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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.