I Entropy and Expansion in a Simple Universe

[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.

 

[PeterDonis]

Consider an expanding universe of infinite extent containing only a single particle.

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

 

[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.

 

[Arman777]

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.

I think it wouldn't
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 shouldn't be any change in entrophy.

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

 

[Grinkle]

@Arman777 Thanks for the response.

1-Is one particle universe make sense ?

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

Theres just one particle.Theres only one state.

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.

 

[PeterDonis]

Consider a single particle in a closed system with volume V

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

 

[kimbyd]

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 shouldn't be any change in entrophy.

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

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.

 

[PeterDonis]

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

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.

Eventually, due to the cosmological constant, there will be either one or zero particles within any given cosmological horizon

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.

 

[kimbyd]

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.

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

 

[Grinkle]

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

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.

It's similar to what the patch within a single cosmological horizon would eventually look like.

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

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.

The issue there is that it's not at all clear that that's the correct thing to do.

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.

 

[PeterDonis]

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.

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

 

[Grinkle]

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

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 anyone position defines a single state, and all possible states of the interiors of a given sphere have the same thermodynamic properties.

 

[PeterDonis]

I have 2 ideally insulated/insulating sphere's, one of volume V and the other of volume 2V

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).

the entropy of space

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.

 

[durant35]

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.

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?

 

[Grinkle]

But I don't know what "the entropy of space" is.

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.

 

[Grinkle]

An expanding universe has no boundary corresponding to the ideally insulated/insulating spheres

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.

 

[Bandersnatch]

Eventually, due to the cosmological constant, there will be either one or zero particles within any given cosmological horizon.

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

 

[PeterDonis]

I meant the entropy of the space inside the sphere

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.

Does the entropy continue to increase without bound?

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.

 

[Arman777]

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 anyone position defines a single state, and all possible states of the interiors of a given sphere have the same thermodynamic properties.

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.

 

[kimbyd]

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

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.

 

[Bandersnatch]

proton decay

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

 

[kimbyd]

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

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.

 

[Grinkle]

matter will still approach a state of one particle per horizon, only much more slowly.

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?

 

[kimbyd]

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?

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

 

[Grinkle]

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

I was referring back to post 20.

 

[kimbyd]

I was referring back to post 20.

Yes, if protons don't decay then they would eventually become isolated as well.

However I think this is pretty unlikely due to the symmetry argument I started earlier.

 

[Grinkle]

However I think this is pretty unlikely due to the symmetry argument I started earlier.

FWIW, I find contemplating an end point of minimal complexity more aesthetically pleasing than an endpoint with some basic particles and some compound particles. I know you weren't arguing from an aesthetic perspective, but in any case proton decay feels to me like something that 'ought' to be correct.

 

[kimbyd]

FWIW, I find contemplating an end point of minimal complexity more aesthetically pleasing than an endpoint with some basic particles and some compound particles. I know you weren't arguing from an aesthetic perspective, but in any case proton decay feels to me like something that 'ought' to be correct.

I can see that. But aesthetic arguments are rarely a good barometer for the truth.

 

[furiobas]

The solution to Einstein's equations for a single particle in otherwise empty space should be the Schwarzschild's metric if the particle is not rotating and uncharged or other solutions if rotation, charge or both are present.
None of these solutions are expanding, but flat at infinity.

 

[Bruce Barron]

There is no such as an infite universe, and if there were, I don't think one particle would do anything or could make any difference in the entropy

 

[kimbyd]

The solution to Einstein's equations for a single particle in otherwise empty space should be the Schwarzschild's metric if the particle is not rotating and uncharged or other solutions if rotation, charge or both are present.
None of these solutions are expanding, but flat at infinity.

For this thread, we're mostly talking about a universe with a cosmological constant. Technically, this would result in a de Sitter-Schwarzschild metric:
https://en.wikipedia.org/wiki/De_Sitter–Schwarzschild_metric

In practice, however, the mass of the particle will be completely and utterly overwhelmed by the cosmological constant, so the particle's gravitational field will be irrelevant.

 

[PeterDonis]

There is no such as an infite universe

Yes, there is; our current best-fit model of our universe is spatially infinite.

 

[Mark Harder]

Theres only one state.So I think there shouldn't be any change in entrophy.

The location of the particle is a variable that describes the state it is in. Therefore, in a volume > zero, there are many locations and therefore many states.

 

[Mark Harder]

The universe is a closed system, therefore it can't exchange heat or mass with any other system (of which there are none, since we're talking about the universe here). Since the single particle can't have any interaction with any other particle and its excluded volume in such a large system is effectively zero, the system you are describing is an ideal gas. So, let's think in terms of an ideal gas in a sealed adiabatic container fitted with a piston or some other means of changing the volume. Assuming the temperature, T>0, and reversibility of the volume change, then for such a change dS = δq /T = 0. So the entropy does not depend on the volume and an increase in volume does not change the entropy. Your universe being an ideal gas, the energy also does not depend on the volume.

