# I Entropy of a simple universe

1. May 3, 2017

### Grinkle

Consider an expanding universe of infinite extent containing only a single particle. Does the entropy of this universe increase over time due to expansion? If it makes any difference in being a sensible question, consider an expanding universe with N particles where N is a known, finite number.

2. May 3, 2017

### Staff: Mentor

How do you know there is such a thing? What solution of the Einstein Field Equation are you using?

3. May 4, 2017

### Grinkle

@PeterDonis I can't just contrive any situation that might occur to me and expect that it makes sense? ;-) I get it - my question might as well be asking about the properties of unicorn horns.

I'll try to think of a better grounded way to get help on what I am pondering. As always, I do appreciate your responses.

4. May 4, 2017

### Arman777

I think it wouldnt
1-Is one particle universe make sense ?
2- If it makes sense then, in one particle universe the system never changes its properities.Like there is no change in tempature or simply in the "information".Theres no energy dissipation/transformation etc.Theres just one particle.Theres only one state.So I think there shouldnt be any change in entrophy.

I am an undergrad student,This is just an opinion.

5. May 4, 2017

### Grinkle

@Arman777 Thanks for the response.

I suspect a one particle expanding universe does not make sense.

I think I might pose the question better like so -

Consider a single particle in a closed system with volume V, and a second closed system, again with one particle but with volume 2V. Is the entropy of the 2nd system greater than the entropy of the first system?

I propose that to an outside observer, the answer is yes, because the 2nd system has more thermodynamically equivalent states than the first system.

I don't think a single particle system has any meaningful temperature property, but maybe I am wrong about that. If I am wrong, then I need to also say that both systems are at the same equilibrium temperature.

Since there is only one particle, I don't know how to assign meaning to entropy observed inside the system. I don't think a zero particle observer makes any sense.

6. May 4, 2017

### Staff: Mentor

What determines the "volume" of the system if there is only a single particle?

7. May 4, 2017

### kimbyd

This one-particle universe is very similar to the eventual fate of our own universe, assuming we have a cosmological constant.

Eventually, due to the cosmological constant, there will be either one or zero particles within any given cosmological horizon. Once this occurs, then the area within each cosmological horizon will eventually reach its ground state, which means any properties you might want to measure within that horizon would no longer change with time at all (for the moment I'm neglecting the fact that measurement would also be impossible because there also couldn't be any observers). This means that the entropy within the horizon is a constant.

However, what does this mean for the area outside the horizon? Here we run into a conundrum: ostensible there are still other particles in the far future of this universe, each in a different horizon. Over time, the universe could be divided into more and more non-overlapping volumes each within its own cosmological horizon, and more and more of those without any particles at all. Ostensibly that would represent an entropy that continues to increase forever without bound.

The issue there is that it's not at all clear that that's the correct thing to do. It might be valid to only consider one cosmological horizon at a time, and the degrees of freedom representing the rest of the universe would then be holographically encoded on that horizon. This reduces the infinite universe to a strictly finite universe, one which eventually reaches a ground state and becomes absolutely static, having a constant entropy. I know Andreas Albrecht has been investigating this kind of finite universe over the last few years, as it solves a number of really difficult mathematical problems of trying to treat the universe as infinite.

8. May 4, 2017

### Staff: Mentor

It's similar to what the patch within a single cosmological horizon would eventually look like. But that's not the entire universe; the model that makes this prediction includes an infinite number of such patches, not just one.

Aside from the difficulty given above, there is another problem with this. The OP is not talking about a universe that eventually includes just one particle. He is talking about a universe that always includes just one particle. If there is a spacetime that meets this description, it is not de Sitter spacetime, which is the one that is being used in the models you describe.

9. May 4, 2017

### kimbyd

I'm aware. I was trying to bring the situation described down to earth, so to speak.

10. May 4, 2017

### Grinkle

I think volume and entropy and for that matter any thermodynamic property for such a system could only be defined by an observer outside the system.

These are very helpful comments for me. I didn't realize that the history of reaching a 1 particle per horizon state matters in terms of whether the OP question makes any sense.

I think this is then the answer to my question?

I posted this hoping the discussion would help me understand why there exists a maximum entropy condition for the holographic principle to hold at the cosmological level. Its hard for me to get my head around, especially given that the space inside a finite event horizon is at maximum entropy, and the holographic principle seems to hold for that.

