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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?)

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  • $\begingroup$ Related: physics.stackexchange.com/q/3436/2451 $\endgroup$ – Qmechanic Oct 23 '16 at 5:06
  • $\begingroup$ I was not aware that photons have any mass... $\endgroup$ – MauganRa Oct 24 '16 at 6:07
  • $\begingroup$ 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. $\endgroup$ – Joseph Hirsch Oct 25 '16 at 0:27
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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.

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  • $\begingroup$ 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. $\endgroup$ – Joseph Hirsch Oct 24 '16 at 22:49
  • $\begingroup$ @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). $\endgroup$ – Mark H Oct 25 '16 at 1:03
  • $\begingroup$ 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. $\endgroup$ – Mark H Oct 25 '16 at 1:03
  • $\begingroup$ Can light be bent in a circular path by gravity without there being a black hole? $\endgroup$ – Joseph Hirsch Oct 25 '16 at 1:31
  • $\begingroup$ @JosephHirsch Neutron stars possibly work as well: 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. $\endgroup$ – Mark H Oct 25 '16 at 1:59
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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.

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  • $\begingroup$ But if the photon were bent in a circular path around its own gravity well, wouldn't it have zero net momentum too? $\endgroup$ – Joseph Hirsch Oct 23 '16 at 14:46
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    $\begingroup$ @JosephHirsch: I have never heard of a structure like that. This doesn't necessarily mean it cannot exist since my knowledge of this area is limited. However I'm fairly sure that any such structure would be unstable just as the circular orbit around a black hole is unstable. $\endgroup$ – John Rennie Oct 23 '16 at 15:22

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