Perpetual Student:

W.D.Clinger:...

At the event horizon, therefore, dr=0 (for outgoing light) or dr=-2 dτ (for ingoing light). The coordinate speed of outgoing light is dr/dτ=0, and the coordinate speed of ingoing light is dr/dτ=-2.

Outgoing light isn't making any outward progress at all. Looks like a black hole to me.

Since I am unable to follow the mathematical argument here, please indulge me to ask a question in (non-mathematical) layman's terms.

**Ziggurat** answered your non-mathematical (scientific) question.

The trouble you're having with the mathematical argument is likely to be my fault.

The steps of that argument go as follows:

- The Painlevé-Gullstrand metric is equal to the familiar Schwarzschild metric outside the event horizon.
That's exercise 25 in the "Black holes" thread. Given exercise 24, the proof of exercise 25 is just a trivial calculation using high school algebra.

- The r, θ, and φ coordinates of the Painlevé-Gullstrand metric are identical to the r, θ, and φ coordinates of the Schwarzschild metric (by exercise 25).
- The Painlevé-Gullstrand metric satisfies Einstein's field equations for empty space.
(I didn't prove that, because the proof is both tedious and uncontroversial. Even Crothers agrees that the familiar Schwarzschild metric satisfies Einstein's field equations for empty space. Combined with exercise 25, that implies that the Painlevé-Gullstrand metric satisfies Einstein's field equations outside the event horizon. The coefficients of the metric are analytic in the coordinates, so the Painlevé-Gullstrand metric must satisfy the field equations at the event horizon as well. That's all we need for the following steps.)

- According to general relativity, light follows null geodesics.
- For null geodesics, ds=0.
- For radial geodesics, which are directed entirely toward or entirely away from the center of spherical symmetry, the conventional spherical coordinates θ (elevation) and φ (azimuth) don't change.
- For radial geodesics, therefore, dθ=dφ=0.
- At the event horizon, r=2m so β=1.
- Substituting ds=dθ=dφ=0 and β=1 into the Painlevé-Gullstrand metric, we find that null radial geodesics at the event horizon satisfy
- That means one of the two factors must be zero.
- That means dr=0 or dr=-2dτ.
- That means dr/dτ=0 or dr/dτ= - 2.
- That means the world line for inwardly directed radial light has coordinate velocity dr/dτ= - 2 at the event horizon. (The radial coordinate for inwardly directly radial light at the event horizon is decreasing as the time coordinate τ advances into the future.)
- That means the world line for outwardly directed light has coordinate velocity dr/dτ=0 at the event horizon. (The radial coordinate for outwardly directly radial light at the event horizon isn't changing.)
- That means a photon that's emitted radially outward at the event horizon never escapes beyond the event horizon. It stays forever at the event horizon.
- That also means a photon that's emitted radially inward at the event horizon proceeds rapidly toward the central singularity, contra Crothers and
**Farsight**.

If you don't understand how one of those steps follows from preceding steps, or you don't understand how one of those steps is derived, please let me know which steps require elaboration.