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Incompressibility of fluid $\Rightarrow \exists\psi: \mathbf{u}=(u,v)=(\psi_y,-\psi_x)$

Where $\mathbf{u}$ is the velocity of the fluid

Could someone explain why $\mathbf{u} \cdot \nabla \psi=0 \Rightarrow \psi$ is constant on streamlines

Many thanks in advance

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up vote 1 down vote accepted

Suppose $\gamma (t)$ is a streamline, and consider $\psi (\gamma (t))$. We want to show that this is constant, so consider $\frac{d}{dt}\psi (\gamma (t))$. By the chain rule, this is $\nabla \psi \cdot \gamma ' (t)$. Since $\gamma (t)$ is a streamline, its derivative is the velocity $\mathbf{u}$. So $\frac{d}{dt}\psi (\gamma (t))=\nabla \psi \cdot \mathbf{u}=0$, so $\psi (\gamma (t))$ is constant.

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That makes so much more sense! thank you very much. – Freeman Mar 27 '12 at 12:14

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