### Polarized Light

#### The Mirror

I describe polarized light according to Maxwell here.
I describe here some details of a polarizing filter.
This was all known in Maxwell’s time and confirmed by observation.
Maxwell’s equations are differential equations and in that spirit we define a polarizing filter by posing boundary conditions in the plane z=0 thru which the the light passes.
That plane is a conductor and thus a mirror.
There are massless charges that can move freely in the plane and produce a current but, constrained to the plane and thus E_{x} = E_{y} = 0 and the electric field is (0, 0, E_{z}) at (x, y, 0).
The physics might be that there are many charge carriers there that are mobile within the plane.
The carriers cancel net charge and ρ=0.
A solution with this boundary condition is:

For z<0

- E = (sin(z − ct) + sin(z + ct), 0, 0)
- B = (0, 0, 0).

For z>0:
For z<0 we see another reflected light stream moving in the negative z direction and the light is entirely reflected.
The magnet fields of the two streams entirely cancel; we have a standing wave.
It is dark for z>0.
I must spell out the induced current in the z=0 plane!!
Perhaps I need the integral equations to cope with the infinite current density on the z=0 plane.

The integral form of ∇×B = ∂E/∂t + J is:

In these expressions Σ is a bounded 2D surface and ∂Σ is its boundary which is a closed loop.
**B** is the magnetic field and **E** is the electric field.
d**S** is a differential vector orthogonal to the surface Σ.
*μ*_{0} and *ε*_{0} are both one here.
“d*ℓ*” is a differential vector element of the boundary of Σ.

For Σ we choose a small square which crosses the z=0 plane.
Its corners are at:

{(−0.01, 0, −0.01), (−0.01, 0, 0.01), (0.01, 0, 0.01), (0.01, 0, −0.01)}.

**B**∙d*ℓ* is 0 everywhere and thus the too left side of this equation.
The vector field **J** is the current in the z=0 plane.
**J** is orthogonal to d**S** and this term is 0 too.
Just now it seems I don’t need no stinking mirror.
I have made a mistake somewhere.

#### The perfect polarizing filter

Now we assume charges constrained in the z=0 plane and free to move in the x direction.
The condition is now:

E(x, y, 0) = (0, E_{y}, E_{z}).