# $\nabla \frac 1 {|\vec{r}-\vec{a}|} = -\frac {\vec{r}-\vec{a}} {|\vec{r}-\vec{a}|^3}$?

This may be an extremely silly question, but ...

Given $\nabla \dfrac 1 {|\vec{r}|} \equiv -\dfrac {\vec{r}} {|\vec{r}|^3}$, is $\nabla \dfrac 1 {|\vec{r}-\vec{a}|} = -\dfrac {\vec{r}-\vec{a}} {|\vec{r}-\vec{a}|^3}$ also true (given that $\vec{a}$ is a constant)? If so, is there a proof?

I do know that given $\vec{r} = r\hat{r}$, there is also an identity $\nabla f(r) \equiv \dfrac{\vec{r}}{r}\space f'(r)$, but I couldn't figure out if that will come handy here.

Are there other identities of this type?

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## 1 Answer

Compute the derivative of the expression in component form:

$$\nabla_j \frac 1 {|\vec{r}-\vec{a}|}=\frac{\partial}{\partial x^j}\left(\sum_{i=1}^n(x^i-a^i)^2\right)^{-\frac{1}{2}}.$$

http://www.wolframalpha.com/input/?i=D[Sum[%28x[i]-a[i]%29^2%2C{i%2C1%2C3}]^%28-1%2F2%29%2Cx[1]]

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Thanks @NickKidman ... – Avijit Mar 7 '13 at 10:38
@Avijit: You obtain a simple generalization of the result if, instead of $-1/2$, you plug in $-m/2$ as exponent. – NikolajK Mar 7 '13 at 14:59