How can I evaluate $\partial_i \partial_j (1/r)$?

Question

I want to show that $\tfrac{\partial}{\partial x_i} \tfrac{\partial}{\partial x_j} \left(\tfrac{1}{r}\right)=\tfrac{1}{r^3}\left(\tfrac{3}{r^2}x_i x_j-\delta_{ij}\right)$ where $r=\sqrt{x_1^2+x_2^2+x_3^2}$. How can I do this? May I write the left hand side as $\tfrac{\partial}{\partial x_i} \tfrac{\partial}{\partial x_j} \left(\tfrac{1}{\sqrt{x_k x_k}}\right)$?

Answer 1

Think I got it now.

$\frac{\partial}{\partial x_j} (1/r)=\frac{-x_k \delta_{kj}}{(x_{k'}x_{k'})^{3/2}}=\frac{-x_j}{(x_{k'}x_{k'})^{3/2}}$

$\Rightarrow \frac{\partial}{\partial x_i} \frac{\partial}{\partial x_j} \frac{1}{r}=\frac{\partial}{\partial x_i}\left[\frac{-x_j}{(x_{k'}x_{k'})^{3/2}}\right] =\lbrace\textrm{Quotient rule}\rbrace=\frac{-\delta_{ij} (x_{k'}x_{k'})^{3/2}+3x_j x_{k'}\delta_{ik'}(x_{k'} x_{k'})^{1/2} }{(x_{k'}x_{k'})^{3}}=\frac{1}{r^3} \left(\frac{3}{r^2}x_i x_j -\delta_{ij}\right)$.

Answer 2

You can calculate the gradient ($g$) and Jacobian ($J$) of the function ($f$) using standard vector notation. $$\eqalign{ r^2 &= x\cdot x \cr 2\,r\,dr &= 2\,x\cdot dx \cr dr &= \frac{x\cdot dx}{r} \cr\cr f &= \frac{1}{r} \cr df &= -\frac{dr}{r^2} = -\frac{x}{r^3}\cdot dx \cr\cr g &= \frac{\partial f}{\partial x} = -\frac{x}{r^3} \cr\cr dg &= \frac{3x\,dr}{r^4}-\frac{dx}{r^3} \cr &= \frac{3xx\cdot dx}{r^5}-\frac{dx}{r^3} \cr &= \frac{3xx-r^2I}{r^5}\cdot dx \cr\cr J &= \frac{\partial g}{\partial x} = \frac{3xx-r^2I}{r^5} \cr }$$