1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook