1. ## Spherical Polar Coordinates

Right bit of a horrible looking one here:

We've got the following:

$\mathbf{u} = U\left(1-\frac{a^3}{r^3}\right)\cos{\theta}\;\mathbf{\hat{r }} - U\left(1+\frac{a^3}{2r^3}\right)\sin{\theta}\;\mat hbf{\hat{\theta}}$ where $a$ and $U$ are constants.

We have already shown that $\nabla \cdot \mathbf{u} = 0$, and because of this we have a vector field $\mathbf{A}$ s.t. $\mathbf{u} = \nabla \times \mathbf{A}$. So assuming that $\mathbf{A} = A(r,\theta)\hat{\mathbf{\phi}}$:

Find 2 equations for $A(r,\theta)$, determine $A(r,\theta)$ and verify that they're consistent.

Right, so does the fact that $\mathbf{A} = A(r,\theta)\hat{\mathbf{\phi}}$ mean that $\mathbf{A}$ is of the form $(0,0,\phi)$? This sort of makes sense.

If so, then do we just complete the cross product and then have two differential equations, essentially $\frac{\partial \phi}{\partial r}$ and $\frac{\partial \phi}{\partial \theta}$.

If so, how do I go about formulating these two equations, I don't seem to be having much luck..

2. Originally Posted by craig
Right bit of a horrible looking one here:

We've got the following:

$\mathbf{u} = U\left(1-\frac{a^3}{r^3}\right)\cos{\theta}\;\mathbf{\hat{r }} - U\left(1+\frac{a^3}{2r^3}\right)\sin{\theta}\;\mat hbf{\hat{\theta}}$ where $a$ and $U$ are constants.

We have already shown that $\nabla \cdot \mathbf{u} = 0$, and because of this we have a vector field $\mathbf{A}$ s.t. $\mathbf{u} = \nabla \times \mathbf{A}$. So assuming that $\mathbf{A} = A(r,\theta)\hat{\mathbf{\phi}}$:

Find 2 equations for $A(r,\theta)$, determine $A(r,\theta)$ and verify that they're consistent.

Right, so does the fact that $\mathbf{A} = A(r,\theta)\hat{\mathbf{\phi}}$ mean that $\mathbf{A}$ is of the form $(0,0,\phi)$? This sort of makes sense.

If so, then do we just complete the cross product and then have two differential equations, essentially $\frac{\partial \phi}{\partial r}$ and $\frac{\partial \phi}{\partial \theta}$.

If so, how do I go about formulating these two equations, I don't seem to be having much luck..

Calling $A_r, ~A_{\theta}, ~A_{\phi}$ the components of A, we have

$\displaystyle u = \hat{r} \frac{1}{r~sin( \theta )} \left [ \frac{\partial}{\partial \theta} \left ( sin( \theta)A_{\phi} \right ) - \frac{\partial A_{\theta}}{\partial \phi} \right ] + \hat{\theta} \left [ \frac{1}{r~sin( \theta )} \frac{ \partial A_r}{\partial \phi} - \frac{1}{r} \frac{\partial}{\partial r} \left ( r A_{\phi} \right ) \right ] + \hat{\phi} \frac{1}{r} \left [ \frac{\partial}{\partial r} \left ( r A_{\theta} \right ) - \frac{\partial A_r }{\partial \theta} \right ]$

This gives
$\displaystyle u = \hat{r} \frac{1}{r~sin( \theta )} \frac{\partial}{\partial \theta} \left ( sin( \theta ) A_{\phi} \right ) - \hat{ \theta } \frac{1}{r} \frac{\partial}{\partial r} \left ( r A_{\phi} \right )$

Eventually leading to the two equations
$\displaystyle \frac{\partial A_{\phi}}{\partial \theta} + cot( \theta )A_{\phi} = U \left ( r - \frac{a^3}{r^2} \right )~cos( \theta )$

and
$\displaystyle \frac{\partial A_{\phi}}{\partial r} + \frac{1}{r}A_{\phi} = U \left ( 1 + \frac{a^3}{2r^3} \right ) ~sin( \theta )$

-Dan

3. I can help with this one, at least.

$\displaystyle \frac{\partial A_{\phi}}{\partial r} + \frac{1}{r}A_{\phi} = U \left ( 1 + \frac{a^3}{2r^3} \right ) ~sin( \theta )$

Multiply through by r:
$\displaystyle r\frac{\partial A_{\phi}}{\partial r} + A_{\phi} = U \left ( r + \frac{a^3}{2r^2} \right ) ~sin( \theta )$

$\displaystyle \frac{\partial }{\partial r} \left ( r A_{\phi} \right ) = U \left ( r + \frac{a^3}{2r^2} \right ) ~sin( \theta )$

$\displaystyle r A_{\phi} = \int \left [U \left ( r + \frac{a^3}{2r^2} \right ) \right ]~dr \cdot sin( \theta ) + \Theta ( \theta )}$

$\displaystyle r A_{\phi} = \frac{1}{2}Ur^2 ~sin( \theta ) - \frac{a^3U}{2r}~sin( \theta ) + \Theta ( \theta )}$

where $\Theta ( \theta )$ is an an arbitrary function of $\theta$. (Notice that it is not also a function of $\phi$. This is a restriction on the original statement on the function $A(r,~ \theta)$.)

