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Find the gradient of 1/mod{r-r'}

  1. Dec 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Find [tex]\nabla\left( \dfrac{1}{\left| \vec{r}-\vec{r'}\right| }\right)[/tex]

    2. Relevant equations



    3. The attempt at a solution[tex]
    \left| \vec{r}-\vec{r'}\right| =\sqrt{(x-x^\prime)^2 + (y-y^\prime)^2 + (z-z^\prime)^2}[/tex]
    and so therefore the derivative of the scalar would be 0. Of course then it seems trivial. So perhaps I need to use the chain rule, but I don't know what to make my substitution for. I can't seem to escape the fact that the bottom always ends up being a scaler.

    So I seem a bit stuck. Any advice on where to go onwards would be appreciated. Thank you.
     
  2. jcsd
  3. Dec 5, 2012 #2

    haruspex

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    I don't understand how you get 0. What do you get for
    [tex]\frac{\partial}{\partial x}((x-x^\prime)^2 + (y-y^\prime)^2 + (z-z^\prime)^2)^{-1/2}[/tex]?
    Yes, you need to use the chain rule. You can break it up as finely as you like, but a reasonable choice would be
    [tex]\frac{\partial f}{\partial x} \frac{\partial f^{-1/2}}{\partial f}[/tex]
    where [tex]f = (x-x^\prime)^2 + (y-y^\prime)^2 + (z-z^\prime)^2[/tex]
     
  4. Dec 6, 2012 #3
    Thank you.

    Okay, so doing that I get [tex]
    \frac{\partial}{\partial x}\left(\frac{1}{\left| \vec{r}-\vec{r'}\right|}\right) = \frac{x^{\prime}-x}{\sqrt{(x-x^\prime)^2 + C}^3}
    [/tex]

    by treating x' as a constant.Does that seem along the right path?
     
  5. Dec 6, 2012 #4

    haruspex

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    Sort of. I assume C here stands for (y-y')2 + (z-z')2, but why are you choosing to represent that with C? Suggests to me you are thinking of it as a constant, and so playing a different role from the (x-x')2 part. Yes, you had treat it as a constant for the purposes of ∂/∂x, but having done that it tells you the value of the derivative at any point (x,y,z), so they should still be considered variables on an equal footing. So I would suggest representing the whole square root part using the original modulus formula involving r.
    What do you then get for the whole grad (∇) vector?
     
  6. Dec 6, 2012 #5
    I got this:

    [tex]\nabla \left(\frac{1}{\left| \vec{r}-\vec{r'}\right|}\right) = \frac{ (x^\prime-x) \vec{i} + (y^\prime-y) \vec{j} + (z^\prime-z)\vec{k}} {\sqrt{(x-x^\prime)^2 + (y-y^\prime)^2 + (z-z^\prime)^2}^3}[/tex]

    by adding the x, y and z parts of the same form as above.
     
    Last edited: Dec 6, 2012
  7. Dec 6, 2012 #6

    haruspex

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    Two things wrong with that. First, I assume you meant ∇ on the left, not ∂/∂x. Secondly, you shouldn't add the three terms up. ∇ is a vector, namely (∂/∂x, ∂/∂y, ∂/∂z), so the answer should be a vector consisting of those three terms.
     
  8. Dec 6, 2012 #7
    Yes, just typos. Corrected as above.
     
  9. Dec 6, 2012 #8

    haruspex

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    Looks good.
     
  10. Dec 6, 2012 #9
    Thank you for your assistance :).
     
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