# Are there actually any photons in orbit in the photon sphere of a black hole?

My intuition says 'no'. I know from my knowledge of spaceflight that every orbit includes the point at which delta-v was last spent; you can't use only a cannon on the ground to put something into a stable orbit because the last point delta-v was applied is on the ground.

I would think a similar limitation would apply to the photon sphere as well. Or am I way off and photons really can get stuck in the photon sphere?

• I'd guess that in between photons which narrowly escape and photons which narrowly fall in, are precise trajectories which orbit as many times as you want while they "decide" whether to escape or fall in, depending on how precise you can make their trajectory. Commented Mar 7 at 1:28
• Light has diffraction and cannot stay on any trajectory for too long. Commented Mar 7 at 7:22
• An interesting variant to this question, since I agree with all of the answers about external photons/particles, is whether the virtual pair production that separates by the event horizon could, either directly or through interactions, cause a lucky particle to orbit or a photon to be in the photon sphere. Commented Mar 7 at 22:44
• Also note that your stated knowledge of spaceflight does not involve general theory of relativity effects, which imply you won't follow a closed elliptical orbit as you might expect. Commented Mar 8 at 13:09
• @SethRobertson “the virtual pair production that separates by the event horizon” - As we observe from outside, nothing can ever cross the horizon. So no pair production can be separated by the horizon for as long as the external universe exists (forever). Commented Mar 8 at 23:49

Yes and no. ;)

In this answer, I'll only discuss photon trajectories in the Schwarzschild metric. Trajectories around Kerr black holes are more complicated. I'll use units such that the Schwarzschild radius equals $$1$$.

Schwarzschild photon trajectories can do things that aren't possible in Newtonian gravity. In particular, a fast body in Newtonian gravity gets deflected in a hyperbolic trajectory. However, in GR, a photon can make one or more loops around a black hole (or neutron star) before travelling on.

We can describe the trajectory in terms of the impact parameter $$b$$: the closest distance from the trajectory to the centre of the BH if there were no gravity.

The critical impact parameter is $$b_0 = 3\sqrt3/2 \approx 2.59807621$$. If $$b>b_0$$ then the photon is deflected. If $$b the photon will fall below the photon sphere and cross the event horizon. In either case, the photon can make a number of loops around the BH. The closer its $$b$$ is to $$b_0$$, the more loops it makes. So if $$b$$ is very close to $$b_0$$ it can make an arbitrarily large number of loops.

Here's a trajectory with $$b=3.62001264240871876$$, which gets deflected by $$60°$$.

The greenish circle at $$1.5$$ is the photon sphere. The radius of the dashed circle is $$b_0$$.

Here's a trajectory with $$b \approx 2.59807621533240768089$$, which gets deflected by $$1140°$$. So it's similar to the previous trajectory, except that it does 3 loops before escaping.

Both of these trajectories were calculated using Carlson's algorithm for evaluating elliptic integrals, with high precision arithmetic. (The dots on the trajectories are computed using elliptic integrals, the connecting curves are cubic Bézier curves).

I have further details and diagrams in this answer, and this more recent answer.

So on first analysis, a photon trajectory with impact parameter $$b=b_0$$ feeds into the photon sphere and stays there indefinitely. However, that orbit is unstable. The slightest perturbation will throw the photon out of the photon sphere. So indefinite photon sphere orbits are purely theoretical. They can only happen in a universe with no perturbing influences.

But on further analysis, the photon cannot orbit indefinitely even in those ideal conditions. The photon-black hole system radiates gravitational radiation, so the orbiting photon is doomed to fall out of the photon sphere. Of course, the power of that radiation is ridiculously tiny, even compared to the energy of a single photon. But it is non-zero, so we can't neglect it.

As I mention in the 1st linked answer above, in Divergent reflections around the photon sphere of a black hole, (published in Scientific Reports, a Nature Portfolio journal), Albert Sneppen gives a nice equation for the number of loops (originally given as Eq. 269 in Chandrasekhar's The Mathematical Theory of Black Holes). Let $$b=b_0+\delta$$. If this trajectory makes $$n$$ loops then the trajectory with $$b=b_0+\delta e^{-2\pi}$$ is almost identical except it makes $$n+1$$ loops. This applies to both trajectories with $$\delta>0$$, which escape, and to trajectories with $$\delta<0$$, which end up plunging into the event horizon. The trajectory with $$b=2.6013402014052278804$$ is deflected by almost exactly $$360°$$, i.e., it makes one loop before continuing on its original path.