-Dan

4. The other is easier than I thought.

$\displaystyle \frac{\partial A_{\phi}}{\partial \theta} + cot( \theta )A_{\phi} = U \left ( r - \frac{a^3}{r^2} \right )~cos( \theta )$

Again use an integrating factor, $sin(\theta)$ this time, and you will eventually get
$\displaystyle A = - \frac{1}{2} U \left ( r - \frac{a^3}{r^2} \right )~cos( \theta )~cot( \theta ) + \left ( \frac{1}{4} U \left ( r - \frac{a^3}{r^2} \right ) + R(r) \right )~csc( \theta )$

Where R(r) is an arbitrary function of r.

-Dan

5. Wow thanks for that!

Ahh so it looks like I was on the right track with the cross product then.

How do I go about verifying that they are consistent, would this involve partially differentiating the first $A_{\phi}$ wrt $\theta$ and vice versa, showing that you get the same values for $\partial A$?

Thanks again for that great reply!

6. Originally Posted by craig
Wow thanks for that!

Ahh so it looks like I was on the right track with the cross product then.

How do I go about verifying that they are consistent, would this involve partially differentiating the first $A_{\phi}$ wrt $\theta$ and vice versa, showing that you get the same values for $\partial A$?

Thanks again for that great reply!
I'm thinking they want you to show that, for specific values of $\Theta$ and R, the two solutions can be put into the same form.

-Dan

Edit: Or perhaps that $\nabla \times A$ is the same for both solutions.

7. Sorry could you just explain how you got to:

$\displaystyle \frac{\partial A_{\phi}}{\partial \theta} + cot( \theta )A_{\phi} = U \left ( r - \frac{a^3}{r^2} \right )~cos( \theta )$

Equating the $\hat{r}$ coefficients from the cross product and the original equation for $u$ gives:

$\frac{1}{r~sin( \theta )} \frac{\partial}{\partial \theta} ( sin( \theta)A_{\phi} ) = U ( 1 - \frac{a^3}{r^3} )~cos( \theta )$

Thanks again for the help.

8. Originally Posted by craig
$\frac{1}{r~sin( \theta )} \frac{\partial}{\partial \theta} ( sin( \theta)A_{\phi} ) = U ( 1 - \frac{a^3}{r^3} )~cos( \theta )$
$\displaystyle \frac{1}{r~sin( \theta )} \left ( cos( \theta)A_{\phi} + sin( \theta ) \frac{\partial A_{\phi}}{\partial \theta } \right ) = U ( 1 - \frac{a^3}{r^3} )~cos( \theta )$

$\displaystyle \frac{1}{r}~cot( \theta )} A_{\phi} + \frac{1}{r}~\frac{\partial A_{\phi}}{\partial \theta } \right ) = U ( 1 - \frac{a^3}{r^3} )~cos( \theta )$

$\displaystyle cot( \theta )} A_{\phi} + \frac{\partial A_{\phi}}{\partial \theta } \right ) = U ( r - \frac{a^3}{r^2} )~cos( \theta )$

-Dan

9. Ahh of course! Not sure why I missed that last night...

10. Sorry could you just clarify something for me:

Originally Posted by topsquark
$\displaystyle \frac{\partial A_{\phi}}{\partial \theta} + cot( \theta )A_{\phi} = U \left ( r - \frac{a^3}{r^2} \right )~cos( \theta )$

Again use an integrating factor, $sin(\theta)$ this time, and you will eventually get
$\displaystyle A = - \frac{1}{2} U \left ( r - \frac{a^3}{r^2} \right )~cos( \theta )~cot( \theta ) + \left ( \frac{1}{4} U \left ( r - \frac{a^3}{r^2} \right ) + R(r) \right )~csc( \theta )$

Where R(r) is an arbitrary function of r.

-Dan
Using the integrating factor:

$\displaystyle \frac{\partial}{\partial \theta} \left ( ~sin( \theta ) A_{\phi}\right ) = U \left ( r - \frac{a^3}{r^2} \right )~cos( \theta )~sin( \theta )$

Which if we integrate both sides wrt $\theta$ gives

$\displaystyle ~sin( \theta ) A_{\phi} = U \left ( r - \frac{a^3}{r^2} \right ) \left (\frac{-1}{2} ~cos^2( \theta ) + R(r) \right )$

Not sure where you went from here?

Sorry for all the questions.