From special relativity, we know that the infinitesimal spacetime interval $ds^2=0$ for lightlike paths. Extending to general relativity, this still holds true if we think of $ds^2$ as the distance between two points on the manifold.

However, in Sean Carroll's GR book it was emphasized that $$ds^2=g_{\mu\nu}dx^\mu dx^\nu$$ is really the metric tensor $\textbf{g}$ in component basis form, i.e. $ds^2\equiv\textbf{g}=g_{\mu\nu} (dx^\mu\otimes dx^\nu)$.

If we see $ds^2$ as the metric tensor, how can we come to the conclusion that the metric tensor is zero for all paths travelled by light?

  • $\begingroup$ $ds^2$ is not the metric tensor, that's the line-element. $g_{\mu\nu}$ is the metric tensor. $\endgroup$
    – robphy
    Mar 16, 2021 at 3:29
  • $\begingroup$ @robphy $g_{\mu\nu}$ is the component of the metric tensor. The author says $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ is the metric tensor. On pg.71 it was said that "$dx^\mu$ in the line element is really a basis dual vector". $\endgroup$
    – TaeNyFan
    Mar 16, 2021 at 3:46
  • $\begingroup$ Understanding the difference between timelike and spacelike separations $\endgroup$
    – mmesser314
    Mar 16, 2021 at 3:51
  • $\begingroup$ The first sentence of 2.5 says "The metric tensor is such an important object in curved space that it is given a new symbol, $g_{uv}$". The paragraph including (2.40) says "...introduced the line element as $ds^2=\eta_{\mu\nu}dx^\mu dx^\nu$..." (Unfortunately, it then says "it becomes natural to use the terms "metric" and "line element" interchangeably. This is loose talk... but note this last sentence does not say "metric tensor".) $\endgroup$
    – robphy
    Mar 16, 2021 at 4:19
  • $\begingroup$ @robphy In section 2.6, he treats $ds^2=g_{\mu\nu} dx^\mu dx^\nu$ as the metric tensor and used it to act on two vector arguements. $\endgroup$
    – TaeNyFan
    Mar 16, 2021 at 4:23

1 Answer 1


Sadly this is overloaded notation and $ds^2$ means something different in those two cases.

  1. Very often, as you point out, $ds^2$ is a symbol that refers to the metric tensor, so $ds^2=g_{\mu\nu} {\rm d}x^\mu dx^\nu$, where ${\rm d}x^\mu$ is a set of basis one forms.

  2. When people write $ds^2=0$ to refer to a null geodesic, what they really mean is that there is a curve $x^\mu(\lambda)$ parameterized by $\lambda$ which satisfies \begin{equation} g_{\mu\nu} \frac{{\rm d}x^\mu}{{\rm d}\lambda} \frac{{\rm d}x^\nu}{{\rm d}\lambda} = 0 \end{equation} The "$ds^2$" in $ds^2=0$ in this usage refers to the left hand side of this equation, multiplied by some finite by small change in the parameter $(\Delta \lambda)^2$. In other words, $ds^2$ here is referring to the spacetime interval between two arbitrarily close points (so that we can think of them as both being in the same tangent space).

Both are really short hand notations. The first is more often than not what is meant in advanced texts and papers in theoretical physics, whereas the second one is typically used in special relativity textbooks.

  • $\begingroup$ Sean Carrol uses $ds^2=0$ as a tensor equation, where $ds^2$ is the metric tensor (section 2.6, pg 76-77). What puzzles me is that even if we know the line element $ds^2=0$, why does this imply that the metric tensor $ds^2=0$? $\endgroup$
    – TaeNyFan
    Mar 16, 2021 at 3:55
  • $\begingroup$ @TaeNyFan The metric tensor is not zero. Indeed this cannot be the case because it would correspond to a singularity -- for example, the metric $ds^2=0$ in cosmology (FRW-spacetimes) occurs at the Big Bang singularity, where the Riemann tensor is infinite. But you know that null geodesics exist in Minkowski space, which is not singular. The statement is that the spacetime interval along null geodesics is zero, that is what is given by point 2 in my answer. It is unfortunate that Carroll uses $ds^2$ in these two different ways without being clear about the distinction. $\endgroup$
    – Andrew
    Mar 16, 2021 at 12:28

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