My question is related to this older question at stackexchange - physics.

How much energy a photon needs to create a black hole ?

My question:

How close to the speed of light a gamma photon source has to move towards an observer such that the frequency (energy)  of the Doppler shifted photon measured by the observer would be sufficient to create a black  hole ? In other words, what is the Doppler factor?


2 Answers 2


An electromagnetic plane wave solution to general relativity will be an electromagnetic plane wave solution in every reference frame you boost to. No black hole will form. you don't just need a specific energy density, you need a stress-energy tensor that satisfies certain conditions, and there is no reference frame where the $T_{tt}$ component is the only nonzero component, because photons carry momentum, and are null in every reference frame. You can try to localize your wave state by superposing different momenta, but this both will not be a "single photon" in any sense, and the localization property will not be lorentz invariant (which makes sense from the perspective of Poisson brackets):


In any reasonable interpretation of this question, no black hole forms in any reference frame, barring some exotic semiclassical quantum gravity effect.

Also, photons are a quantum mechanical concept that doesn't really exist in general relativity, so talking about them int his discussion requires care.


Actually i think there will be no black hole no matter how fast that source moves.

First, from the point of view of source itself, there won't be any Doppler shift at all, hence no black hole. And remember principle of special relativity:

The laws of physics don’t change, even for objects moving in inertial (constant speed) frames of reference.

If one says that there is no black hole, other observers have to come to same conclusion, because it makes no sense to have a black hole in one frame, while other frame doesn't see it! (e.g by Hawking's radiation or experiments such as observing other frames time rate)

Second, source of gravitation itself is stress energy tensor in general relativity, and if i recall correctly it's invariant under change of observers.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Apr 19, 2019 at 20:33

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