I am currently looking to calculate the static structure factor of a computer-generated sphere packing I have been referring this paper as well as numerous other online sources to try and understand how to accomplish this.

It is my understanding that the pair correlation function $g$ of a homogeneous collection of point objects (such as the centers in a sphere packing) can be defined with the statement that $\rho g(\mathbf{r}-\mathbf{r}_{0})dV$ is the probability of finding an object in in arbitrarily small volume $dV$ at position $\mathbf{r}$ given there is an object at position $\mathbf{r}_{0}$, where $\rho$ is the global number density of the system. This implies that $g$ is a unitless quantity, and that $g(\mathbf{r})=1$ corresponds to a Poisson distribution.

I have read in several places that the structure factor $S(\mathbf{k})$ is related to the pair correlation function by the equation $$S(\mathbf{k})=1+\rho\bar{h}(\mathbf{k})$$ where $\bar{h}(\mathbf{k})$ is the Fourier transform of $h(\mathbf{r})=1-g(\mathbf{r})$, which I assume is also unitless.

I haven't been able to find a good explanation of the above relation. It seems to me that if $\rho$ signifies the global number density then $S(\mathbf{k})$ would depend on the units used to measure to measure the size of the system. Is there some kind of unitless density or am I misinterpreting some other part of the equation?

My knowledge of this type of statistical mechanics is still somewhat rudimentary, so any explanation or background information would be appreciated. Thanks.


The $g(\mathbf r_1, \mathbf r_2)$ is defined as

$$g(\mathbf{r}) = \frac{\rho^{(2)}(\mathbf{r}_1,\mathbf{r}_2)}{\rho^{(1)}(\mathbf{r}_1) \rho^{(1)}(\mathbf{r}_2)}$$


$$\rho^{(n)} (\mathbf r_1, \dots, \mathbf r_n) = \frac{N!}{(N-n)!} \frac 1 Z \int e^{-\beta V} d \mathbf r^{(N-n)}$$

If the system is homogeneous,

$$\rho^{(1)} (\mathbf r) = \rho \ \ \ \ \text{(bulk density)}$$

so that

$$g(\mathbf{r}) = \frac{\rho^{(2)}(\mathbf{r}_1,\mathbf{r}_2)}{\rho^2}$$

and if the system is also isotropic,

$$g(\mathbf r_1, \mathbf r_2) = g(\mid \mathbf r_1-\mathbf r_2\mid) = g(r)$$

So we can interpret $g(r)$ as the probability to find a particle in a volume $d \mathbf r$ around a chosen particle, and $g(r) \rho d \mathbf r$ as the average number of particles in the volume $d \mathbf r$.

Now, it can be shown that

$$\rho g(\mathbf r) = \frac 1 N \langle \sum_{i\neq j} \delta (\mathbf r + \mathbf r_i - \mathbf r_j) \rangle $$

The structure factor is defined as

$$S(\mathbf k) = \frac 1 N \langle \sum_{i,j} e^{-i \mathbf k \cdot (\mathbf r_i - \mathbf r_j)} \rangle$$

so that you have

$$S(\mathbf k) = 1+\frac 1 N \langle \sum_{i\neq j} \int d \mathbf r e^{-i \mathbf k \cdot \mathbf r} \delta (\mathbf r + \mathbf r_i - \mathbf r_j) \rangle = 1+\rho \int d \mathbf r e^{-i \mathbf k \cdot \mathbf r} g(r) = 1 + \rho \tilde g(\mathbf k)$$

where $\tilde {(.)}$ is the Fourier transform.

$h$ is defined as


So that

$$S(\mathbf k) = 1 + \rho \tilde h(\mathbf k) + (2 \pi)^3 \rho \delta(\mathbf k)$$

which for $\mathbf k \neq \mathbf 0$ becomes

$$S(\mathbf k) = 1 + \rho \tilde h(\mathbf k)$$

For a more complete exposition, I would suggest Theory of Simple Liquids by Hansen and McDonald.

The confusion arises from the fact that, while $h(r)$ is a dimensionless quantity, its Fourier transform $\tilde h(\mathbf k)$ is not: it has the dimension of a volume.

  • $\begingroup$ Thanks For the derivation. The issue seems to be that I was using a discrete Fourier-Bessel transform on h(r), which apparently differs from the continuous Fourier transform by a factor with appropriate dimensions. $\endgroup$ – Kajelad Jul 24 '16 at 18:19
  • $\begingroup$ @Kajelad You're welcome. If this answer solved your doubt, please consider accepting it. $\endgroup$ – valerio Jul 25 '16 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.