I was studying the equation of motion for the probability density function of the position coordinates of the Brownian particles, also known as the Smoluchowski Equation (SE).

Particularly, I came across:

$$\frac{\partial}{\partial t} \rho (r,t) = D_o [\nabla^2 \rho (r,t) + \beta \nabla \rho (r,t) \int dr' [\nabla V (|r-r'|)] \rho (r',t) g(r,r',t)]$$

Which is the SE form for interacting particles.


1) $\rho (r,t)$ is the probability density function.

2) $D_o$ is the diffusion coefficient.

3) $$\nabla = \frac{\partial}{\partial x}$$

4) $$\beta = \frac{1}{k_\mathrm BT}$$

5) $V (|r-r'|)$ is the potential.

6) $g(r,r',t)$ is the pair correlation function.

I got curious and wanted to verify this equation using dimensional analysis.

We know that:

$$[D_o] = \frac{L^2}{T}$$

$$[\beta] = \frac{T^2}{ML^2}$$

$$[\rho] = \frac{1}{L T}$$

$$[\nabla] = \frac{1}{L}$$

$$[g] = \frac{1}{L^2T}$$

Note that dimensions of $\rho$ come from the fact that the integral of the probability density function over its entire support is 1. I used the same reasoning with $g$.

Based on this information I got on the left hand side:


My struggle is triggered by the second term on the right hand side of the equation, as I got as the final result:

$$\frac{1}{LT^2} = \frac{1}{LT^2} + \frac{1}{L^6T^3}$$

I think I may be missing something related to the dimensions of the correlation function...

  • $\begingroup$ Are you working in three spatial dimensions, or one? The units of $\rho$ don’t look like a 3D or a 1D probability density to me. $\endgroup$
    – G. Smith
    Jan 5 '19 at 3:08
  • $\begingroup$ The gradient operator produces a vector, but it isn’t getting contracted with anything. $\endgroup$
    – G. Smith
    Jan 5 '19 at 3:11

The pair correlation function is defined to be dimensionless. The probability density is defined such that its integral over all space is unity for any time $t$. So, if you are thinking only of one spatial dimension, both sides are $L^{-1}T^{-1}$, while in three dimensions both sides are $L^{-3}T^{-1}$.

Actually, in the context that I have seen this version of the Smoluchowski equation (papers by Jan Dhont and colleagues on dynamics in colloidal systems), $\rho(r,t)$ is the single particle density, not the single particle probability density. The difference is simply that the integral of $\rho(r,t)$ over all space gives the total number of particles in the system, $N$, rather than unity. This does not affect the dimensional analysis at all, but I thought I should clarify.

  • 1
    $\begingroup$ @JD_PM please note my second paragraph, added for clarity after you accepted the answer. $\endgroup$
    – user197851
    Jan 5 '19 at 15:38

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