User:Kupirijo/Stokes stream function

Vorticity
The vorticity is defined as:
 * $$\mathbf{\omega} = \nabla \times \mathbf{v} = \nabla \times \nabla \times \mathbf{\psi}$$
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!Derivation of vorticity $$\omega$$ of Stokes stream function
 * Consider the vorticity as defined by
 * $$\mathbf{\omega} = \nabla \times \mathbf{v}$$
 * $$\mathbf{\omega} = \nabla \times \mathbf{v}$$

From the definition of the curl in spherical coordinates:

\begin{align} \omega_r     &= {1 \over r\sin\theta}\left({\partial \over \partial \theta} \left( v_\phi\sin\theta \right) - {\partial v_\theta \over \partial \phi}\right) \boldsymbol{\hat r} \\ \omega_\theta &= {1 \over r}\left({1 \over \sin\theta}{\partial v_r \over \partial \phi} - {\partial \over \partial r} \left( r v_\phi \right) \right) \boldsymbol{\hat \theta} \\ \omega_\phi  &= {1 \over r}\left({\partial \over \partial r} \left( r v_\theta \right) - {\partial v_r \over \partial \theta}\right) \boldsymbol{\hat \phi} \\ \end{align} $$ First we notice that the $$r$$ and $$\theta$$ components are equal to 0. Secondly we substitute $$v_r$$ and $$v_\theta$$ into $$\omega_\phi$$ then we get:

\begin{align} \omega_r     &= 0\,{\hat r} \\ \omega_\theta &= 0\,{\hat \theta} \\ \omega_\phi  &= {1 \over r}\left({\partial \over \partial r} \left( r \left(-\frac{1}{r \sin\theta}\frac{\partial\psi}{\partial r}\right) \right) - {\partial \over \partial \theta}\left(\frac{1}{r^2 \sin\theta}\frac{\partial\psi}{\partial \theta}\right)\right) \boldsymbol{\hat \phi} \\ \end{align} $$ If we do the algebra:

\begin{align} \omega_\phi  &= {1 \over r}\left(-\frac{1}{\sin\theta}\left({\partial \over \partial r} \left(\frac{\partial\psi}{\partial r}\right)\right) - \frac{1}{r^2 }{\partial \over \partial \theta}\left(\frac{1}{\sin\theta}\frac{\partial\psi}{\partial \theta}\right)\right) \boldsymbol{\hat \phi} \\ &= {1 \over r}\left(-\frac{1}{\sin\theta}\left(\frac{\partial^2\psi}{\partial r^2}\right) - \frac{\sin\theta}{r^2 \sin\theta}{\partial \over \partial \theta}\left(\frac{1}{\sin\theta}\frac{\partial\psi}{\partial \theta}\right)\right) \boldsymbol{\hat \phi} \\ &= -\frac{1}{r\sin\theta} \left(\frac{\partial^2\psi}{\partial r^2} + \frac{\sin\theta}{r^2}{\partial \over \partial \theta}\left(\frac{1}{\sin\theta}\frac{\partial\psi}{\partial \theta}\right)\right) \boldsymbol{\hat \phi} \\ \end{align} $$ If we do the calculation we find that the vorticity vector is equal to:
 * }
 * $$\mathbf{\omega} = (0,0, -\frac{1}{r\sin\theta} \left(\frac{\partial^2\psi}{\partial r^2} + \frac{\sin\theta}{r^2}{\partial \over \partial \theta}\left(\frac{1}{\sin\theta}\frac{\partial\psi}{\partial \theta}\right)\right))$$

If we define the new operator $$E = \frac{\partial^2}{\partial r^2} + \frac{\sin\theta}{r^2}{\partial \over \partial \theta}\left(\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\right)$$ then we have
 * $$\mathbf{\omega} = (0,0, -\frac{E\psi}{r\sin\theta})$$