11. May 4, 2017

### Staff: Mentor

So how are the observers in your hypothesis in post #5 defining the volume?

12. May 4, 2017

### Grinkle

As an embedded volume in a larger space. I have 2 ideally insulated/insulating sphere's, one of volume V and the other of volume 2V, and I calculate the volumes because know the radii of the 2 spheres. I further know there is exactly one particle in each sphere.

My reasoning is that the the entropy of space inside the larger sphere is greater than the entropy of the space inside the smaller sphere because the particle in the larger sphere has more potential positions it can be in, and the particle being in any one position defines a single state, and all possible states of the interiors of a given sphere have the same thermodynamic properties.

13. May 4, 2017

### Staff: Mentor

This makes the case you are considering different from the case of an expanding universe. An expanding universe has no boundary corresponding to the ideally insulated/insulating spheres, and the presence of that boundary is essential in your reasoning about the entropy (see below).

What does this mean? The particle inside the sphere has states it can be in, and the number of those states does increase with the volume of the sphere, because the boundary of the sphere confines the particle (note that you have to consider both position and momentum in this calculation, since the states are states on phase space, not configuration space). But I don't know what "the entropy of space" is.

14. May 4, 2017

### durant35

Interesting. What do you mean by the term 'one or zero' particles and how are they equivalent in regarding the space of the far future vacuum as empty?

Also, how would the outside of the horizon differ from the inside? The particles and the radiation would still get diluted away - no matter what horizon we're talking about, right?

Last edited: May 4, 2017
15. May 4, 2017

### Grinkle

I meant the entropy of the space inside the sphere - I am not sure how else to describe the system. The absent 'the' in 'the space' was a typo.

16. May 4, 2017

### Grinkle

That is where my poorly posed OP came from. Does the entropy continue to increase without bound? Maybe the question stop being defined after one reaches a density of 1 or 0 particles per horizon.

17. May 4, 2017

### Bandersnatch

Wait. With cosmological constant you should end up with isolated bound systems rather than single particles. Did you mean quintessence rather than cosmological constant?

18. May 4, 2017

### Staff: Mentor

Still doesn't make sense. I think you need to spend some time with a statistical mechanics textbook that discusses the microphysical definition of entropy.

If there is no boundary confining the particle, then there isn't a "volume" associated with it. So it's not clear how to even define the phase space of the system.

19. May 4, 2017

### Arman777

To keep the particle in that volume we need boundries,as PeterDonis said.

So we know that even in the changes in the volume can increase the entropy of the system (or decrease), but this case is true for the ideal gases.I found an equation or describes the equation.
http://hyperphysics.phy-astr.gsu.edu/hbase/Therm/entropgas.html

Our questions is just for a single particle situation.So this equation will not hold.But I think we can think a system like this.Think an electron and we put it in a very small container (very very very small) so that from uncertanity principle we know its position well so its momentum/energy should increase due to QM effects. When we expend the volume of the container, we are increasing the position uncertanity so momentum/energy of an electron should decrease.So I believe this change in the energy can cause in the change in entropy.

Note:The problem could be this , we do the change in the volume in isothermal situation so this may somehow affect our particle I mean I am not sure about the details of the system.

20. May 5, 2017

### kimbyd

That happens first. Then, over time that matter decays through proton decay or falls into black holes which then evaporate. Eventually there's nothing left but photons, neutrinos, and electrons/positrons that are flowing freely.

21. May 6, 2017

### Bandersnatch

That's speculative, though. Protons don't decay in the standard model.

22. May 6, 2017

### kimbyd

Proton decay is demanded by the observed fact of baryogenesis. The time-inverse of that process will necessarily result in proton decay, though it may be hugely suppressed if, for instance, it requires a very massive intermediate particle (just as the weak nuclear force is suppressed).

Even if this reasoning ultimately proves to be incorrect, which seems highly unlikely, matter will still approach a state of one particle per horizon, only much more slowly.

23. May 6, 2017

### Grinkle

If protons don't decay, then is there reason to believe protons won't become isolated in their own horizon without any black hole to fall into?

24. May 7, 2017

### kimbyd

I don't understand your question. A single proton would be one particle in the horizon.

25. May 7, 2017

### Grinkle

I was referring back to post 20.