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.


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 $S$)


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



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.

  • $\begingroup$ +1 Thanks a lot, this unblocked me. I started differently, but then got stuck. I did not think about using distributions to continue, so now I guess I can continue. However I have a few comments. 1st: I think using Maxwell distribution for metal at low temperatures (below $10 kK$) is usually not a good approximation. Better is to use Fermi-Dirac statistics. 2nd) I am unsure about the part where you replace the chemical potential by the electrostatic potential, in the exponential. That seems odd to me. 3rd) The part where you consider the proton density also troubles me. Holes could be almost $\endgroup$ – thermomagnetic condensed boson Sep 17 '18 at 17:55
  • $\begingroup$ as mobile as electrons. But overall I think you paved the way to go, even though some details are not quite correct. $\endgroup$ – thermomagnetic condensed boson Sep 17 '18 at 17:57
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    $\begingroup$ I wrote this answer in the middle of class so there might be some errors indeed. But I'm happy to help! $\endgroup$ – Gabriel Golfetti Sep 17 '18 at 18:16
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    $\begingroup$ The chemical potential comment is a good one. If we used it instead of the electrostatic potential, the $V_0$ parameter probably wouldn't even show up. I didn't do quantum because solving things in classical electromagnetism along with quantum distributions feels a little off to me. $\endgroup$ – Gabriel Golfetti Sep 17 '18 at 18:19
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    $\begingroup$ By the way, the proton density is just the density of the underlying positive ions in the lattice. Holes are just lack of electrons, but the ions stay in place. That's why the proton density is homogeneus while the electron density can vary (leading both to electrons and holes) $\endgroup$ – Gabriel Golfetti Sep 17 '18 at 19:06

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