# Confusion about Lorentz invariance of scalar product

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?

• 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.
Commented Mar 2, 2023 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 $f_{\mu \nu}$? Commented Mar 2, 2023 at 17:11

• 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? Commented Mar 2, 2023 at 16:03
• 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 Commented Mar 2, 2023 at 17:23