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
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.
