How does the number of microstates affect the entropy of an isolated system?

[Rap]

Yes - Look at the Wikipedia article "Sackur-Tetrode equation" (https://en.wikipedia.org/wiki/Sackur–Tetrode_equation).

You can get into trouble defining cells which are of the order of the size of a gas molecule, etc. because you are entering the quantum realm, and your conclusions can be wrong or nonsensical. The Sackur-Tetrode equation approaches things from a somewhat simplistic quantum viewpoint and gives a very good estimate of the entropy of an ideal monatomic gas, even down to close to absolute zero. Another good article is the thermal wavelength (https://en.wikipedia.org/wiki/Thermal_de_Broglie_wavelength) which tells you when you are entering the quantum realm.

 

[Buzz Bloom]

If I understand the question correctly, one should search for the "Sackur-Tetrode formula" or "Sackur-Tetrode equation".

The Rap approaches things from a somewhat simplistic quantum viewpoint and gives a very good estimate of the entropy of an ideal monatomic gas, even down to close to absolute zero

Hi @Lord Jestocost and @Rap:

Thank you both for your posts. Wikipedia gives the Sackur-Tetrode equation in the following form:

S/kN = ln V + (3/2) ln (2πemkT) - 3 ln h - (1/N) ln N!
≈(5/2) + ln ( (V/N) + ln (2πmkT/h²)^(3/2) )
where V is the volume of the gas, N is the number of particles in the gas, ..., k is Boltzmann's constant, m is the mass of a gas particle, h is Planck's constant, and ln is the natural logarithm, [and T = temperature kelvin].​

The approximation is based on Stirling's:

ln N! ≈ N ×(-1+ln N).​

(One detail I don't get with confidence is whether "e" is intended to be the base of the natural logarithms. )

If we take into account that k, m, and k are natural constants, and N is an assumed constant for both systems (1) and (2), the approximation for S can be written in the following form:

S = C₁ + C₂ × ln ( C₃ × V × T^3/2 ).​

where C₁, C₂, and C₃ are constants.
Since V = const × R³, if the entropy S₁ and S₂ (corresponding to system (1) and (2) respectively), are equal, then

R₁ × T₁^1/2 = R₂ × T₂^1/2.​

This was not what I was expecting. I was expecting:

R₁ × T₁ = R₂ × T₂​

because the corresponding universe models requires this in order to avoid a conflict with the second law of thermodynamics. I will explain this further in a later post about the universe models.

Regards,
Buzz

 

[Buzz Bloom]

Hi @Grinkle, @anorlunda, @jartsa, @Stephen Tashi, @Nugatory, @Mark Harder, @Lord Jestocost, and @Rap:

Thank you all for participating in this thread. I was able to learn from all of you enough so that I could properly complete this post.

Below is my description of a hypothetical GR based universe as a thermodynamic system. I am a bit shaky regarding technical nomenclature, so I will try to be extra careful in explaining the properties of this universe.
1. The shape of the hypothetical universe system is nearly exactly a finite 3D hypersphere. That is, the ratio of the density of stuff (matter and energy, including dark stuff) to the critical density is represented by

Ω_S >1.
https://en.wikipedia.org/wiki/Shape_of_the_universe#Curvature_of_the_universe​

I use the subscript "s" (for stuff) here to distinguish this ratio from sum of the four Ω ratios in the Friedmann equation.
2. The Friedmann equation

https://en.wikipedia.org/wiki/Friedmann_equations​

For a scale factor of a = 1, H = H₀, so

Ω = Ω_R + Ω_M + Ω_k + Ω_Λ = 1.​

Now, Ω_k is not related to stuff. It is related to the curvature of the universe. So,

Ω_S = Ω_R + Ω_M +M_Λ​

and

Ω = Ω_S + Ω_k = 1​

Since for the closed universe,

Ω_S > 1,​

this means that

Ω_k < 1.​

3. Calculating the radius of curvature

https://www.physicsforums.com/threads/radius-of-curvature-of-universe.240763/​

The radius of curvature for the universe corresponding to a = 1 is

R_a=1 = c/( H₀ sqrt(-Ω_k) )​

4. For the purpose of this hypothetical universe, I am interested in the calculation of total entropy in the universe for the values a = 1 and a = 2. I assume the following initial conditions:

Ω_M << 1
Ω_R ≈ 1+|Ω_k|
Ω_k = - c² / (R_a=1² × H₀²)
Ω_Λ << 1​

Thus the hypothetical universe expansion and contraction effects are assumed to be almost entirely depending only on the radiation and curvature, the matter and dark energy contributions being negligible. The time it takes for the hypothetical universe the expand from a=1 to a=2 depends on a value assumed for H₀, and it can be calculated, but I will omit that calculation since I assume it to be irrelevant.
5. I choose initial conditions such that at a=2 the universe will stop expanding and begin its collapse. This requires that

