# Einstein summation notation for magnetic dipole field

by mmpstudent
Tags: dipole, einstein, field, magnetic, notation, summation
 P: 16 I can do this derivation the old fashioned way, but am having trouble doing it with einstein summation notation. Since $\vec{B}=\nabla \times \vec{a}$ $\vec{B}=\mu_{0}/4\pi (\nabla \times (m \times r)r^{-3}))$ $4\pi \vec{B}/\mu_{0}=\epsilon_{ijk} \nabla_{j}(\epsilon_{klm} m_{l} r_{m} r^{-3})$ $=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_{j}m_{l}r_{m}r^{-3}$ here is where I am stumbling. My professor has for the next step $=m_{l}(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})r^{-3} \delta_{jm}-3 r_{m}\hat{r}_{j}r^{-4})$ but I don't really know how to get to that step
 Emeritus Sci Advisor PF Gold P: 9,522 You may be interested in the The LaTeX guide for the forum. Link.
P: 16
 Quote by Fredrik You may be interested in the The LaTeX guide for the forum. Link.
You were too fast. Was trying to get it to work just needed to delete the spaces in brackets I guess.

 Emeritus Sci Advisor PF Gold P: 9,522 Einstein summation notation for magnetic dipole field My first thought is that he's using the product rule for derivatives to evaluate ##\partial_j## acting on a product.
 C. Spirit Sci Advisor Thanks P: 5,661 First off, tell your professor that he is horribly butchering Einstein notation. Seriously, what was written down misses the entire point of the notation. Anyways, ##\frac{4\pi}{\mu_{0}}B^{i} = \frac{4\pi}{\mu_{0}}\epsilon^{ijk}\partial_{j}A_{k} = \epsilon^{ijk}\epsilon_{klm}m^{l}[r^{-3}\partial_{j}r^{m} + r^{m}\partial_{j}(r^{-3})]## hence ##\frac{4\pi}{\mu_{0}}\epsilon^{ijk}\partial_{j}A_{k} = 2\delta^{[i}_{l}\delta^{j]}_{m}m^{l}[r^{-3}\partial_{j}r^{m} + r^{m}\partial_{j}(r^{-3})]##. Now, ##\partial_{j}r^{m} = \delta^{m}_{j}## and ##\partial_{j}(r^{-3}) = -3(-r^i r_{i})^{-5/2}r_{k}\partial_{j}r^{k} = -3r^{-4}\hat{r}_{j}## giving us ##\frac{4\pi}{\mu_{0}}\epsilon^{ijk}\partial_{j}A_{k} = 2\delta^{[i}_{l}\delta^{j]}_{m}m^{l}[r^{-3}\delta^{m}_{j} -3r^{-4}\hat{r}_{j}r^{m}]## as desired. EDIT: By the way, in the above it should be ##(r^i r_{i})^{-5/2}## not ##(-r^i r_{i})^{-5/2}##; I've gotten too used to General Relativity xD.
 P: 16 O wow, thanks. that makes much more sense now.
 P: 16 Do you know of any materials online that would give more written out examples of such derivations with Einstein summation? I just need more practice
C. Spirit