How much energy does a photon need to form a black hole? 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.
 A: This question is unanswerable using current loop quantum gravity, which does not yet have a completely consistent way to couple gravity to matter.
A: 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.
A: 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.
