Electrostatics of non-equilibrium situations

Question

In footnote 1 of Chapter 13 of Ashcroft and Mermin, they remark that

The only case we shall discuss in which the local equilibrium distribution is not the uniform equilibrium distribution (13.1) (with constant $T$ and $\mu$) is when the spatially varying temperature $T(\mathbf{r})$ is imposed by suitable application of sources and/or sinks of heat, as in a thermal conductivity measurement. In that case, since the electronic density $n$ is constrained (electrostatically) to be constant, the local chemical potential will also depend on position, so that $\mu(\mathbf{r}) = \mu_{equilib}(n, T(\mathbf{r}))$. In the most general case the local temperature and chemical potential may depend on time as well as position.

Note that (13.1) is simply the Fermi function with a local temperature and local chemical potential, in general.

My question is why can we say that $n$ is electrostatically constrained to be constant? In a measurement of thermal conductivity, we generally have an open circuit and in this case I suppose I can argue that there are therefore no macroscopic fields (and so $n$ must be constant, else there would be macroscopic fields). But if I am not at open circuit, why must this be the case? For example, if we are analyzing the Thomson effect, then I have both a temperature gradient and a current (not at open circuit). Is $n$ constant throughout my sample in this case too?

Answer 1

A varying electronic density would imply an accumulation of charge, since we assume the background density of positive charge (ions) is constant. Moreover, let me show that we cannot have a static accumulation of charge, that is charge density $\rho \ne 0$ with no time dependency $\frac{\partial \rho}{\partial t} = 0$.

From the continuity equation we have $$\nabla \cdot \mathbf{j} = - \frac{\partial \rho}{\partial t} = 0,$$ where $\mathbf{j}$ is the current density. Ohm's law state that $$ \mathbf{j} = \sigma \mathbf{E},$$ where $\sigma$ is the the conductivity (which we assume to be uniform) and $\mathbf{E}$ is the electric field. We take the divergence of both sides and get $$ \nabla \cdot \mathbf{j} = \sigma \nabla \cdot \mathbf{E}.$$ Now the left hand side vanishes, as we showed above, and the right hand side is proportional to $\rho$ by Gauss's law.

Note that we use Ohm's law in order to formulize the notion that charge is moved by electric field. This formulation also allows us to see when the statement is false, i.e. in which situations we can get $\rho \ne 0$:

1. For an insulator $\sigma = 0$ and indeed we can have charge distribution while electrostatic is still valid (strictly speaking the only case).
2. When $\sigma$ is not uniform we get an accumulation of charge given by $$ \rho = \frac{\mathbf{E} \cdot \nabla \sigma}{\sigma}.$$

Answer 2

Following Ashcroft and Mermin, let's use Gaussian units. Ampere's law is \begin{equation}\tag{1} \nabla \times \vec{H} = \frac{4\pi}{c}\vec{J}_f + \frac{1}{c}\frac{\partial\vec{D}}{\partial t}. \end{equation} If the electric field is static, the partial derivative of $\vec{D}$ vanishes. Therefore, \begin{equation}\tag{2} \nabla\cdot\vec{J}_f = 0. \end{equation} Equation of continuity then tells us that $\rho_e = -en$ is constant. Here $\rho_e$ is the electronic charge density and $n$ is the number density. We do not need a constitutive relation like Ohm's law to come to this conclusion.

About constancy of $n$, not just time independence: A slightly more rigorous explanation for constancy of $n$ beyond the elementary reasoning in the comments is to look at the distribution function in presence of a dc electric field. It is given by equation (13.22) in the book and is \begin{equation}\tag{3} g(\vec{k}) = g^0(\vec{k}) - e\vec{E}\cdot\vec{v}(\vec{k})\tau(\mathcal{E}(\vec{k}))\left(-\frac{\partial f}{\partial\mathcal{E}}\right). \end{equation} As you see, $g$ depends solely on $\vec{k}$ and not on $\vec{r}$. The number density is just the integral of $g$ over the $k$-space with an appropriate normalising factor.