Statistical mechanics, confused about an approximation and limits of integration

[fluidistic]

Homework Statement

Hello, I tried to solve a problem on my own and then I looked up a solution on the web, and I realize that it seems that I goofed. The problem statement can be found at http://www.hep.fsu.edu/~reina/courses/2012-2013/phy5524/homework/solutions/hw5_sol.pdf (Problem 1, part a).

Homework Equations

Partition function and Helmholt free energy expressions.

The Attempt at a Solution

O[/B]n page 2 of the document, it is written that $Z_1(T,V)=C\int d \vec R d\vec r \exp \left ( - \frac{\beta K |\vec r|^2}{2} \right )$ so far so good. Despite having only 1 "integral sign", it's really 6 integrals to perform.
Now according to the solution on the web, $Z_1(T,V)=C V \left ( \frac{2\pi kT}{K} \right )^{3/2}$.
It means that the solution considers that the triple integral with $d\vec R$ is equal to the volume of the container of the gas, where $\vec R =\frac{\vec r_1 +\vec r_2}{2}$; I still have to digest this.
But what really annoys me is the other triple integral, namely $\int d\vec r \exp \left ( - \frac{\beta K |\vec r|^2}{2} \right )=\left ( \frac{2\pi kT}{K} \right )^{3/2}$
Where $\vec r =\vec r_1-\vec r_2$. The result the solution provide implies that the limits of integration are negative and positive infinity... On my draft I considered a cubic box of lengths $V^{1/3}$ so that the result of that triple integral contained the error function (erf).
And I do not see any argument as to why I could assume that the limits of integrations can be considered as infinite. Is it because the lengths of the sides of the box are huge compared to $|\vec r|$?
Ah no... it must be that the integrand is almost 0 for small values of r and onward... okay... and so the approximation is valid. Is this correct?

The other problem I have with the solution given is when they obtain the Helmholtz free energy as $F\approx NkT \{ \ln (N) -1 -\ln [Z_1(T,V)] \}$ and they just sit on that value.
But since N is enormous, isn't better to reduce F to approximately $NkT \{ \ln (N) -\ln [Z_1(T,V)] \}$?
Why do they keep the "1"... is it because it's multiplied by NkT ? Still that doesn't make sense to me... it's ridiculous small compared to the natural logarithm of N...

 

[mark726]

Hello, thank you for sharing your attempt at the solution. It seems like you have some valid concerns and questions about the solution provided. Let's address them one by one.

Firstly, you are correct in your understanding that the triple integral with $d\vec{R}$ is equal to the volume of the container. This is because $\vec{R}$ represents the center of mass of the two particles, so integrating over all possible values of $\vec{R}$ gives the volume of the container.

You are also correct in your understanding that the limits of integration for the other triple integral, $\int d\vec{r} \exp \left ( - \frac{\beta K |\vec{r}|^2}{2} \right )$, can be considered as infinite due to the fact that the integrand becomes very small for small values of $\vec{r}$. This is known as the "thin shell" approximation, where we only consider the contribution from particles that are close to each other (i.e. within a thin shell), as the contribution from particles that are far apart is negligible.

As for your question about the Helmholtz free energy, the solution provided is indeed a valid approximation. The "1" term that you are questioning is actually the constant term in the Taylor expansion of the natural logarithm. When we have a very large number like N, this constant term becomes insignificant compared to the logarithmic term, so it can be neglected. However, in some cases, it may be useful to keep this term for the sake of completeness.

I hope this helps to clarify your doubts. Keep up the good work in trying to understand the solution and do not hesitate to ask for further clarification if needed. Good luck with your future problem solving!