# Homework Help: Gradient of 1/r

1. Sep 7, 2007

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 $$\vec{P}$$ viewed at $$\vec{r}$$.

He then states:

$$\nabla ' \left( \frac{1}{r} \right) = \frac{ \hat{r}}{r^2}$$

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

2. Relevant equations

3. The attempt at a solution

Following from the definition of the gradient,

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

So I guess all would be well as long as
$$\frac{\partial x}{\partial x'} \hat{x}' = - \hat{x}$$
However, this isn't clear to me at the moment

2. Sep 7, 2007

### christianjb

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

3. Sep 7, 2007

### christianjb

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
4. Sep 7, 2007

### christianjb

The 'pro' method is to memorize the useful formula

You can prove this using the methods above.

5. Sep 7, 2007

### cristo

Staff Emeritus
Why not use the gradient operator for spherical polar coordinates, noting that f is a function of r alone?
$$\nabla f=\hat{r}\frac{\partial f}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial f}{\partial \theta}+\hat{\varphi}\frac{1}{r\sin\theta}\frac{\partial f}{\partial\varphi}$$

6. Sep 7, 2007

Thanks for the help so far.

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

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,
$$\nabla f = \hat{r} \frac{\partial f}{\partial r}$$
And presumably I can extend this to my case by changing the co-ordinate system so that
$$\nabla ' f = \hat{r}' \frac{\partial f}{\partial r'}$$
Now substituting 1/r for f,
$$\nabla ' \left( \frac{1}{r} \right) = \hat{r}' \frac{\partial}{\partial r'} \left(\frac{1}{r} \right) = \frac{- \hat{r}'}{r^2} \frac{\partial r}{\partial r'}$$

So that this time it appears that I require
$$\hat{r} = - \frac{\partial r}{\partial r'} \hat{r}'$$
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.

7. Sep 7, 2007

### cristo

Staff Emeritus
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!)

8. Sep 7, 2007

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 $$\vec{P}$$. We want to know what the potential is due to this blob.

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

Where $$\mathcal{R}$$ 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
$$\vec{p} = \vec{P} d \tau'$$ in each volume element $$d \tau'$$ 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 $$(\vec{r}')$$

To me, this isn't very clear but I reckon that we've got an origin. We want the potential at point $$\vec{r}$$ from the origin. Now we have a dipole at position $$\vec{r} '$$. And we are told that our point is at $$\vec{ \mathcal{R} }$$ from the dipole.
So that surely $$\vec{\mathcal{R}} = \vec{r} - \vec{r}'$$

9. Sep 7, 2007

### christianjb

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.