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The Seebeck effect is really interesting and the reasoning behind the phenomenon does seem to make sense, that is, that electrons move from the hotter areas to the colder areas. What I don't understand is the need for two conductors. I don't see why electrons couldn't move around in a circle of copper wire. In fact, wouldn't a strand of straight copper wire produce voltage for a brief moment as the available electrons move (but they can't come go back, so no more voltage)?

I've spent a fair bit of time researching and it seems like the reason is something to do with how electrons diffuse between the different materials which causes a potential difference, that would otherwise be cancelled out when using the same metal, but I don't quite understand the reasoning behind this. I would greatly appreciate if the need for dissimilar conductors was explained.

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  • $\begingroup$ electron flow induced by thermal gradient in a homogeneous metal is called the Thomson effect, see en.wikipedia.org/wiki/Thermoelectric_effect#Thomson_effect $\endgroup$
    – hyportnex
    Commented May 21, 2018 at 19:16
  • $\begingroup$ @hyportnex I'm not sure your statement is accurate. The Thomson effect manifests itself as a heating/cooling due to a non vanishing current (and a non vanishing thermal gradient). It does not produce a current, as far as I know. What do you think? $\endgroup$ Commented Sep 18, 2018 at 20:45

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You don’t need two dissimilar metals, but you do need some way to break the symmetry of the system to produce a current in one direction or the other. Heating the junction of two different metals is just a particularly good way to do this since the asymmetry can be strong, leading to a strong effect.

Think about it this way: If you heat the middle of a single, long wire, which way would the current flow? Based on the sign of the Seebeck coefficient (which depends on the details of the band structure and the doping level), electrons will all flow away from or toward the heat on both sides of the hot spot. If everything is totally symmetric, the currents to the left and right simply cancel at all times.

The trick with dissimilar metals is that they have different Seebeck coefficients, so electrons naturally flow away/toward heat better in one metal than the other, breaking the symmetry. This gives a net current which you can measure.

But you can get a current in other asymmetric situations, too. For example, I have personally measured a Seebeck current arising from laser heating near the junction of two thicknesses of the same metal. It was a thin film of gold (25 nm), contacted on both ends by thicker gold (200 nm). Laser illumination of one contact resulted in a DC current, and illumination of the other contact resulted in a current in the opposite direction. The current flipped sign from one contact to the other because the direction of the asymmetry in the circuit switched (thick metal toward +, or thick metal toward -).

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  • $\begingroup$ Nice answer. Do you think this argument is applicable to pn-junctions as well? $\endgroup$
    – V.F.
    Commented May 21, 2018 at 17:22
  • $\begingroup$ Thanks! In Understanding Physics Asimov explains the effect as depending on the different contact potential between dissimilar metals. How does that play a role? $\endgroup$ Commented Jun 15, 2019 at 21:00
  • $\begingroup$ Why heat the "middle" of a single wire? Why not heat the end of a single wire -- connect a multimeter to both the hot and cold end -- and measure a current ? $\endgroup$
    – mumtaz
    Commented Feb 18, 2022 at 14:14
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    $\begingroup$ @mumtaz it’s usually impractical to have a multimeter right at the heated point. So, you’re generally going to have some length of wire on both sides of the heat. Then what matters is how far away from the heat is the wire still elevated with respect to equilibrium temperature? If the wires entering each side of the voltmeter are the same temperature, and they are the same wire, then the system is symmetric, and you don’t get a Seebeck current. $\endgroup$
    – Gilbert
    Commented Feb 18, 2022 at 15:58
  • $\begingroup$ @Gilbert ah thanks .. got it. $\endgroup$
    – mumtaz
    Commented Feb 18, 2022 at 17:07

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