H_a=2 = 0,​

which implies

Ω_R = 4 × |Ω_k|.​

This combined with the above relationship between Ω_R and Ω_k implies that

Ω_k = -1/3,​

and

Ω_R = 4/3.​

6. It is assume the universe between a=1 and 1=2 to be filled with H₂ gas and photons in thermodynamic equilibrium. The relevant variables for calculating entropy changes depend only on R (radius of curvature) and T (temperature) as discussed in my previous post. Here the useful forms are:

R_a=2 = 2 R_a=1
T_a=2 = (1/2) T_a=1​

The difference between entropy at a = 2 and at a = 1 is

S_a=2 - S_a=1 = C₂ × ln( (R_a=2/R_a=1)³ × (T_a=2/T_a=1)^3/2 )
= C₂ × ln( 8 × (1/8)^1/2 ) = C₂ × (1/2) ln 8 ≈ C₂ × 1.0397​

CONCLUSION
Let t* be the time between a=1 and a=2. Then

t' (> t_a=2) = t_a=1 + 2 t*​

is the time when a is again =1. Therefore as time increases from t_a=2 to t', entropy decreases from S_a=2 back again to S_a=1. This seems to violate the second law.
However,

https://www.physicsforums.com/threads/entropy-and-contracting-universe.943975/
Post #10 by @kimbyd
Gravity changes entropy dramatically. In fact, nobody knows how to actually calculate the entropy for a generic gravitational system. We know a few extreme cases (like black holes), but that's about it.​

[Underlining above is mine for highlighting purpose.] So, maybe the apparent violation of the second law really does not really happen because of the assumed hypothetical universe system has a definite gravitational component.

Regards,
Buzz

 

[kimbyd]

Hi @Grinkle, @anorlunda, @jartsa, @Stephen Tashi, @Nugatory, @Mark Harder, @Lord Jestocost, and @Rap:

Thank you all for participating in this thread. I was able to learn from all of you enough so that I could properly complete this post.

Below is my description of a hypothetical GR based universe as a thermodynamic system. I am a bit shaky regarding technical nomenclature, so I will try to be extra careful in explaining the properties of this universe.
1. The shape of the hypothetical universe system is nearly exactly a finite 3D hypersphere. That is, the ratio of the density of stuff (matter and energy, including dark stuff) to the critical density is represented by

Ω_S >1.
https://en.wikipedia.org/wiki/Shape_of_the_universe#Curvature_of_the_universe​

I use the subscript "s" (for stuff) here to distinguish this ratio from sum of the four Ω ratios in the Friedmann equation.
2. The Friedmann equation

https://en.wikipedia.org/wiki/Friedmann_equations​

View attachment 227275
For a scale factor of a = 1, H = H₀, so

Ω = Ω_R + Ω_M + Ω_k + Ω_Λ = 1.​

Now, Ω_k is not related to stuff. It is related to the curvature of the universe. So,

Ω_S = Ω_R + Ω_M +M_Λ​

and

Ω = Ω_S + Ω_k = 1​

Since for the closed universe,

Ω_S > 1,​

this means that

Ω_k < 1.​

3. Calculating the radius of curvature

https://www.physicsforums.com/threads/radius-of-curvature-of-universe.240763/​

The radius of curvature for the universe corresponding to a = 1 is

R_a=1 = c/( H₀ sqrt(-Ω_k) )​

4. For the purpose of this hypothetical universe, I am interested in the calculation of total entropy in the universe for the values a = 1 and a = 2. I assume the following initial conditions:

Ω_M << 1
Ω_R ≈ 1+|Ω_k|
Ω_k = - c² / (R_a=1² × H₀²)
Ω_Λ << 1​

Thus the hypothetical universe expansion and contraction effects are assumed to be almost entirely depending only on the radiation and curvature, the matter and dark energy contributions being negligible. The time it takes for the hypothetical universe the expand from a=1 to a=2 depends on a value assumed for H₀, and it can be calculated, but I will omit that calculation since I assume it to be irrelevant.
5. I choose initial conditions such that at a=2 the universe will stop expanding and begin its collapse. This requires that

H_a=2 = 0,​

which implies

Ω_R = 4 × |Ω_k|.​

This combined with the above relationship between Ω_R and Ω_k implies that

Ω_k = -1/3,​

and

Ω_R = 4/3.​

6. It is assume the universe between a=1 and 1=2 to be filled with H₂ gas and photons in thermodynamic equilibrium. The relevant variables for calculating entropy changes depend only on R (radius of curvature) and T (temperature) as discussed in my previous post. Here the useful forms are:

R_a=2 = 2 R_a=1
T_a=2 = (1/2) T_a=1​

The difference between entropy at a = 2 and at a = 1 is

S_a=2 - S_a=1 = C₂ × ln( (R_a=2/R_a=1)³ × (T_a=2/T_a=1)^3/2 )
= C₂ × ln( 8 × (1/8)^1/2 ) = C₂ × (1/2) ln 8 ≈ C₂ × 1.0397​

CONCLUSION
Let t* be the time between a=1 and a=2. Then

t' (> t_a=2) = t_a=1 + 2 t*​

is the time when a is again =1. Therefore as time increases from t_a=2 to t', entropy decreases from S_a=2 back again to S_a=1. This seems to violate the second law.
However,

https://www.physicsforums.com/threads/entropy-and-contracting-universe.943975/
Post #10 by @kimbyd
Gravity changes entropy dramatically. In fact, nobody knows how to actually calculate the entropy for a generic gravitational system. We know a few extreme cases (like black holes), but that's about it.​

[Underlining above is mine for highlighting purpose.] So, maybe the apparent violation of the second law really does not really happen because of the assumed hypothetical universe system has a definite gravitational component.

