# Isentropic fluid: cross product of gradients is zero, why?

In the vorticity equation we have the baroclinic term of the form: $$\frac{ {\nabla}\rho}{\rho}\times\frac{ {\nabla}{P} }{\rho}.$$ Why does it go to zero for isentropic flow?

I understand that, if the flow is barotropic, the above term vanishes. However, an isentropic (reversible $$dS=Q$$ + adiabatic $$Q=0$$, i.e. $$dS=0$$) flow is more general, in the sense that the pressure depends on both the density and temperature.

• Pressure can depend on both density and temperature, but in the way to keep entropy constant. You'd better write $P(\rho, s)$, to realize what happens. See the answer below Commented Sep 27, 2022 at 21:52
• Commented Jan 16, 2023 at 10:45

Write pressure as a function of density and specific entropy, $$P(\rho, s)$$, so that its differential reads

$$dP = \left( \dfrac{\partial P}{\partial \rho}\right)_s d \rho + \left( \dfrac{\partial P}{\partial s}\right)_{\rho} d s$$

$$\nabla P = \left( \dfrac{\partial P}{\partial \rho}\right)_s \nabla \rho + \left( \dfrac{\partial P}{\partial s}\right)_{\rho} \nabla s$$.
For isentropic flow, with uniform entropy, $$s = \overline{s}$$, $$ds = 0$$, and thus
$$\nabla P = \left( \dfrac{\partial P}{\partial \rho}\right)_s \nabla \rho = c^2(\rho, \overline{s}) \nabla \rho$$,