Falling angle of a photon near the event horizon

I have been doing simulations using Mathematica of light and matter paths (plotted in the $$r, \phi$$ Schwarzschild coordinates) around a black hole, in the Schwarzschild metric. This was in order to practice my learning of GR. I was able to accurately reproduce the Mercury perihelion shift and the light deviation close to the Sun.

One of the results I obtained is that a beam of light emitted perpendicular to the radial direction in the Schwarzschild metric, just under the photon sphere radius (in my example at (1.5 – 0.00001)$$\times R_s$$, spirals down to $$R_s$$ (Schwarzschild radius) and crosses the horizon at an angle of $$\sin^{-1}[2/\sqrt{27}]$$ above the horizon itself.

I made my simulation in units where $$c=1$$ and $$R_s=1$$. But I think this has no incidence on the falling angle.

On another hand, I have read after intensive search on this topic in this forum, that for a photon to escape the event horizon, when starting very close above $$R_s$$, it should necessarily be emitted radially, otherwise it will fall back onto the black hole.

My questions are :

• can my simulation be correct regarding the falling angle?
• and ok, what is the physical reason for the falling angle not being the same as the escaping one?

• Why do you think these two situations should be symmetric? If the photon is below the photon sphere and you've emitted it with no radial velocity it isn't the same as a photon being emitted purely radially above the Schwarzschild radius. Commented Nov 6, 2023 at 16:21
• In what coordinate system are you showing us your results. In Schwarzschild coordinates light does not cross the horizon (in either direction). Commented Nov 6, 2023 at 19:45
• As a new user, I am not sure how to answer to correctly to everyone ... so apologies in advance. I was thinking that the situation should be symmetric because my program basically use geodesics equations in Schwarzschild coordinates, and I was unable to produce a non symmetric trajectory for the light. Now I think I understand, that the situation is indeed symmetric in the coordinates system I was using, but the pure radial angle was associated to a stationary observer at the emission point near the horizon. Thank you for the time taken for the explanations. Daniel Commented Nov 7, 2023 at 18:30
• Schwarzschild coordinates are defined in terms of an observer "at infinity", where spacetime is flat. I have some relevant info here: physics.stackexchange.com/a/552874/123208 Note that the r coordinate of a point is defined in terms of the area (or circumference) of the sphere centred on the BH passing through that point. Commented Nov 8, 2023 at 1:15
• Sure, $t\to\infty$ as $r\to r_s$, but we can side-step that issue by eliminating $t$ and expressing the trajectory in terms of $\phi$ and $r$. (Actually, the equations are nicer if we use the parameter $u=r_s/r$). I have some examples & diagrams: physics.stackexchange.com/a/680961/123208 & physics.stackexchange.com/a/766180/123208 Commented Nov 8, 2023 at 1:29

If the plot is in Schwarzschild coordinates then it appears to be correct. Ity should also be symmetric in the sense that light geodesics are reversible - the light should be able to follow them in either direction. The confusion over the limiting angle at which light can escape is just due to confusion over in which frame of reference the angle is measured.

I think the maths is the following: The equations of motion written in Schwarzschild coordinates are $$\frac{dr}{dt} = \pm \frac{b}{r_s}\left(1 - \frac{r_s}{r}\right)\left( \frac{r_s^2}{b^2} - \left(1-\frac{r_s}{r}\right)\frac{r_s^2}{r^2}\right)^{1/2}\ ,$$ and $$r\frac{d\phi}{dt} = \frac{b}{r}\left(1 - \frac{r_s}{r}\right)\$$ where $$r_s$$ is the Schwarzschild radius and $$b$$ is the impact parameter of the light (the ratio of its angular momentum to its energy, as measured far from the black hole).

If we construct a right angled triangle then the angle $$\theta$$ the trajectory makes with the radial direction is given by $$\theta = \tan^{-1}\left[\frac{r d\phi/dt}{dr/dt}\right] = \tan^{-1}\left[\frac{r_s}{r}\left( \frac{r_s^2}{b^2} - \left(1-\frac{r_s}{r}\right)\frac{r_s^2}{r^2}\right)^{-1/2}\right]\ .$$

Thus if you have a trajectory with $$b$$ just larger than $$3\sqrt{3}r_s/2$$, then at the photon sphere we have $$r=1.5r_s$$ and $$\theta = \pm \pi/2$$. Then, when $$r=r_s$$, $$\theta = \tan^{-1}(3\sqrt{3}/2)=68.9^{\circ}$$ (and not $$\sin^{-1}$$?). This appears to match your plot and I see something similar using the GRorbits software (see below; note, the axes are in geometrised units where $$r_s=2M$$).

Starting just inside the photon sphere - the time-reversal of this would not escape.

Note that the time reversal of this trajectory (and the one you plotted), won't quite escape the black hole. A time-reversed escaping trajectory would start just above the photon sphere angled slightly inwards. Here is one I produced in GR orbits, but you can see that the angle at which it meets the event horizon is imperceptibly different.

Starting just outside the photon sphere - light would escape if time reversed

The notion that in order to escape, light must be emitted almost radially from just above the event horizon is not correct in Schwarzschild coordinates but it is correct when considered in the coordinate system of the local inertial frame of an observer hovering just above the event horizon. For the full details of that you can see https://physics.stackexchange.com/a/787543/43351

And I hope, to finally make this clear, I show one further plot from GRorbits. This shows the trajectory of light launched from $$r=1.001r_s$$ that just escapes from the black hole (with $$b = 2.5980r_s$$). The red arrow shows the direction in which this light is launched in the frame of reference of the observer at $$r=1.001r_s$$. You can see that this is almost radial (the angle is 4.7 degrees to the radial), but the angle that the light trajectory makes to the event horizon is almost identical to those shown in the pictures above and is distinctly non-radial.