• I think much more important than any radiation produced, is the fact that a wavepacket has a finite size (unless it has infinite energy, in which case the radiation produced become very relevant). In the short wave length limit, the time scale for the number photons on the light ring to decrease by a factor e is roughly $\sqrt{3}3GM/c^3$ this works out to roughly a tenth of a millisecond for a 4 solarmass BH. Commented Mar 6 at 16:21
• @TimRias Interesting! So in my notation, which I borrowed from Albert Sneppen's article, that mean lifetime is only $b_0/c$. Heuristically, I expected it to be a bit longer... Commented Mar 6 at 21:15
• Note that Sneppen's article was published in Scientific Reports not Nature (which would have been astonishing since there really isn't anything new in that article, it is nicely presented though.). Commented Mar 7 at 8:37
• Note that the formula you quote from that article is Eq. 269 in Chandrasekhar's The Mathematical Theory of Black Holes. Commented Mar 7 at 8:52
• Thanks, @Tim. Updated. Commented Mar 7 at 9:01

An intuitive reason why a spacecraft cannot enter into a circular orbit without accelerating itself is that a time-reversed trajectory is also a valid trajectory. Thus, if an unaccelerated spacecraft could enter a circular orbit, then an unaccelerated spacecraft already on a circular orbit could spontaneously depart from that orbit. We know this cannot happen: Newtonian circular orbits are stable.

However, photon orbits in the photon sphere are unstable, meaning that small deviations are exponentially amplified over time. So a photon on an (arbitrarily close to) circular orbit does indeed depart from that orbit, either escaping the system or falling into the black hole. And so, by time-reversing that picture, a photon arriving from far away can indeed enter into an (arbitrarily close to) circular orbit.

Of course, the instability of these orbits also means that a photon is highly unlikely to remain "stuck" in the photon sphere for long.

The photon sphere is a mathematical construct. Infinitely thin.

• If a photon dips below, it's gone. And if it just an epsilon outside while not going inwards, it will leave for good. The photon needs to be precisely on the sphere.
No uncertainty in height (=location) is allowed.
• Likewise, if the photon is moving downwards at the photon sphere, it's gone. And if it's just moving ever so slightly upwards, it'll leave for good. The photon needs to move precisely along the sphere.
No uncertainty in direction (=momentum) is allowed.

Now, photons are quantum objects, so photons have an uncertainty in location and momentum. So either a photon is not on the photon sphere, or it is not moving along the photon sphere. In either case, it's not going to stay there.

You can't use your Newtonian intuition. The effective potential of light in the Schwarzschild metric has a local maximum, which does not exist in Newtonian gravity. If light arrives at $$1.5r_s$$ with exactly the right ratio of energy to angular momentum such that it has no inward motion, then in principle it can enter a circular orbit.

The magic number is an impact parameter of $$3\sqrt{3}r_s/2$$ and the closer to this value, the more "orbits" the light will make before either falling inwards or heading outwards to infinity.

• Light arriving with the exact impact parameter is conceptually impossible. A light trajectory in a static spacetime is time reversible. If light arrives from outside of the photon sphere, then upon the reversal it would leave to the outside. And vice versa. In no case light would stay on the photon sphere even if neglecting diffraction. The only conceptual possibility for light to stay there is if this light originates on this sphere or changes the direction at the entry point, such as by reflecting off a mirror hovering exactly at the photon sphere. Commented Mar 8 at 23:42
• @safesphere You can get arbitrarily close to $1.5 r_s$ and execute an arbitrarily large number of orbits. Commented Mar 9 at 8:55
• The impact parameter is just a distance. In the geometric optics approximation, mathematically, nothing stops you from having the exact critical impact parameter. And yet this produces a physically meaningless outcome with a contradiction of escaping “after” the “infinite number” of loops that would take the infinite time to complete. In reality diffraction takes care of this dilemma. Commented Mar 9 at 17:25
• @safesphere write an answer, I won't be editing mine. Commented Mar 9 at 19:06