The Maxwell Equations in the Light Gauge: QED?

The E and B Fields, and the Gauge with No Name
The Maxwell Equations in the Light Gauge

What makes a theory non-classical?  Use an operational definition: a classical approach neatly separates the scalar and vector terms of a quaternion.  Recall how the electric field was defined (where {A, B} is the even or symmetric product over 2, and [A, B] is the odd, antisymmetric product over two or cross product).

E = vector(even((d by dt, Del), (phi, -A))) = (0, - Grad phi - A dot)

B = odd((d by dt, Del), (phi, A)) = (0,  Curl A)

The scalar information is explicitly discarded from the E field quaternion.  In this notebook, the scalar field that arises will be examined and shown to be the field which gives rise to gauge symmetry.  The commutators and anticommutators of this scalar and vector field do not alter the homogeneous terms of the Maxwell equations, but may explain why light is a quantized, transverse wave.

The E and B Fields, and the Gauge with No Name

In the previous notebook, the electric field was generated differently from the magnetic field, since the scalar field was discard.  This time that will not be done.

E = even((d by dt, Del), (phi, -A)) = (phi dot - div A, - A dot - Grad phi)

B = odd((d by dt, Del), (phi, A)) = (0,  Curl A)

What is the name of the scalar field, d phi/dt - Del.A which looks like some sort of gauge?  It is not the Lorenz or Landau gauge which has a plus sign between the two.  It is none of the popular gauges: Coulomb (Del.A = 0), axial (Az = 0), temporal (phi = 0), Feynman, unitary...

[special note: I am now testing the interpretation that this gauge constitutes the gravitational field.  See the section on Einstein's Vision]  

The standard definition of a gauge starts with an arbitrary scalar function psi.  The following substitutions do not effect the resulting equations.

phi goes to phi prime = phi - psi dot

A goes to A prime = A + Grad psi

This can be written as one quaternion transformation.

(phi, A)  goes to  (phi prime, A prime) =  (phi, A) + (- phi dot, Grad psi)

The goal here is to find an arbitrary scalar and a 3-vector that does the same work as the scalar function psi.  Let

p = - phi dot  and  Alpha = Grad psi

Look at how the gauge symmetry changes by taking its derivative.

(d by dt, Del) acting on (- phi dot, Grad psi) = (-  Laplacian psi-psi double dot, Curl Del psi - Del phi dot + Del phi dot) = (p dot - div Alpha, 0)

This is the gauge with no name!  Call it the "light gauge".  That name was chosen because if the rate of change in the scalar potential phi is equal to the spatial change of the 3-vector potential A as should be the case for a photon, the distance is zero.

The Maxwell Equations in the Light Gauge

The homogeneous terms of the Maxwell equations are formed from the sum of both orders of the commutator and anticommutator.

even((d by dt, Del), odd((d by dt, Del), (phi, A))) + odd((d by dt, Del), even((d by dt, - Del), (phi, -A))) =

= (- div Curl A , - Curl  Grad phi) = (0, 0)

The source terms arise from of two commutators and two anticommutators.  In the classical case discussed in the previous notebook, this involved a difference.  Here a sum will be used because it generates a simpler differential equation.

odd((d by dt, Del),odd((d by dt, Del), (phi, A))) - even((d by dt, Del), even((d by dt, - Del), (phi, -A))) =

= (phi double dot + Laplacian phi , - A double dot +  Curl (Curl A) - Del  divA)

= (phi double dot + laplacian phi , - A double dot -  Laplacian A) =  4 pi (rho, J)

Notice how the scalar and vector parts have neatly partitioned themselves.  This is a wave equation, except that a sign is flipped.  Here is the equation for a longitudinal wave like sound.

Omega double dot -  Laplacian Omega = 0

The second time derivative of w must be the same as Del^2 w.  This has a solution which depends on sines and cosines (for simplicity, the details of initial and boundary conditions are skipped, and the infinite sum has been made finite).

Omega = Sum from n = 0 to infinity of cosine (n pi t) sine (n pi R)

Omega double dot - Laplacian Omega = 0

Hit w with two time derivatives, and out comes -n^2 pi^2 w.  Take Del^2, and that creates the same results.  Thus every value of n will satisfy the longitudinal wave equation.

Now to find the solution for the sum of the second time derivative and Del^2.  One of the signs must be switched by doing some operation twice.  Sounds like a job for i!  With quaternions, the square of a  normalized 3-vector equals (-1, 0), and it is i if y = z = 0 .  The solution to Maxwell's equations in the light gauge is

Omega = Sum from n = 0 to infinity of cosine (n pi t) sine (n pi R V)

if   V squared = -1, then Omega double dot +  Laplacian Omega = 0

Hit this two time derivatives yields -n^2 pi^2 w.  Del^2 w has all of this and the normalized phase factor V^2 = (-1, 0).  V acts like an imaginary phase factor that rotates the spatial component.  The sum for any n is zero (the details of the solution depend on the initial and boundary conditions).


The solution to the Maxwell equations in the light gauge is a superposition of waves--each with a separate value of n--where the spatial part gets rotated by the 3D analogue of i.  That is a quantized, transverse wave.  That's fortunate, because light is a quantized transverse wave.  The equations were generated by taking the classical Maxwell equations, and making them simpler.

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