I Value of beta in Boltzmann Statistics taking degeneracy into account

JohnnyGui

Summary
It is concluded in MB Statistics that $\beta = -\frac{1}{k_BT}$ But this is based on Boltzmann Statistics when degeneracy of quantum states is not taken into account. If it is taken into account, the equation to solve for $\beta$ gives a different value. Is this correct or am I misunderstanding something?
Hello,

The relationship between entropy $S$, the total number of particles $N$, the total energy $U(β)$, the partition function $Z(β$ and a yet to be defined constant $β$ is:
$$S(\beta)=k_BN \cdot \ln(Z(\beta)) - \beta k_B \cdot U(\beta)$$
$$\frac{dS}{d\beta} = -k_B\beta \cdot \frac{dU}{d\beta}$$
And since $dS = \frac{dU}{T}$, this means that $\beta = -\frac{1}{k_BT}$. This is the known derived value for $\beta$ for the Maxwell-Boltzmann Distribution. Source starting from sheet 40

However, this derivation does not take the degeneracy of quantum states into account. If it does then $S(β)$ would have an extra parameter in its formula. If the number of quantum states of an energy level is $gj$, then this wold be:
$$S = k_BN \cdot \ln(Z(\beta)) - k_B\beta\cdot U(\beta) + k_B\cdot \sum^n_{j=1}\bigg[\ln(g_j)\cdot \frac{N}{Z(\beta)} \cdot e^{\beta E_j}\bigg]$$
And since $\ln(g_j) = \ln\big(\frac{N}{Z(\beta)}e^{\beta E_j}\big)- \beta E_j - \frac{N}{Z(\beta)}$, this would eventually give me:
$$\frac{dS}{d\beta} = k_B\cdot\left( - \beta\cdot\frac{dU}{d\beta} + \frac{U(\beta)\cdot N}{Z(\beta)} - U(\beta) \right)$$
Derivation under Part II (I asked about the derivation there and someone helped me with it)
But using again $dS = \frac{dU}{T}$, this relationship doest not give $\beta = -\frac{1}{k_BT}$.

Is it permitted for $β$ to have a different value than $-\frac{1}{k_BT}$ when quantum states is taken into account or am I misunderstanding something here?

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PeterDonis

Mentor
am I misunderstanding something here?
I suspect you must be, since other derivations that take degeneracy into account still obtain $\beta = \frac{1}{k_B T}$ (the sign is just a convention). For example, see the derivation from the microcanonical ensemble here:

JohnnyGui

I suspect you must be, since other derivations that take degeneracy into account still obtain $\beta = \frac{1}{k_B T}$ (the sign is just a convention). For example, see the derivation from the microcanonical ensemble here:

I was about to derive it based on that link but before starting I noticed that the wiki is stating that:
$$\ln(W) = \sum_{j=1}^n \bigg[n_j \cdot \ln(g_j) - n_j \cdot \ln(n_j) + n_j \bigg]$$
While sheet 13 in this lecture states:
$$\ln(W) = N \cdot \ln(N) - N + \sum_{j=1}^n \big[n_j \cdot \ln(g_j)\big] -\sum_{j=1}^n \big[n_j \cdot \ln(n_j)\big] + \sum_{j=1}^n [n_j]$$
They don't seem to be the same equation

PeterDonis

Mentor
They don't seem to be the same equation
That's right. Can you see why? (Hint: the equation at the top of slide 13 in the lecture also appears in the Wikipedia article. But the Wikipedia article does not take the log of that equation; it takes the log of a different equation, further down the article, and explains the reason why.)

PeterDonis

Mentor
They don't seem to be the same equation
Also note that, if you remove the first two terms from the RHS of the equation from slide 13 of the lecture, they are the same equation. And those first two terms are the same whether there is degeneracy or not. So this difference between the two equations should not lead to any difference in the final formula due to degeneracy.

JohnnyGui

That's right. Can you see why? (Hint: the equation at the top of slide 13 in the lecture also appears in the Wikipedia article. But the Wikipedia article does not take the log of that equation; it takes the log of a different equation, further down the article, and explains the reason why.)
The Wiki corrected the equation for the Gibbs Paradox while the lecture has not?

Also note that, if you remove the first two terms from the RHS of the equation from slide 13 of the lecture, they are the same equation. And those first two terms are the same whether there is degeneracy or not. So this difference between the two equations should not lead to any difference in the final formula due to degeneracy.
Yes, I indeed noticed that. Is it because those constant terms will disappear either way because the derivative of $\ln(W)$ will eventually be taken for the further derivation?

Regardless of the further derivation, since the lecture does not remove those two terms based on the Gibbs paradox, doesn't this mean that one of the two sources is incorrect regarding the equation for $\ln(W)$ itself?

stevendaryl

Staff Emeritus
I think there's a confusion from the very first post about whether you are talking about total entropy, or entropy per particle. Similarly, whether your $Z(\beta)$ means the partition function for the many-particle system, or the partition function for a single particle.

