I'm studying this right now as well and had the same question. I think the answer has to do with the concept of "universality class" from statistical mechanics. Basically, I think the idea is that two Lagrangians that are invariant under the same symmetries will flow to the same fixed point in...
Someone sent me a private message asking about a better version of these videos. It won't let me respond so I'll just respond here for everyone. Anyway, I don't know of better videos, but after awhile, you can start to see what he writes on the board. Especially in some of the later videos...
Sidney Coleman's http://www.physics.harvard.edu/about/Phys253.html. I went through these over and over during my QFT class last year. I really feel like he has been my teacher.
Yeah J dot E is power per unit volume. That's usually the starting point for deriving expressions for the poynting vector and the energy density in the field.
After the initial motion, there will be no emf. Charge will build up on both ends, causing an electric field inside the conductor such that E + v x B = 0. Thus, the net force on each charge will be zero. Therefore, the emf will be zero as well since it's the integral of the force per unit charge.
In Fourier space, you are referring to the equation
\left[ \nabla^2 +\mu(\omega)\epsilon(\omega) \omega^2 \right] \mathbf E(\mathbf r, \omega)=0
To find out what that equation is in terms of time, you would have to take its inverse Fourier transform. Then, the functions of \omega will become...
The justification is the result. Multiplying both sides of an equation by the same thing doesn't need any justification since it is a valid algebraic manipulation.
Perhaps you mean more motivation or intuition? Well, Force x Velocity = Power = rate of change of energy, and the first equation...
Our classical ideas align more closely with expectation values anyway. It's definitely possible for any number of electrons in the universe to have the same expectation value of momentum.