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The Schwarzchild spacetime has Killing vectors that result in the existence of two conserved quantities for test particles, interpreted as energy per unit mass and angular momentum per unit mass. For a massless particle, these are both infinite, but we can talk about their finite ratio $L/E=r^2A^{-1}d\phi/dt$. Here the units are chosen so that the Schwarzschild radius is 1, and $A=1-1/r$.

For $r>1$, we have $|L/E| \le rA^{-1/2}$, for straightforward reasons that are easy to understand by analogy with Newtonian gravity. For larger values of $L/E$, due to the centrifugal barrier, we get trajectories that never penetrate to values of $r$ this small.

What's confusing me is how to find a bound on $|L/E|$ for $r<1$, i.e., for a photon inside the horizon. Suppose I set up a tangent vector in $t-r-\phi$ coordinates, arbitrarily setting the $r$ component to $-1$, which I can do because the choice of an affine parameter is arbitrary, making it possible to rescale a tangent vector by any factor. If I express the $\phi$ component in terms of the other variables, I get a tangent vector $(\dot{t},-1,r^{-2}A(L/E)\dot{t})$, where the dot represents differentiation with respect to the affine parameter. If I now take the norm of this vector and set it equal to zero, I get $\dot{t}^{-2} = A^2(1-r^{-2}A(L/E)^2)$, which fixes the other two components of the vector. The compoments come out real, because $A<0$.

This all seems to work fine, suggesting that there is no bound on $L/E$ for $r<1$. But that seems to be a contradiction, because we could use the geodesic equation to extrapolate backward, out through the horizon, and then we would have an arbitrarily large value of $L/E$ at $r>1$ -- where we already know that there is a bound. (Of course the Schwarzschild coordinates break down at the horizon, but there is nothing in principle to stop us from switching to some other chart as we pass through the horizon, then switching back.)

Can anyone help me figure out what's wrong with my analysis?

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  • $\begingroup$ For sufficiently large $L/E$ there should be some back-reaction: one cannot have a photon with so much angular momentum that the total black hole plus photon would be a super-extremal Kerr hole when viewed from the outside. But this seems to be a different kind of constraint than the one in the question. $\endgroup$ – Anders Sandberg Jul 18 '18 at 6:46
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There is, indeed, no lower bound on $|L/E|$ for a photon inside the Schwarzschild black hole. To see that, we could simply note that it is possible to have photons with $E=0$ (and finite $L$) inside. The trajectory of such a photon in an equatorial plane would be given by constant Schwarzschild time $t=\mathrm{const}$ and equation $$ 0 = \frac{dr^2}{1-\frac{1}r}+r^2d\phi^2.$$ It is simply an equation for a cardioid that touches $r=1$. In the maximally extended manifold this trajectory starts in the past singularity, touches the horizon and ends in the future singularity. In spacetime diagram of Kruskal–Szekeres coordinates, the projection of this trajectory is simply vertical line passing through the origin ($X=0$ in wikipedia notation).

So the error of OP's analysis is the assumption that every null geodesic in the inside of Schwarzschild metric could be extended into $r>1$ region. In addition to the trajectory considered above, examples of nonextendable into $r>1$ region geodesics are given by null geodesics passing through the "antiworld" (the left half of K–S or Penrose diagram).

Incidentally, the equation $|L/E| \le rA^{-1/2} $ is not much of a "bound" for the outside trajectories by itself. By choosing the turning point of the trajectory very close to the horizon we could make $A$ arbitrarily small and thus the r.h.s. arbitrarily large. Trajectories that would saturate this bound only briefly emerge from the inside of black hole achieve some the maximum od radial coordinate slightly larger than $1$ and fall back inside.

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  • $\begingroup$ Thanks, this is very helpful. With hindsight, what I was really searching for was a characterization of photons that could reach $r$ from null infinity. I'll have to think about this some more, and see if I can come up with a better characterization. $\endgroup$ – Ben Crowell Jul 18 '18 at 12:21

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