Post by: chiralSPO on 11/06/2015 16:12:31
Is there a frequency above which a single photon's gravitational field results in a black hole? Or could there exist a high enough concentration of energetic photons that together create a black hole? Would this black hole still travel at the speed of light?
Part of how I got to this question was thinking about the result of the impact of two black holes that are identical except that one was formed from matter, and the other from antimatter. My understanding is that the apparent masses would just add together whether or not we imagine that an annihilation event occurred. Either the degenerate matter is somehow incapable of annihilation, or if the mass of the black holes is actually completely converted into energy, it is still confined within the event horizon, and exerts the same gravitational effects as the mass before annihilation. (no information leaves the event horizon, so we cannot distinguish between the impact of two matter black holes and one matter plus one anitmatter)
Post by: syhprum on 11/06/2015 17:54:05
Post by: PmbPhy on 11/06/2015 17:57:21
Quote from: chiralSPO
As I understand it, although a photon has no rest mass, it does have a gravitational field that is a function of the energy (frequency) of that photon.The following is my understanding of the related physics: Although a single photon has zero mass and thus no zero momentum frame, the same can't be said for two photons moving in opposite directions. The invariant mass of the two photons is non-zero and there is a frame of reference in which the total momentum is zero. If, in this frame, two photons of sufficient energy "collide" they will form a black hole.
Quote from: chiralSPO
Is there a frequency above which a single photon's gravitational field results in a black hole? Or could there exist a high enough concentration of energetic photons that together create a black hole? Would this black hole still travel at the speed of light?If such a frequency did exist then it'd be possible to change to a frame of reference where the energy/frequency is too small and therefore no black hole exists in that frame. The existence of a black hole is invariant so it follows that the answer to your question is no.
Post by: chiralSPO on 11/06/2015 20:32:42
The reference to an anti matter BH presumably refers to one that has formed by the collapse of an massive anti matter star, after the BH has formed is it possible to determine whether the original star was matter or anti matter ?
It should not be possible to tell an "antimatter black hole" from any other type of black hole. I first stated:
...two black holes that are identical except that one was formed from matter, and the other from antimatter...
from there on I lazily referred to an "antimatter black hole" and a "matter black hole." I appologize for any confusion caused by my laziness.
Post by: chiralSPO on 11/06/2015 20:36:44
If such a frequency did exist then it'd be possible to change to a frame of reference where the energy/frequency is too small and therefore no black hole exists in that frame. The existence of a black hole is invariant so it follows that the answer to your question is no.
Aha! That is a good way of explaining this, and certainly one I would not have thought of. Thank you.
The following is my understanding of the related physics: Although a single photon has zero mass and thus no zero momentum frame, the same can't be said for two photons moving in opposite directions. The invariant mass of the two photons is non-zero and there is a frame of reference in which the total momentum is zero. If, in this frame, two photons of sufficient energy "collide" they will form a black hole.
So, in this case, with two (or n pairs of) photons traveling in opposite directions, no frame of reference will result in a lower observed aggregate energy?
Post by: jeffreyH on 11/06/2015 20:56:23
EDIT: So each photon would have
Post by: PmbPhy on 12/06/2015 02:19:52
Post by: lightarrow on 13/06/2015 18:28:06
So, in this case, with two (or n pairs of) photons traveling in opposite directions, no frame of reference will result in a lower observed aggregate energy?What does it mean?
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Post by: lightarrow on 13/06/2015 18:35:35
If, in this frame, two photons of sufficient energy "collide" they will form a black hole.It also depends on how large are the photons and I have no idea of that.
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Post by: PmbPhy on 14/06/2015 06:13:53
Photons are point particles. However the probability of where they are is not a point function.If, in this frame, two photons of sufficient energy "collide" they will form a black hole.It also depends on how large are the photons and I have no idea of that.
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Post by: lightarrow on 14/06/2015 09:48:25
If photons really were Point particles, the Energy density would be very high and two opposite (common) beams of light should easily have the possibility to form Black holes. Or we have been so Lucky that none of the photons smashed exactly head-on yet?Photons are point particles. However the probability of where they are is not a point function.If, in this frame, two photons of sufficient energy "collide" they will form a black hole.It also depends on how large are the photons and I have no idea of that.
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Post by: evan_au on 14/06/2015 11:46:24
The wavelength of light gets smaller as the energy gets higher.
When the energy gets high enough to form a black hole, the photon's wavelength gets vanishingly small, so there is little likelihood that they will collide.
I suspect that long before you got to black-hole energies, the photons would turn into a shower of other particles.
Post by: PmbPhy on 14/06/2015 12:03:25
Quote from: lightarrow
If photons really were Point particles, the Energy density would be very high and two opposite (common) beams of light should easily have the possibility to form Black holes.That's a classical argument which you're trying to use in quantum mechanics so it's invalid argument. Electrons, photons etc. behave the laws of quantum field theory. See
http://www.physlink.com/Education/AskExperts/ae191.cfm
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But, another reason that the electron is not considered a black hole, even assuming that its radius is infinitely small, is that it obeys the laws of quantum field theory. Normally, when one speaks about black holes, one is talking about them in terms of Einstein's theory of general relativity. No one is sure how nature merges Einstein's theory with quantum field theory. So we aren't really sure if the idea of a black hole makes sense on distance scales as small as the (possible) radius of the electron.. Our best idea to unify general relativity with quantum field theory is an idea called string theory, but string theory still appears to be a long way from being put to any experimental tests.If I were you I'd ignore the second response. I consider it to be quite flawed.
Post by: jeffreyH on 14/06/2015 14:08:29
would have M equaling x number of photons. How would you ever prove this? Would you want to? The vortex around these objects would be intense. I wouldn't want to be anywhere near one.
Post by: lightarrow on 14/06/2015 17:49:25
It's not used in classical mechanics only, even in QM. Let's say, rather, that we don't have a quantum theory of gravity so our reasoning is speculation (my reasoning as well as your).Quote from: lightarrowIf photons really were Point particles, the Energy density would be very high and two opposite (common) beams of light should easily have the possibility to form Black holes.That's a classical argument which you're trying to use in quantum mechanics so it's invalid argument.
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Electrons, photons etc. behave the laws of quantum field theory. SeeBut even that answer can have some value, it brings an interesting viewpoint.
http://www.physlink.com/Education/AskExperts/ae191.cfmQuoteBut, another reason that the electron is not considered a black hole, even assuming that its radius is infinitely small, is that it obeys the laws of quantum field theory. Normally, when one speaks about black holes, one is talking about them in terms of Einstein's theory of general relativity. No one is sure how nature merges Einstein's theory with quantum field theory. So we aren't really sure if the idea of a black hole makes sense on distance scales as small as the (possible) radius of the electron.. Our best idea to unify general relativity with quantum field theory is an idea called string theory, but string theory still appears to be a long way from being put to any experimental tests.If I were you I'd ignore the second response. I consider it to be quite flawed.
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Post by: lightarrow on 14/06/2015 18:07:48
The probability of finding a photon in a particular place is related to its wavelength.Wavelength is related to how light (em waves) behaves, but from this it's not possible to extrapolate a photon's dimension. If you have a plane wave, it's infinitely large, but it can have as small wavelength as you want.
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The wavelength of light gets smaller as the energy gets higher.So it only depends on energy? But even little masses (and so little energies) can form black holes, if the dimensions are little enough.
When the energy gets high enough to form a black hole,
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