<p>Consider the phase space of the universe as a box with hypervolume <span class="math-container">$V$</span> and divide its phase space into volume <span class="math-container">$h^6$</span> boxes. That gives us <span class="math-container">$N=V/h^6$</span> boxes. If we put in <span class="math-container">$k$</span> particles we get <span class="math-container">$S=\binom{N}{k}$</span>  possible states. One can bound this as <span class="math-container">$N^k/k^k\leq S \leq N^k/k!$</span>. So if we assume random states occur each <span class="math-container">$\tau$</span> seconds, the time until a particular state shows up is somewhere between <span class="math-container">$\tau N^k/k^k$</span> and <span class="math-container">$\tau N^k/k!$</span>.</p>
<p>Now, the observable universe is about <span class="math-container">$V=10^{80}$</span> cubic meters and has about <span class="math-container">$k=10^{84}$</span> particles (mostly photons). So <span class="math-container">$N=10^{160}/(10^{-36})^6=10^{376}$</span> (where I assumed the phase space volume is the 3D volume squared). So the time is between <span class="math-container">$\tau (10^{376})^{10^{84}}/(10^{84})^{10^{84}}$</span> and <span class="math-container">$\tau (10^{376})^{10^{84}}/(10^{84})!$</span>.  So the timescale is roughly <span class="math-container">$10^{2.92\times 10^{86}}$</span>: at this point we can drop the <span class="math-container">$\tau$</span> since to a good approximation measuring the time in Planck seconds or Hindu kalpas give the same answer - double exponentials are big.</p>
<p>The big bang might not correspond to a single state, so the actual recurrence time is shorter. But it is still doubly exponential.</p>
<p>The fundamental problem is that space is expanding and that particles are lost over the cosmological horizon. This also causes a very weak horizon radiation of finite temperature that in principle can add new particles. This hardly replenishes the universe enough.</p>

CC BY-SA 4.0
2023-07-06T02:24:32.487Z
Thanks, exactly what I was looking for, although I was hoping the number would be so large it would be impractical to represent with exponentials.
Thomas
CC BY-SA 4.0
2023-07-16T08:56:05.553Z
@Thomas - Well, double exponentials are bad enough. I have never seen any physics problem produce anything in need of up-arrow notation or the other super-exponential functions.
Anders Sandberg
CC BY-SA 4.0
2023-07-05T08:30:14.027Z
2023-07-05T08:30:14.027Z
Anders Sandberg
<p>Going off of the Boltzmann brain thought experiment, approximately how many years from now would there be a 50% chance that all the particles in the Universe would randomly assemble themselves into another state of very low entropy similar to what was present at the beginning of the Big Bang?</p>

CC BY-SA 4.0
2023-07-05T05:05:11.993Z
I'd say it would never happen-- expansion of space would imply that particles in the universe (after evaporation of black holes) could never again be causally connected.  But my understanding may be wrong...
antlersoft
<entropy>
How many years will it take before there's a 50% chance that all particles would have assembled into another Big Bang?
Thomas
# How many years will it take before there's a 50% chance that all particles would have assembled into another Big Bang?

(Asked by: username_0 on 2023-07-05)

Going off of the Boltzmann brain thought experiment, approximately how many years from now would there be a 50% chance that all the particles in the Universe would randomly assemble themselves into another state of very low entropy similar to what was present at the beginning of the Big Bang?

**Tags:** entropy
> **Comments:**
> **username_1**: I'd say it would never happen-- expansion of space would imply that particles in the universe (after evaporation of black holes) could never again be causally connected.  But my understanding may be wrong...  

---

## 1 answers

### Answer by username_2 on 2023-07-05. Score: 4 | Accepted answer

Consider the phase space of the universe as a box with hypervolume $V$ and divide its phase space into volume $h^6$ boxes. That gives us $N=V/h^6$ boxes. If we put in $k$ particles we get $S=\binom{N}{k}$ possible states. One can bound this as $N^k/k^k\leq S \leq N^k/k!$. So if we assume random states occur each $\tau$ seconds, the time until a particular state shows up is somewhere between $\tau N^k/k^k$ and $\tau N^k/k!$.

Now, the observable universe is about $V=10^{80}$ cubic meters and has about $k=10^{84}$ particles (mostly photons). So $N=10^{160}/(10^{-36})^6=10^{376}$ (where I assumed the phase space volume is the 3D volume squared). So the time is between $\tau (10^{376})^{10^{84}}/(10^{84})^{10^{84}}$ and $\tau (10^{376})^{10^{84}}/(10^{84})!$. So the timescale is roughly $10^{2.92\times 10^{86}}$: at this point we can drop the $\tau$ since to a good approximation measuring the time in Planck seconds or Hindu kalpas give the same answer - double exponentials are big.

The big bang might not correspond to a single state, so the actual recurrence time is shorter. But it is still doubly exponential.

The fundamental problem is that space is expanding and that particles are lost over the cosmological horizon. This also causes a very weak horizon radiation of finite temperature that in principle can add new particles. This hardly replenishes the universe enough.

> **Comments:**
> **username_0**: Thanks, exactly what I was looking for, although I was hoping the number would be so large it would be impractical to represent with exponentials.  
> **username_2**: @Thomas - Well, double exponentials are bad enough. I have never seen any physics problem produce anything in need of up-arrow notation or the other super-exponential functions.  

---

