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Laplace Operator in Polar Coordinates

  1. Oct 5, 2008 #1

    dav2008

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    1. The problem statement, all variables and given/known data
    Compute [tex] \nabla \cdot \nabla f[/tex] in polar coordinates.


    2. Relevant equations
    [​IMG]

    3. The attempt at a solution
    It seems like a straightforward dot product yields
    [tex] \nabla \cdot \nabla f = {\partial^2 f \over \partial \rho^2}
    + {1 \over \rho^2} {\partial^2 f \over \partial \theta^2}
    + {\partial^2 f \over \partial z^2 }[/tex]
    since the 3 basis vectors are mutually orthogonal.

    This is obviously not the correct expression, which should be:
    [tex]{1 \over \rho} {\partial \over \partial \rho}
    \left( \rho {\partial f \over \partial \rho} \right)
    + {1 \over \rho^2} {\partial^2 f \over \partial \theta^2}
    + {\partial^2 f \over \partial z^2 [/tex]

    Where is that first term coming from?

    It seems like I'm ignoring something simple when calculating the dot product. What am I failing to take into account?
     
  2. jcsd
  3. Oct 5, 2008 #2

    gabbagabbahey

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    It comes from the fact that the divergence of a vector [itex]\vec{v}[/itex] in cylindrical coordinates is

    [tex]\vec{\nabla} \cdot \vec{v}=\frac{1}{\rho} \frac{\partial}{\partial \rho} (\rho v_{\rho})+\frac{1}{\rho} \frac{\partial v_{\theta}}{\partial \theta}+ \frac{\partial v_{z}}{\partial z}[/tex]

    not just

    [tex]\vec{\nabla} \cdot \vec{v}= \frac{\partial v_{\rho}}{\partial \rho}+\frac{1}{\rho} \frac{\partial v_{\theta}}{\partial \theta}+ \frac{\partial v_{z}}{\partial z}[/tex]
     
  4. Oct 5, 2008 #3

    dav2008

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    ok, but why is it incorrect to compute the dot product:

    [tex]\left(\frac{\partial }{\partial \rho}\mathbf{e}_\rho+
    \frac{1}{\rho}\frac{\partial }{\partial \theta}\mathbf{e}_\theta+
    \frac{\partial }{\partial z}\mathbf{e}_z\right) \bullet \left(
    \frac{\partial }{\partial \rho}\mathbf{e}_\rho+
    \frac{1}{\rho}\frac{\partial }{\partial \theta}\mathbf{e}_\theta+
    \frac{\partial }{\partial z}\mathbf{e}_z\right) [/tex]

    as

    [tex] {\partial^2 \over \partial \rho^2}
    + {1 \over \rho^2} {\partial^2 \over \partial \theta^2}
    + {\partial^2 \over \partial z^2 }[/tex]

    What assumptions in taking the dot product are incorrect?

    Edit: I guess I'll just write it in terms of the Cartesian base vectors and work it out that way. I see now that the dot products of the unit vectors don't work out to just be 1 or 0.
     
  5. Oct 5, 2008 #4

    dav2008

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    Hmm I wrote er and eθ in terms of the cartesian base vectors, did the dot product, and it still comes out the same way...which I guess is expected.
     
  6. Oct 5, 2008 #5

    Ben Niehoff

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    You need to remember to differentiate the unit vectors also. For example,

    [tex]\frac{\partial}{\partial \rho} \hat e_{\rho} \cdot \frac{\partial}{\partial \rho} \hat e_{\rho} = \frac{\partial^2}{\partial \rho^2} \hat e_{\rho} + \frac{\partial}{\partial \rho} \frac{\partial \hat e_{\rho}}{\partial \rho}[/tex]

    It just so happens that

    [tex]\frac{\partial \hat e_{\rho}}{\partial \rho} = 0[/tex]

    so this particular term does not contribute. However, the [itex]\theta[/itex] term will give you

    [tex]\frac{\partial \hat e_{\theta}}{\partial \theta}[/tex]

    which is nonzero; this should give you your missing term (you will have to collect terms and undo a product rule to get the exact form needed).
     
  7. Oct 5, 2008 #6

    dav2008

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    Ahh there it is. Thanks.

    Edit: Do you mean [tex]
    \frac{\partial}{\partial \rho} \hat e_{\rho} \cdot \frac{\partial}{\partial \rho} \hat e_{\rho} = \frac{\partial^2}{\partial \rho^2}} + \hat e_{\rho} \cdot \frac{\partial}{\partial \rho} \frac{\partial \hat e_{\rho}}{\partial \rho}
    [/tex]
     
    Last edited: Oct 5, 2008
  8. Oct 6, 2008 #7

    Ben Niehoff

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    Right, yes. I forgot to take the dot product. :P
     
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