# Photonic black holes

"Can a photon turn into a black hole?" - usually the answer to this question is - it can't, because it has zero rest mass. However, when we derive the Schwarzchild Metric initially the $$2M$$ term just remains a constant (assuming $$G=c=1$$), we fine tune it in Newtonian limit to equate it to $$2M$$.

Can $$M$$ actually be the total energy, instead of the rest mass - allowing a photon to be a black hole? If not, why?

• A number of comments removed. Comments are for clarifying or improving the post they're attached to. Comments are not for partial answers, nor for discussion of other comments that are partial answers, nor for personal remarks. Knock it off, y'all.
– rob
Commented Jul 3 at 14:31

For the sake of argument, let us simplistically take the photon energy $$E=hf$$, and see what kind of black hole we get.

The Schwarzschild radius of a given parcel of energy $$E$$, which has an equivalent mass-energy $$m=E/c^2$$, is

$$r=\frac{2G}{c^2}m= \frac{2G}{c^2}\left(\frac E {c^2} \right) =\frac{2G}{c^4} E$$

Real physicists will see this as an act of barbarity, but let's roll with it.

And so if we plug in $$E=hf$$, this radius for our hypothetical "photonic black hole" (black hole with the equivalent energy of a single photon), is therefore

$$r=\frac{2Gh}{c^4}f$$

This is the radius within which all of the given energy must be confined to generate the spacetime corresponding to a black hole.

Now, the approximate spatial extent of a real photon is its wavelength

$$\lambda = \frac c f$$

In order for our black hole formation condition to be met, we require $$\lambda < r$$. Let us compute these for a typical gamma ray photon, on the high-energy side of photon energy:

$$f = 10^{20}~ \rm Hz$$

$$\lambda = 3\cdot 10^{-12} \rm ~ m$$

$$r = 1\cdot 10^{-57} \rm ~ m$$

So we see, the energy of the gamma ray photon, being "smeared out" over a region on the order of its wavelength, is not even remotely localized enough to fit within a Schwarzschild radius, to form a black hole. This is part of the reason we know that fundamental particles cannot be miniscule black holes flying around.

If we set $$\lambda < r$$ and solve for the frequency required to achieve this condition, we get:

$$f > \sqrt{\frac {c^5}{2Gh} } \sim 5\cdot 10^{42} \rm ~ Hz$$

This is on the same order as the Planck Frequency (inverse of the Planck Time), and so would require a Quantum Theory of Gravity to describe. Bear in mind, such an individual photon would have an energy (assuming we could even apply $$E=hf$$) of about $$\rm 10~GJ$$, or about the total energy used by every household in the USA over a year – in a single photon, if such a thing could be. But this is, very roughly, the energy scale photon that would be required to have any chance of being comparable to a black hole, per our current theories.

This order of magnitude argument, while blindly equating different types of energies from Quantum Mechanics and Relativity in a way real physicists would rightly balk at, gives an idea why an individual photon cannot constitute something resembling a black hole.

• Good analysis about single-photon case. Maybe it's worth to mention in the post that if we replace energy of single into multiple photons, like $\sum E = nhf$, then we really could get a Kugelblitz type BH. Commented Jul 2 at 21:23
• Good point. And the analysis shows that $n$ has to be on the order of $10^{45}$ or greater before photons in gamma ray range would be able to form a Kugelblitz Commented Jul 2 at 22:17
• Yes. Or in other words, to create a Kugelblitz one needs about $10^{15}$ units of $10 ~PW$ lasers directed into same spot. Or the most powerful quasar found with radiation output power on the order $10^{41}~W$ collected and focused light. Surely, trademark "made@home" doesn't apply for this type of "product" :-D Commented Jul 3 at 7:07

The existing answers don't mention a crucial fact, which is that the $$M$$ in the Schwarzschild metric is an invariant parameter, whereas a photon's energy is frame dependent. So a single, free photon cannot form a black hole. Loosely speaking its energy is cancelled out by its momentum in the stress-energy-momentum tensor, which is the source of spacetime curvature.

A pair of photons, or a photon in a box, could indeed form a black hole under the right circumstances, because the system as a whole possesses an invariant minimum energy ("rest mass"). But as others have pointed out, the energy required is substantial.

"According to electromagnetic theory, the rest mass of photon in free space is zero and also photon has non-zero rest mass, as well as wavelength-dependent. The very recent experiment revealed its non-zero value as $$10^{-54}$$ kg ( $$5.610 \times 10^{-25}$$ MeV $$c^{-2}$$ ) ." So, if you consider at rest, then no black hole would be formed. However, if you want to apply the Schwarzschild metric, you first assume the photon or the body is at rest, which is a big mistake and impractical. So the frank answer is no.

But let's consider it that we apply it for some time, irrespective of the condition.

Then,

$$R_s=\frac {2GM}{c^2}$$

Which on plugging values gives

$$R_s=1.48317778×10^{-73}$$

As you can notice, it is far below the planck length. So, applying GR is not correct.

I hope this helps you.

• From the same article- So photon has no fixed real mass like other particles and objects, it can be zero and have real; imaginary value again. If photon changes it's rest mass according to the situation (like neutrinos changes their flavor on the fly),- then it's not correct to choose one specific instance of photon mass into the Schwarzschild radius formula. Because somebody could equally choose zero and say that it's impossible to get a BH from photon. Commented Jul 2 at 18:24
• Definitely correct. But please recheck the assumptions. There is no black hole for stationary photon. If we were forced to take another mass because the question says such. Then, a wrong hypothetical condition arises. Which I explained but definitely again is not correct. Commented Jul 3 at 2:05

A photon in a box has mass and pressure. Put it in an impossibly small box and it'll theoretically make a black hole.

• The mass-energy for the black hole comes from the box, though, doesn't it? Do we even have to put a photon inside? Isn't the quantum mechanical zero point energy already diverging as the size of the box goes to zero? If I were a theoretician I would now probably make a holographic argument that the entire history of the black hole was already contained on (in) the boundary (conditions) of the box. Is that a poor guess? Commented Jul 2 at 0:37
• A single photon in a massless box is problematic. When the photon hits a side of the box and reflects, conservation of momentum cannot be fulfilled. (Not my downvote.) Commented Jul 2 at 4:57
• @safesphere Yes, I glossed over the fact that the box needs to be flexible and have some mass. The theory of the proton is that it's a box containing quarks. Quarks have little rest mass, but within the box they have plenty of kinetic energy. And, as FlatterMann points out, the field has energy, too. All that energy makes the proton a relatively massive particle. Commented Jul 2 at 11:25