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As the wavelength of a photon shrinks, its energy rises, and so its mass rises (using $E=hc/\lambda$ and $m=E/c^2$). On calculating the Schwarzschild radius for a photon based on its mass derived from those two equations, I found that the Schwarzschild radius of the photon will be equal to $\lambda/2\pi$ in one instance, when the wavelength of the photon equals $2\pi$ times the Planck's length:

$$\lambda=2\pi \times \mathscr{L}_P\implies r_s=\frac{\lambda}{2\pi}$$

where $r_s$is the Schwarzschild radius and $\mathscr{L}_P$ is the Planck's length.

In other words, a photon with a wavelength $\lambda=2\pi \times \mathscr{L}_P$ would gravitationally trap itself in a circular orbit with a radius equal to the plank length. A photon in a circular path with diameter of $2\pi \mathscr{L}_P$ would have a gravity well that would trap itself at the corresponding radius of the plank length (with an orbital path diameter of $\lambda=2\pi \times \mathscr{L}_P$). Has this been discussed as a conceptual mechanism as to why the plank length is a lower limit on potential allowed wavelengths, and the resolution of the universe (that a photon with a wavelength of $2\pi \mathscr{L}_P$ in a circular path with a diameter equal to that wavelength would in fact be the definition of a black hole?)

Mark H
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Joseph Hirsch
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  • Related: http://physics.stackexchange.com/q/3436/2451 – Qmechanic Oct 23 '16 at 05:06
  • I was not aware that photons have any mass... – MauganRa Oct 24 '16 at 06:07
  • I know photons have no rest mass. Also I have been told that individual photons have no mass, but groups of photons do, BUT I think that a photon in a circular path around a gravity well would have mass for the same reason that groups of photons do, that the center of gravity defines a rest frame. – Joseph Hirsch Oct 25 '16 at 00:27

3 Answers3

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A photon of sufficiently small wavelength would not become a black hole. To see this, consider two observers: one who measures the photon as having a Planck-scale wavelength, and another that is traveling at high speed in the same direction as the photon. This second observer will observe the photon to have a wavelength that is longer than that measured by the first observer due to Doppler shifting. The second observer will conclude that the photon does not have enough energy to create a black hole. All observers will agree on whether a black hole exists or not, so the only consistent conclusion is that no black hole forms.

We can also reason in the reverse. A photon of visible light obviously doesn't have enough energy to create a black hole. Otherwise, light bulbs would be dangerous black hole generators. However, due to Doppler shifting, there is a frame of reference traveling at high speed towards the photon in which that photon has a much larger amount of energy--large enough to create a black hole if that was possible.

Be careful when assigning importance to Planck-scale measurements. We do not know if the Planck units have any physical significance. It is not known if spacetime is continuous or not, and if it's not, we have no reason to think that the "resolution" of spacetime is at all related to the Planck length. Just as a counter example, the Planck mass is about 20 micrograms--a small amount, but one that is handled all the time by pharmacists and far larger than any fundamental particle mass.

