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

[Ashiataka]

Homework Statement

Find \nabla\left( \dfrac{1}{\left| \vec{r}-\vec{r'}\right| }\right)

Homework Equations

The Attempt at a Solution

<br /> \left| \vec{r}-\vec{r&#039;}\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.

 

[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

 

[Ashiataka]

Thank you.

Okay, so doing that I get <br /> \frac{\partial}{\partial x}\left(\frac{1}{\left| \vec{r}-\vec{r&#039;}\right|}\right) = \frac{x^{\prime}-x}{\sqrt{(x-x^\prime)^2 + C}^3}<br />

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

 

[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?

 

[Ashiataka]

I got this:

\nabla \left(\frac{1}{\left| \vec{r}-\vec{r&#039;}\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.

 

[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.

 

[Ashiataka]

Yes, just typos. Corrected as above.

 

[haruspex]

Looks good.

 

[Ashiataka]

Thank you for your assistance :).