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We switch gears to discuss the normal-state Nernst signal vN. In zero field, a thermal gradient along x produces a diffusion of charge carriers (assume holes). Because of the boundary condition J_x = 0 in a finite sample, the pile-up of holes on the right leads to an E field that opposes further increase in charge imbalance. The observed thermopower S is defined as E_x/|grad T|.
Reasoning from the Lorentz deflection of the holes in an applied magnetic field, it seems plausible that the normal-state Nernst signal is of the order S*tan(\theta). However, it is actually an order of magnitude smaller because of the following cancellation. The total electric current density J (in an infinite sample) has a piece from the E field and a piece from the applied gradient, viz.
J = \sigma . E + \alpha . (-grad T)
where \sigma is the conductivity tensor and \alpha the Peltier conductivity tensor. Both tensors develop an off-diagonal term linear in H. Since E_x is always directed against (-grad T) in a finite sample, the two off-diagonal currents are opposite as shown in the top-right inset. In a Boltzmann-eqn. treatment (Sondheimer 1949), the cancellation is exact if the relaxation time \tau is independent of energy \epsilon. As shown below, deviation from perfect cancellation produces the observed Nernst signal E_y.

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