Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.