Light launched (almost) radially from $$1.001r_s$$ that just escapes the black hole. The arrow shows the launch vector (4.7 degrees to the radial) in the local frame of reference of the light source.

• I think that there must be a $(g_{rr}) ^{-1/2}$ multiplier in $\tan^{-1}$ argument in your equation for $\theta$ because the proper radial distance is $(g_{rr}) ^{1/2}dr$. Commented Nov 7, 2023 at 10:50
• @A.V.S. but the plot is in Schwarzschild coordinates (I think), so isn't my angle correct in terms of a trajectory in a Schwarzschild $r, \phi$ diagram? Commented Nov 7, 2023 at 11:08
• Yes, and I agree with the last sentence of your answer in its current version. But the sentence before that raises doubts in me, since I would think that “being emitted radially” is a coordinate independent property. Commented Nov 7, 2023 at 11:31
• @A.V.S. I am busy for the next couple of hours but I am going to work it out in both coordinate systems. The OP's plot is correct and I confirmed using the GRorbits software. Looks like 69 degrees to me. It isn't exactly radial until $r=r_s$ but there can't be a shell observer at $r=r_s$. Commented Nov 7, 2023 at 11:37
• @A.V.S. see physics.stackexchange.com/a/787543/43351 Commented Nov 7, 2023 at 17:08

Q: "when starting very close above the horizon, it should necessarily be emitted radially, otherwise it will fall back onto the black hole."

This is correct.

Q: "what is the physical reason for the falling angle not being the same as the escaping one?"

Because you used

Q: "a beam of light emitted perpendicular to the radial direction"

so the result can't be radial, that would be the case if the light came from infinity to where the time reversed ray should escape to. Your ray started transversal below the photon sphere, not radial at infinity, so why would you expect something else in reverse?

Q: "can my simulation be correct regarding the falling angle?"

Seems about right, with my simulator and your initial conditions the result looks similar, although your ray crosses the horizon at an other position, it should be at $$\phi \approx 674°$$ but that might be a numerical issue on your end (I used 32 bit precision and implicit Runge Kutta with difference order 20):

The numbers show the crossing at $$\rm r = 2$$ where $$\rm \dot{r} \approx -1/\surd 3, \ \dot{\phi} \approx 3/4$$ $$\text{(G=M=c=1)}$$.

• This is correct.” - No, it is frame dependent, as already explained in the earlier answer and comments. Commented Nov 9, 2023 at 3:49
• @safesphere - If you know of some coordinates in which it isn't radial let me know which ones, until then I'll stand my ground. In all the coordinates I know of, a zero angular momentum worldline is radial, so I don't know what exotic frames you're talking about. Commented Nov 9, 2023 at 4:59
• I think your answer will confuse the OP even more. I agree that it does not matter if you transform the $t$ coordinate to plot the trajectory, but it does matter if you transform the spatial coordinates. You say "this is correct" but then don't explain to the OP why that statement can be true and why all the things you've said in comments can be true also. The fact that the OP has chosen a light ray that just wouldn't escape the black hole if reversed appears irrelevant to me. They could have plotted one starting just outside the photon sphere and it would look roughly the same. Commented Nov 9, 2023 at 8:19
• @Yukterez Why zero angular momentum? What are you talking about? The angle of the trajectory crossing the horizon is far from radial on your own diagram. And per ProfRob’s comment above, a minor increase of the original orbit would let the reversed ray escape without tangibly changing the angle. Also the question is about trajectories, not worldliness. You are confusing trajectories with geodesics. Commented Nov 9, 2023 at 16:50
• @safesphere asked: "Why zero angular momentum?" - I'm not talking about rotation around your own axis, I mean the orbital angular momentum. If you send a photon radially outwards while you cross the horizon it won't fall into the singularity, but if you give it an angle it will. Under Newton it doesn't matter in what direction you try to escape, but under Schwarzschild it depends: above the photon sphere rp=1.5rs the photon can escape in the transverse direction, but below rp the angle needs to get smaller, up to zero in the limit r→rs, where the escape vector needs to be radial. Commented Nov 9, 2023 at 17:56

Answering the question related to the falling angle of a photon in the event horizon: applying the Schwarzschild metric, if $$L_{impact}$$ is the angle between the event horizon and the trajectory of the photon entering, with $$b$$ the impact parameter, we have $$\tan L_{impact}=\frac{R_s}{b}$$. For $$b=b_{crit}=\frac{3\sqrt{3}}{2}R_s$$ (limit case for a photon to impact a black hole), $$L_{impact}=\arctan\frac{2}{3\sqrt{3}}$$ that is about $$21.1^\circ$$ or $$68.9^\circ$$ for the complementary angle.

The angle of emission is the same for the same value of $$b$$ : $$\tan L_{emission}=\frac{R_s}{b}$$, (which is to be expected under the principle of the inverse return of light).

Answering the question related to the exiting of the event horizon: if a light source is located immediately outside of the event horizon, a photon emitted by this source can escape the event horizon if the impact parameter $$b$$ is less than $$b_{crit}$$. This means that the photon can escape even if its trajectory is not radial ( $$b=0$$).