- #1
mmpstudent
- 16
- 0
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
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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