4
$\begingroup$

Cross-posted from the PhysicsSE.

I am reading Theory of Simple Liquids by Hansen and McDonald, and they in chapter 3, they describe the density-density correlation for a simple liquid in the grand canonical ensemble. This is how they have defined it: $$H^{(2)}(r,r') = \langle [\rho (r) - \langle \rho (r) \rangle ][ \rho (r') - \langle \rho (r') \rangle ]\rangle \tag{1}$$ $$ = \rho ^{(2)}(r,r') - \rho ^{(1)}(r)\rho ^{(1)}(r') + \rho ^{(1)}(r) \delta (r-r').\tag{2}$$ This is my understanding of where the terms in the RHS comes from. I understand that the two-particle density $\rho^{(2)}$ term arises from: $$\langle \rho (r) \rho (r') \rangle\tag{3}$$ in the RHS.

The second term in the RHS comes from expanding the the product and taking like terms together.

$$-\rho ^{(1)}(r)\rho ^{(1)}(r') = -\langle \rho(r) \langle \rho (r') \rangle \rangle - \langle \rho(r') \langle \rho (r) \rangle \rangle + \rho ^{(1)}(r) \rho ^{(1)}(r'),\tag{4}$$ where $\rho^{(1)}(r)$ is the average single-particle density at $r$, for a homogeneous liquid.

But I still do not get why that $\delta$-function exists there. Is the $\delta$-function just there to say that if the two particles are in the same spot ($r=r'$), the density correlation is... maximized? How does the $\delta$ fall out of the averaging, mathematically?

I would appreciate any advice you have for me!

$\endgroup$

1 Answer 1

3
$\begingroup$

While deriving this formula in Ch. 3 of your source, the author references equation 2.5.13, \begin{equation} \rho_N^{(2)}\left(\mathbf{r}, \mathbf{r}^{\prime}\right)=\left\langle\sum_{i=1}^N \sum_{j=1}^N{}^{'} \delta\left(\mathbf{r}-\mathbf{r}_i\right) \delta\left(\mathbf{r}^{\prime}-\mathbf{r}_j\right)\right\rangle.\tag{2.5.13} \end{equation} The author calls $\rho^{(2)}\left(\mathbf{r},\mathbf{r}'\right)$ an analogue of the above.

Notice the second sum has a prime indicating it is missing a set of terms where $i = j$. Those missing terms are

\begin{equation} \left\langle\sum_{i=1}^{N}\delta\left(\mathbf{r} - \mathbf{r}_{i}\right)\delta\left(\mathbf{r}' - \mathbf{r}_{i}\right)\right\rangle = \left\langle\sum_{i=1}^{N}\delta\left(\mathbf{r} - \mathbf{r}_{i}\right)\right\rangle\delta\left(\mathbf{r} - \mathbf{r}'\right) =\rho^{(1)}\left(\mathbf{r} \right)\delta\left(\mathbf{r} - \mathbf{r}'\right), \end{equation} where the final equality comes from Eq. (3.1.2) in the source. Anytime $\mathbf{r} \neq \mathbf{r}'$, the whole thing vanishes due to the two coupled $\delta$-functions.

$\endgroup$
5
  • $\begingroup$ +1. Just to be clear: the equation that you labeled 2.5.13 in your answer is also 2.5.13 in the book? I'm a bit confused because your sentence seems to suggest that 2.5.13 in the book is an "analogue" of your 2.5.13. $\endgroup$ Jan 31, 2023 at 4:19
  • $\begingroup$ I edited the comment for clarity. $\endgroup$
    – Hayden S
    Jan 31, 2023 at 4:35
  • 1
    $\begingroup$ This is great! Thank you for spelling it out! $\endgroup$
    – megamence
    Jan 31, 2023 at 16:04
  • $\begingroup$ I am trying to write in words what the second equation you have written means. Put simply, $\rho (r,r')$ is a measure of how likely it is for two identical particles to be at $r$ and $r'$, averaging (integrating) out the positions of other particles. What I am struggling to see is how I can go from the left-most term in the second equation, to the one in the middle. Why does the $\delta (r-r')$ term come OUTSIDE the ensemble averaging brackets? $\langle \rangle$? $\endgroup$
    – megamence
    Jan 31, 2023 at 17:24
  • $\begingroup$ To begin with, in the second $\delta$ function, you can clearly switch out $\mathbf{r}_{i}$ for $\mathbf{r}$ without affecting the overall sum, as $\mathbf{r} = \mathbf{r}' = \mathbf{r}_{i}$ or the term drops out. So then you must think about whether averaging over the sum of products of $\delta$ functions is any different than averaging over the sum of one and multiplying by the other. In either case if $\mathbf{r} \neq \mathbf{r}'$, the entire average drops to zero. Only if $\mathbf{r} = \mathbf{r}'$ does the average become finite, and both options will simply multiply all terms by 1. $\endgroup$
    – Hayden S
    Jan 31, 2023 at 17:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .