# Why is the Schwarzschild metric equal to 0 for light at the event horizon?

Carroll & Ostlie's Introduction to Modern Astronomy and Astrophysics provides a derivation on why light and information is frozen at the event horizon:

However I noticed that they equated $$ds=0$$ for light at the event horizon. But we know that the Schwarzschild metric is undefined at the event horizon (How to derive the Schwarzschild radius? , look at the answer with the most upvotes) because of a division by $$0$$. So qualitatively or quantitatively speaking, why can we say that $$ds=0$$ at the event horizon even though the Schwarzschild metric is undefined at that point (coordinate singularity)? Thank you!

It's true that there are null geodesics that stay permanently on the event horizon, but the derivation in the textbook is wrong. Schwarzschild coordinates don't cover the event horizon, so you can't use them to analyze what happens there. $$r=R_s$$ is, as you said, a coordinate singularity; neither light nor anything else can be frozen in time or do anything else at $$r=R_s$$, because it isn't actually a part of the manifold.

The answer to that other question (direct link) is also wrong: it says

This is a coordinate singularity, and that's what defines the event horizon.

but the event horizon most definitely is not defined as the place where there's a coordinate singularity in some silly man-made coordinate system. The event horizon is the boundary between the region from which you can reach future infinity and the region from which you can't, which is independent of coordinates.

The coordinate singularity problem is peculiar to Schwarzschild coordinates and is not shared by other popular coordinate charts for the Schwarzschild geometry, such as Eddington-Finkelstein infalling coordinates: $$ds^2 = (1-R_s/r)(c\,dt)^2 - 2 (R_s/r)\,c\,dt\,dr - (1+R_s/r)\,dr^2 - r^2 (dθ^2 + \sin^2 θ\,d\phi^2)$$

If you do the same calculation as the textbook, setting $$r=R_s$$, $$ds=0$$, and $$dθ=d\phi=0$$, you'll find $$dr/dt\in\{-c,0\}$$, indicating that outgoing null geodesics are "frozen" at this radius while ingoing null geodesics go right through it, as you'd expect of a one-way surface. $$r=R_s$$ in Eddington-Finkelstein coordinates actually is the event horizon. (In Schwarzschild coordinates you instead get $$dr/dt\in\{0\}$$, implying that ingoing and outgoing geodesics are both frozen and never separate even though they're moving away from each other at the speed of light, which is another sign that there's something wrong with those coordinates.)

• +1. Thanks for pointing a subtle point in the first paragraph. But in above Eddington Finkelstein, the coordinate velocity of all outgoing radial null geodesics at whatever r (not just at the horizon but at any r) is turning out to be zero. Why is that happening... why is the velocity of outgoing rays 0 at any r.. May 9 at 8:41
• @Shashaank It isn't, unless I wrote it down wrong. The speeds should be $-c$ and $c(1-R_s/r)/(1+R_s/r)$. May 9 at 15:39
• "The event horizon is the boundary between the region from which you can reach future infinity and the region from which you can't, which is independent of coordinates" could I clarify what you mean by "future infinity"? Thank you for your insightful response May 10 at 10:09

Think of plain old Minkowski spacetime in two dimensions, with line element $$ds= -c^2 dt^2 + dx^2$$. Since light is massless, its "proper time" as quantified by $$ds$$ is just zero. If we plug that into the line element, we get the equation $$(\frac{dx}{dt})^2 = c^2$$ which gives you the standard speed of light.

Now, this is exactly what they have done in the text, except for the Schwarzchild metric. Notice that the metric itself is undefined at the Schwarzchild radius (at least in these coordinates,) but the coordinate speed of light you get by solving for $$\frac{dr}{dt}$$ in the equation $$ds=0$$ is well-defined there, and it evaluates to zero.

• The coordinate speed that you get from this calculation makes no sense, though: the ingoing and outgoing speeds are both zero, implying that light beams moving in opposite directions never separate, which is just wrong. In coordinate systems that aren't singular at the horizon, you get different ingoing and outgoing speeds. May 9 at 6:58
• @benrg “the ingoing and outgoing speeds are both zero, implying that light beams moving in opposite directions never separate, which is just wrong” - Not wrong at all, but exactly what the entire universe observes from outside. May 16 at 3:48