In the derivation of the momentum transport equation in Kirkwood's paper (https://aip.scitation.org/doi/10.1063/1.1747782), I am stuck at a particular point in the derivation. The rate of change of the expectation value of a dynamical quantity is derived (using Liouville's theorem and the assumption that $\gamma$ doesn't explicitly depend on time) and is given as: $$\frac{\partial\langle\gamma;f\rangle}{\partial t} = \sum_{k=1}^{N} \Big \langle\Big(\frac{p_{k}}{m_{k}}\cdot\nabla_{R_{k}}\Big)p_{k}\delta(R_{k}-r)-\nabla_{R_{k}}U\cdot\nabla_{p_{k}}\gamma ; f\Big\rangle $$ where $\gamma$ is any time dynamical variable and $f$ is the phase space distribution/density. $U$ is the interparticle potential. $p_{k}$ is the momentum of the $k$th particle and $R_{k}$ is the position of the $k$th particle. Now consider $$\gamma = \sum_{i=1}^{N} = p_{i}\delta(R_{i} - r)$$ The first term in the first equation in the paper is given as $$\Big(\frac{p_{k}}{m_{k}}\cdot\nabla_{R_{k}}\Big)p_{k}\delta(R_{k}-r) = - \nabla_{r} \cdot \Big[\frac{p_{k}p_{k}}{m_k}\delta(R_{k}-r)\Big].$$ I tried to arrive at the above identity, using index notation; $$\frac{p^{\alpha}_{k}}{m_{k}}\nabla^{\alpha}_{R_k}p^{\beta}_{k}\delta(R_k-r)$$ Using the identity $$\nabla_{x^{\prime}} f(x-x^{\prime} )=- \nabla_{x} f(x-x^{\prime}) $$ we get, $$=-\frac{p^{\alpha}_{k}}{m_{k}}p^{\beta}_{k}\nabla^{\alpha}_{r}\delta(R_k-r) + \frac{p^{\alpha}_{k}}{m_{k}}\delta(R_k-r)\nabla^{\alpha}_{R_k}p^{\beta}_{k} $$ Is the above expression correct? I am not quite sure how to proceed beyond this point or even if I am doing it right. Thanks for the answers.

  • $\begingroup$ Could you please address the basis of that equation (where you saw it, what your hypothesys are, and so on)? $\endgroup$ – FGSUZ Jul 21 '18 at 18:19
  • $\begingroup$ @FGSUZ Please check $\endgroup$ – Rohan Mittal Jul 21 '18 at 18:39

OK, I think that I have figured out what is being asked here. Not everyone will have access to the famous Irving-Kirkwood paper, unfortunately. The equation which seems to be at issue is (5.3) which is typeset in an ugly way over several lines in the journal. The kinetic term in the equation is $$ \left(\frac{\mathbf{p}_k}{m_k}\cdot \mathbf{\nabla}_{\mathbf{R}_k}\right)\mathbf{p}_k \delta(\mathbf{R}_k-\mathbf{r}) $$ The quantity in parentheses is a scalar dot product. As you suggest, the cleanest way to approach this is by explicitly writing out the indices and using the Einstein convention to sum over repeated Greek indices $$ \left(\frac{p_k^\alpha}{m_k}\nabla_{\mathbf{R}_k}^\alpha\right) p_k^\beta\delta(\mathbf{R}_k-\mathbf{r}) = \left(\frac{p_k^\alpha p_k^\beta}{m_k}\right) \nabla_{\mathbf{R}_k}^\alpha \delta(\mathbf{R}_k-\mathbf{r}) . $$ The gradient has been moved, since it operates only on functions of $\mathbf{R}_k$, i.e. the delta function in this case. Again as you suggest, we can use the identity to change it to a derivative with respect to $\mathbf{r}$: $$ -\left(\frac{p_k^\alpha p_k^\beta}{m_k}\right)\nabla_{\mathbf{r}}^\alpha \delta(\mathbf{R}_k-\mathbf{r}) $$ and it doesn't matter if we place the gradient at the front, since it doesn't act on the momenta $$ -\nabla_{\mathbf{r}}^\alpha \left(\frac{p_k^\alpha p_k^\beta}{m_k}\right) \delta(\mathbf{R}_k-\mathbf{r}) $$ Remembering the Einstein convention, we have omitted the explicit $\sum_{\alpha}$ symbol, but it is understood that we are summing over $\alpha$ here. With that in mind, this is the corresponding term on the right hand side of eqn (5.3) in the paper, except that they write it in dyadic notation: $$ -\mathbf{\nabla}_{\mathbf{r}} \cdot \left(\frac{\mathbf{p}_k \mathbf{p}_k}{m_k}\right) \delta(\mathbf{R}_k-\mathbf{r}) $$ The juxtaposition $\mathbf{p}_k \mathbf{p}_k$ is taken to mean a $3\times 3$ matrix with indices $\alpha\beta$ and the dot product is a contraction of the first index $\alpha$ with the index on the gradient vector.

So your method was essentially correct, but the answer that you wanted to be checked seems to have a mysterious extra term in it.

Just to put this in context for people who come across this question. The aim of the derivation is to take the time derivative of the momentum density and show that it may be written as the spatial divergence of something: the (negative of the) stress tensor. The momentum density is expressed as a field, in terms of particle momenta, using the Dirac delta function. The time derivative may be written in terms of derivatives, with respect to particle positions and momenta, using the Liouville equation. This part of the derivative leads to the ideal gas or kinetic part of the stress tensor; the other part depends on intermolecular forces. The derivation may be found in several textbooks. In another answer the book by DJ Evans and GP Morriss "Statistical Mechanics of Nonequilibrium Liquids" was mentioned. My own preference is for JP Hansen and IR McDonald's "Theory of Simple Liquids", where Fourier-transformed densities are defined, thus avoiding some of the tiresome business of delta functions.

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  • $\begingroup$ I kinda get it. In equation 2, you shifted the $p^{\beta}_{k}$ and claimed that the gradient operator only acts on the delta function , but doesn’t the momentum of the $k$ th particle also depend on the time derivative of $R_{k}$? Or did you shift it because the $\beta$ component of the momentum has no dependence on the $\alpha$ component of the gradient operator for $\alpha \neq \beta$. Thanks. Sorry if this is a silly question to ask. $\endgroup$ – Rohan Mittal Jul 23 '18 at 22:50
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    $\begingroup$ It certainly isn't $\alpha\neq\beta$ because we are summing over all $\alpha=x,y,z$, one of which will be $\beta$. No, it is because the particle coordinates $\mathbf{R}_k$ and momenta $\mathbf{p}_k$ are taken to be independent variables, defining the $6N$-dimensional phase space. Any dynamical variable depends on some or all of these variables and the usual rules of calculus apply: if it is independent of $R_k^\alpha$, say, then differentiating it with respect to $R_k^\alpha$ will give zero. None of this changes because of the existence of an equation of motion. $\endgroup$ – user197851 Jul 24 '18 at 7:27
  • $\begingroup$ That clears it up. Thank you. I had been doubting this thing about the independence of $R_{k}$ and $p_{k}$. The derivations fell into place when I had assumed this independence but couldn’t be too sure of it. Thanks a lot for clearing it up. This was exactly one of my concerns when I posted the question. $\endgroup$ – Rohan Mittal Jul 24 '18 at 7:55
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    $\begingroup$ OK, I've made a couple of minor edits to my answer, to try and clarify these points in the light of your comments, for other readers. $\endgroup$ – user197851 Jul 24 '18 at 8:18

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