There's one worm in the ointment, as far as I can see. All the above assumes that there is no non-uniform field in the container for which a gradient will constitute a force, therefore pressure, of its own. To my untutored mind, this is what our relativists are referring to: If there's only one particle there's no gravitational field that the particle can sense. As soon as you include more than one particle - with nonzero mass - then the particles will experience their mutual gravitational attraction. The pressure of the system, compared to a perfect vacuum (which doesn't exist, I know, but neither does the system we're talking about, so..), will be negative. So it would take work to separate the particles, i.e. to increase the volume of the system. The energy of the system will depend on the volume. So our assumption of ideality goes out the window. For laboratory-scale systems, gravitation is ignored, making ideal gases approachable. On the other hand, you didn't specify what the mass of the particle is, so it could be humongous and though it is only one particle, space curvature might make a difference. If the universe expands, will the curvature have to change? If so, will that change the energy or entropy? As to that, I have no idea; but whatever the answers, thermodynamics will still apply.

 

[durant35]

Yes, there is; our current best-fit model of our universe is spatially infinite.

Isn't that just because the curvature has not been detected yet and for simplification of calculations?

I mean, it's still an open question if the universe is spatially finite or infinite. Whatever the flat-lambda model says should be compatible even if the universe was a finite, closed system.

 

[PeterDonis]

Isn't that just because the curvature has not been detected yet

You say "yet". How do you know spatial curvature will ever be detected?

Our best current model is spatially flat because that's the model that best fits the data we have.

it's still an open question if the universe is spatially finite or infinite

Only in the sense that the error bars in our measurements cannot conclusively rule out the possibility that the universe is spatially closed. But we have no positive evidence of spatial curvature. And at some point, as the error bars continue to narrow, we might be able to rule out spatial closure as a possibility, if future measurements continue in the pattern of the data we have now.

Whatever the flat-lambda model says should be compatible even if the universe was a finite, closed system.

No, it won't; if we ever get positive evidence that the universe is not spatially flat (which we do not currently have--see above), we will have to change our best fit model; it will no longer be the flat lambda CDM model. It might be a spatially closed lambda CDM model, but that's still a different model.

 

[PeterDonis]

the system you are describing is an ideal gas

No, it isn't. A single particle is not a continuous fluid. Also, there is no piston in the "single particle in an otherwise empty universe" model. As has already been pointed out, such a model is either flat Minkowski spacetime (if the single particle has negligible mass) or Schwarzschild spacetime (if the single particle has non-negligible mass). If you want to reason about entropy, you need to do it using one of those two models; the model you used is neither.

 

[Mark Harder]

No, it isn't. A single particle is not a continuous fluid.

That's not the ideal gas model with which I'm familiar. The statistical mechanics of an ideal gas assumes that it consists of particles in thermal, i.e. ramdom, motion. That can't be a continuous fluid. The particles must occupy zero volume and do not participate in any energetic interaction with other particles. A very unreal situation, but one that can be approximated experimentally by helium and some other gases, especially at low concentrations that minimize inter-particle collisions.

 

[durant35]

You say "yet". How do you know spatial curvature will ever be detected?

Well, I don't but the same can be said about you and the claim that it won't. At this point it is pointless to claim anything about it.

Only in the sense that the error bars in our measurements cannot conclusively rule out the possibility that the universe is spatially closed. But we have no positive evidence of spatial curvature. And at some point, as the error bars continue to narrow, we might be able to rule out spatial closure as a possibility, if future measurements continue in the pattern of the data we have now.

What if only our region is flat and the rest is not?

You seem to be pretty optimistic about the potential extrapolations of a measurement that is local and inconclusive about the greater scale. Sure, it can be said that there's no evidence in my claim but your claims are incredibly strong wrt to available measurements.

No, it won't; if we ever get positive evidence that the universe is not spatially flat (which we do not currently have--see above), we will have to change our best fit model; it will no longer be the flat lambda CDM model. It might be a spatially closed lambda CDM model, but that's still a different model.

What would be different?

 

[PeterDonis]

At this point it is pointless to claim anything about it.

I wasn't. You were the one making an implicit claim when you used the word "yet". If you had left out that word your statement would just have been a description of our current evidence, which is fine.

What if only our region is flat and the rest is not?

Then we will have to change models--assuming we could observe the non-flatness outside "our region" at some point.

your claims are incredibly strong wrt to available measurements

My "claims" are just a description of our current best fit model. It is our current best fit model because it best fits all of our available evidence--more precisely, it is the simplest model which fits all of our available evidence. It is the simplest because it assumes that the rest of the universe that we can't see looks similar to the part of it that we can see. Your questions (such as "what if only our region is flat and the rest is not?") assume a more complex universe in which the region we can see is somehow special, different from the rest. If we ever get positive evidence that that's the case, then yes, we'll need to revise our model, but in the meantime Occam's Razor tells us to use the simplest model that fits the data. So that's what we do. If that's "incredibly strong wrt to available measurements", then so are practically all of our current scientific theories.

What would be different?

Um, the universe would be spatially closed and have positive spatial curvature, instead of being spatially flat and having zero curvature? Is it not obvious that those are different models?

 

[PeterDonis]

The statistical mechanics of an ideal gas assumes that it consists of particles in thermal, i.e. ramdom, motion.

No, the statistical mechanics justification for using the ideal gas model assumes that the gas consists of particles in random thermal motion, and then shows how, given that assumption (plus a few others, such as zero interaction between the particles except elastic collisions on contact), you can ignore the motions of the individual particles and just model the gas as a continuous fluid with a few simple thermodynamic properties. And for that justification to work, you need a very large number of particles; just one won't do.