I like to know why photon with sufficient energy can avoid being pulled into the event horizon? Most light will get trapped and forms into photon sphere while some can escape, isn't light supposed to be geodesic also more energy does not equal more speed particularly in this case of light... so what's give?

I read in the Wikipedia article "Photon sphere:"

As photons approach the event horizon of a black hole, those with the appropriate energy avoid being pulled into the black hole by traveling in a nearly tangential direction known as an exit cone. A photon on the boundary of this cone does not possess the energy to escape the gravity well of the black hole. Instead, it orbits the black hole. These orbits are rarely stable in the long term.

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    $\begingroup$ Who says energy matters for light to escape an event horizon? Light will get trapped regardless of energy, no? $\endgroup$
    – Allure
    Commented Apr 19, 2019 at 8:54
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    $\begingroup$ I like to know why photon with sufficient energy can avoid being pulled into the event horizon? Where did you get the idea that this was true? Please edit the question to explain why you would think this. $\endgroup$
    – user4552
    Commented Apr 19, 2019 at 12:06

3 Answers 3


Thanks for editing the question to provide clarification. The material you quote from the Wikipedia article is simply wrong. Note that the article has a template at the top warning that "This article needs attention from an expert on the subject." The trajectory of a photon near a black hole has nothing to do with its energy. The geodesic equation has uniquely defined solutions for a given initial position and initial four-velocity. Two photons with different energies will still have the same trajectories if they start at the same position and with the same initial four-velocity.

I've edited out the incorrect material from the WP article, and left comments on the article's talk page.


The general relativity states that according to Schwarzschild metrics, the "fate" of a photon in respect of a black hole depends only on the mass of the black hole and one parameter $ b $ of positive or zero value known as the impact parameter, and such that: $$ b^2=\frac{c^2l^2}{\varepsilon^2} $$ with $ c $ speed of light in vacuum, $ l $ and $ \varepsilon $ respectively angular momentum and energy of the photon, which are invariants along its geodesic path.

Without taking into account the trivial case $ b=0 $ which means no deflection and a radial trajectory ($ l=0 $) to the event horizon, the critical value $ b_{crit} $ defined as: $$ b_{crit}=3\sqrt{3}\frac{GM}{c^2} $$ with $ G $ gravitational constant and $ M $ mass of the black hole, leads to three possible cases, for a photon coming from infinity:

$ 1 $ - if $ b>b_{crit} $, the photon is deflected and escapes the black hole,

$ 2 $ - if $ b=b_{crit} $, the photon is deflected and moves to an unstable orbit of radius $ \frac{3GM}{c^2} $ around the black hole (photon sphere),

$ 3 $ - if $ b<b_{crit} $, the photon is deflected and enters in the event horizon of the black hole.

So the photon will escape the black hole only in case $ 1 $ which means $ b^2=\frac{c^2l^2}{\varepsilon^2}>b_{crit}^2=27\frac{G^2M^2}{c^4} $ or $$ \frac{l^2}{\varepsilon^2}>27\frac{G^2M^2}{c^6} $$

In conclusion, the "fate" of the photon depends not only of its energy but depends on the ratio angular momentum / energy: the higher it is, the more chance the photon has of escaping the black hole.

Hoping to have answered your question,

Best regards.


The difference between escaping photons and trapped photons is the direction of their geodesic. If you consider a 4 dimensional blackhole,which really exists in our universe, the spacetime position of photon is represented by coordinates (t,r,θ,φ) where θ and φ express the angular position around the blackhole. And let's suppose photons are at the slightly outside of the event horizon.

A photon which travels in the purely radial direction (when time goes, r is increasing but θ and φ are unchanged) escape from the blackhole (because it is slightly outside the event horizon). But A photon which travels in the angular direction can stay with respect to radial position. This is unique phenomenon of Einstein gravity in some sense, because in Newtonian gravity it depends only on its energy whether a particles can escape from the gravity.(And I suppose this is why you mentioned the energy of light.) In Einstein gravity, photons go through the same null geodesic regardless of its energy, defined as initial condition of geodesic equation.

In the case of 2 dimensional blackhole, which is frequently used for simplicity, there is no angular direction, so there is no trapped photon except for the case that it is exactly on the event horizon. This is easily checked by writing Penrose diagram.

enter image description here

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    $\begingroup$ The OP's question is about energy. You haven't said anything about energy. $\endgroup$
    – user4552
    Commented Apr 19, 2019 at 12:07
  • $\begingroup$ @BenCrowell that might be because the answer to OP's question has nothing to do with energy as you correctly mentioned in your answer. I think Takumi gives a good explanation. $\endgroup$ Commented Apr 19, 2019 at 17:57

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