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How to deduce rot and div formulas under spherical or cylinderical coordinat

  1. Oct 4, 2005 #1
    [SOLVED] How to deduce rot and div formulas under spherical or cylinderical coordinat

    It is said that in general, these formulas are:

    if [tex]
    dl_1=h_1du_1, dl_2=h_2du_2, dl_3=h_3du_3
    [/tex]

    then
    [tex]\nabla \Psi=\frac{1}{h_1}\frac{\partial\Psi}{\partial u_1}\vec{e}_1 + \frac{1}{h_2}\frac{\partial\Psi}{\partial u_2}\vec{e}_2 + \frac{1}{h_3}\frac{\partial\Psi}{\partial u_3}\vec{e}_3
    [/tex]

    with the first sentence, which above is easy to be understood, but how to deduce the two formulas below?

    [tex]\nabla\cdot\vec{f}=\frac{1}{h_2h_3}[\frac{\partial(h_2h_3f_1)}{\partial u_1} + \frac{\partial(h_3h_1f_2)}{\partial u_2} + \frac{\partial(h_1h_2f_3)}{\partial u_3}]
    [/tex]

    [tex]\nabla\times\vec{f}=\frac{1}{h_2h_3}[\frac{\partial(h_3f_3)}{\partial u_2} - \frac{\partial(h_2f_2)}{\partial u_3}]\vec{e}_1 + \frac{1}{h_3h_1}[\frac{\partial(h_1f_1)}{\partial u_3} - \frac{\partial(h_3f_3)}{\partial u_1}]\vec{e}_2 + \frac{1}{h_1h_2}[\frac{\partial(h_2f_2)}{\partial u_1} - \frac{\partial(h_1f_1)}{\partial u_2}]\vec{e}_3
    [/tex]
    ==========================
    some one says that
    [tex]
    h_1=\sqrt{(\frac{\partial x_1}{u_1})^2+(\frac{\partial x_2}{u_1})^2+(\frac{\partial x_3}{u_1})^2}
    [/tex]
    and etc.
    But I don't know how to deduce this too.

    thanks a lot!
     
    Last edited: Oct 4, 2005
  2. jcsd
  3. Oct 4, 2005 #2

    robphy

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  4. Oct 5, 2005 #3
    I know that [tex]h_1=\sqrt{(\frac{\partial x_1}{u_1})^2+(\frac{\partial x_2}{u_1})^2+(\frac{\partial x_3}{u_1})^2}[/tex]
    is just the formula(6) in http://mathworld.wolfram.com/CurvilinearCoordinates.html now.

    and by using the limit definitions of the divergence, I can deduce the formula like below:
    [tex]

    \nabla\cdot\vec{f} = \lim_{\delta V\rightarrow 0} \frac{1}{\delta V}\oint\vec{f}\cdot\vec{n}dS

    [/tex]
    [tex]

    = \frac{1}{h_1h_2h_3du_1du_2du_3}[(h_2h_3f_1)^{'}-(h_2h_3f_1)]du_2du_3 + [(h_1h_3f_2)^{'}-(h_1h_3f_2)]du_2du_3 +[(h_1h_2f_3)^{'}-(h_1h_2f_3)]du_2du_3

    [/tex]
    [tex]

    = \frac{1}{h_1h_2h_3du_1du_2du_3}\cdot[\frac{\partial(h_2h_3f_1)}{\partial u_1} + \frac{\partial(h_1h_3f_2)}{\partial u_2} +\frac{\partial(h_1h_2f_3)}{\partial u_3}]du_1du_2du_3

    [/tex]
    [tex]

    = \frac{1}{h_1h_2h_3}[\frac{\partial(h_2h_3f_1)}{\partial u_1} + \frac{\partial(h_1h_3f_2)}{\partial u_2} +\frac{\partial(h_1h_2f_3)}{\partial u_3}]

    [/tex]

    but the formula of rot seems to be hard because it's different definition form, I am puzzled...
     
    Last edited: Oct 5, 2005
  5. Oct 5, 2005 #4

    robphy

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    [tex]\left(\nabla\times\vec{f}\right) \cdot \hat n
    = \lim_{\delta A\rightarrow 0} \frac{1}{\delta A}\oint\vec{f}\cdot d\vec{s}[/tex]
    finds the component of the curl (rot), parallel to [tex]\hat n[/tex].

    The calculation follows a similar strategy to the one for the divergence.
    Consider the case when [tex]\hat n=\hat e_1[/tex].
    Consider a small oriented rectangle in the [tex]e_2e_3[/tex]-plane, with unit normal [tex]\hat e_1[/tex] and area [tex](h_2du_2)(h_3du_3)[/tex]. Now do the line-integral... analogous to the surface integral you did for the divergence.
     
  6. Oct 10, 2005 #5
    IC, thanks to robphy!
     
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