Thursday, March 4, 2010

The Scalar Field

In the previous post, we saw that a homogeneous paravector wave provided an exact description of the Maxwell's equations, except there is an added scalar field, which is on the same footing as the Electric and Magnetic fields. This scalar field does not appear in standard electromagnetic theory, and we are going to find out why.

To begin, we will define the electro-magnetic bi-paravector F in terms of the potential A

A = ( V/c, A )

F = cA

F = ( ∂V/∂t + cA, - ∂A/∂t - V + ic ( × A) )

We can make the following definitions for the fields

E = -∂A/∂t - V

B = × A

φ = (1/c) ∂V/∂t + A

In terms of these fields the electro-magnetic bi-paravector is given by

F = ( cφ, E + icB )

If you want to recover the orthodox version of the electro-magnetic field tensor, then just take the vector portion of this, ignoring the scalar portion. For now, we are not going to ignore the scalar portion, so we can see what physical implications it has.

To begin with, we see that the scalar field is a Lorentz invariant. It is the same in all reference frames.

The vector fields are gauge invariant. We can express the classical concept of a gauge transformation as follows

A' → A + Χ

F' → F + cΧ

Now, if we want to get full gauge invariance, we should expect the second term to vanish. You can show that since Χ is a scalar the vector portion of the second term vanishes identically, therefore the E and B fields are trivially gauge invariant.

We can achieve gauge invariance of the scalar field, only in the case where the scalar Χ belongs to a restricted set of functions which obey the condition

Χ = 0

This expression reduces to the scalar homogeneous wave equation. The idea of a restricted gauge invariance provides for some non-trivial gauge transformations, which end up being valid, even in the context of the traditional, unrestricted, gauge principle.

For instance, consider a paravector Ψ, which is not the gradient of a scalar, but which does satisfy the expression

Ψ = 0

Such a paravector can be used as a gauge function, in the sense that

A' → A + Ψ

F' → F

Such a gauge function ends up being non-trivially gauge invariant for the vector fields. This means that the vector fields are gauge invariant, but it is due to the form of the gauge function, not because of mathematical identities.

We will escalate the concept of gauge invariance to include the non-trivial gauge functions as well. The higher principle that we used to introduce the non-trivial gauge functions requires that we restrict the traditional scalar gauge functions to only include solutions to the homogeneous wave equation. This refinement resides within the limitations of traditional electro-magnetic theory, though it may require us to rethink the physical meaning behind various gauge transformations.

For instance, the Lorenz gauge

(1/c) ∂V/∂t + A = 0

If you remember the definition of the scalar field, you see that applying the Lorenz gauge is the same as making the statement

φ = 0

Since the derivatives of the scalar field are equated with the source terms, we see that setting this field to zero, or any constant for that matter, has the physical meaning of having no source terms. Luckily, this is the case where the Lorentz gauge is employed.

We will continue discussing the scalar field in the next post, where we will ask the question, "what force does the scalar field apply to a particle?"

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, θR + iθI )

θ = (1/c) ∂ψ/∂t + ψ

θR = -(1/c) ∂ψ/∂t - ψ

θI = × ψ

Now we apply the second gradient to the bi-paravector

Θ∂ = ( θ, ΘR + iΘI )( (1/c) ∂/∂t, - ) = 0

( (1/c) ∂θ/∂t - θR - iθI, (1/c)∂θR/∂t + (i/c) ∂θI/∂t - θ + i × θR - × θI ) = 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

θR = (1/c) ∂θ/∂t

θI = 0

× θR = - (1/c) ∂θI/∂t

× θI = -θ + (1/c) ∂θR/∂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

θR → cE


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

ΨA = ( 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.

Gradient Operator

I think some of the physics that is encoded in the algebra manifests itself mainly in the derivatives. For instance, the true meat of the complex algebra is the analytic functions. Can we come up with some type of space-time version of an analytic function? I've made a few attempts, and I've seen other attempts, but in general I'm just not satisfied. There is no parallel that is as clean and clear as the analytic functions of a complex variable. The troubles with derivatives in the algebraic representation of space-time all boil down to non-commutativity. I'll keep searching.

In the mean time, let's discuss a derivative operator, which I believe is a valid operator, though I wish it would arise in a more clean fashion. Since I am not satisfied with any particular derivation of this operator, I will just introduce it without derivation: the algebraic gradient

= ∂μeμ

Here the notation μ makes reference to contra-variant (raised index) partial derivative with respect to the μth coordinate. Remember that a standard partial derivative is defined with the index lowered.

μ = ∂/∂xμ

If we use the minkowski metric to raise the index, we must change the sign on the spatial terms of the gradient.

= (1/c) ∂/∂t e0 - ∂/∂x e1 - ∂/∂y e2 - ∂/∂z e2

Now, in tensor algebras, we use index balancing in order to construct relativistically significant quantities. In this algebra we don't deal with the component indexes much, but we still need to be concerned with making relativistically significant quantities.

We can use the fact that the gradient transforms like a para-vector. Thus, if we want the gradient to act on a paravector, it needs to do so like this


Which results in a biparavector.

If the gradient acts on a biparavector it must do so like this

B∂ or B

Which results in a paravector. Note that the gradient acting from the right is a distinct operation due to non-commutativity.

And if the gradient acts on a spinor, it must do so like this


which results in another spinor. I only know that this expression is relativistically significant, but since I haven't yet seen this operation in action, I have no physical intution as to its significance.