This is the theory of light that immediately emerged from Maxwell’s equations of electricity and magnetism. Michael Farady had already produced this theory in a somewhat qualitative fashion. Maxwell provided a precise set of differential equations that matched Faraday’s notions, and also led to an explicit formulation for light waves. Here is such a formulation.

First we recapitulate some of the vector notations due to Gibbs and still commonly taught. We denote 3D vectors by upper case latin letters or triples of 3 reals: (3, 4, 5).

(p, q, r)×(x, y, z) = (qz–ry, rx–pz, py–qx)
(p, q, r)∙(x, y, z) = px + qy + rz
In these definitions p, q, and r might be all reals, or all differential operators, such as ∇ = (∂/∂x, ∂/∂y, ∂/∂z). With this notation we can write Maxwell’s equations, in suitable units, thus:

∂E/∂t = –∇×B
∇∙E = ρ
∂B/∂t = ∇×E
∇∙B = 0

In these equations:

ρ = 0 in this exercise.

Consider the electromagnetic field:
E = (sin(z – ct), 0, 0)
B = (0, sin(z – ct), 0).
We claim that this represents a beam of polarized light moving in the z direction, with intensity = 1, wave length = 2π, spread over all space, and lasting forever.

We compute ∇×B for this field:
∇×B = (∂0/∂y–∂sin(z – ct)/∂z, ∂0/∂z–∂0/∂x, ∂sin(z – ct)/∂x–∂0/∂y)
= (–cos(z – ct), 0, 0).
Note that this is exactly –∂E/∂t as required by Maxwell’s equations, considering that we have chosen units where c=1.
∇×E = (∂0/∂y – ∂0/∂z, ∂sin(z – ct)/∂z – ∂0/∂x, ∂sin(z – ct)/∂x – ∂0/∂y)
= (0, cos(z – ct), 0) = ∂B/∂t

That this proposed field solves the differential equations shows that the waves will persist as described.

Symmetries

The symmetries of this field are generated by the following:
(x, y, z, t) → (x–a, y–b, z–ct, ct) for arbitrary reals a, b, c.
(x, y, z, t) → (–x, –y, –z, t); B → –B; E → –E

The presumed symmetries of space must also be complied with. The natures of Gibb’s notation ensures that equations written in that notation is properly symmetric. If we shift our solution in the z direction by π/2 and rotate it by π/2 about the Z axis we get:
E = (0, cos(z – ct), 0)
B = (–cos(z – ct), 0, 0).

The equations are linear and the sum of two solutions also solves the equations. Considering that eib = cos b + i sin b the sum is:
E = (e, f, 0)
B = (–f, e, 0)
where e + if = ei(z – ct).
Here e and f are real numbers but i(z – ct) is complex.

This solution reveals a helical symmetry:
(x, y, z, t) → (x (cos p) + y (sin p), – x (sin p) + y(cos p), z + p, t)
This is circularly polarized light.