# Why is a hard sphere gas correlated?

In stat mech we calculated the radial distribution function (a.k.a. pair correlation function) for a classical gas by using perturbation theory for the BBGKY hierarchy. (I could post more details of the calculation if you want, but it is a rather long winded but standard perturbation theory type calculation.) The result we got was

$$g_2 (r) = \mathrm{e}^{-u(r)/T} \left[ 1 + n_0 \int\mathrm{d}^3r'\ f(r')f(|\vec{r}-\vec{r}'|)\right],$$

where $u(r)$ is the interaction potential, $T$ is the temprature, $n_0$ is the density and $f(r)=\mathrm{e}^{-u(r)/T}-1$ is the Mayer function. $g_2$ roughly measures the probability of finding two particles seperated by a distance $r$. $n_0$ is the small parameter of the perturbation theory.

If you then apply this result to hard sphere (infinite repulsive potential of diameter $a$), you get this:

Now it makes perfect sense that $g_2$ is zero for $r<a$. Also the asymptote to one at large $r$ is part of the definition of $g_2$, meaning that particles are uncorrelated at large distances. The problem is the peak at $r\sim a$ which implies that you are more likely to find particles clustered together, despite the complete absence of any attractive forces! Why is that?

Our lecturer seems to think it is because when two particles collide they stop, then bounce, hence spending more time in the vicinity of each other than for an ideal gas. But this seems dubious because perfect hard sphere collisions are instantaneous. I can imagine three possibilities:

• This argument could be formalized as a limit of soft sphere scattering and is the correct explanation of the correlation,
• there is some other (presumably entropic) explanation,
• the correlation doesn't exist - the perturbation theory gives a qualitatively wrong picture (seems unlikely in this case).

So what is it?

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Is there misunderstanding here? For $|r|<a/2$, $u=\infty$, so $f(r)=-1$. The convolution integral in the equation of the two sphere larger than $|r|>a$ have no overlap, so it should be 0 and $g_2(r)=1$ –  hwlau Aug 29 '13 at 9:09
@hwlau We covered that in lecture. $a$ is the diameter of the spheres but the radius of the Mayer function, so the Mayer functions overlap for $a<r<2a$. Counter-intuitive yes! But it's because the seperation between the centres of hard sphere particles is twice the radius when they touch, so $u(r)\to\infty$ at $r=a$ rather than $a/2$. –  Michael Brown Aug 29 '13 at 9:26

If two particles are close to each other, there is more space for the rest of the particles to move. This gives rise to an effective entropic attraction between the particles because when looking at two particles for different separations while "tracing out" over the degrees of freedom of the rest of the system, the entropy of the rest is higher when the two tagged particles you are looking at are close to each other.

In fact at high density, you should also observe oscillations in the g(r) and not a single peack. The width of the bumps in these oscillations is related to the particle size.

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How exactly does this work? The volume excluded by the particles is constant, unless the separation is less than $2a$, in which case particles are excluded between the two particles, i.e. you can't fit a third particle between them... This effect would seem to go the other way than you say! –  Michael Brown Aug 29 '13 at 9:43
@MichaelBrown: I guess the point it that if the particles are less than two diameters apart, then the closer they are the smaller the gap between them is, and so the more room there is for other particles. Basically, looking at the possible locations of the centers of each particle, each of them excludes other particles from a sphere of radius $a$ around it. But if you bring two particles closer than $2a$ apart, then their exclusion radii overlap, leaving more room for other particles elsewhere. –  Ilmari Karonen Aug 29 '13 at 10:32
@IlmariKaronen Aaahhh... that seems to be making sense. I'd have to do another little calculation to make sure this is the complete effect, but it sounds plausible. Thank you. :) –  Michael Brown Aug 29 '13 at 10:34
Yes limari Karonen's explanation is right. This entropic effect exists also if you consider a gas of hard spheres in a box with hard walls. The one particle density that you will find will be higher at the walls than in the middle of the box because it is simply more likely for them to be there as they optimize the available volume for the other particles and hence the entropy of the system –  gatsu Aug 29 '13 at 11:07

I think gatsu's right: it's because of an entropic attraction resulting from the fact that two spheres whose centers are less than $2a$ apart leave more room for other spheres. To see why this happens, it may help to draw a picture:

Here, the blue spheres all have radius $\frac12a$. The gray dashed circle around each sphere has radius $a$, and shows the region from which the centers of other spheres are excluded.

Since the spheres A and B are more than $2a$ apart (measured from their centers), the total volume from which they together exclude other spheres simply equals the sum of their respective exclusion volumes. Spheres B and C, however, are closer than $2a$ from each other, and so their exclusion zones partially overlap, leaving more room for sphere A and any other spheres in the gas.

Looking only at pairwise interactions, this explanation predicts that the quasi-attractive interaction between the particles should appear only at distances less than $2a$. This matches your graph, which looks flat at $r/a > 2$. Of course, a more through analysis would also consider clusters of three or more nearby particles, which ought to produce some effects at longer distances (at least at sufficiently high gas densities), but presumably your perturbation analysis also neglected these higher-order effects.

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I see now this is the right track (nice picture by the way!). Indeed the integral is just the volume of the overlap between two spheres of radius $a$, so the $n_0 \int \cdots$ term just counts the average number of particles excluded. So this is obviously the number that matters, but why it should contribute to the correlation function in precisely the way it does (without say a factor of two or anything like that) is still a bit mysterious to me... This has been really helpful however. –  Michael Brown Aug 29 '13 at 12:07
This is a good answer. Comparable enough to gatsu's answer but the picture really helps. Hope you don't mind that I accepted gatsu's instead because he has the lower score. :) Thanks anyway though. –  Michael Brown Aug 30 '13 at 1:19
@MichaelBrown: No, of course I don't mind. My answer is really just an extension of gatsu's, anyway. –  Ilmari Karonen Aug 30 '13 at 14:03