# Piezoelectric effect in terms of classical thermodynamics

Non-volumetric work associated with the increase in polarization in a dielectric is equal to $$dW = E dP$$, where $$E$$ and $$P$$ are vectors of the electric field and the total polarization in the dielectric (for a homogeneous sample, $$P$$ is equal to the electric dipole moment per unit volume times volume).

The question is how to show that if in a substance an increase in polarization due to compression is observed, it follows from phenomenological thermodynamics laws that introducing the same substance into the electric field $$E$$ will cause its mechanical deformation (the so-called piezoelectric effect).

• Review Maxwell relations and relate $\partial P/\partial p$ (the increase in polarization for an increase in pressure $p$) to $\partial V/\partial E$ (the increase in volume for an increase in electric field). Commented Jun 18 at 18:09

I do not believe that this can be proven because the question is, in essence, asking to prove that some process, here volumetric work coupled to polarization, is reversible. One can instead start with Bronsted's work-heat axiom or an equivalent formulation of thermodynamics in which the Gibbs-Duhem equation is extended to cover the process between the equilibrium endpoints and write that $$\delta S \Delta T+\delta P\Delta E-\delta V \Delta p=T\sigma \tag{1}.$$ In this Eq 1, $$\sigma \ge 0$$ is the internally generated entropy (zero iff the process is reversible) while $$\delta S, delta P, \delta V$$ denote the entropy, polarization and volume that during the process will have been transported through $$\Delta T, \Delta E, -\Delta p$$ temperature, electric field and pressure drops, resp., between the equilibrium endpoints.
If you now assume that the process is actually reversible, $$\sigma = 0$$ and isothermal, $$dT=0$$, then you have the two work-processes, volumetric and polarization, in balance, which what you were asking for: $$dT = 0 \cap \sigma = 0:P\Delta E-V\Delta p=0 \tag{2}$$