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D/dx in spherical coordinates

  1. Aug 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi. I have a simple question. Is it true that [tex] \frac{\partial r}{\partial x} = (\frac{\partial x}{\partial r})^{-1} [/tex] ?

    Because I'm having some trouble with the conversion between rectangular and spherical coordinates.

    2. Relevant equations

    [tex] x = r cos \phi sin \theta [/tex]

    [tex] y = r sin \phi sin \theta [/tex]

    [tex] z = r cos \theta [/tex]

    [tex] r = \sqrt{x^2+y^2+z^2} [/tex]

    3. The attempt at a solution

    It is easy to show that
    [tex] \frac{\partial r}{\partial x} = cos \phi sin \theta [/tex]

    However, we see that
    [tex] (\frac{\partial x}{\partial r})^{-1} = (\frac{\partial (r cos \phi sin \theta)}{\partial r})^{-1}= \frac{1}{cos \phi sin \theta} [/tex]

    and these are clearly not equal.

    What am I missing? :confused:
     
  2. jcsd
  3. Aug 27, 2009 #2

    tiny-tim

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    Hi raul_l! :smile:

    (have a curly d: ∂ :wink:)
    No, it doesn't work for partial derivatives, because they depend on what the other (unwritten) coordinates are.

    ∂r/dx keeps y constant, but ∂x/dr keeps θ constant …

    and keeping y and θ constant aren't the same! :wink:
     
  4. Aug 27, 2009 #3
    Yes, I suspected that. Thanks.

    But it's weird that if I multiply both sides by [tex] \frac{\partial x}{\partial r} [/tex] I get
    [tex] \frac{\partial r}{\partial x} \frac{\partial x}{\partial r} = 1 [/tex]
    and that seems mathematically correct.
     
  5. Aug 27, 2009 #4
    You made a little mistake.:wink:

    [tex]\frac{dx}{dr}=cos \phi sin \theta[/tex]

    Proof.

    [tex]
    r = \frac{x}{cos \phi sin \theta}
    [/tex]

    [tex]dr=\frac{(x)'cos \phi sin \theta - x(cos \phi sin \theta)'}{cos^2 \phi sin^2 \theta} dx[/tex]

    [tex]dr=\frac{cos \phi sin \theta}{cos^2 \phi sin^2 \theta}dx[/tex]

    [tex]dr=\frac{1}{cos \phi sin \theta}dx[/tex]

    [tex]\frac{dr}{dx}=\frac{1}{cos \phi sin \theta}[/tex]

    Is this right, mate :smile: ?
     
    Last edited: Aug 27, 2009
  6. Aug 27, 2009 #5

    tiny-tim

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    Sorry, chuck :redface:

    not unless you're using those weird (x,θ,φ) coordinates. :rolleyes:
     
  7. Aug 27, 2009 #6
    Just you need to prove that

    [tex] \frac{dr}{dx} = (\frac{dx}{dr})^{-1} [/tex]

    [tex](\frac{dx}{dr})^{-1}=(cos \phi sin \theta)^{-1}=\frac{1}{cos \phi sin \theta}=\frac{dr}{dx}[/tex]

    I can do science me. :smile:
     
  8. Aug 27, 2009 #7
    njama, there's a difference between [tex] \frac{dr}{dx} [/tex] and [tex] \frac{\partial r}{\partial x} [/tex].

    Correct me if I'm wrong but I think the left side of what you wrote is equal to
    [tex] \frac{dr}{dx} = \frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} \frac{d y}{d x} + \frac{\partial r}{\partial z} \frac{d z}{d x} [/tex]

    while right side is equal to
    [tex] (\frac{dx}{dr})^{-1} = (\frac{\partial x}{\partial r} + \frac{\partial x}{\partial \phi} \frac{d \phi}{d r} + \frac{\partial x}{\partial \theta} \frac{d \theta}{d r})^{-1} [/tex]

    and it doesn't look like these would be equal.
     
    Last edited: Aug 28, 2009
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