# How to use the calculate velocity correlation from VanHove correlation?

I am trying to derive an expression for the velocity correlation function $C(t) = \langle \mathbf{V}(t) \mathbf{V}(0)\rangle$ from the VanHove correlation:

$$G(\mathbf{r},t) = \frac{1}{N}\left( \sum_{i,j} \langle \delta(\mathbf{r}- \mathbf{r}_i(t) +\mathbf{r}_j(0) \rangle \right)$$

However I am not sure how to get the time dependency and the $\dot{\mathbf{r}}(t)$ from the VanHove correlation. The average $\langle\rangle$ is over the ensemble.

I tired to figure out how to derive a term for velocity autocorrelation from the original paper, but without any luck.

Probably by now you have found the answer elsewhere, but for the benefit of others who may come across this question, here is my answer.

The link to the original paper seems to be broken. However the problem you describe is fairly well known in the field of liquid state physics, and the lecture notes that you refer to give a good introduction to the relevant correlation functions, including the van Hove function. Certainly this is explained in Theory of Simple Liquids by JP Hansen and IR McDonald (Academic Press, 4th edition). That is a comprehensive graduate-level text which doesn't pull its punches: the liquids may be simple, but the theory isn't!

There are two problems in making this connection. Firstly, as you have noted, the van Hove function relates to the positions of the atoms, not their velocities. Therefore, we must anticipate that it will need to be time-differentiated (twice, in fact).

The second problem is that the velocity correlation function involves the correlation of the velocity of a particle $i$ at time $0$ with the velocity of the same particle $i$ at time $t$. However, the van Hove function involves cross correlations of the positions of different atoms $i\neq j$ as well as positions of the same atom $i=j$. These are respectively the distinct part and the self part of $G(\mathbf{r},t)$, as defined in the notes that you cite. Formally, the velocity correlation function may be linked to the self part. If you are attempting to connect it to the full $G(\mathbf{r},t)$, you need to develop a justification for neglecting the cross correlations.

The self part of the van Hove function is defined $$G_s(\mathbf{r},t) = \frac{1}{N}\left\langle \sum_{i=1}^N \delta[\mathbf{r}+\mathbf{r}_i(0)-\mathbf{r}_i(t) ] \right\rangle = \left\langle \delta[\mathbf{r}+\mathbf{r}_i(0)-\mathbf{r}_i(t) ] \right\rangle$$ where I have chosen one arbitrary atom $i$ since they are all equivalent.

It is convenient to Fourier transform this in space, to give the self-part of the intermediate scattering function (so-called because of its importance in inelastic neutron scattering experiments): $$F_s(\mathbf{k},t) = \int G_s(\mathbf{r},t) \exp(-i\mathbf{k}\cdot\mathbf{r}) d\mathbf{r} = \left\langle \exp(i\mathbf{k}\cdot\mathbf{r}_i(t)) \, \exp(-i\mathbf{k}\cdot\mathbf{r}_i(0)) \right\rangle .$$ For simplicity, we will define $\mathbf{k}$ to lie along the $z$-axis: $$F_s(k,t) = \left\langle \exp(ikz_i(t)) \, \exp(-ikz_i(0)) \right\rangle .$$ Then we differentiate with respect to time, twice. Let $v_i$ be the $z$-component of the velocity of particle $i$. $$\frac{d^2}{dt^2}F_s(k,t) = -k^2\left\langle v_i(t) \exp(i k z_i(t)) \, v_i(0) \exp(-i k z_i(0)) \right\rangle .$$ A subtle trick has been used on the right hand side: the so-called dot-shifting trick. Because of stationarity $\langle \dot{A}(t) B(0)\rangle = -\langle A(t) \dot{B}(0)\rangle$ where $A$ and $B$ are any variables, and the dot denotes time differentiation. This is how we are able to bring down $v_i(t)$ from the first exponential, and $v_i(0)$ from the second one.

The derivation is completed by dividing both sides by $-k^2$ and taking the limit $k\rightarrow 0$: $$-\lim_{k\rightarrow 0} \frac{1}{k^2}\frac{d^2}{dt^2}F_s(k,t) = \left\langle v_i(t) v_i(0) \right\rangle .$$ This is the answer.

In some circumstances one can make the Gaussian approximation, namely that the self part of the van Hove function is a Gaussian function of $\mathbf{r}$ at all times. This is true at short times (free-particle limit) and at long times (diffusion limit), but has been shown to be inaccurate (by about 10 percent) in typical liquids at intermediate times. If we nonetheless assume that it is true, then the intermediate scattering function is a simple function of the mean-squared displacement $$F_s(k,t) = \exp\left[-\frac{1}{6}k^2 \langle |\mathbf{r}_i(t)-\mathbf{r}_i(0)|^2\rangle\right]$$ where now $k$ represents the magnitude of the $\mathbf{k}$-vector. This can be related straightforwardly to the velocity correlation function.

• Thank you! This is great. Any chance you add a comment if it's applicable in the theory of liquid crystal? – 0x90 Jun 19 '18 at 16:16
• In a liquid crystal the van Hove function, and the velocity correlation function, will both lose their spherical symmetry. For example, in the nematic phase, with the director along the $z$-axis, they will both have cylindrical symmetry about that axis: motion in the $x$ and $y$ directions will be equivalent, but motion in the $z$ direction will be different. However, I don't think that this will affect this particular relationship. It should go through for each of the $x$, $y$ and $z$ components separately, pretty much unchanged. – user64968 Jun 19 '18 at 16:44