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Average of the internal energy of a system

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I am struggling to get the right value for $Z_1$, which is $\beta^{\frac{-5}{2}}$ multiplied by constants. Once you have the value for the one particle partition function it is just about applying the definition of average energy.

$$\langle E \rangle = -\frac{ \partial \ln(Z_1)}{\partial \beta}$$

I got :

$$ \langle E \rangle = \frac{9}{4\beta}$$

The right answer is though:

$$\langle E \rangle = \frac{5}{2\beta}$$

NOTE: I know this is more a mathematical issue and I have already asked for this question in MSE, but I could not get the right value for < E > yet. If you want to have a look:

https://math.stackexchange.com/questions/2941009/solving-a-partition-function-in-polar-coordinates

More information about the partition function and more definitions if needed be. Please let me know if you need information that is not stated:

$$\beta = \frac{1}{k_B T}$$

asked Oct 5 in Physics Problems by JD_PM (490 points)
edited Oct 5 by JD_PM
Your first explanation is based on the degrees of freedom but I am thinking about the particles. They do have potential energy, that is why I am confused as I directly relate $\frac12 kT$ to kinetic energy and do not see how the Equipartition Theorem takes into account the potential energy...
I cannot see any potential energy mentioned in the Rigid Rotor problem.

Perhaps the container is in a gravitational field, in which case there should be gravitational PE of $kT$ per molecule. However, this assumes that the molecules can gain $kT$ of gravitational PE relative to the bottom of the container.

At room temperatures $kT$ is much larger than the gravitational PE which the molecules can gain in rising from the bottom to the top of the container. So the molecules do not actually have the potential to store this amount of gravitational PE while confined to a container of laboratory dimensions. For this reason it is neglected during calculations.

The same caveat applies for springs : If the spring cannot extend freely to the point at which the KE is zero, then the average PE stored in this oscillation is less than $\frac12 kT$.
Yes I assumed the container is in a gravitational field. Oh I see that GPE can be neglected, I understand your point now, thanks.
@Sammy_gerbil may you have a look at my answer? By the way, what can I do to fix MathJax issues?
@Sammy_gerbil problem solved. I would rather select your answer because it provides a physical insight. If you do not matter, I would like to keep posting questions related to topics that you may have not studied in a while but in which you can provide physics knowledge. I will do research in mathematical methods and post a second answer afterwards if needed be. What do you think?

2 Answers

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Best answer

Setting the mathematics aside, the solution suggested by Count Iblis in a comment to your question in Mathematics SE, is convincing and physically intuitive. It uses the Equipartition of Energy Theorem. This states that when a system is in thermal equilibrium at temperature $T$, each degree of freedom has an average kinetic energy of $\frac12 kT$.

In the present case there are 5 degrees of freedom per diatomic molecule : 3 orthogonal directions of translational motion, and 2 orthogonal axes of rotation. Because the atoms within the diatomic molecule are point particles there is no rotation about the axis joining them.

So the result is very simply obtained : the average energy per molecule is $5\times \frac12 kT=\frac52 kT$.


If the bond between the atoms acted as a spring, so that the molecule could vibrate as an harmonic oscillator, there would be another $\frac12 kT$ from the kinetic energy and also $\frac12 kT$ from the potential energy, making $\frac72 kT$ in total.

As you have suggested there ought also to be some average potential energy associated with vertical motion in the gravitational field. Because this is linear in the displacement variable $z$ ($U=mgz$), rather than quadratic like the potential for the harmonic oscillator ($U=\frac12 kz^2$), the average energy is $kT$ instead of $\frac12 kT$. In general, for a potential function proportional to $z^p$ the average energy would be $\frac{1}{p}kT$.

However, the gas is confined to a laboratory-sized container which prevents molecules from exploring the full range of the potential energies from $0$ to $kT$. At room temperature $kT$ is very much larger than the difference in gravitational PE for typical diatomic gas molecules such as nitrogen and oxygen. An extremely tall container would be required. For laboratory-sized containers and atmospheric gases the capacity of the gravitational field for storing energy can be neglected. But if the particles were much larger, such as sediment or colloid particles suspended in a liquid, the gravitational PE would become significant even for small containers.

answered Oct 9 by sammy gerbil (26,096 points)
selected Oct 11 by JD_PM
0 votes

The first detail to take into account is that the factor $\lambda_t$ does not appear until we have integrated over the momentum and position (and it will appear if we want to), thus it should not appear before integrating over it. Note that in Cartesian coordinates the one particle partition function is (this is its general form):

In this case the integral has to be arranged in polar coordinates so :

Note $\hbar^5$ plays a dimensional role and it is powered to 5 because there are 5 momentum degrees of freedom.

Let's start our integral. Firstly, the integral over position $\vec{Q}$ (3D) which gives the volume V:

$$\int d\vec{Q} = V$$

Secondly, the integral over momentum $\vec{P}$ (3D) which gives:

$$\int d\vec{P} e^{\frac{-\beta \vec{P}^2}{2m}} = (\frac{2 \pi m}{\beta})^{3/2}$$

Note it is the momentum over the 3 components x, y, z.

Thirdly we integrate over $p_\theta$:

$$\int dp _{\theta} e^{\frac{-\beta p_{\theta}^2}{2I}} = (\frac{2 \pi I}{\beta})^{1/2}$$

Fourthly and fifthly we integrate over $\theta$ and $p_\phi$

Finally we integrate over $ \phi$:

$$\int_{0} ^{2 \pi} d \phi = 2 \pi$$

We just have to multiply:

$$Z_1 = \frac{4 \pi V}{\hbar^5} \frac{2 \pi I}{\beta} (\frac{2 \pi m}{\beta})^{3/2} = \beta ^{-5/2} * consts$$

Then we just have to compute the following:

$$\langle E \rangle = -\frac{ \partial \ln(Z_1)}{\partial \beta} = \frac{5}{2\beta}$$

NOTE: I wrote $\lambda_t$ would appear if we wanted to because you can get the answer without using the definition:

$$\lambda_t = h \sqrt{\frac{\beta}{2 \pi m}}$$

If we were to use it, our $Z_1$ would look like this:

$$Z_1 = \frac{4 \pi V}{\hbar^2 \lambda^3} \frac{2 \pi I}{\beta}$$

answered Oct 8 by JD_PM (490 points)
edited Oct 11 by JD_PM
I have tried to fix the formatting error, but without success. I have sent a message to Kenshin (Super Administrator) to have a look at it.

Your answer has moved the problem (how to derive $Z_1 \propto \beta^{-5/2}$ mathematically) a little further forward, but has not solved it. I suggest that you post a message in the Problem Solving chatroom asking John Rennie if he can help.
I will do it, thanks
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