How Does Equation (23) Apply to Classical Systems with Total Energy E?

[A-fil]

I've got a few problems that I got stuck on. I'll add more problems in this thread as time passes.

Homework Statement

3.5 Show that Equation (23) holds for any classical system that has a total energy of the form
E = \sum_{i=1}^{N}{\bf{ϵ}(\bf{v}_{i},\bf{r}_i)}
where ϵ(v_i,r_i) is any function of (v_i,r_i). Obtain an expression for the particle partition function in this case.

Homework Equations

Equation (22): f(E) = D_N Z^{-1}e^{-\beta E}

Equation (23): Z = \delta^{N}
where the quantity δ is known as the molecular (or particle) partition function.Z = D_N \int_{v}\int_{-\infty}^{\infty}d^{3}rd^{v_N}e^{-\beta E}

The Attempt at a Solution

I should be able to assume that N is fixed. Then D_N could be assumed to be 1 [length-velocity^-3N].

\delta = Z^{1/N} = ( \int_{v}\int_{-\infty}^{\infty}d^{3}r d^{3}v_N e^{-\beta E})^{1/N}= \int_{v}\int_{-\infty}^{\infty}d^{3}r d^{3} v_N e^{- \beta \frac{E}{N}}

Now, is there something about classical systems that assures the following? Because this is the answer according to my book:

\frac{E}{N}=\frac{\sum_{i=1}^{N}{\bf{ϵ}(\bf{v}_{i},\bf{r}_i)}}{N}=\bf{ϵ}(\bf{v},\bf{r})

Thankful for all the help I can get.

 

[henry_m]

Hi A-fil,

You haven't used all the information that's been given to you. Your strategy should be to evaluate Z given the form of E. Bear in mind that you need to integrate over all the degrees of freedom; how many are there?

You seem to be trying to claim that
\left(\int \mathrm{d}x f(x)\right)^{1/N}=\int \mathrm{d}x f(x)^{1/N}.
Try a few simple examples and convince yourself that this doesn't make sense.

 

[A-fil]

Hehe, the integral thing was a classical case of me being tired. Thanks for replying!

So, I get to this:
Z = \int_{V}d^{3} r_{1} ... \int_{-\infty}^{\infty}d^{3}v_{N} e^{-\beta E} = \int_{V}d^{3} r_{1} ... \int_{-\infty}^{\infty}{d^{3}v_{N} e^{\sum_{i=1}^{N}{-\beta \bf{ϵ}(\bf{v}_i,\bf{r}_i)}}} = \prod_{i=1}^{N}\int_{v}d^{3} r_{i}\int_{-\infty}^{\infty}{d^{3}v_{i} e^{-\beta \bf{ϵ}(\bf{v}_i,\bf{r}_i)}}

Now, how do I go from this to?

(\int_{V}d^{3} r \int_{-\infty}^{\infty}d^{3}v e^{-\beta \bf{ϵ}(\bf{v},\bf{r})})^N

 

[Mute]

In your integral in the product, r_i and v_i are dummy variables - you can relabel them however you like. That includes getting rid of the label i. Since the only thing you have that depends on i is the dummy variables, the integrals are independent of i after the relabelling. So, what do you get?

 

[A-fil]

Ohh, thanks! That would give me what I'm looking for (see below). However, I still don't fully understand why. Is it because we integrate over all of the room V and for all speeds v, so whatever inital energy particle number i had doesn't really matter, meaning that we can say that all particles had a certain energy (in this case ϵ(v,r)? I really want to get an intuitive feel for what a partition function is.


\delta = \int_{V}d^{3} r \int_{-\infty}^{\infty}d^{3}v e^{-\beta \bf{ϵ}(\bf{v},\bf{r})}

By the way, I have a few other questions in the same field (from the same book). Should I post them here or in a new thread?

 

[Mute]

The reason is the energy function \epsilon(\mathbf{v},\mathbf{r}) is the same for every particle, and all particles are independent. Since the particles are independent (no interactions between them), your 6N-dimensional integral factor into a product of N independent 6-dimensional integrals. Then, since every particle feels the same \epsilon, all N integrals are the same.

Some ways the result could change: suppose instead of all particles being the same, all particles were different (say they all have different masses). Then \epsilon would be different for every particle. The particles are still independent, so the 6N-dimensional integral still factors into N 6-dimensional integrals, but the 6 dimensional integrals are no longer the same. You would thus find your partition function to be

Z = \prod_{i=1}^N \delta_i
where
\delta_i = \int_V d^3r \int_{-\infty}^\infty d^3v e^{-\beta \epsilon_i(\mathbf{v},\mathbf{r})}.

Another way the result could be different is if there were interactions between the particles. In this case you generally can't factor the 6N-dimensional integral into independent integrals because the interaction couples all of the integrals together. The partition function can't usually be calculated exactly in such cases. Some sort of approximation scheme is generally needed.

If you have further questions related to the current question, you can continue to post in this thread. However, if you have separate questions, you are free to post a new thread - that way people will see that it has not been answered and you are more likely to get responses.