_x\cdot\hat{e}_t=\hat{e}_t\cdot\hat{e}_x=0$$ and by definition of metric $\hat{e}_x\cdot\hat{e}_x=g_{xx},\hat{e}_t\cdot\hat{e}_t=g_{tt}, \hat{e}_x\cdot\hat{e}_t=g_{xt}=g_{tx}$.

We also want

$$\hat{e}_{x'}\cdot\hat{e}_{t'}=\hat{e}_{t'}\cdot\hat{e}_{x'}=0$$

Where prime refers to some other inertial coordinate with axes parallel to (ct,x) coordinates.

The bases vector transformation, which can be derived from Lorentz transformations gives:

$$\hat{e}_{x'}=\gamma(\hat{e}_{x}+\beta\hat{e}_{t})$$ $$\hat{e}_{t'}=\gamma(\hat{e}_{t}+\beta\hat{e}_{x})$$

Thus demanding $\hat{e}_{t'}\cdot\hat{e}_{x'}=0$ gives

$$\gamma^2((\hat{e}_{x}+\beta\hat{e}_{t}))\cdot(\hat{e}_{t}+\beta\hat{e}_{x})=0$$ $$-\hat{e}_{x}\cdot\hat{e}_{x}=\hat{e}_{t}\cdot\hat{e}_{t}=\varphi=g_{tt}$$

Therefore our metric becomes

$$\varphi\begin{pmatrix}1 & 0\\\ 0 & -1 \end{pmatrix}$$

Here we set $\varphi=1$ (or $-1$) for simplicity.

GedankenExperimentalist
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