# Gravitational collapse of a photon cloud

The Schwarzschild solution shows that a spherically symmetric, static fluid will undergo gravitational collapse if too much mass-energy is concentrated together (i.e. if the fluid's radius is less than the Schwarzschild radius for its total mass-energy).

If a very large number of photons in a spherical arrangement ar travelling towards a central point, which all arrive at the same time, can gravitational collapse occur, or does the non-staticity due to movement allow one to avoid the gravitational collapse that would otherwise occur according to the Schwarzschild solution?

Tl;dr: Can a large number of photons create a black hole?

• "The Schwarzschild solution shows that a spherically symmetric, static fluid will undergo gravitational collapse" - A collapsing fluid is not static, a static fluid is not collapsing. – Alfred Centauri Jul 3 '14 at 0:39
• A better way to say it is that a static fluid at high densities must have infinite pressure, and thus cannot exist, and so therefore gravitational collapse when a large amount of matter is placed at the same location. – Brian Rushton Jul 3 '14 at 0:44
• Well, my answer was deleted so: 1) photons do not form a cloud, they each run away in a straight line with velocity c and do not interact with each other. 2) the heisenber uncertainty principle has to be taken into account when thinking of overlapping photons in a small volume. 3) there exist fusion reactors using lasers in a sphere concentrating at the center in order to create the plasma energy for fusion. No danger of gravitational collapse has been envisaged as far as a I know.en.wikipedia.org/wiki/Inertial_confinement_fusion – anna v Jul 3 '14 at 15:49

Yes, you can form a black hole from just photons. There is a fairly simple solution that captures exactly what you are describing, given on pg. 83 of Eric Poisson's GR lecture notes: http://www.physics.uoguelph.ca/poisson/research/agr.pdf.

Basically, the setup is that you have a thin sphere of energy density collapsing at the speed of light, which could be taken to be your spherical arrangement of photons. Inside, the metric is flat space,

$$ds^2 = -dt^2_-+dr^2 + r^2 d\Omega^2,$$

and outside the sphere it is the Schwarzschild solution

$$ds^2 = -fdt^2_++\frac1fdr^2+r^2d\Omega^2,$$

with $f=1-2M/r$, and $M$ constant. Here $t_-$ and $t_+$ don't necessarily coincide at the boundary of the sphere, but the spatial coordinates $(r,\theta, \phi)$ do coincide at the boundary. From the inside, the surface of the photon shell is located at $t_-=v_--r$, while from the outside the boundary is $t_+=v_+-r_*$, where $v_-, v_+$ are constants and $r_*=r+2M\ln(r/2M-1)$ is the tortoise coordinate.

The notes go on to calculate the surface energy density and pressure, and unsurprisingly find that the energy density is $\mu = M/4\pi r^2$ and the pressure is zero.

It's also worth noting that you don't really need a large number of photons for this to work: as long as you have them collapsing in a perfectly spherical, very thin shell, they will form a black hole once the shell becomes smaller than the Schwarzschild radius. Fewer (or less energetic) photons will simply make a smaller black hole.

Note that as a practical matter, you would be limited by how thin you can actually make the shell. In order for the thin shell approximation to work, you need the actual width of the shell (or say the photon wavepackets) to be much less than the Schwarzschild radius of the eventual black hole that will form. For laboratory lasers, the Schwarzschild radius will be way too tiny to ever make a black hole using this setup.

• If the pressure is zero, why is this not a shell of dust? – Jerry Schirmer Jul 3 '14 at 15:39
• The Heisenberg Uncertainty principle has to be taken into account, the volume cannot go to zero for photons, so the energy density has to be high. See my comment to the question – anna v Jul 3 '14 at 15:52
• @JerrySchirmer I think it is a shell of dust, just null dust. It's a collection of photons in a spherical configuration following radial null geodesics. – asperanz Jul 3 '14 at 20:12
• @annav I updated my answer to address your point. – asperanz Jul 3 '14 at 20:17