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Partial derivative in spherical coordinates

  1. Oct 12, 2011 #1
    I am facing some problem about derivatives in spherical coordinates

    in spherical coordinates:
    x=r sinθ cos[itex]\phi[/itex]
    y=r sinθ sin[itex]\phi[/itex]
    z=r cosθ

    and
    r=[itex]\sqrt{x^{2}+y^{2}+z^{2}}[/itex]
    θ=tan[itex]^{-1}[/itex][itex]\frac{\sqrt{x^{2}+y{2}}}{z}[/itex]
    [itex]\phi[/itex]=tan[itex]^{-1}[/itex][itex]\frac{y}{x}[/itex]

    [itex]\frac{\partial x}{\partial r}[/itex]=sinθ cos[itex]\phi[/itex]
    then [itex]\frac{\partial r}{\partial x}[/itex]=[itex]\frac{1}{sinθ cos \phi }[/itex]

    but if i calculate directly from r:
    [itex]\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}[/itex]
    substitute:
    =[itex]\frac{r sinθ cos \phi }{r}[/itex]
    = sinθ cos[itex]\phi[/itex]

    Why do the results are different? what i did wrong?


    From https://www.physicsforums.com/showthread.php?t=63886
    not this case is the second case? but why the inverse still not true?
     
    Last edited: Oct 12, 2011
  2. jcsd
  3. Oct 12, 2011 #2

    mathman

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    ∂r/∂x is defined for constant x and y.
    ∂x/∂r is defined for constant θ and φ.

    There is no reason that they should be reciprocal.
     
  4. Oct 12, 2011 #3

    HallsofIvy

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    You mean "for constant y and z" don't you?

     
  5. Oct 13, 2011 #4

    mathman

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    Correct - my typo.
     
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