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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.

Quillo
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martín canullán
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    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 – basics Sep 27 '22 at 21:52
  • Related: https://physics.stackexchange.com/q/116779/226902 https://physics.stackexchange.com/q/439508/226902 https://physics.stackexchange.com/q/745716/226902 – Quillo Jan 16 '23 at 10:45

1 Answers1

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

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

and its gradient

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

For isentropic flow, with uniform entropy, $s = \overline{s}$, $ds = 0$, and thus

$\nabla P = \left( \dfrac{\partial P}{\partial \rho}\right)_s \nabla \rho = c^2(\rho, \overline{s}) \nabla \rho$,

the gradient of pressure field is proportional to (aligned with) the gradient of the density field in every point of the space. Taking the vector product of two proportional vectors, you get zero.

basics
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