If you're talking about the total entropy and partition function, then the relationship is:

$S = k (ln(Z(\beta)) + \beta U(\beta))$

There is no factor of $N$.

JohnnyGui

I think there's a confusion from the very first post about whether you are talking about total entropy, or entropy per particle. Similarly, whether your $Z(\beta)$ means the partition function for the many-particle system, or the partition function for a single particle.

If you're talking about the total entropy and partition function, then the relationship is:

$S = k (ln(Z(\beta)) + \beta U(\beta))$

There is no factor of $N$.
Does choosing entropy per particle or total entropy depend on whether or not you include degeneracy?

If so, it is not only the factor of $N$ that changes my derivation for $S$ when degeneracy is taken into account, so I don't think that's the cause of my misunderstanding.

BTW, the lecture also includes the factor of $N$.

PeterDonis

Mentor
The Wiki corrected the equation for the Gibbs Paradox while the lecture has not?
Yes.

I indeed noticed that. Is it because those constant terms will disappear either way because the derivative of $\ln(W)$ will eventually be taken for the further derivation?
Yes.

since the lecture does not remove those two terms based on the Gibbs paradox, doesn't this mean that one of the two sources is incorrect regarding the equation for $\ln(W)$ itself?
The lecture is not correcting for the Gibbs paradox; whether you think that makes it "incorrect" depends on what you think the lecturer was trying to accomplish. Since, as you have agreed, the correction for the Gibbs paradox makes no difference once you take the derivative of $\ln(W)$, the lecturer might simply have wanted to keep things simple by not mentioning the Gibbs paradox since it makes no difference to the result the lecture was intended to communicate.

stevendaryl

Staff Emeritus
Okay, I think I understand what's going on. In the source http://hep.ph.liv.ac.uk/~hock/Teaching/StatisticalPhysics-Part1-Handout.pdf

$Z$ is definitely the partition function for a single particle. It's defined as:

$Z(\beta) = \sum_i e^{-\beta \varepsilon_i}$

where $\varepsilon_i$ is the single-particle energy level for state $i$.

But in this formula, the index $i$ ranges over all possible states, not over all possible energy levels. So there is no degeneracy factor $g$ involved. This formula is not for calculating the number of particles with energy $e_i$, it's for calculating the number of particle in state $i$. This formula allows for the possibility that two different states might have the same energy.

So the Wikipedia article is doing a sum over energy levels, while the handout is doing a sum over states. So you can't directly compare them. In the approach taken in the handout, the degeneracy does not enter into the calculation. It would enter if instead of wanting to know the number of particles in state $i$, you wanted to know $n(\varepsilon)$, the number of particles with energy $\varepsilon$.

PeterDonis

Mentor
the Wikipedia article is doing a sum over energy levels, while the handout is doing a sum over states
And if what you are interested in is the correct formula for $\beta$, as in the OP of this thread, then what you want is the sum over energy levels, because $\beta$ comes in as a Lagrange multiplier for the energy (to enforce the constraint that the total energy is conserved). So the Wikipedia article's method is the correct method for finding a formula for $\beta$ in the presence of degeneracy.

JohnnyGui

Okay, I think I understand what's going on. In the source http://hep.ph.liv.ac.uk/~hock/Teaching/StatisticalPhysics-Part1-Handout.pdf

$Z$ is definitely the partition function for a single particle. It's defined as:

$Z(\beta) = \sum_i e^{-\beta \varepsilon_i}$

where $\varepsilon_i$ is the single-particle energy level for state $i$.

But in this formula, the index $i$ ranges over all possible states, not over all possible energy levels. So there is no degeneracy factor $g$ involved. This formula is not for calculating the number of particles with energy $e_i$, it's for calculating the number of particle in state $i$. This formula allows for the possibility that two different states might have the same energy.

So the Wikipedia article is doing a sum over energy levels, while the handout is doing a sum over states. So you can't directly compare them. In the approach taken in the handout, the degeneracy does not enter into the calculation. It would enter if instead of wanting to know the number of particles in state $i$, you wanted to know $n(\varepsilon)$, the number of particles with energy $\varepsilon$.
So what you're saying is that the summation in the handout sums up states with the same energies along with different energy levels? Such that they're counting in degeneracy as well but in a different manner than the wiki?
If so, then that's what I was thinking too, until I read, starting from sheet 21, that they're treating the numbers of $j$ only as different energy levels. Other sources with the same derivation as the handout also only speak of summing up different energy levels.

And if what you are interested in is the correct formula for ββ\beta, as in the OP of this thread, then what you want is the sum over energy levels, because ββ\beta comes in as a Lagrange multiplier for the energy (to enforce the constraint that the total energy is conserved). So the Wikipedia article's method is the correct method for finding a formula for ββ\beta in the presence of degeneracy.
If the handout is counting in degeneracy as well but in a different manner (summing up states with the same energy as I think @stevendaryl is saying), doesn't this mean that the handout shouldn't come up with the correct value for $\beta$, since you're saying that the Wiki is the correct way to derive $\beta$ using $g_j$ as degeneracy?