Regards,
Buzz

Classically-speaking, the entropy of a comoving volume of radiation is constant.

You can see this here:
https://en.wikipedia.org/wiki/Photon_gas

The entropy of a photon gas is proportional to $V T^3$. As volume increases proportional to $a^3$, and temperature decreases proportional to $1/a$, the entropy of the expanding (or contracting) volume does not change. This shouldn't really be a surprise since a uniform expanding (or contracting) photon gas doesn't change over time in any way other than the change in redshift (or blueshift) of the photons that make it up.

 

[Buzz Bloom]

Classically-speaking, the entropy of a comoving volume of radiation is constant.

Hi @kimbyd:

Thanks for your post.

I am confused by your comment since it restricts the issue to photons. The discussion I referenced was about a comoving system containing H₂ gas in a state of thermodynamic equilibrium. I mentioned that photons were also included, since there was a uniform temperature, but I ignored calculating the entropy of the photons because the use of the

Sackur-Tetrode equation

in my post #32 produces a clear increase in the H₂ gas entropy during expansion.

Do you disagree with the conclusion that (1) the fact that the finite hypothetical universe expansion and contraction is governed by gravitational effects, and (2) your comment that such systems can not have their entropy calculated, together avoid the violation of the second law during contraction?

Regards,
Buzz

 

[Grinkle]

a comoving volume of radiation

When one says comoving volume of radiation, is it the centroid of the volume that is comoving? I'm not confident I understand what that is meaning - can photons themselves be co-moving?

 

[kimbyd]

When one says comoving volume of radiation, is it the centroid of the volume that is comoving? I'm not confident I understand what that is meaning - can photons themselves be co-moving?

A comoving volume is a volume that increases in size along with the expansion. Think a hypothetical, finite-size cube whose sides scale along with the expansion factor.

In the case of the photon gas, the number of photons within a comoving volume is a constant. Photons exit the volume, but just as many photons enter it. The number density of photons decreases as the universe expands, and they also redshift as it expands.

 

[kimbyd]

Hi @kimbyd:

Thanks for your post.

I am confused by your comment since it restricts the issue to photons. The discussion I referenced was about a comoving system containing H₂ gas in a state of thermodynamic equilibrium. I mentioned that photons were also included, since there was a uniform temperature, but I ignored calculating the entropy of the photons because the use of the

I mentioned photons because your argument assumes radiation domination. Thus photons make up nearly all of the entropy of the system.

I'm pretty sure that as long as clustering isn't happening, though, any non-relativistic gases will also expand in an adiabatic fashion. Without clustering, the full behavior of the system at any point in time is determined solely by the temperature and density of the gases. Because the temperature and density are purely functions of the expansion, the expansion is trivially a reversible process. Thus it must be adiabatic.

With clustering, things get messy and nobody knows how to calculate the entropy. So in the case of the above, my bet is that there genuinely is something wrong with your calculations.

Edit:
One possible resolution, without looking into it in detail, is that you have to include both the H₂ gas and the photon gas in the calculation together, with changes in temperature of the H₂ gas causing energy to be dumped into or extracted from the photon gas.

 

[Grinkle]

With clustering, things get messy and nobody knows how to calculate the entropy.

Is this because a clumped system is not at equilibrium, or for some other (or maybe additional) reasons?

 

[Buzz Bloom]

In the case of the photon gas, the number of photons within a comoving volume is a constant.

Hi @kimbyd:

I started another thread

https://www.physicsforums.com/threads/how-to-calculate-the-number-of-photons.950409​

to seek some education about photons, and @Charles Link was helpful in leading me to the reference

https://en.wikipedia.org/wiki/Photon_gas.​

It is clear from the equation for N (the average number of photons) that the number does not change. It is also clear from the equation for S (entropy) that the entropy of the photons does not change as the hypothetical universe expands.

Am I correct that in a equilibrium mixture of H₂ gas and photons, the entropy of the mixture is the sum of he entropy of the gas and the entropy of the photons? If that is correct, then the entropy of the mixture increases (using these photon gas formulas photons together with the Sackur-Tetrode equation for the gas) as the hypothetical universe expands, and decreases as it contracts as discussed in my post #33.

So it becomes necessary to have a specific reason why this does not violate the second law. I tend to like the reason based on your statement which I quoted about gravity at the end of post #33. What do think about this?

Regards,
Buzz