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Math Help - 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:

    \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..

    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:

    \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..

    Thanks in advance
    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
    Last edited by topsquark; March 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 \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
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  4. #4
    Forum Admin topsquark's Avatar
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    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
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  5. #5
    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 A_{\phi} wrt \theta and vice versa, showing that you get the same values for \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 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.
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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 \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 )

    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
    \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
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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 \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.
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