Solenoidal and irrotational vector field

  • #1
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Homework Statement


I am to prove (using the equations for gradient, divergence and curl in spherical polar coordinates) that vector field $$\mathbf{w}=w_{\psi}(r,\theta)\hat e_{\psi}$$ is solenoidal, find $$w_{\psi}(r,\theta)$$ when it's irrotational and find a potential in this case.

Homework Equations




The Attempt at a Solution


For vector field to be solenoidal, divergence should be zero, so I get the equation:

$$\nabla\cdot\mathbf{w}=\frac{1}{r\sin\theta}\frac{\partial w_{\psi}(r,\theta)}{\partial \psi}=0$$

For a vector field to be irrotational, the curl has to be zero. After substituting values into equation, I get:

$$\cos\theta\cdot w_{\psi}+\frac{\partial w_{\psi}}{\partial \theta}\cdot \sin\theta=0$$
and
$$w_{\psi}+\frac{\partial w_{\psi}}{\partial r}\cdot r=0$$.

Is it right? How to proceed?
 

Answers and Replies

  • #2
andrewkirk
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For vector field to be solenoidal, divergence should be zero, so I get the equation:

$$\nabla\cdot\mathbf{w}=\frac{1}{r\sin\theta}\frac{\partial w_{\psi}(r,\theta)}{\partial \psi}=0$$
How did you derive that formula? Are you sure you didn't just use ##\frac{\partial w^r}{\partial r}+
\frac{\partial w^\theta}{\partial \theta}+\frac{\partial w^\phi}{\partial \phi}##? If I recall correctly that formula for divergence only works for Cartesian coordinates, in which case it won't be correct for spherical coordinates. I think you either need to convert to Cartesian coordinates and calculate divergence as ##\frac{\partial w^x}{\partial x}+
\frac{\partial w^y}{\partial y}+\frac{\partial w^z}{\partial z}##, or else add correction terms to the calculation in spherical coordinates to allow for the change in the coordinate vectors as the position moves.
 

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