Pinned with hard homework problem

In summary, the problem involves finding the solution to the Liouville equation for a noninteracting gas of N particles in a cubic box with periodic boundary conditions. The Hamiltonian is independent of position and the probability density at time t=0 is given. The formal solution to the Liouville equation is of the form e^{-Lt}\rho\left ( \mathbf{p} ,\mathbf{q}\right,0 ) and the boundary conditions can be easily accounted for. The problem can be simplified by considering only one particle and the given integral does not seem to be necessary for solving the problem.
  • #1
lamq_31
1
0

Homework Statement


I am having problems with the following problem: "For a noninteracting gas of N particles in a cubic box of volume [itex]V = L^3 [/itex] where L is the length of the side of the box, find the solution, [itex]\rho (\mathbf{p}^{3N},\mathbf{q}^{3N},t) [/itex], of the Liouville equation at time t, where [itex] \mathbf{p}^{3N} = \left ( \mathbf{p_{1},...,\mathbf{p_N}} \right ) [/itex] and [itex] \mathbf{q}^{3N} = \left ( \mathbf{q_{1},...,\mathbf{q_N}} \right ) [/itex]. Assume periodic bondary conditions, and that the probability density at time t=0 is given by:


Homework Equations


[tex] \rho \left ( \mathbf{p}^{3N},\mathbf{q}^{3N},t \right ) = \left ( \frac{\sqrt{\pi}}{2L} \right )^{3N}\prod_{i=1}^{3N}e^{-p_{i}^{2}/2m}sin\left ( \pi q_{i}/L \right ) [/tex]
with
[itex] 0\leq q_{i}\leq L [/itex]

The following equation may be helpful:
[itex] \int _{0}^{L}dx \sin \left ( \pi q/L \right )\ln \left [ \sin\left ( \pi q/L \right ) \right ] = \frac{L}{\pi}\left ( 2 - \ln 2\right ) [/itex]

The Attempt at a Solution


So far, I started from the Liouville equation:
[itex] \frac{\partial \rho}{\partial t} = -\sum _{i}^{3N}\left [ \frac{\partial \rho}{\partial q}\frac{\partial H}{\partial p} + \frac{\partial \rho}{\partial p}\frac{\partial H}{\partial q}\right ] [/itex]
Since the gas is noninteracting, I can assume that the Hamiltonian is:
[itex]H = \sum _{i}^{3N}\frac{p_{i}^2}{2m} [/itex]
So, I arrive to the equation:
[itex]\frac{\partial \rho}{\partial t} = -\sum _{i}^{3N}\frac{p_{i}}{m}\frac{\partial \rho}{\partial q} [/itex]
I have read in a few sources that the formal solution to the Liouville equation is of the form:

[itex] \rho\left ( \mathbf{p},\mathbf{q},t \right ) = e^{-Lt}\rho\left ( \mathbf{p} ,\mathbf{q}\right,0 ) [/itex]

And, in this case:
[itex] L =\sum _{i}^{3N}\frac{p_{i}}{m}\frac{\partial }{\partial q} [/itex]
However , at this point I don't know how to proceed because I don't know how explicitly include boundary conditions. (Actually I am chemist and I am not used to solving partial differential equations).
Thank you very much for your help
 
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  • #2
Hello lamq_31!

I think this problem may be easier than it at first appears - here's a few comments:

I think the boundary value business is probably trivial - you just have to imagine filling up all of space with little cubes of side L, all of which have the same initial probability density as your original cube. Because the Hamiltonian is independent of position, the probability density will continue to be periodic in q at any future time.

Next comment is that all the particles are doing exactly the same thing, so you can just think about what one particle is doing. That reduces your degrees of freedom from 3N to 3!

So then you write down your Liouville equation for one particle (for what it's worth there should be a - sign in the middle of that equation). That should tell you what you expected anyway - that the probability fluid in phase space just drifts along in q-space with speed determined by its momentum.

What I don't understand is why they've given you that integral - it's an integral for the entropy of the system (which stays constant). I don't think it's needed to solve the problem at hand.

:confused:
 

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