# Hard sphere gas

1. Jun 14, 2014

### WannabeNewton

Hi guys. Consider the problem of calculating the multiplicity (phase space volume) of N hard sphere gases each of whose center of mass is confined to a volume V. The spheres themselves have volume $\omega$ and do not interact with one another in equilibrium time scales. Then $\Omega \propto V_S \int d^3q_1...d^3q_N$ where $V_S$ is the volume of the 3N-sphere defined by $\sum \frac {p_i^2}{2m} =E$.
Now the standard treatment then claims $\int d^3q_1...d^3_N = V (V -\omega)... (V-(N-1)\omega)$ by adding in one particle into the box at a time e.g. the first particle has access to volume V while the second has access to volume $V-\omega$ due to the presence of the first. But this claim doesn't fully make sense to me, the reason being this claim calculates the configuration space volume sequentially for each particle i.e. we first just have one particle and calculate $\int d^3 q_1$ to get V and then calculate $\int d^3q_2$ to get $V-\omega$ since the second particle is put in after the first and so has an accessible volume lessened by that of the first and so on. However in reality the phase space volume must be calculated assuming all particles are in the box simultaneously as we are in a 3N dimensional configuration space consisting of the center of mass coordinates of all the particles. This means we must correlate the restrictions on the allowed coordinates of one center of mass to the restrictions on those of the others simultaneously in the volume integral. We cant calculate this configuration space volume by pretending theres first only one particle with access to the full volume V and then treat it as just a volume $\omega$ sitting in the box when a second particle is added and doing the integral for this particle in a manner completely uncorrelated to the set of allowed dynamical coordinates of the first and so on. I hope my confusion is somewhat clear as I am not sure how to properly articulate it. So how then does one justify calculating the configuration space volume integral in this manner? Is it some kind of approximation? If so could someone explain the approximation in detail? Thanks in advance!

2. Jun 14, 2014

### Jano L.

Yes, it is approximation. It is an analogy to the task :

Find the number of arrangements of $N$ balls into $M$ departments.

There we calculate this way: first we have to place one ball somewhere, we have $M$ possibilities. Then we have to place second ball, for that we have $M-1$ possibilities. And so on... in the end we have $M.(M-1)...(M-(N-1))$ possibilities of arrangement. Although we put the balls one by one, it is clear that we have thus calculated all possibilities - there are not any others.

Similarly for the phase volume, only this time, the idea that each ball decreases the volume available by the same value is only an approximation, based on the idea that all balls are isolated from each other so that the spheres of excluded volume (of radius $2R$) are not connected. For dense arrangement of the balls , they will begin to connect and addition of new ball will not decrease the available volume by as much as $4/3 \pi (2R)^3$, but less since part of the sphere is already excluded by the other balls.

3. Jun 15, 2014

### WannabeNewton

Hi Jano, thanks for the reply. Actually what I'm confused about is more primitive than that which you elucidated. What I'm looking for is a formal argument justifying going from the exact volume integral $\int_{V^N} d^3 q_1...d^3 q_N$ with the constraint $|q_i - q_j| \geq 2R$, to the approximate integral $\int _V d^3 q_1 \int _{V - \omega}d^3 q_2...\int _{V - (N-1)\omega}d^3 q_N$. Do you know of such a formal argument?

Last edited: Jun 15, 2014
4. Jun 15, 2014

### Jano L.

Sorry, I do not know. It seems like a difficult math problem.

5. Jun 16, 2014

### WannabeNewton

That's alright Jano, I appreciate the help you already gave. Perhaps you can help me understand how one transitions from the full integral to the approximate integral at least intuitively?

