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I am a bit confused about the way that Lorentz invariance of the scalar product $A^\mu g_{\mu\nu}B^\nu$ is proved. Usually, the proof would go like this (see also e.g. this Physics SE question). The transformed product is $A^{'\mu}g_{\mu\nu}B^{'\nu} = g_{\mu\nu}\Lambda^{\mu} _{\ \ \ \rho}A^\rho\Lambda^{\nu} _{\ \ \ \sigma}B^\sigma = A^\rho g_{\rho\sigma}B^\sigma $, since we define Lorentz transformations by the condition $\Lambda^{\mu} _{\ \ \ \rho}g_{\mu\nu}\Lambda^{\nu} _{\ \ \ \sigma}=g_{\rho\sigma}$.

However, if we say that the metric $g_{\mu\nu}$ is a tensor, should not the transformation of the scalar product also involve a transformation of the metric, i.e. $A^{'\mu}g' _{\mu\nu}B^{'\nu}$? This quantity would be automatically invariant, for any kind of transformation, since lower indices transform with the inverse transformation. In other words, the quantity $A^\mu g_{\mu\nu}B^\nu$ would be automatically invariant, without any conditions on the transformation actually being a Lorentz transformation, since it is a fully contracted tensor. What am I missing here?

Qmechanic
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Quercus Robur
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    I think your question is conflating two coordinate-transforming issues. In matrix terms for brevity, after special relativity worked out the consequences of Lorentz ($\Lambda^Tg\Lambda=g$) transformations, general relativity looked at diffeomorphism ($\Lambda^Tg'\Lambda=g,,ds^2=g_{\mu\nu}dx^\mu dx^\nu=g'_{\mu'\nu'}dx^{'\mu'}dx^{'\nu'}$) transformations. This is part of the reason we write $g$ as $\eta$ in SR. – J.G. Mar 02 '23 at 11:31
  • Thanks for your comment! Would you agree though, that the invariance of the Minkowski metric $\eta$, if we say it is a tensor, has no bearing on the Lorentz invariance of $A^\mu \eta_{\mu \nu}B^\nu$, which is Lorentz invariant simply by virtue of being fully contracted? This is what I'm getting at with my question. The non-trivial statement is the invariance of the metric. The invariance of $A^\mu \eta_{\mu \nu}B^\nu$ however has nothing to do with the invariance of $\eta$. Surely $A^\mu f_{\mu \nu}B^\nu$ is also Lorentz invariant for an arbitrary tensor $f_{\mu \nu}$? – Quercus Robur Mar 02 '23 at 17:11

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Since the Minkowski metric is Lorentz invariant, both viewpoints can be taken. Nevertheless, your second argument also applies for more general coordinate transformations: it holds for the diffeomorphisms we use in General Relativity, for example.

Níckolas Alves
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  • Thanks for your answer! Ok, I agree that the Minkowski metric is Lorentz invariant (which is a non-trivial statement). Would you agree though, that the invariance of the Minkowski metric, if we say it is a tensor, has no bearing on the Lorentz invariance of $A^\mu \eta{\mu\nu} B^\nu$, which is Lorentz invariant simply by virtue of being fully contracted? – Quercus Robur Mar 02 '23 at 16:03
  • @QuercusRobur Both arguments are correct. I personally prefer your version, which is also the one we would use in GR. I find it way more intuitive. However, it is also possible to argue using the "Lorentz trick". I don't see it as a wrong argument. – Níckolas Alves Mar 02 '23 at 17:21
  • Your argument is also way more general. For example, it shows that $A^{\mu} \eta_{\mu\nu} B^{\nu}$ is still invariant under other changes of coordinates that are not Lorentz transformations. In this sense, I do agree it gets more to the bottom of things, so to speakl – Níckolas Alves Mar 02 '23 at 17:23
  • Alright, thanks! – Quercus Robur Mar 02 '23 at 17:54