 Kelvin thermoelectric relation

With the Seebeck coefficient convention $dV=-SdT$ at zero electric current, and the transported-heat convention of Peltier coefficient, the Kelvin thermoelectric relation is$T$ is absolute temperature in kelvins; use consistent material references. This scalar linear response relation assumes local thermal equilibrium and reciprocal transport without time-reversal-breaking fields or order. Onsager reciprocal relations supply the symmetry; conservation of energy alone is insufficient. A magnetic field requires a reversed-field reciprocity comparison.^[ref]

In particular, the A-minus-B thermocouple coefficient in Seebeck effect is $S_A-S_B$, whereas absorbed junction heat for conventional current A to B uses $S_B-S_A$. There is no contradiction: terminal voltage order and crossing direction are separate choices. Multiplying a measured voltage by an absolute temperature without specifying these choices does not determine junction heating.

A short reciprocity calculation makes the factor $T$ visible. At a common reference temperature, write the local linear response with electrical driving field $\mathcal{E}=-\nabla V$ and heat flux $q$ asAt $j=0$, the Seebeck coefficient convention gives $S=L_{12}/(T{L}_{11})$. At $\nabla T=0$, the Peltier coefficient is $q/j=L_{21}/L_{11}$. Thus $L_{21}=L_{12}$ gives $\Pi =TS$. The chosen forces are conjugate to the fluxes in the local entropy production, $j\mathcal{E}/T+q(-\nabla T)/T^2$; arbitrary rescaling of just one force would spoil this argument.

The relation is local. Varying $S(T)$ permits distributed Thomson effect heating, with $\tau =TdS/dT$; the constant-coefficient model below omits it. Endpoint substitutions into a constant-property formula can therefore break its energy accounting.