Furthermore, the handout is not applying the correction for the Gibbs paradox and yet it concludes the correct value for $\beta$ by using its "incorrect" equation for $\ln(\Omega)$ through the formula of entropy $S$. This puzzles me as well.

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PeterDonis

Mentor
in this formula, the index $i$ ranges over all possible states, not over all possible energy levels.
I'm not sure this is the intent of that handout. The handout in the text right below equation (41) on sheet 37, which is the equation for $Z$ that you give, says the summation ranges over all energy levels, not over all states. It never mentions degeneracy at all.

• JohnnyGui

stevendaryl

Staff Emeritus
I'm not sure this is the intent of that handout. The handout in the text right below equation (41) on sheet 37, which is the equation for $Z$ that you give, says the summation ranges over all energy levels, not over all states. It never mentions degeneracy at all.
Hmm. Maybe for simplification they are only considering non-degenerate energy levels?

PeterDonis

Mentor
Maybe for simplification they are only considering non-degenerate energy levels?
That's my suspicion, yes.

JohnnyGui

I have figured it out. @PeterDonis @stevendaryl

The way I included degeneracy was correct but I made some subtle mistakes during substitution of some parameters.

The formula for $\ln(\Omega)$ when degeneracy is taken into account is:
$$\ln(\Omega)= N \cdot \ln(N) - N - \sum^n_{j=1}[n_j \cdot \ln(n_j) - n_j] + \sum^n_{j=1} [\ln(g_j) \cdot n_j]$$
Substituting $n_j = g_j \cdot \frac{N}{Z} \cdot e^{\beta E_j}$ (I erroneously left out the $g_j$ before during this substitution) along with rewriting, splitting the summations and simplifying eventually gives me:
$$\ln(\Omega) = N \cdot \ln(Z) - \beta U$$
Which is the exact equation as when degeneracy is not taken into account, and thus I get the same value for $\beta$ when taking the derivative of $S = k_B \cdot \ln(\Omega)$ and putting it next to the equation of entropy $dS = \frac{dU}{T}$.

EDIT: The following is incorrect, they are different.
Apparently, correcting for the Gibbs paradox is not necessary. I used the "incorrect" $\ln(\Omega)$ formula to substitue $n_j$ and the only differences with the Wiki's final formula for $\ln(\Omega)$ are the signs in front of $\alpha$ ($= \ln(\frac{N}{Z}$)) and $\beta$ which are just conventions, since you can either add or substract Lagrange Multipliers.

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• stevendaryl

PeterDonis

Mentor
Apparently, correcting for the Gibbs paradox is not necessary.
If all you're interested in is the correct relationship between $S$ and $\beta$, yes. You figured out why in post #6.

JohnnyGui

If all you're interested in is the correct relationship between $S$ and $\beta$, yes. You figured out why in post #6.
No, this time I'm talking about the formula $\ln(\Omega)$ itself when $n_j$ in that formula is substituted by $g_j \cdot \frac{N}{Z} \cdot e^{\beta E_j}$. Whether it's the Wiki's stated formula for $\ln(\Omega)$ or that of the handout, it always simplifies to $\ln(\Omega) = N \cdot \ln(Z) - \beta U$. There is no derivative this time that makes the first two terms of the handout's formula obsolete.

Hold on, I think I got it wrong, they're different...

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PeterDonis

Mentor
this time I'm talking about the formula $\ln(\Omega)$ itself
Well, as we've already seen, correcting for the Gibbs paradox certainly does make a difference in that formula. The difference just doesn't matter when you take the derivative.

Note that the reason for taking the derivative is to find the distribution of particles that maximizes $\ln(\Omega)$, i.e., that maximizes entropy. Also note that in the Wikipedia article, the derivative is taken with respect to $N_i$ (for each value of $i$), not $\beta$.

JohnnyGui

Well, as we've already seen, correcting for the Gibbs paradox certainly does make a difference in that formula. The difference just doesn't matter when you take the derivative.

Note that the reason for taking the derivative is to find the distribution of particles that maximizes $\ln(\Omega)$, i.e., that maximizes entropy. Also note that in the Wikipedia article, the derivative is taken with respect to $N_i$ (for each value of $i$), not $\beta$.
Indeed, just double checked and they're different. When using the same signs for $\alpha$ and $\beta$:
Handout's formula: $\ln(\Omega) = N \cdot \ln(Z) - \beta U$
Wiki's formula: $\ln(\Omega) = N - \ln(\frac{N}{Z})\cdot N + \beta U$

"Value of beta in Boltzmann Statistics taking degeneracy into account"

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