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})...
I forgot I didn't post the whole problem in the beginning. Disregard the complex conjugate part. This might be helpful need to chug thru this now. Thanks
yeah I guess, but i was hoping to be able to take it back into a form that would have exponentials again. I wonder if me dropping the complex conjugate terms in the prior steps be the reason why its not simplifying the way I want it to. I will try with them included i guess
i understand that the second term is 1
but my attempt at reducing the first term
|1+e^{ik \Delta cos\theta}|=|(1+cos(k \Delta cos\theta))+i sin(k \Delta cos\theta)|
\sqrt{(1+cos(k \Delta cos\theta))^{2}+sin^{2}(k \Delta cos\theta)}
=\sqrt{2+2cos(k \Delta cos\theta)}
in which I'm stuck
I have an EM problem (michelson interferometerish) where I have a term that I need to reduce. It is
|1+e^{ik \Delta cos\theta}|^{2}+| e^{ik \Delta sin\theta}|^{2}
I have foiled it and squared the last term but is there something that I am missing. I am multiplying it by a large matrix and...
Can anyone suggest a book that has a ton of examples using einstein summation? I feel behind most of my class in regards to the notation. It just takes me too long to do problems.
unless I am seeing this completely wrong, the left side (first line up above) and the right side (on the last line) is twice the left side when I add them together
I was thinking maybe I had to express the right side like this
\epsilon_{ijk} (\nabla f)_{j} v_{k} + f \epsilon_{klm} \partial_{l}...
prove the identity $$\nabla\times(f\cdot\vec{v})=(\nabla f) \times \vec{v} + f \cdot \nabla \times \vec{v}$$
I can do the proof with normal vector calculus, but I am in a tensor intensive course and would like to do this with
einstein summation notation, but am having some trouble since I am...