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Light Propagation in a Transparent Dielectric Material

The goal here is the demonstrate electromechanically how a wave behaves when it enters a dielectric material after exiting a vacuum. Asking why a light wave slows down when it enters a dielectric of permittivity `epsilon` is the same as asking why Maxwell's wave equation contains the parameter `n^2=epsilon/epsilon_0` where where `epsilon_0` is the permittivity of vacuum and `n>1` is the index of refraction. The use of `n` in Maxwell's wave equation results in the wave phase speed being reduced by factor `1/n` in the dielectric. That also results in the energy antinodes in the dielectric being spaced at distance 1/n closer together than in the vacuum.

Calculation of the wave energy density in the dielectric

An important fact is that in free space (vacuum) the relation between the electric intensity, `E_0`, and magnetic intensity, `H_0`, is

`E_0/H_0=sqrt(mu_0/epsilon_0)`

where `mu_0` is the magnetic permeability of vacuum and `epsilon_0` is the permittivity of vacuum. Now, when the wave enters a dielectric medium, the magnetic intensity does not change but the electric intensity,`E`, follows the same equation:

`E/H_0=sqrt(mu_0/epsilon)`

where `epsilon` is the permittivity of the medium. For a high frequency wave we can write

`epsilon=n^2epsilon_0`

where `n` is the index of refraction of the dielectric medium. Therefore we can write:

`E/H_0=sqrt(mu_0/(n^2epsilon_0))=(E_0/H_0)/n`

so `E` becomes

`E=E_0/n`

In the case of a radiation wave, the electric field is "loaded" or reduced by the dielectric material.

Wave Energy Conservation

The rate of energy flow in medium `epsilon` has to be the same as in medium `epsilon_0`. Energy flow of a plane wave must be measured as `"Joule"sm^(-2)sec^(-1)`. The Poynting vector magnitude is `S=EH="Watts"m^(-2)="Joules"m^(-2)sec^(-1)`. In `epsilon_0` `S_0=E_0H_0` while in `epsilon` we have `S=EH_0=(E_0/n)H_0` which is less than `S_0`. In fact `S=S_0/n`. This is the "bare field" power per unit area. But we must also consider the energy associated with the motion of the electron clouds of the atoms of the dielectric. Since `S` has units watts`meter^(-2)` and the energy density has units Joules`m^(-3)`, we must divide `S` by a speed `msec^(-1)` to obtain wave energy density. An appropriate equation for the total energy density associated with the wave is

`U=(EH_0)/v=(E_0H_0)/(nv)=(E_0H_0)/c`

from which we find that

`nv=c`
`v=c/n`

This shows that the rate of wave energy flow is constant if `v=c/n`. and this applies to both vacuum and dielectric material. The total energy density induced by the wave also includes the mechanical energy of the moving dipoles.

Origin of `epsilon_0`, the Permittivity of Vacuum

In order to calculate what would be the speed of light in vacuum, Maxwell needed to have independent values for the parameters `epsilon_0` and `mu_0`. He could obtain `epsilon_0` from the Coulomb force between two charged bodies and that is the measurement I will discuss here. By Coulomb's Law the repelling force between two charges, `q_1` and `q_2` (coulombs) is measured to be

`bbF=bbhatr(q_1q_2)/(4piepsilon_0r^2)`      

where `r` is the distance between the charge centers and `bbhatr` is a unit vector in the `r` direction.

Derivation of the Force Equation

For the derivation we will make one of `Q` much larger than the other `q` i.e.

`bbF=bbhatr(Qq)/(4piepsilon_0r^2)`      

Now we invoke Gauss's law for the electric field, `bbE`, of `Q`.

`epsilon_0intbbE*dbbs=Q`      

where the integral is over the surface of a sphere surrounding `Q` and `dbbs` is an outward vector surface increment of the sphere and the centered `*` dot represents the dot product of `bbE` with `dbbs`. This equation for `bbE` is also valid in a dipolar medium where `epsilon` can replace `epsilon_0`. Obviously, if `bbE` is parallel to `dbbs` everywhere the the integral is just `4pir^2|bbE|`, where `r` is the distance from the sphere center, so the equation becomes:

`epsilon_04pir^2|bbE|=Q`      

and

`bbE=Qbbhatr/(4piepsilon_0r^2)`      

The force on `q` due to this electric field is `qbbE` and becomes:

`F=bbhatr(Qq)/(4piepsilon_0r^2)`      

Measuring `epsilon_0` in Vacuum

After measuring `|bbF|` we can compute the value of `epsilon_0` as

`epsilon_0=(qQ)/(4pi|bbF|r^2)`      

To measure `bbF` due to charges, one might resort to placing known changes `q` on each of two spheres just as was done for measuring the gravitational constant `G`. A big problem with this measurement is being sure of the value of `q`. Charging the spheres is done by providing a known current for a known time. But other parts of the apparatus may have charge too.

Explanation of the Controls

Dielectric Constant Slider

For this animation the dielectric constant `epsilon` is the square of the index of refraction, `n`. Increasing `epsilon` using its slider has 3 effects:

1:It decreases the amplitude by the factor 1/`sqrt(epsilon)` of the wave in the dielectric in Canvas 1.

2: It decreases the spacing between waves and wave crests shown in Canvases 1 and 2 by the factor 1/`sqrt(epsilon)`.

3:It decreases the wave and wave crest speed by the factor 1/`sqrt(epsilon)` of the wave in the dielectric in Canvases 1 and 2.

Note that in both Canvases, when a wave or wave crest is reaches the boundary in the vacuum/dielectric, a new wave or wave crest always is born in the dielectric/vacuum. That is much easier to see in Canvas 2 than Canvas 1 because of the amplitude differences in Canvas 1.

Number of Random Dipoles Slider

This slider changes the number of randomly placed dipoles that react to the moving wave in Canvas 1. If you decrease the number of dipoles to, say, 200 and check the Single Step box and then press Start successively while watching a single dipole, you can see that the dipoles' rotation angles are such that they tend to cancel the electric field of the wave. This canceling effect reduces the wave's electric field by the factor `1/sqrt(epsilon)`.

Explanation of Graphics

Canvas 1 shows the relative motion and amplitude of the wave in both vacuum and dielectric material. To avoid the complexity of having wave reflections at the interfaces between vacuum and dielectric I have chosen to have applied a 1/4th wave antireflection coating (see darker layers at both of these interfaces). The reflections that might have occurred bring into play the Fresnel equations which are an entirely different subject. Thus the graphics are explained more than adequately under "Explanation of the Controls"