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Gradient of 1/r

  1. Sep 7, 2007 #1
    1. The problem statement, all variables and given/known data

    This is from Griffiths' Intro to Electrodynamics. He is discussing the field of a polarized object of dipole moment per unit volume [tex] \vec{P} [/tex] viewed at [tex] \vec{r} [/tex].

    He then states:

    [tex] \nabla ' \left( \frac{1}{r} \right) = \frac{ \hat{r}}{r^2} [/tex]

    Where [tex] \nabla ' [/tex] denotes that the differentiation is with respect to the source co-ordinates [tex] \vec{r}' [/tex]

    2. Relevant equations

    3. The attempt at a solution

    Following from the definition of the gradient,

    [tex] \nabla ' \left( \frac{1}{r} \right) = \frac{-1}{r^3} \left[ x \frac{ \partial x}{\partial x'} \hat{x}' + y \frac{\partial y}{\partial y'} \hat{y}' + z \frac{\partial z}{\partial z'} \hat{z} \right]

    So I guess all would be well as long as
    [tex] \frac{\partial x}{\partial x'} \hat{x}' = - \hat{x} [/tex]
    However, this isn't clear to me at the moment
  2. jcsd
  3. Sep 7, 2007 #2
    The easiest method is to use the chain rule in r

    d/dx = d r^2/dx * d/d r^2

    r^2 = x^2+y^2+z^2

    d r^2/dx = 2x

    so d/dx= 2x d / dr^2 = 2x dr/dr^2 d/dr = (x/r) d/dr

    d/dx (1/r) = -(x/r) 1/r^2
  4. Sep 7, 2007 #3
    Well, just to complete the above:

    grad (1/r) = (d/dx xhat + d/dy yhat +d/dz zhat) (1/r)
    = -x/r^3 xhat -y/r^3 yhat -z/r^3 zhat
    = -(x,y,z)/r^3=-rhat/r^2

    apologies for being too lazy to latex this.

    (edit- corrected dumb mistakes)
    (edit again- corrected dumb corrections. Hopefully this is right now. I had to write it down)
    Last edited: Sep 7, 2007
  5. Sep 7, 2007 #4
    The 'pro' method is to memorize the useful formula

    grad (f(r))= rhat df(r)/dr

    You can prove this using the methods above.
  6. Sep 7, 2007 #5


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    Why not use the gradient operator for spherical polar coordinates, noting that f is a function of r alone?
    [tex]\nabla f=\hat{r}\frac{\partial f}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial f}{\partial \theta}+\hat{\varphi}\frac{1}{r\sin\theta}\frac{\partial f}{\partial\varphi}[/tex]
  7. Sep 7, 2007 #6
    Thanks for the help so far.

    Christianjb, what you've written is correct, but it proves
    [tex] \nabla \left( \frac{1}{r} \right) = \frac{- \hat{r}}{r^2} [/tex]

    I'm cool with this, but my problem is that Griffiths is differentiating with respect to a different co-ordinate system (hence the prime on the gradient operator), which seems to cause one to lose a minus sign. He calls this the source coordinates and they are integrated over since we are not dealing with a point charge.

    Christo, so if I use the spherical gradient, ignoring angular parts,
    [tex] \nabla f = \hat{r} \frac{\partial f}{\partial r} [/tex]
    And presumably I can extend this to my case by changing the co-ordinate system so that
    [tex] \nabla ' f = \hat{r}' \frac{\partial f}{\partial r'} [/tex]
    Now substituting 1/r for f,
    [tex] \nabla ' \left( \frac{1}{r} \right) = \hat{r}' \frac{\partial}{\partial r'} \left(\frac{1}{r} \right) = \frac{- \hat{r}'}{r^2} \frac{\partial r}{\partial r'} [/tex]

    So that this time it appears that I require
    [tex] \hat{r} = - \frac{\partial r}{\partial r'} \hat{r}' [/tex]
    in order to be in agreement with Griffiths.

    This is kinda neater than what I first posted with individual components, but I'm still not sure why it's true.
  8. Sep 7, 2007 #7


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    Staff Emeritus
    Science Advisor

    Ok, I didn't notice the prime. What's the relationship between the primed and the unprimed coordinates? It may turn out that you cannot ignore the angular parts-- just because f is a function only of r it doesn't mean that f is a function of only r'.

    (I don't have the text, so am relying solely on what you write here!)
  9. Sep 7, 2007 #8
    Yeah, I was beginning to realise that was the problem. The relationship between the co-ordinate systems isn't explicitly stated.

    What we've got is the usual arbitrary blob in space, which has a dipole moment per unit volume [tex] \vec{P} [/tex]. We want to know what the potential is due to this blob.

    For a simple dipole [tex] \vec{p} [/tex] we have
    V (\vec{r}) = \frac{1}{4 \pi \epsilon_0 } \frac{\hat{\mathcal{R}} \cdot \vec{p}}{\mathcal{R}^2}

    Where [tex]\mathcal{R}[/tex] is the vector from the dipole to the point at which we are evaluating the potential.

    So in our case we have a dipole moment
    \vec{p} = \vec{P} d \tau'
    [/tex] in each volume element [tex] d \tau' [/tex] so the total potential is:

    V(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \int_\mathcal{V} \frac{ \hat{\mathcal{R}} \cdot \vec{P}( \vec{r}') }{\mathcal{R}^2} d \tau '

    Now he states that
    \nabla ' \left( \frac{1}{\mathcal{R}} \right) = \frac{\hat{\mathcal{R}}}{\mathcal{R}^2}
    where the differentiation is with respect to the source coordinates [tex] (\vec{r}') [/tex]

    To me, this isn't very clear but I reckon that we've got an origin. We want the potential at point [tex] \vec{r} [/tex] from the origin. Now we have a dipole at position [tex] \vec{r} ' [/tex]. And we are told that our point is at [tex] \vec{ \mathcal{R} } [/tex] from the dipole.
    So that surely [tex] \vec{\mathcal{R}} = \vec{r} - \vec{r}' [/tex]
  10. Sep 7, 2007 #9
    It's a simple sign change when you diff wrt the axis coordinates.

    In 1D- moving the origin 1 unit to the left has the effect of increasing all x values by 1. Thus the signs are reversed.
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