I Entropy and Expansion in a Simple Universe

[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.