I am facing a conceptual difficulty while reading about black-hole(BH) horizons. Consider for simplicity a Schwarzchild(SW) BH, with a null hypersurface $\zeta$ denoting its horizon at $r=2M$ and let $g_{\alpha\beta}$ denote the metric. Finally suppose $K^{\alpha}$ is a Killing vector for this space. Then, we know that $C = g_{\alpha\beta}\dot{x}^\alpha K^\beta$ is a conserved charge/quantity of this space.

MY QUESTION: Since $g_{\alpha\beta}$ (in the $(t,r,\theta,\phi)$ coordinate) blows up at $r=2M$, I am having trouble understanding whether the quantity $C$ is defined on $\zeta$ i.e. is $C$ still conserved at the hypersurface $\zeta$?.

A mathematical approach would be better. If anyone can find relevant papers please point them out (I couldn't find any that discuss this issue at the horizon $\zeta$). Also it would be better if someone can generalise for a non-SW BH and answer the same question.

  • $\begingroup$ So as long as the horizon is not a physical singularity, I can still use the Killing fields to calculate $C$. Is this correct ? @JohnRennie $\endgroup$
    – Lelouch
    Jul 14, 2020 at 7:51

1 Answer 1


The quantity $C = g_{\alpha\beta}\dot{x}^\alpha K^\beta$ is a scalar invariant and is therefore independent of the coordinate system used.

Since coordinate systems exist that are not singular at the event horizon (for example Kruskal Szekeres) you could calculate $C$ at the horizon using these coordinates. The quantity is not affected by a change of coordinates.

This can be done at the horizon because it is only a coordinate singularity not a curvature singularity. At a curvature singularity all coordinate systems are singular.


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