Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 0 down vote accepted

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]]

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.