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

1. Dec 5, 2012

### Ashiataka

1. The problem statement, all variables and given/known data
Find $$\nabla\left( \dfrac{1}{\left| \vec{r}-\vec{r'}\right| }\right)$$

2. Relevant equations

3. The attempt at a solution$$\left| \vec{r}-\vec{r'}\right| =\sqrt{(x-x^\prime)^2 + (y-y^\prime)^2 + (z-z^\prime)^2}$$
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. Dec 5, 2012

### haruspex

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

3. Dec 6, 2012

### Ashiataka

Thank you.

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

by treating x' as a constant.Does that seem along the right path?

4. Dec 6, 2012

### haruspex

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?

5. Dec 6, 2012

### Ashiataka

I got this:

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

by adding the x, y and z parts of the same form as above.

Last edited: Dec 6, 2012
6. Dec 6, 2012

### haruspex

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.

7. Dec 6, 2012

### Ashiataka

Yes, just typos. Corrected as above.

8. Dec 6, 2012

Looks good.

9. Dec 6, 2012

### Ashiataka

Thank you for your assistance :).