Invariant line element in GR

Question

In my General Relativity course, we had argued that even in a curved metric space, light follows a trajectory that obeys $ds^2 = 0$. The reasoning we were given is as follows:

The line element in any metric tensor $g_{\mu\nu}$ is $$ds^2 = d\overline{x}^T g_{\mu\nu} d\overline{x}.$$

If the metric is symmetric, then it can be diagonalised by an orthogonal transformation $Q$ at every point. So if $\overline{x} = Q\overline{x'}$, then

$$ds^2 = d\overline{x'}^T g'_{\mu\nu} d\overline{x'},$$

where $g'$ is diagonal. If the eigenvalues ordered from top to bottom are $\lambda_i$, then we subject our space to a scaling transformation $$x'_i = \frac{x''_i}{\sqrt{|\lambda_i|}},$$ so that all our eigenvalues become $+1$ or $-1$ depending on their initial sign.

If our original metric was Lorentzian, then the result of our transformations is the Minkowski metric at the point.

We then argued that since $ds^2 = 0$ for light in the Minkowski metric and $ds^2$ is invariant under transformations, it must be so at every point in our original metric as well. I don't understand how the line element $ds^2$ is invariant under scaling transformations. Is it invariant under disproportionate scaling of coordinates?

If not, how do we prove that $ds^2 = 0$ for light in all metrics?

Answer 1

Note that will all the operations you have described, both the metric and the coordinates are changed. In particular, both are changed such that $ds^2$ is left unchanged. Let's look at this in detail starting from your rescaling: $$ x^\prime_i=\frac{x_i^{\prime\prime}}{\sqrt{\lambda_i}}. $$ (I will suspend the summation convention in what follows...coordinates should be specified with upper indices if we were using the summation convention anyway.) So we write $$ ds^2=\sum_{\mu\nu}g_{\mu\nu}^\prime dx^{\prime\,\mu}\otimes dx^{\prime\,\nu}=\sum_{\mu\nu}g^\prime_{\mu\nu}\frac{x^{\prime\prime\,\mu}\otimes x^{\prime\prime\, \nu}}{\sqrt{\lambda_\mu}\sqrt{\lambda_\nu}}. $$ Assuming we have diagonalized $g_{\mu\nu}^\prime$, as you have done, then we see the denominators combine to just be some $\lambda_\mu$, which divides the corresponding diagonal element of $g_{\mu\nu}^\prime$ and hence rescales the eigenvalues as you have described (the elements of a diagonal matrix are the eigenvalues).

You will note that in the above arguments we never changed $ds^2$, we just changed the variables we used to describe it.

As a side note, let me also point out that the diagonalization you have described is not quite right, though the end result remains the same. If you were to apply a transformation $x=Qx^\prime$, the metric would not transform as described. The simplest way to see this is to note that for the factors $dx$, we do not have $dx=d(Qx^\prime)=Qdx^\prime$ since the differential $d$ will generally hit the $Q$ as well since, unless we are actually in Minkowski space, $Q$ need not be a constant.