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Thread: Spherical Polar Coordinates

  1. #1
    Super Member craig's Avatar
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    Spherical Polar Coordinates

    Right bit of a horrible looking one here:

    We've got the following:

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

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

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

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

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

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

    Thanks in advance
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by craig View Post
    Right bit of a horrible looking one here:

    We've got the following:

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

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

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

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

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

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

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

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

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

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

    -Dan
    Last edited by topsquark; Mar 20th 2011 at 02:26 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    I can help with this one, at least.

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

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

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

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

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

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

    -Dan
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    Forum Admin topsquark's Avatar
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    The other is easier than I thought.

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

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

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

    -Dan
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    Super Member craig's Avatar
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    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 $\displaystyle A_{\phi}$ wrt $\displaystyle \theta$ and vice versa, showing that you get the same values for $\displaystyle \partial A$?

    Thanks again for that great reply!
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by craig View Post
    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 $\displaystyle A_{\phi}$ wrt $\displaystyle \theta$ and vice versa, showing that you get the same values for $\displaystyle \partial A$?

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

    -Dan

    Edit: Or perhaps that $\displaystyle \nabla \times A$ is the same for both solutions.
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  7. #7
    Super Member craig's Avatar
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    Sorry could you just explain how you got to:

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

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

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

    Just not sure how you've got your answer from the above?

    Thanks again for the help.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by craig View Post
    $\displaystyle \frac{1}{r~sin( \theta )} \frac{\partial}{\partial \theta} ( sin( \theta)A_{\phi} ) = U ( 1 - \frac{a^3}{r^3} )~cos( \theta ) $
    $\displaystyle \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 \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 \displaystyle cot( \theta )} A_{\phi} + \frac{\partial A_{\phi}}{\partial \theta } \right ) = U ( r - \frac{a^3}{r^2} )~cos( \theta ) $

    -Dan
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  9. #9
    Super Member craig's Avatar
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    Ahh of course! Not sure why I missed that last night...
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  10. #10
    Super Member craig's Avatar
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    Sorry could you just clarify something for me:

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

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

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

    -Dan
    Using the integrating factor:

    $\displaystyle \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 $\displaystyle \theta$ gives

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

    Not sure where you went from here?

    Sorry for all the questions.
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