March 29, 2021. Black holes leak energy into the environment and ultimately evaporate. In this post, I’ll explore two applications of evaporation: estimating the life expectancy of the universe, and showing that the LHC cannot produce dangerous black holes.
Contents
1. Introduction
In 1974, Stephen Hawking discovered that black holes glow. This means that, over time, a black hole loses energy and eventually disappears altogether. We’re going to use this to do some fun stuff, first guessing the time until entropic heat death, and secondly, giving a heuristic argument that the Large Hadron Collider (LHC) does not produce dangerous microscopic black holes. We won’t derive anything in detail, but rather, slap estimates together to get an order-of-magnitude vibe for what’s going on. Our key formula will be that, for a black hole of mass $M$, the evaporation timescale is roughly
\[t_\text{evap} \sim \frac{G^2M^3}{\hbar c^4} \sim 10^{67} \text{ years} \left(\frac{M}{M_\odot}\right)^3,\]where $M_\odot = 2\times 10^{30} \text{ kg}$ is the mass of the sun, and $G$ is Newton’s constant, $c$ the speed of light, and $\hbar$ is Planck’s constant. I’ll explain these constants if and when we need them. Now for our applications!
2. The end of the universe
Careful observation of the night sky reveals that everything is flying apart at an accelerating rate. Eventually, every galaxy (or any other system clumped together by gravity) will be moving away from other systems faster than light. At this point, it has entered its cosmological horizon, from which it can no longer send messages. We can loosely estimate how long it will take us to fall into our cosmological horizon as follows. The bigger gravitationally bound structure we are part of is the local group, a collection of galaxies a few million light years across. Around 13 million light years away, we encounter the M94 group, which is not gravitationally bound to us.
So, we can estimate “time to horizon” as the time it takes the M94 group to move at the speed of light, relative to us. To do this, we can use Hubble’s law. This says that an object a distance $d$ away recedes from us at a speed $v$ given by
\[v = H_0d, \quad H_0 \approx 70 \, \frac{\text{km/s}}{\text{Mpc}} = 7 \times 10^{-11} \, \frac{c}{\text{ly}},\]where $H_0$ is the Hubble parameter, and we have converted from the conventional units (km/s per megaparsec) to even weirder units (light speed units per light year, or inverse years). To see how long it takes M94 to reach light speed, we technically need to do calculus, but I’m lazy and we won’t. Instead, I simply claim that things move away exponentially quickly, with distance and speed
\[d(t) = d_0 e^{H_0t}, \quad v(t) = H_0 d_0 e^{H_0t} = H_0 d(t).\]So, to figure out when M94 reaches light speed, we note that (if now is $t = 0$) then
\[d_0 = 13 \times 10^6 \text{ ly},\]and we want $v(t) = c$. Thus, the time until we fall into our horizon obeys $c = H_0 d_0 e^{H_0t_\text{hor}}$, or
\[t_\text{hor} = \frac{1}{H_0}\log\left(\frac{c}{H_0 d_0}\right) = \frac{10^{11}}{7}\log\left(\frac{10^{11}}{7 \cdot (13 \times 10^6)}\right) \text{ y} \approx 100 \text{ Gy}.\]In $100$ billion years, the universe will get radically smaller. But this is not the end! The local group is a huge place, and lots of interesting things can still happen. But according to the second law of thermodynamics, entropy always increases, reducing the amount of useful energy in a system until all that’s left is a cool, uniform goo in thermal equilibrium.