
#1
Apr3013, 05:32 PM

P: 16

I can do this derivation the old fashioned way, but am having trouble doing it with einstein summation notation.
Since [itex]\vec{B}=\nabla \times \vec{a}[/itex] [itex]\vec{B}=\mu_{0}/4\pi (\nabla \times (m \times r)r^{3}))[/itex] [itex]4\pi \vec{B}/\mu_{0}=\epsilon_{ijk} \nabla_{j}(\epsilon_{klm} m_{l} r_{m} r^{3})[/itex] [itex]=(\delta_{il}\delta_{jm}\delta_{im}\delta_{jl})\partial_{j}m_{l}r_{m}r^{3}[/itex] here is where I am stumbling. My professor has for the next step [itex]=m_{l}(\delta_{il}\delta_{jm}\delta_{im}\delta_{jl})r^{3} \delta_{jm}3 r_{m}\hat{r}_{j}r^{4})[/itex] but I don't really know how to get to that step 



#2
Apr3013, 05:37 PM

Emeritus
Sci Advisor
PF Gold
P: 9,018

You may be interested in the The LaTeX guide for the forum. Link.




#3
Apr3013, 05:43 PM

P: 16





#4
Apr3013, 05:52 PM

Emeritus
Sci Advisor
PF Gold
P: 9,018

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.




#5
Apr3013, 06:20 PM

C. Spirit
Sci Advisor
Thanks
P: 4,941

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. 



#6
Apr3013, 06:35 PM

P: 16

O wow, thanks. that makes much more sense now.




#7
Apr3013, 06:55 PM

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



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