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I was wondering if it is possible for a single photon to form a black hole if it has a small enough wavelength.If so, what would this wavelength be? I came across this question because I am reading about loop quantum gravity, and have heard that it is 'untestable'. I know that the loops making up spacetime in loop quantum gravity are thought to have a size in the order of $10^{-35}$m, and I was wondering whether making a photon with a wavelength small enough to investigate this would form a black hole.

I am aware of the two posts here, but neither seem to really answer my question because they refer to several photons: Can a black hole be formed by radiation? and https://physics.stackexchange.com/questions/107207/photons-and-black-holes.

Thank you.

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    $\begingroup$ How would you reconcile this with Lorentz invariance? The photons emitted by the Sun have an arbitrary small wavelength in the frame of observers who move close enough to the speed of light. $\endgroup$
    – Count Iblis
    Commented Oct 16, 2014 at 20:21
  • $\begingroup$ @CountIblis: LQG quantizes area at the planck scale, so I would assume that you lose Lorentz invariance at similar length scales. $\endgroup$
    – Zo the Relativist
    Commented Oct 16, 2014 at 20:41
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    $\begingroup$ @JerrySchirmer: No, LQG does not have to violate Lorentz invariance. See Rovelli, 2010, “Loop quantum gravity: the first twenty-five years,” arxiv.org/abs/1012.4707 . "I want to stress the fact that loop gravity does not imply a violation of Lorentz invariance. In particular, the naive argument, often heard, that a minimal length is incompatible with Lorentz invariance is wrong, because it disregards quantum theory." $\endgroup$
    – user4552
    Commented Oct 16, 2014 at 23:08
  • $\begingroup$ It seems plausible that two photons colliding from different directions could form a black hole, but it's hard to imagine how energy and momentum could be balanced in converting one lightspeed, massless entity into a less-than-lightspeed, massive entity. $\endgroup$
    – S. McGrew
    Commented Apr 8, 2019 at 17:55

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This question is unanswerable using current loop quantum gravity, which does not yet have a completely consistent way to couple gravity to matter.

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  • $\begingroup$ Hi Jerry- thanks for your reply. I know loop quantum gravity still has a long way to go and is early on in its development, but setting it aside, do you know if a high energy photon would form a black hole? And if so, what energy would it have to have? Thanks :) $\endgroup$
    – Meep
    Commented Oct 16, 2014 at 20:10
  • $\begingroup$ @21joanna12: there is no answer in LQG yet. The theory, as it currently exists, is a theory of gravity alone. There is no matter content in it, and thus, no concept of "a photon" in a full, quantum sense. $\endgroup$
    – Zo the Relativist
    Commented Oct 16, 2014 at 20:40
  • $\begingroup$ @JerrySchirmer ok, just forget LQG, use string theory or the least assuming model. Is the comment made by Count Iblis the correct answer? and why it would be different if you have more than one photon? (in that case the answer was that it is indeed possible) $\endgroup$
    – user65081
    Commented Oct 16, 2014 at 20:58
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Could it have been: $\large l_p = 1.61619926^{-35}m \small \quad \text{ (Planck Length)}$

At any rate, to create a black hole, you simply need enough energy density in a single area that its escape velocity (the speed at which the sums of $E_k$ and $E_p$ are $0$) is larger than the speed of light. As you should know, the photon has no mass. However, its momentum and energy are contributed to the mass of black holes curving spacetime in such a way that the escape velocity of light is not high enough to escape it. The energy of a photon, in accordance with its frequency is: $$E=hf=pc$$ E=Energy
h=Planck's Constant
f=Frequency
p=Momentum
c = the speed of light constant

The Schwarzschild radius is the radius at which the mass of something would cause its escape velocity to be equal to the speed of light, $c$. The equation is $$R = 2GM/c^2$$ Since the photon has no mass, we can rearrange Einsteins mass/energy equivaleny equation in the sense of mass as follows: $$ E^2=(mc^2)^2+(pc)^2\\ E^2=m^2c^4+p^2c^2\\ \frac{E^2}{c^4+p^2c^2}=m^2\\ m=\sqrt\frac{E^2}{c^4+p^2c^2}\\ m=\frac{E}{c^2+pc} $$ I calculate that the wavelength would be: $$ R=\frac{2G{\frac {E}{c^2+pc}}}{c^2} \\ R=\frac{2GEc^2}{c^2+pc} \\ R=\frac{2GEc}{p} \\ R=\frac{2Ghfc}{p} \\ f=\frac{Rp}{2Ghc} \\ \lambda = \frac{1}{f} = 2\cdot G\cdot h \cdot c \cdot \frac{1}{Rp} $$

Now this is just me tinkering around with some equations. I've probably done some illegal operations here, but in the interest of my amusement, that's the answer I came up with.

Thanks to @Hypnosifl for recommending the far simpler (possibly illegal) equation: $$R=\frac{2Ghf}{c^4}$$ which doesn't require us to know momentum.

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    $\begingroup$ Since p is an unknown, don't you want to remove it from the equation? Since any bound system will behave as if its inertial mass is proportional to total energy (including kinetic and potential along with the rest mass of its parts) over c^2, couldn't you just plug in hf/c^2 in place of M in the Schwarzschild radius equation and solve for f? Like you said this may be an illegal move, but at least it would give you an equation for frequency/wavelength that doesn't require already knowing the momentum. $\endgroup$
    – Hypnosifl
    Commented Oct 16, 2014 at 21:24
  • $\begingroup$ As a last step, you could substitute the Planck length, L_p = sqrt[hG/(2pi*c^3)], in for R and solve for f, which could be a guesstimate for the frequency that would be required to form a Planck-sized black hole. The equation would suggest that photons with even higher frequencies could form larger black holes, but this might not be physically meaningful in a theory of quantum gravity as it would probably mean an energy density higher than the Planck density. I get f = 0.032c/L_p this way, and since the wavelength = c/f (not 1/f as you wrote), this gives wavelength L_p/0.032 $\endgroup$
    – Hypnosifl
    Commented Oct 16, 2014 at 21:52
  • $\begingroup$ (and given the questionable nature of this procedure I wouldn't take the coefficient of 1/0.032 arising from the factor of 2pi too seriously, the point is that this back-of-the-envelope calculation suggests that to form a Planck-sized black hole, a photon would need a wavelength pretty close in terms of orders of magnitude to the Planck length) $\endgroup$
    – Hypnosifl
    Commented Oct 16, 2014 at 21:57
  • $\begingroup$ In the third line, should it not be $m^2=\frac{E^2-p^2c^2}{c^4}$? $\endgroup$
    – Meep
    Commented Oct 17, 2014 at 10:16
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    $\begingroup$ You can have a photon of arbitrarily large kinetic energy by just changing the reference frame. $\endgroup$
    – Anixx
    Commented Dec 7, 2018 at 21:33
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The minimum mass for a black hole is called Planck mass, which is when the Compton wavelength less than Schwarzschild radius. According to wikipedia, the value is 2.4e15TeV comparing to 14TeV that LHC can produce.

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    $\begingroup$ Which Wiki entry? $\endgroup$
    – Kyle Kanos
    Commented Apr 8, 2019 at 17:46

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