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Average of the internal energy of a system

1 vote

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:


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, 2018 in Physics Problems by Jorge Daniel (686 points)
edited Oct 5, 2018 by Jorge Daniel
I agree : this is more of a mathematical question. You have already got an answer on Mathematics SE. I think it would be better to ask the user who posted that answer to explain how to get the correct result.
Actually I asked Lecuippus for details in the resolution of the integral but got no answer... That is why I looked for help again. Should I make another question on MSE? My main problem is solving the integral and after applying the subsequent constants getting < E >.
Your question is a bit too complicated for me. I am very rusty on statistical mechanics. I don't understand what Lecuippus means by "add in the various constants" but I don't think that will make any difference to the answer.

The fact that you have already accepted the answer in Mathematics SE gives no incentive for Leucippus to answer any follow-up questions, and no incentive for anyone else to post a more complete answer. I suggest that you withdraw the "accepted" status, and ask specifically in the question, as well as in the comments to Leucippus, how to get $5/2\beta$ instead of $3/2\beta$. If you are able to you could also consider offering a bounty.

If that does not work then yes try posting another question.

The comment by Count Iblis provides the simplest, most physical solution. It is the answer which I would give if I knew what I was doing.
Okey thanks for your advises.
I would rather not hide this question and post the solution once I have it if that is OK for you.
Yes that is ok. You can post your own solution.

Incidentally, I don't see how you are getting $Z_1 \propto \beta^{9/4}$. I think you should get a factor of $\beta^{-3/2}$ from the definition of $\lambda_T$ and another factor of $\beta^{-3/2}$ from the integral. The product of these factors is the sum of the indices, which is $\beta^{-6/2}=\beta^{-3}$. But this is still not the correct answer to the question.
My answer is wrong. You are right saying that from the definition of $\lambda_T$ we get $\beta^{-\frac{3}{2}}$. Yes, that makes sense but there has to be something left (in theory a factor of $\beta^{\frac{1}{2}}$ is missing) which results in getting $\beta^{-\frac{5}{2}}$, the right answer. Otherwise the official result may be incorrect.
I think the official answer must be correct. The explanation of Count Iblis using the Theorem of Equipartition of Energy is simple and convincing : there are 5 degrees of freedom, and each degree of freedom holds an average energy of $\frac12 kT$. The calculation must arrive at the answer of an average energy of $\frac52 kT$ per molecule.
Ohh I see this is based on one of the theorems of statistical mechanics; 'the mean value of the kinetic energy for any motion at the absolute temperature $T$ is $\frac{1}{2} k T$ for each independent motion'. But what about the potential and other kind of energies? I mean, I have the understanding $\frac{1}{2} k T$ is just the mean value for the kinetic energy.
A degree of freedom can be a translational or rotational motion, which has only kinetic energy. It can also be a vibration/oscillation motion, which exchanges energy between kinetic and potential. Regardless of whichever type/mode of motion it is, the Equipartition Theorem says there is on average $\frac12 kT$ of energy in that mode.
I have not quite answered your question. Potential energy counts as another mode in which energy can be stored. So two masses joined by a spring can have oscillation energies which are kinetic and potential. This oscillator has on average $\frac12 kT$ of kinetic energy and $\frac12 kT$ of potential energy - a total of $kT$.

Actually the situation is more complicated than that. In general, if the potential energy is proportional to $x^s$ where $x$ is the displacement, then at equilibrium the average of this kind of PE is $\frac{1}{s}kT$. For springs $s=2$ so this works out the same as for kinetic energy. But for gravitational or electrostatic PE $s=-1$ so the average in these cases would be $kT$. See https://en.wikipedia.org/wiki/Equipartition_theorem.
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

1 vote
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, 2018 by sammy gerbil (28,806 points)
selected Oct 11, 2018 by Jorge Daniel
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, 2018 by Jorge Daniel (686 points)
edited Oct 11, 2018 by Jorge Daniel
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