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Both Seebeck effect and Peltier effect we have the need of two dissimilar conductors to break the symmetry of the system to produce a current. But, why in the Thomson effect we just need one conductor, that is, a homogeneous conductor? Shouldn't there be a balance of charges as well? What is the big difference between these effects? And, it seems that in the Thomson effect we need to have a temperature gradient and a current flow at the same time, so, it is possible generate an electric current through Thomson effect? I would really appreciate if someone could answer these questions for me.

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Take a single material in a wire loop shape, and place a fire under part of it. As you pointed out, the Seebeck effect wouldn't be able to create a current, because even though there would be a Seebeck voltage between any 2 points in the circuit, the total emf would be worth 0 V. Because the emf from hot to cold would exactly cancel the one from cold to hot.

The Peltier effect is quite different, it appears when there are dissimilar materials (or a single inhomogeneous material, though this bears another name, but it is exactly the same mathematically), and it is a generation/absorption of heat right at the interface of the 2 materials, which by the way, adds up onto a surface Joule effect (due to the resistance interface). The Peltier effect arises because someone has put a current in the loop, it does not create a current per se.

Similarly for the Thomson effect, which, as you point out, requires both an electric current and a thermal gradient to exist. It is a heat released/absorbed at every single point in the material where both $I$ and $\nabla T$ exist. It does not create a current per se. You actually need to input a current for it to exist.

If you want an electric current, you need the quantity $\vec J_e =-\sigma \nabla \overline{\mu}-\sigma S \nabla T$ not to vanish. $\mu$ is the electrochemical potential which generally reduces to $eV$, i.e. a quantity proportional to the electrostatic potential (from which you derive the voltage across 2 points).

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  • $\begingroup$ Thank you for answer, do you know some book about thermoelectrics that you could recommend to me? $\endgroup$
    – Brising
    Mar 22, 2022 at 12:32

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