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Cylindrical Coordinates -Div/Curl/Grad

  1. Mar 4, 2006 #1
    We just started a section dealing with div/curl/grad in different orthogonal systems... before I get started doing problems involving these operations I wanted to make sure I am dealing with these operation correctly. Our first homework problem is as follows:

    In cylindrical coordinates compute:
    (1) [tex] \nabla \theta [/tex]
    (2) [tex] \nabla r^4 [/tex]
    (3) [tex] \nabla \cdot i_r [/tex]
    (4) [tex] \nabla \times \nabla \cdot \theta [/tex]
    ...

    The definition for [tex] \nabla [/tex] in cylindrical coordinate is:
    [tex]\nabla f = \frac{\partial f}{\partial r} \hat r + \frac{1}{r}\frac{\partial f}{\partial \theta} \hat \theta + \frac{\partial f}{\partial z} \hat z [/tex]

    So...
    (1) I first take [tex] \frac{\partial \theta}{\partial r} [/tex] which equals 0. Next I take [tex] \frac{1}{r}\frac{\partial \theta}{\partial \theta} [/tex] which equals [tex]\frac{1}{r}\hat \theta [/tex], etc...

    so for problem (1)
    [tex] \nabla \theta = \frac{1}{r}\hat \theta [/tex]

    (2)
    [tex] \nabla r^4 = 4r^3 \hat r [/tex]

    (3)
    [tex] \nabla \cdot i_r = \frac{1}{r} [/tex] from the definition of [tex] div [/tex] in cylindrical coordinates:
    [tex] \nabla \cdot \vec F = \frac{1}{r}\frac{\partial}{\partial r} (r F_r) + ... [/tex]

    (4)
    [tex] \nabla \cross \nabla \theta [/tex]
    Well from (1) I have [tex] \nabla \theta = \frac{1}{r}\hat \theta [/tex]

    The general definition of curl is:

    [tex] \nabla \times \vec F = \frac{1}{abc} \left| \begin{array}{ccc} a\hat i_1 & b\hat i_2 & c\hat i_3 \\ \partial u & \partial v & \partial w \\ aF_1 & bF_2 & cF_3 \end{array} \right| [/tex]
    where: a=1, b=r, c=1, u=r, v=theta, w=z

    Then plugging in the values I get:

    [tex] \nabla \times \frac{1}{r}\hat \theta = \frac{1}{r} \left| \begin{array}{ccc} \hat r & r\hat \theta & \hat z \\ \partial r & \partial \theta & \partial z \\ 0 & r(\frac{1}{r}) & 0 \end{array} \right| = 0 [/tex] which is equal to 0, as it should...


    So does everything look sound? Is my thought process here ok? I\d appreciate any help.
     
    Last edited: Mar 4, 2006
  2. jcsd
  3. Mar 4, 2006 #2

    benorin

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    Homework Helper

    (1) and (2) are fine, What is [tex]i_r[/tex] in (3)? and, about (4), I get

    [tex] \nabla \times \nabla \cdot \theta = \nabla \times (\nabla \theta) = \nabla \times \left(\frac{1}{r}\hat \theta\right)
    = \left| \begin{array}{ccc} \hat r & \hat \theta & \hat z \\ \frac{\partial}{\partial r} & \frac{1}{r}\frac{\partial}{\partial \theta} & \frac{\partial}{\partial z} \\ 0 & \frac{1}{r} & 0 \end{array} \right| = -\frac{1}{r^2} \hat z [/tex]

    But I am not sure: what do you think?
     
  4. Mar 4, 2006 #3
    Woops sorry I didn't define [itex] i_r [/itex]. That is the books notation for a unit vector for the [itex] r [/itex] component. So [itex] i_r = e_r = \hat r [/itex].

    And for (4), isn't a standard identity that:
    [tex] \nabla \times \nabla \cdot \vec F = \vec 0 [/tex] for any orthogonal coordinate system?

    EDIT:
    wait sorry... I looked that up. Its div curl F = 0

    Well the book says 0 for (4).
     
    Last edited: Mar 4, 2006
  5. Mar 4, 2006 #4

    benorin

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    I tried

    [tex] \nabla \times \nabla = \left| \begin{array}{ccc} \hat r & \hat \theta & \hat z \\ \frac{\partial}{\partial r} & \frac{1}{r}\frac{\partial}{\partial \theta} & \frac{\partial}{\partial z} \\ \frac{\partial}{\partial r} & \frac{1}{r}\frac{\partial}{\partial \theta} & \frac{\partial}{\partial z} \end{array} \right| = 0\hat r - 0\hat \theta +\left[\frac{\partial}{\partial r}\left( \frac{1}{r} \frac{\partial}{\partial \theta}\right) -\frac{1}{r}\frac{\partial ^2}{\partial \theta \partial r}\right] \hat z = \left[ \left(-\frac{1}{r^2} \frac{\partial}{\partial \theta}+\frac{1}{r}\frac{\partial ^2}{\partial \theta \partial r}\right) -\frac{1}{r}\frac{\partial ^2}{\partial \theta \partial r}\right] \hat z= -\frac{1}{r^2}\frac{\partial}{\partial \theta} \hat z [/tex]

    which gives the same result as before, but here the order of--I digress: operators get applied rather than multiplied--application of the operators is crucial, for if the other order is taken, there is no product rule and the cross product is identically 0. But your book says it's 0.
     
  6. Mar 4, 2006 #5

    benorin

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    Homework Helper

    I found a definition I could better grasp, your initial calculation is indeed correct! My bad.
     
  7. Mar 4, 2006 #6
    No. No bad at all. I really appreciate you looking over it! I was not confident with what I was doing. I didn't feel comfortable applying the operators in different coordinate systems due to the order of the vector components. I HATE dealing in: [tex] F_1 \hat r + F_2 \hat \theta + F_3 \hat z [/tex] notation (unit vector notation ?). I'd much rather handle it in this type of notation: [tex] \left[ \begin{array}{c} F_1 & F_2 & F_3 \end{array} \right] [/tex]


    But, like anything... I'll just have to do it a few times until I'm comfortable.

    Anyways, thanks benorin.
     
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