In the following derivation of the vorticity equation, I do not understand how $\nabla \cdot v=0$ implies $\frac{1}{\rho^2}\nabla \rho \times \nabla p=0$.
We start with the Euler equation $$\frac{\partial \vec{v}}{\partial t}+(\vec{v}.{\nabla})\vec{v}=-\frac{1}{\rho}{\nabla}p + \vec{F},$$
where $\vec{v}$, $p$, $\rho$ and $\vec{F}$ are velocity, pressure, density, and body force, respectively. Now, we take the curl of this equation and obtain:
$$\frac{\partial \vec{w}}{\partial t}={\nabla}\times(\vec{v}\times\vec{w})+\frac{1}{\rho^2}\nabla \rho \times \nabla p,$$
Where, $\vec{w}=\nabla \times \vec{v}$ is the vorticity field. For incompressible fluids we have that ${\nabla \cdot \vec{v}}=0$ and the vorticity equation should become:
$$\frac{\partial \vec{w}}{\partial t}= {\nabla}\times(\vec{v}\times\vec{w}).$$

I don't understand how $\nabla \cdot v=0$ implies $\frac{1}{\rho^2}\nabla \rho \times \nabla p=0$. Is this generally true or there is some extra hypothesis (other than $\nabla \cdot v=0$) that is needed to obtain the above vorticity equation?

  • $\begingroup$ Are you considering a "barotropic" fluid? (i.e. $P=P(\rho)$). In this case, the gradients of $\rho$ and $P$ are parallel and their cross-product vanishes, en.wikipedia.org/wiki/Barotropic_fluid . The same is valid for isentropic flow: physics.stackexchange.com/q/729637/226902 $\endgroup$
    – Quillo
    Commented Jan 16, 2023 at 10:47
  • $\begingroup$ No, I am not considering barotropic fluid. Just incompressible fluid. $\endgroup$
    – bhoutik
    Commented Jan 16, 2023 at 11:25
  • 1
    $\begingroup$ Are you considering a special case of an incompressible fluid, namely constant density? (in this case $\nabla \rho $ is trivially zero). However, in the general incompressible case the pressure has a specific meaning, try to have a look at physics.stackexchange.com/q/507526/226902 and physics.stackexchange.com/q/427000/226902, it may help. $\endgroup$
    – Quillo
    Commented Jan 16, 2023 at 11:38
  • 1
    $\begingroup$ I agree with @Quillo . This is one thing I could never understand. The usual assumption that goes along with the incompressibility assumption is that density doesn't depend on space and is constant all along the flow. That allows us to write down the continuity as follows: $\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \cdot \vec u) = \frac{\partial \rho}{\partial t} + \nabla \rho \cdot \vec u + \rho ( \nabla \cdot \vec u) = \frac{\partial \rho}{\partial t} + \rho ( \nabla \cdot \vec u) = 0$ because $\nabla \rho = 0$ due to $\rho = invar(\vec x)$. $\endgroup$ Commented Jan 16, 2023 at 16:43
  • $\begingroup$ In fluid mechanics, when we say that the flow is incompressible, we usually imply that the density is constant. Which, in fact, is more restrictive than just "incompressible". But nonetheless, it is assumed that everybody understands that the density is taken constant. I understand that's not rigorous and can be annoying. But that's how it is. $\endgroup$ Commented Jan 16, 2023 at 16:47

1 Answer 1


Acheson = Acheson, D.J.: Elementary Fluid Dynamics, Oxford: Clarendon Press, 2005.
Falkovich = Falkovich, G: Fluid mechanics, 2nd ed., Cambridge: Cambridge University Press, 2018.
The vorticity equation
The barotropic version: $\frac{D \pmb{\omega}}{Dt}=(\pmb{\omega}\cdot\nabla)\pmb{u}$ [Acheson, p.17, (1.25)].
Remark. The incompressible, constant density [Acheson, p.6, §1.3, (ii)] version is a special case of the barotropic version.
The generalized version: $\frac{D}{Dt}(\frac{\pmb{\omega}}{\rho})=(\frac{\pmb{\omega}}{\rho}\cdot \nabla)\pmb{u}-\frac{1}{\rho}\nabla (\frac{1}{\rho})\times \nabla p$ [Acheson, p.25, (1.39)].
I. The barotropic version is a special case of the generalized version.
$\nabla p=p'(\rho)\nabla \rho$.
$\nabla (\frac{1}{\rho})=-\rho^{-2}\nabla \rho$.
Consequently, $\nabla (\frac{1}{\rho})\times \nabla p=\pmb{0}$.
$\frac{D}{Dt}(\frac{\pmb{\omega}}{\rho})=\frac{1}{\rho}\frac{D\pmb{\omega}}{Dt}$ [we assume that $\rho$ is constant in $t$].
II. Proof of the generalized version.
$\frac{\partial \pmb{u}}{\partial t}+\pmb{\omega}\times \pmb{u}+\frac{1}{2}\pmb{u}^2=-\frac{1}{\rho}\nabla p-\nabla \chi$ [Acheson, p.24, l.$-$1].
Taking the curl of the above equality, we have
$\frac{\partial \pmb{\omega}}{\partial t}+\nabla\times (\pmb{\omega}\times \pmb{u})=-\nabla\times (\frac{1}{\rho}\nabla p)$
$=-\nabla (\frac{1}{\rho})\times \nabla p$ [Acheson, p.348, (A.8)]. Thus,
$\frac{\partial \pmb{\omega}}{\partial t}=\nabla\times (\pmb{u}\times \pmb{\omega})-\nabla (\frac{1}{\rho})\times \nabla p$ ($*$).
If we substitute [Falkovich, p.17, (1.20)] into $\frac{\partial \pmb{\omega}}{\partial t}$ given in [Falkovich, p.17, $-$11], we obtain [Falkovich, p.17, (1.22)]. However, if we substitute ($*$) instead into $\frac{\partial \pmb{\omega}}{\partial t}$ given in [Falkovich, p.17, $-$11], we obtain [Acheson, p.25, (1.39)].
Remark. With the generalized version at hand, we may easily recognize what conditions we should impose on its hypothesis in order to get a specialized version.
This answer is excerpted from §1.12.(A), Remark 4 in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/quantum-mechanics.

  • $\begingroup$ What does this add wrt the comments? Not a criticism, just to know if I'm missing something. Thanks :) $\endgroup$
    – Quillo
    Commented Feb 20, 2023 at 8:59
  • $\begingroup$ The proofs of various versions are interlinked. The generalized version may clarify the key idea and provide the big picture better. $\endgroup$ Commented Feb 21, 2023 at 10:28
  • $\begingroup$ The question is: "I don't understand how ∇⋅v=0 implies ∇ρ×∇p=0." From a concrete point of view, how does this answer the specific question? $\endgroup$
    – Quillo
    Commented Feb 21, 2023 at 10:54
  • $\begingroup$ Proof in I shows that $\nabla (\frac{1}{\rho})\times \nabla p=\pmb{0}$ is derived from the barotropic hypothesis $p=p(\rho)$ rather than $\nabla\cdot\pmb{u}=0$. The remark before I shows that the incompressible, constant density case is a special case of the barotropic case. In general, $\nabla (\frac{1}{\rho})\times \nabla p\neq \pmb{0}$. $\endgroup$ Commented Feb 22, 2023 at 0:49
  • $\begingroup$ Thank you for confirming the picture that was sketched in the comments, +1 for me. $\endgroup$
    – Quillo
    Commented Feb 22, 2023 at 0:53

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