einstein summation notation for magnetic dipole field


by mmpstudent
Tags: dipole, einstein, field, magnetic, notation, summation
mmpstudent
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#1
Apr30-13, 05:32 PM
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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
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Fredrik
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#2
Apr30-13, 05:37 PM
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mmpstudent
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#3
Apr30-13, 05:43 PM
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Quote Quote by Fredrik View Post
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.

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Apr30-13, 05:52 PM
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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.
WannabeNewton
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#5
Apr30-13, 06:20 PM
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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.
mmpstudent
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#6
Apr30-13, 06:35 PM
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O wow, thanks. that makes much more sense now.
mmpstudent
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#7
Apr30-13, 06:55 PM
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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
WannabeNewton
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Apr30-13, 07:01 PM
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Quote Quote by mmpstudent View Post
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
I honestly can't think of any online resources off of the top of my head because I got used to the notation when learning special relativity (the text used was Schutz).


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