The Wave Equation

Note: I usually prefer to use the nabla symbol (∇) to represent the vector portion of the gradient. If you are using IE and can see the nabla symbol as an upside down triangle rather than a box, then you are lucky :(

I suggest using firefox, or another browser. Sorry microsoft, I love ya, but I need my math notation.

We are going to construct the homogeneous wave equation, for a paravector wavefunction. The reason for doing this is because I have a hunch that it may be possible to show that this homogeneous wave equation is a natural result of the definition of the derivative with respect to a paravector variable, but so far this hunch currently has the status of a hypothesis. Therefore, for the sake of my hypothesis we will proceed on this path.

To generate the wave equation, we will operate on a paravector two times with the gradient operator. Begin by letting the gradient operate on an arbitrary paravector that we will call Ψ. Remember that this must be done in a specific way in order to maintain relativistic significance.

Θ = ∂Ψ

This quantity is a bi-paravector. Let's act on this bi-paravector with another gradient operation in a relativistically correct way, and set the result to zero, for a homogeneous wave.

Θ∂ = ∂Ψ∂ = 0

This is the expression for a homogeneous paravector wave. Let's see if this wave equation resembles anything that we have any physical intuition for.

Let's express the gradient and the wavefunction in terms of their scalar and vector parts

∂ = ( (1/c) ∂/∂t, -∇ )

Ψ = ( ψ, ψ )

When we act with the gradient operator, we treat it as if we were multiplying it, in the following manner

∂Ψ = ( (1/c) ∂/∂t, -∇ )( ψ, -ψ )

= ( (1/c) ∂ψ/∂t + ∇•ψ, -(1/c) ∂ψ/∂t - ∇ψ + i∇ × ψ )

For the sake of brevity, rather than writing out all of these terms, we will just consider that the resulting bi-paravector has three parts: a scalar part, a real vector part, and an imaginary vector part.

Θ = ( θ, θ_R + iθ_I )

θ = (1/c) ∂ψ/∂t + ∇•ψ

θ_R = -(1/c) ∂ψ/∂t - ∇ψ

θ_I = ∇ × ψ

Now we apply the second gradient to the bi-paravector

Θ∂ = ( θ, Θ_R + iΘ_I )( (1/c) ∂/∂t, -∇ ) = 0

( (1/c) ∂θ/∂t - ∇•θ_R - i∇•θ_I, (1/c)∂θ_R/∂t + (i/c) ∂θ_I/∂t - ∇θ + i∇ × θ_R - ∇ × θ_I ) = 0

Now, this is a long and hairy equation, which involves four pieces: a real scalar, an imaginary scalar, a real vector and an imaginary vector. Physically speaking, there is a scalar term, a pseudo-scalar term, a vector term, and a pseudo-vector term. If the result of this expression is truly zero, then each of these four peices must independantly equal zero as well. We will set them to zero, and rearrange them slightly

∇•θ_R = (1/c) ∂θ/∂t

∇•θ_I = 0

∇ × θ_R = - (1/c) ∂θ_I/∂t

∇ × θ_I = -∇θ + (1/c) ∂θ_R/∂t

Can you see it yet? The first time I saw this, I about peed my pants.

These equations resemble the Maxwell equations in their differential form. We can make the following substitutions in order to get the equations closer in form to Maxwell's equations

θ_R → cE

θ_I → B

These substitutions constrain the wave function Ψ to take on the role of the electro-magnetic four-potential

Ψ → A = ( V/c, A )

Using these assignments of our wave variables, we can express the wave equation in a from that is very similar to the Maxwell equations.

∇•E = ∂θ/∂t

∇•B = 0

∇ × E = - ∂B/∂t

∇ × B = -∇θ + (1/c²) ∂E/∂t

Now, other than the discovery that the Maxwell equations are a manifestation of a paravector wave equation, there are a few other interesting tidbits you should take note of.

First, there is no magnetic monopole, and there is no way to introduce one without messing up the covariance of the expression.

Second, the source terms are represented by the derivatives of a scalar field. What is this scalar field?

Third, it is the homogeneous wave equation which represents the electro-magnetic field with sources. We usually introduce inhomogeneous terms in order to represent sources.