Assume a photon enters the event horizon of a black hole. The gravity of the black hole will draw the photon into the singularity eventually. Doesn't the photon come to rest and therefore lose it's mass?

  • $\begingroup$ I will let someone with a better knowledge of GR provide a full answer, but I will point out that a photon does not have mass in the first place so cannot lose it. $\endgroup$ Jan 20, 2012 at 17:44
  • $\begingroup$ From reading on this site about photons: they have zero mass at rest, but they acquire mass (therefore effected by gravity) when in motion. $\endgroup$ Jan 20, 2012 at 17:47
  • $\begingroup$ Okay, I can see that. My recollection is that photon mass due to motion is simply refered to in terms of the associated momentum since this is what is actually measured. $\endgroup$ Jan 20, 2012 at 18:16
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    $\begingroup$ Photons have zero mass, period. They do have a nonzero amount of energy, because of their momentum. Also, photons never come to rest; they always move at speed $c$. $\endgroup$
    – David Z
    Jan 20, 2012 at 18:45
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    $\begingroup$ @Ramashalanka: the reason light appears not to move at speed $c$ in materials is the time-averaged effect of interactions with the electron states of the material, not some fundamental change in the physics of EM waves. The important fact is that the medium doesn't create a rest frame for a photon. Also, a theoretical black hole is a (quantum) vacuum, except at the singularity. $\endgroup$
    – David Z
    Jan 20, 2012 at 19:16

2 Answers 2


We don't know what will happen when a photon or any other particle hits a singularity of a black hole. The singularity is a phenomenon of classical general relativity and the singularity is really is an indication that classical general relativity breaks down there. To really understand what happens near a singularity we need a full quantum mechanical version of general relativity. String theory is the best quantum mechanical version of general relativity that we currently have but string theory is not developed enough to give a definitive answer to your question.

  • $\begingroup$ Thanks Frank. I really wish someone with figure out this singularity thing soon. I guess I'll just have to wait for now. $\endgroup$ Jan 20, 2012 at 19:44
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    $\begingroup$ @KentByerley, yes, I am impatiently waiting also :-) $\endgroup$
    – FrankH
    Jan 20, 2012 at 21:17

Meh, no string theory or singularity or unitology etc. Black Holes are accelerators. Every particle is broken down to the lowest level, accelerated to the highest speed and shot out. The path is non-vectorial, because the exitement of such magnitude makes the low-level particles act as "quantum binded particles". Which means, for every exited particle, there is another particle, it communicates. Through this binding efect, both particles and their respective "anti or dark matter" particles interact and lose energy. they slow down and under the rule of conservation of mass, stick together again. Far far away. the thing here is to think of binded particles as identical, reformable parts, rather than individual particles whizzing about.

Thats why general physics and traditional laws of nature break down in a black hole. Because the respective laws only go down as far as neutrons and protons.

Quarks, leptons, bosons, muons, tauonic particles etc.

Once we reach the bottom, we figure out the black holes. Until then, its all classical mechanics...on a cosmic scale :)


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