Mean energy using the partition function

[patrickmoloney]

Homework Statement

System of two energy levels, $E_0$ and $E_1$ is populated by $N$ particles, at
temperature $T$. The particles populate the levels according to the classical
(Maxwell-Boltzmann) distribution law.

(i) Write an expression for the average energy per particle.

Homework Equations

The Attempt at a Solution

The partition function of our system $$z=\sum_s{e^{-\beta E_s}}= e^{-\beta E_0}+ e^{-\beta E_1}$$ where $\beta = \frac{1}{kT}$.
The probability of any number of the $N$ particles being in either system is given by

$$P_0 = \frac{1}{z}e^{-\beta E_0}$$ $$P_1 = \frac{1}{z}e^{-\beta E_1}$$

The average energy $\overline{E}$ is

$$\overline{E}= -\frac{1}{z}\frac{\partial z}{\partial \beta} = \frac{E_0 +E_1}{e^{-\beta E_0}+ e^{-\beta E_1}}$$since $$\frac{\partial z}{\partial \beta }= \frac{\partial}{\partial \beta}(e^{-\beta E_0}+ e^{-\beta E_1}) =-(E_0 + E_1)$$

is this a correct method to the problem?

 

[vela]

You wrote down the partition function for one particle. You need the one for N particles.

 

[patrickmoloney]

You wrote down the partition function for one particle. You need the one for N particles.

$$z= \sum_s{e^{-\beta E_s}}$$ is that what you mean? and then just use this to find $\overline E$

 

[patrickmoloney]

Sorry, I get what you mean now! We haven't covered that in class yet so here goes:

Letting $$z_1= e^{-\beta E_0}+ e^{-\beta E_1}$$

Thus for $N$ particles we can use
$$Z_N = \frac{(z_1)^N}{N!}$$

and plugging in our value for $z_1$ we get

$$Z_N = \frac{(e^{-\beta E_0}+ e^{-\beta E_1})^N}{N!}$$

 

[NFuller]

$$Z_N = \frac{(e^{-\beta E_0}+ e^{-\beta E_1})^N}{N!}$$

Yes, this looks right. However it may be useful to write your definition for the average energy in post #1 in a different way
$$\left<E\right>=-\frac{\partial}{\partial\beta}\text{ln}(Z_{N})$$
This will allow you to make use of Stirling's approximation to get rid of the $N!$ and get rid of the $N$ power.

 

[vela]

There's no need for Stirling's approximation since $N!$ is a constant.

Patrick, you should take another look at your expression for $\partial z/\partial\beta$ as well. You didn't differentiate correctly.

 

[patrickmoloney]

There's no need for Stirling's approximation since $N!$ is a constant.

Patrick, you should take another look at your expression for $\partial z/\partial\beta$ as well. You didn't differentiate correctly.

I understand what you mean now! $$Z_{sp} = (Z_{sp})^N$$ For distinguishable particles.

 

[NFuller]

There's no need for Stirling's approximation since $N!$ is a constant.

What does $N$ being constant have to do with Stirling's approximation? The only requirement is that $N$ be a very large number, which is often assumed in these types of problems.

I understand what you mean now! $$Z_{sp} = (Z_{sp})^N$$ For distinguishable particles.

Even for a classical system the $N!$ is necessary in order for the entropy to be extensive.

 

[patrickmoloney]

What does $N$ being constant have to do with Stirling's approximation? The only requirement is that $N$ be a very large number, which is often assumed in these types of problems.

Even for a classical system the $N!$ is necessary for the entropy to be extensive.

Does that mean I should always use $$Z_N = \dfrac{(Z_1)^N}{N!}$$

 

[NFuller]

Does that mean I should always use $$Z_N = \dfrac{(Z_1)^N}{N!}$$

Not always, but the problem never said the particles were not indistinguishable it only said it was a classical system. I feel like the problem could be more specific but my intuition is that you should keep the $N!$ since leaving it out leads to things like the Gibb's paradox.

 

[vela]

What does $N$ being constant have to do with Stirling's approximation? The only requirement is that $N$ be a very large number, which is often assumed in these types of problems.

Nothing, but if the $\log N!$ term is going to disappear anyway when you differentiate, why bother approximating it in the first place?

 

[NFuller]

Nothing, but if the $\log N!$ term is going to disappear anyway when you differentiate, why bother approximating it in the first place?

I see your point. However I would still argue in favor of including it for pedagogical reasons. If the particles are indistinguishable, writing $Z$ without the $N!$ term is incorrect; if the problem has a second part asking for the entropy or free energy, then the OP would be lead to the wrong answer.

 

[patrickmoloney]

$$\langle E \, \rangle= -\dfrac{\partial \ln Z}{\partial \beta}= \Bigg {(}\dfrac{E_1 e^{-E_1 \beta}}{1 + e^{-E_1 \beta}}\Bigg{)}$$

if we let $E_0 = 0$ be our zero-energy level.

 

[NFuller]

$$\langle E \, \rangle= -\dfrac{\partial \ln Z}{\partial \beta}= \Bigg {(}\dfrac{E_1 e^{-E_1 \beta}}{1 + e^{-E_1 \beta}}\Bigg{)}$$

if we let $E_0 = 0$ be our zero-energy level.

This is the average energy of an individual particle. So if that is what you are looking for, then this is correct. For the average energy of the whole ensemble, you need a factor of $N$.
$$\langle E \, \rangle= -\dfrac{\partial \ln Z_{N}}{\partial \beta}= \Bigg {(}\dfrac{NE_1 e^{-E_1 \beta}}{1 + e^{-E_1 \beta}}\Bigg{)}$$

 

[patrickmoloney]

Yep it's just for one particle. What would happen to the average energy if the temperature $T \to 0$ and $T \to \infty$ I assume that for $T \to \infty$ we would get

$$\langle E \rangle = E_1$$

since $\dfrac{E_1 e^{0}}{1 + e^{0}} = E_1$. For $\beta = \dfrac{1}{kT}$

 

[NFuller]

Close, what is $1+e^{0}$?

 

[patrickmoloney]

Close, what is $1+e^{0}$?

Right! $$\dfrac{E_1}{2}$$

The other one is kind of confusing me. I think as $T \to 0$ we have $\dfrac{-E_1}{kT} \ll 0$. That's all I can kind of deduce from that

 

[NFuller]

As $T\rightarrow0$, $\beta\rightarrow\infty$. So what happens to $e^{-\beta E_{1}}$ and hence $\langle E \rangle$ in this limit?

 

[patrickmoloney]

As $T\rightarrow0$, $\beta\rightarrow\infty$. So what happens to $e^{-\beta E_{1}}$ and hence $\langle E \rangle$ in this limit?

$$e^{-E_1 \beta} = \dfrac{1}{e^{E_1 \beta}}$$

and $\beta \to 0 \implies e^{-E_1 \beta} =0$ But wouldn't that give us the same value $\dfrac{E_1}{2}$??

 

[NFuller]

$\beta \to 0 \implies e^{-E_1 \beta} =0$

I think you mean $\beta \to \infty \implies e^{-E_1 \beta} =0$. With this in mind, what is $E_{1}e^{-\beta E_{1}}$ in this limit?

 

[patrickmoloney]

I think you mean $\beta \to \infty \implies e^{-E_1 \beta} =0$. With this in mind, what is $E_{1}e^{-\beta E_{1}}$ in this limit?

Oh sorry yeah. That gives us $$\langle E \, \rangle= \dfrac{0}{1+0}=0$$

Thank you so much for that!