For simplicity let's say we only have two hard spheres of volume $\omega$ in a box of volume $V$. The full volume integral in phase space satisfies $\Omega(E) \propto \int d^3 q_1 d^3 q_2$ with the constraint that $|q_1 - q_2| \geq 2R$. The constraint rules out all accessible microstates $(q_1,q_2, p_1,p_2)\in V^2 \times S^2$ for which $|q_1 - q_2| < 2R$. In principle then all we would have to do is take $A \equiv \{(q_1,q_2)\in V^2: |q_1 - q_2| < 2R\}$ and find the volume of $V^2 \setminus A$ with the measure $d^3q_1 d^3q_2$ i.e. $\int_{V^2 \setminus A} d^3 q_1 d^3q_2$.

Now this is all pretty formulaic but I'm just trying to see how to get from here to the logical leap that we can put in the hard spheres one after another and calculate the accessible microstates for each hard sphere separately and sequentially. What exactly is the approximation being made in this regard? More precisely, we first calculate $\int _V d^3q_1 = V$ which constitutes all allowed values of $q_1$ since it's the only hard sphere in the box; then we put in the second hard sphere and find all allowed coordinates of it in the presence of the first hard sphere i.e. $\int _{|q_2 - q_1| \geq 2R} d^3q_2$.

My questions are then:

(1) What is the approximation being made in $\int_{V^2 \setminus A} d^3 q_1 d^3q_2 \approx \int _{|q_2 - q_1| \geq 2R} d^3q_2\int_V d^3 q_1$ and why can we make it? If this is what you answered above then I apologize but I'm not seeing the connection too easily.

(2) Why is $\int _{|q_2 - q_1| \geq 2R} d^3q_2 = V - \omega$? My problem is that in making the approximation in (1), we've broken up the configuration space $V^2$ of the entire system, which simultaneously includes all particles, into the physical spaces of each particle separately such that the physical space of each successive particle includes the hard core exclusion of the previous particle. But if we're looking at the physical space of the second particle exclusively, with the presence of the first, then we are looking for all values of $q_2(t)$ inside the box which remains a diameter's length away from the center of mass of the first hard sphere. But in the physical space of the second particle, the first particle is moving around i.e. $q_1 = q_1(t)$; we don't have a static collection of 2-tuples $(q_1,q_2)$ anymore. With the first particle moving around in the space of the second particle, how does one then find all values of $q_2(t)$ inside the box which remains a diameter's length away from $q_1(t)$?

I hope I'm not being too confusing. Thanks!

6. Jun 16, 2014

### Jano L.

$$\int_{[q_1,q_2] \in V^2 \setminus A} d^3 q_1 d^3q_2 = \int_{q_1\in V} \left( \int_{q_2\in V ~\&~ |q_2-q_1|>2R} d^3q_2 \right) d^3 q_1$$
The transition to the right-hand side should be exact. The excluded volume for placement of the ball 2 depends on the position of the ball 1: if the ball 1 is placed right next to the wall, the excluded volume is only half of the ball of radius $2R$, while if the ball 1 is placed in the center of the container, the excluded volume is full ball of radius $2R$. However, if the volume $V$ of the container is much larger than its surface multiplied by $R$, this can be neglected.

Then the approximation can be made in the right-hand side: it is assumed that ball 1 decreases the volume of the space available to 2 by $\omega = \frac{4}{3}\pi (2R)^3$ whatever the position of 1.

Then the inner integral is $V-\omega$, which does not depend on $q_1$ and thus can be removed from the integral over $q_1$, which then is equal to $V$. The resulting value is therefore $(V-\omega)V$.

When there are two balls and third is being placed, there is similar dependence of the excluded volume on the configuration of the system 1+2 already at place - the excluded volume is $2\omega$ only for some configurations of 1+2 (when the balls are both far enough from the wall and from each other (this is what I described at first).
If the volume of the balls already in the container is much smaller than $V$, this can be neglected and placement of $k-$th ball has volume $V-k\omega$ available.

If the above approximations can be made, each new ball decreases volume available by $\omega$, independently of the position of the ball 1. This is so because in this approximation, any possible motion of the balls won't change the unoccupied volume. So their motion is irrelevant for calculating this volume.