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## Electromagnetic (EM) Wave Response to a Single Dipole

This animation will visualize how a EM wave amplitude, `E`, is suppressed and delayed by a single dipole.
Presently the suppression and delay are not linked to the electric field of the wave and
the restoring spring constant, `k`, of the dipole. However we CAN say that the displacement, `deltay`,
of the dipole requires
an energy `U=1/2kdeltay^2` where `k` is the spring constant of the dipole restoring force.
We also know that the electric field energy per unit volume is equal in vacuum to
`(dU)/(dV)=epsilon_0E^2/2` where `epsilon_0` is the dielectric pemitivity of vacuum.
`E(x)` will be reduced for x close to the
dipole, for example

`EdeltaE(x)=1/2(kdeltay^2)gamma^2/((x-x_D)^2+gamma^2)`

where `x_D` is the dipole location and `gamma` is used to keep the denominator
from going to zero.

Why am I providing this animation?
It is standard practice to compute the macroscopic effects of a volume density of dipoles
in a transparent medium without trying to visualize what is happening in the nanoscopic world
of the individual dipoles. The macroscopic results for the delay at frequencies
far below any resonance frequency are really pretty simple:
the dielectric permittivity is

`epsilon=epsilon_0+N*e^2/k`

where `N` is the dipoles per unit volume, `e` is the dipole charge, and `k` is again the dipole
spring constant. Then, of course, the refractive index, n, is just `n=sqrt(epsilon/epsilon_0)`
and the refractive index represents the relative increase in spatial frequency of the wave.
So then one might say that the speed of light in the transparent medium is `c/n` were `c`
is the light speed in vacuum. What really happens is that the EM frequency and light speed in the medium
remain constant and the light is scattered and is delayed very locally at the dipole (as visualized here)
and then proceeds at speed c to the next dipole.
As always, the learner is welcome to hit F12 in Windows to view the code used for this visualization.