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Gradient of (1/r)

  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data

    gradient(1/r) = r(hat) / r^2


    2. Relevant equations
    r = (x-x')i + (y-y')j + (z-z')k
     
  2. jcsd
  3. Sep 11, 2014 #2

    ShayanJ

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    gradient(1/r) = -r(hat) / r^2
     
  4. Sep 11, 2014 #3
    how do you prove that?
     
  5. Sep 11, 2014 #4

    ShayanJ

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    Gold Member

    In spherical coordinates, the gradient of a scalar function f is:
    [itex]\vec\nabla f(r, \theta, \phi) = \frac{\partial f}{\partial r}\hat r+ \frac{1}{r}\frac{\partial f}{\partial \theta}\hat\theta+ \frac{1}{r \sin\theta}\frac{\partial f}{\partial \phi}\hat\phi [/itex].
    And we have [itex] \frac{d}{dr}\frac 1 r=-\frac{1}{r^2} [/itex].
     
  6. Sep 11, 2014 #5
    can this be done in cartesian coordinates?
     
  7. Sep 11, 2014 #6

    HallsofIvy

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    Or, in Cartesian coordinates,
    [tex]\frac{1}{r}= \frac{1}{x^2+ y^2+ z^2}= (x^2+ y^2+ z^2)^{-1/2}[/tex]

    [tex]\left(\frac{1}{r}\right)_x= -\frac{1}{2}(x^2+ y^2+ z^2)^{-3/2}(2x)= -\frac{rcos(\theta)sin(\phi)}{r^3}= -\frac{1}{r^3} (rcos(\theta)sin(\phi))[/tex]

    [tex]\left(\frac{1}{r}\right)_y= -\frac{1}{2}(x^2+ y^2+ z^2)^{-3/2}(2y)= -\frac{rsin(\theta)sin(\phi)}{r^3}= -\frac{1}{r^3} (rsin(\theta)sin(\phi)([/tex]

    [tex]\left(\frac{1}{r}\right)_z= -\frac{1}{2}(x^2+ y^2+ z^2)^{-3/2}(2z)= -\frac{rcos(\phi)}{r^3}= -\frac{1}{r^3} (rcos(\phi))[/tex]

    So that [tex]\nabla \frac{1}{r}= -\frac{1}{r^3}\vec{r}= -\frac{1}{r^2}\frac{\vec{r}}{r}= -\frac{1}{r^2}\hat{r}[/tex]

    Where [itex]\hat{r}[/itex] is the unit vector in the direction of [itex]\vec{r}[/itex].
     
    Last edited: Sep 11, 2014
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