Charge distribution in a heated metal (Seebeck effect) Consider the Seebeck effect taking place in a single metallic rod, where one end is kept at $T_0$ and the other end at $T_0+\Delta T$. The charge distribution will not be homogeneous: the electrons are accumulating mostly on one side (which one depends on the sign of the Seebeck coefficient of that particular metal), while holes accumulate on the other side. My question is, how does one calculate the charge distribution? 
I am particularly interested to know whether the charges accumulate exactly at the ends of the metal, or whether they are spread linearly or exponentially within the metal in the direction of the temperature gradient, for example.
 A: The Seebeck effect can be stated mathematically as follows. The current density $\mathbf j$ in a conductor is related to the potential $V$ and temperature $T$ by
$$\mathbf j=-\sigma(\nabla V+S\nabla T),$$
Where $\sigma$ is the material's conductivity and $S$ is the Seebeck coefficient. Pretty much, this states that the temperature difference creates an electric field $$\mathbf E_{Seebeck}=-S\nabla T.$$
As such, in the stationary state we'd need the following relationship between the potential and the temperature
$$\nabla V=-S\nabla T,$$
Which allows us to write (for constant $S$)
$$V=-ST+V_0.$$
If we now suppose that everything happens in one dimension $x$ and suppose the effect happens between to infinite plates at $x=0$ and $x=L$ as to simplify our calculations and not have to consider field lines escaping from the wire, we can write the following relationship between the charge density $\rho$ and the potential (Poisson's equation)
$$\frac{d^2V}{d x^2}=-\frac{\rho}{\epsilon},$$
$$\frac{d^2T}{d x^2}=\frac{\rho}{S\epsilon}$$
With $\epsilon$ being the permitivitty of the material. If we now suppose the electron charge density is governed by a Boltzmann distribution (stationary thermal state) with respect to electric potential
$$\rho_e=-\rho_0\exp(\frac{eV}{k_BT}),$$
$$\rho_e=-\rho_0\exp(-\frac{eS}{k_B})\exp(\frac{eV_0}{k_BT}).$$
If we also add the (homogenous) proton density 
$$\rho_p=-\frac{1}{L}\int\limits_0^L\rho_e\,dx,$$
Which is also a constant of the material, the charge density becomes
$$\rho=\rho_e+\rho_p$$
$$\rho=\rho_p\left(1-\frac{L\exp(\frac{eV_0}{k_BT})}{\int\limits_0^L\exp(\frac{eV_0}{k_BT})\,dx}\right)$$
The differential equation we need to solve is then 
$$\frac{d^2T}{dx^2}= \frac{\rho_p}{S\epsilon}\left(1-\frac{L\exp(\frac{eV_0}{k_BT})}{\int\limits_0^L\exp(\frac{eV_0}{k_BT})\,dx}\right),$$
Which can be solved numerically with (I think) $V_0$ as a free parameter. You can try it with your favorite software now. 
