# Entropy of a vacuum and heat death of the universe

1. Aug 4, 2013

### Maximise24

According to the third law of thermodynamics, one could argue that a vacuum has zero entropy, since it has only one ground state and a temperature at absolute zero.

However, assuming the accelerated expansion of the universe to result in a 'heat death', i.e. a state of absolute thermal equilibrium with maximum entropy (following from the second law of TD), how is this state different from that of a vacuum? If the universe expands infinitely, energy density will near zero, resulting in absolute zero temperature and the universe basically having one ground state.

Could one not argue that, since the thermal equilibrium of a vacuum is (near-)perfect and since there is little 'usable' information, its entropy is actually at a maximum level?

In short: which thermodynamical definition of a vacuum or vacuum-like state (such as the infinitely expanding universe) is correct?

2. Aug 4, 2013

### kevinferreira

Why would the vacuum have zero entropy? Use the statistical definition
$$S=k_B~log(W)$$
where $W$ is the number of micro-states compatible with a vacuum. Now, you're saying there's only one of these micro-states. What does this mean?

3. Aug 4, 2013

### Maximise24

Since there is no real energy in a vacuum, very little configurations or micro-states are possible. There may be some virtual effects, but can it not be argued that, due to the absence of real energy and temperature, vacuum entropy is, at least, very low? Or would you regard virtual energy as a valid source of micro-states?

4. Aug 4, 2013

### Staff: Mentor

Consider a volume V containg N particles bouncing around in it. If V is large compared to N, that's a pretty decent vacuum. The N particles can be arranged in some number of microstates. Now let's make the vacuum better by increasing V while holding N constant. What happens to the number of available microstates?

5. Aug 4, 2013

### Maximise24

Surely that depends on what effect an increased V has on the particles? If, such as in the case of the expanding universe, particles get pulled away from each other and eventually disintegrate into photons, there will be a lot less interactions and thus less microstates will be possible (increasing entropy). One could say, in the heat death scenario, that, since there are only photons and since all photons are the same, only one microstate will ultimately be possible, namely that of the non-interacting free photon.

Last edited: Aug 4, 2013
6. Aug 4, 2013

### Staff: Mentor

The possibility or not of interaction has nothing to do with the number of microstates.

The end state isn't just a single photon, and not sure why you think it might be. The bolded text suggests that you may have misunderstood the statement that photons are indistinguishable. That's true, but it doesn't mean that they're all the same. It just means that if I have two identical photons in a given microstate and I exchange them, I don't get to count that as a new microstate.

7. Aug 4, 2013

### kevinferreira

Let's consider the macroscopic state "1 particle in a volume V". This is a pretty good vacuum. What is the number of microstates compatible with this?
Counting: the particle can be anywhere within the volume V, so it W has to be proportional to V, in my opinion.

8. Aug 4, 2013

### Maximise24

I did indeed not express myself very well on that point, but - still in the heat death scenario - you cannot deny that the entropy of a sea of photons is very high, at least higher than in any other state (since it is the last macrostate of the universe): this is a direct application of the second law of thermodynamics, which says that entropy always goes up.

If not the expansion of the universe and the associated reduced effect of gravity, then what does, according to you, produce this entropy?

And, related to that, how would the heat death macrostate be different from that of the classical vacuum, which is sometimes said to have zero entropy or indeed none at all?

9. Aug 5, 2013

### Maximise24

Yes, and what conclusions might one draw from that with respect to the entropy of the vacuum? Or do you think that a vacuum has no entropy?

10. Aug 5, 2013

### kevinferreira

I don't know what to consider as 'microstates' when there is nothing to construct microstates with! A classical vacuum, with no particles nor anything, may not even have an assigned entropy! An alternative (instead of letting the number of particles $N$ go to zero) is to demand instead that the relation $N/V$ goes to zero, i.e. large volume/low number of particles...

11. Aug 5, 2013

### Maximise24

But what about the quantum information in the vacuum? If you look at the Casimir effect or virtual energy/particles in general, could you not consider them as information (and thus entropy) in the vacuum?

And, in the inflationary model, what about the vacuum quantum fluctuations that are enlarged by inflation, creating real structures? If an initial vacuum has no entropy at all, how could its fluctuations be the basis for systems that do have entropy?

12. Aug 5, 2013

### Staff: Mentor

The classical vacuum contains zero particles and zero energy. However, in post #4 I suggested that you start with a different concept: You have N particles within a volume V; consider what happens to the number of microstates as you increase V while holding N constant. You can get the density as low as you wish by increasing the volume sufficiently; but no matter how much you increase the volume it will always be an N-particle system, never a classical vacuum.

13. Aug 5, 2013

### Maximise24

You would indeed never get a complete vacuum, but by infinitely expanding the volume you would come as close to it as is possible with an N-particle system (lowering the density and temperature to almost zero), right? So then my question would be: could you compare this high-entropy heat death scenario with an actual vacuum? Might we conclude from this that a classical vacuum also possesses entropy or not?

14. Aug 5, 2013

### Staff: Mentor

There are no quantum fluctuations in the classical vacuum, which is what you've been discussing above; it contains no particles and no energy.

15. Aug 5, 2013

### Maximise24

Okay, a quantum state vacuum, then. I'm trying to refer to the vacuum that is hypothesised in inflationary theory. What about that?

16. Aug 6, 2013

### Maximise24

Anyone? Or is the question a bit too hard? :-)

17. Aug 6, 2013

### ModusPwnd

No energy? Is there somewhere that this is defined? Wiki describes it as a reference vacuum for electromagnetic effects which means it does have energy, doesn't it?

The vacuum at the universe's heat death would have energy and this macrostate would have microstates associated with it. No good?

18. Aug 6, 2013

### Maximise24

Indeed: a "classical vacuum" with completely no energy does not actually exist - there are inevitably quantum effects. I am indeed referring to a vacuum with quantum effects and I was wondering whether it has entropy or not.

19. Aug 6, 2013

### Staff: Mentor

Yes, if I start with a system of N particles and expand the volume that it occupies enough, then the overwhelming majority of its microstates will closely resemble the macrostate known as "classical vacuum". But that doesn't mean doesn't mean that it is a classical vacuum, nor that all of its properties must necessarily approach those of a classical vacuum - and the entropy is one of the properties that is most interestingly different.

20. Aug 6, 2013

### ModusPwnd

I dont even think you need to appeal to quantum effects. A "classical vacuum" is not broken when you shine a light on it, at least not how I conceptualize the phrase. Thats why I was wondering where the phrase's definition comes from.