Mark H
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  • Mark, the photon does not have a direction because it is travelling in a circular path around its own center of gravity. Also, while the plank mass, as far as we know is merely derived from the plank length, c and G, wasn't the plank length itself experimentally derived from black body radiation experiments which demonstrated that photons with wavelength shorter than the plank length don't exist? The doppler effect would allow an unlimited reduction in wavelength for an observer given the right reference frame, but yet we don't get any photons with wavelength shorter than plank length. – Joseph Hirsch Oct 24 '16 at 22:49
  • @JosephHirsch You still have the problem of being able to boost any photon to arbitrarily high energies just by changing the observer's reference frame. If a single photon can turn into a black hole, then every photon would turn into a black hole. You need at least two photons to collide in order to form one (see John Rennie's answer below). – Mark H Oct 25 '16 at 01:03
  • Also, the Planck length was not derived from experiment, but from combining physical constants. The highest energy photon ever observed had one billionth the energy of a photon with a Planck-length wavelength. We have absolutely no data on photons with higher energy. Any supposed limit on the minimum wavelength of a photon is pure speculation. – Mark H Oct 25 '16 at 01:03
  • Can light be bent in a circular path by gravity without there being a black hole? – Joseph Hirsch Oct 25 '16 at 01:31
  • @JosephHirsch Neutron stars possibly work as well: http://adsabs.harvard.edu/full/1993ApJ...406..590N At a quantum level with Planck-energy photons and arbitrarily warped spacetime, who knows. That's exactly the regime where all theories in physics start giving contradictory answers. – Mark H Oct 25 '16 at 01:59
  • At what "point in time" would two observers, moving relative to each other, have to agree that the photon had become a black hole (since their accounting of time is dependent on their relative motion)? – Joseph Hirsch Aug 31 '19 at 19:54
  • Similarly, wouldn't a moving mass achieve a mass in some reference frames that would make it appear to be a black hole, while not in another? Just say we launch a rocket that accelerates until it's mass is great enough to make it a black hole, and another observer is flying next to it. The initial frame of reference wouldn't see it as a black hole because it's so far away that we would have to wait to get a signal back from it. So is it also true that you can't boost a material object up to being a black hole by accelerating it? – Joseph Hirsch Aug 31 '19 at 19:56
  • If two people are moving relative to each other, they can disagree on the wavelength of light coming from a laser far away, but they can't disagree on the wavelength of a single photon because only the observer who interacts with a given photon will be able to observe it-they can't both observe it directly, so for me to call a blue shifted photon a "black hole" it would have to interact with me, and a relatively slower moving observer would see the affects as if I interacted with a black hole. – Joseph Hirsch Aug 31 '19 at 20:32
  • @JosephHirsch Physicists do not think relativistic mass is a useful concept anymore. An object's mass is invariant. A black hole is a black hole in all reference frames. – Mark H Sep 01 '19 at 11:02
  • OK, last question. A black hole is a black hole in all reference frames, but "when?" Say two black holes form, can the observers in different reference frames disagree on which one formed first and both be right? – Joseph Hirsch Sep 01 '19 at 14:33
  • @JosephHirsch Sure. If there is a reference frame where two black holes form simultaneously, then there are other reference frames where the black holes form in different orders. The important point is that if one reference frame sees a black hole form, then all reference frames will eventually see it form. – Mark H Sep 02 '19 at 01:51
  • For every position in any reference frame, every photon will eventually pass that position and leave the light cone of that point and reference frame. – Joseph Hirsch Sep 02 '19 at 03:06
  • @JosephHirsch "If two people are moving relative to each other, they can disagree on the wavelength of light coming from a laser far away, but they can't disagree on the wavelength of a single photon...". It is important to recognise that even with just one observer they are free to do their maths in any reference frame they like (including boosted frames where the photon kinetic energy is different) and they should get the same physical predictions in all of them. You don't need people in relative motion. The kinetic energy related to linear centre-of-mass motion does not generate gravity. – Dast Nov 24 '21 at 14:36
  • @MarkH I don't believe that the issue that a black hole must exist in all reference frames is the problem you make it out to be. If a photon were boosted to the level of a black hole, it could only be observed in ANY reference frame after it has interacted with an object in the reference frame where it is a black hole. Also ff a photon enters my eye, then no observer in reference frame that could have seen it boosted to black hole status would ever be able to observe it. – Joseph Hirsch Nov 30 '21 at 17:36
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I can't improve on Mark's answer, but I would add that while a single ray of light cannot form a black hole, light can form a black hole if multiple rays are focussed onto the same point. This object is known as a Kugelblitz.

The important distinction from the single ray is that in a Kugelblitz the net momentum is zero due to the (approximate) spherical symmetry.

John Rennie
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This is no doubt both mathematically and physically correct. When we construct either a proton or neutron we have “cancelled out” charge components. But what is a photon but oscillating opposite charges? Your instinct is probably correct. These charge components become trapped over their own gravity well an their propagation becomes rotation. Forward momentum is now angular momentum. $m=E/c^2$. Mass is energy in orbit which acquires through this angular momentum, inertia, mass.

jng224
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Caleigh Fisher
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