# Entropy change independent of diffusion coefficient?

Assume we have a system with particle density $n$, such that $n$ fulfils the diffusion equation with $D$ as the diffusion coefficient. Initially $n(\vec{r}, t=0) = n_0\delta^3(\vec{r})$, which implies $$n(\vec{r}, t) = \frac{n_0}{(4 \pi Dt)^{3/2}} \exp \left(-\frac{r^2}{4Dt}\right).$$ If we define the entropy as $$S = -\int d^3r n(\vec{r}, t) \ln \left(n(\vec{r}, t)\Lambda^3 \right),$$ we get that the time derivative of the entropy is given by $$\frac{d S}{dt} = \frac{3n_0}{2t}.$$ However, this expression does not depend on $D$ which seems strange. If we would rescale $D \rightarrow D' = 2D$ and also rescale $t\rightarrow t' = t/2$ we get the same equation for $n(\vec{r}, t)$. Hence we expect the rescaled system to behave exactly as the system before rescaling, but the time evolution is twice as fast. At a specific time $t_0$ we expect the two systems with $D$ and $D'$ to have evolved differently much. Does this imply that the entropy time derivatives for the different systems are different? If so, shouldn't $\frac{d S}{dt}$ depend on $D$?

One way to see why the time derivative of the entropy is independent of the diffusion coefficient is to look at how the entropy changes as $D \mapsto kD$. The entropy is given by

$$S_{kD}(t) = - \int d^3 \vec{r} \frac{n_0}{(4 \pi kDt)^{3/2}} \exp \left(-\frac{r^2}{4kDt}\right) \ln \left( \frac{n_0 \Lambda^3}{(4 \pi kDt)^{3/2}} \exp \left(-\frac{r^2}{4kDt}\right) \right).$$

Let us make a variable substitution in the integral such that $x = r/\sqrt{k}$.

$$S_{kD}(t) = - \int d^3 \vec{x} \frac{n_0 k^{3/2}}{(4 \pi kDt)^{3/2}} \exp \left(-\frac{x^2}{4Dt}\right) \ln \left( \frac{n_0 \Lambda^3}{(4 \pi kDt)^{3/2}} \exp \left(-\frac{x^2}{4Dt}\right) \right).$$

By using $\ln{xy} = \ln{x} + \ln{y}$ we get that

$$S_{kD}(t) = S_D(t) - \int d^3 \vec{x} \frac{n_0}{(4 \pi Dt)^{3/2}} \exp \left(-\frac{x^2}{4Dt}\right) \ln \left( \frac{1}{k^{3/2}} \right)$$

Let us now change variables in the intergal such that $y = x / \sqrt{4Dt}$ then we get

$$S_{kD}(t) = S_D(t) - \int d^3 \vec{y} \frac{n_0}{\pi^{3/2}} \exp \left(-y^2 \right) \ln \left( \frac{1}{k^{3/2}} \right) = S_D(t) + f(k),$$

where $f(k)$ is some function independent of $t$

The conclusion is that the entropy increases by a constant amount for all times when we change the diffusion coefficient. The time derivative of the entropy must there therefore be independent of the diffusion coefficient since

$$\frac{d S_{kD}}{dt} = \frac{d S_D}{dt} \quad \forall k > 0.$$