I Electric field equation

1. Mar 19, 2017

davidge

I was reading a book on Electromagnetism and it's said on deriving the electric field that $$\nabla \frac{1}{|x-x'|} = - \frac{x-x'}{|x-x'|^3}$$ where $|x-x'|$ is the magnitude of the distance between two point charges. I've tried to derive this result and I found that $$\nabla |x-x'| = \frac{x-x'}{|x-x'|}$$ must be true for the first identity to be valid. Is this right?

2. Mar 19, 2017

Orodruin

Staff Emeritus
Yes, it essentially boils down to $\nabla r = \vec e_r$ in spherical coordinates.

3. Mar 19, 2017

davidge

Thanks. Can you point out where I'm wrong?

I'm assuming a Minkowskian metric, so vectors- and covectors- components are equal. So we have $|x-x'| = \sum_i(x^i - x'^{\ i})^2$. I guess we need some formalism to express $x-x'$, because it's a vector. Let's say an arbitrary vector can be written as $V = V^i \partial_i$. Then $x - x' = (x^i - x'^{\ i})\partial_i$. We have then $$\nabla |x-x'| = \frac{x-x'}{|x-x'|} = \sum_i\partial_i[(x^i-x'^{\ i})^2] = \sum_i\frac{(x^i - x'^{\ i})\partial_i}{(x^i - x'^{\ i})^2}$$
If we apply the derivative on the far right-hand-side on $|x-x'|$, we get a bad result, namely $(x^i - x'^{\ i}) = 1$. I guess this can't be right..

4. Mar 19, 2017

Orodruin

Staff Emeritus
Why are you using $|\vec x| = x^i x^i$?? This does not even make sense dimensionally. You are missing some square roots ...

5. Mar 19, 2017

davidge

Oh yea. I got the correct result now. Thank you.