## Wednesday, July 21, 2010

### The "N" Field - part 2

To discuss my thoughts of the N in this section, I will not use Lagrangians, rather I will use Hamiltonians. The Lagrangian (L) is related to the Hamiltonian (H) as follows:

$L=H-\vec{p}\cdot\vec{v}$

The Hamiltonian represents the total energy of a system. The Hamiltonian of a free particle is merely the kinetic energy of the particle.

$H=E=\frac{1}{2}mv^2 = \frac{1}{2}\vec{p}\cdot\vec{v}$

There is a very simple trick to find the Hamiltonian of a charged particle in the influence of an electromagnetic field. The extension can be made by replacing the components of the relativistic four momentum with the components of the canonical four momentum.

$p\rightarrow p+eA$

Where A is the electromagnetic four potential. In terms of space and time pieces, this is

$E\rightarrow E+eV$
$\vec{p}\rightarrow \vec{p}+e\vec{A}$

Making this transformation on the free particle Hamiltonian yields

$E+eV=\frac{1}{2}\left(\vec{p}+e\vec{A}\right)\cdot\vec{v}$

Solving for E gives

$E=E_{free}+e\vec{A}\cdot\vec{v}+eV$

So we started with the energy of a free particle, and transformed the four momentum to get the energy of a particle that is subject to the electromagnetic field.

Start with a free particle. Extend the momentum. End up with interaction terms.

The N field seems to be a similar type of extension - except in this case we aren't extending from four momentum to canonical four momentum, but rather from four potential to "canonical four potential", like this

$A\rightarrow A+\frac{m}{e}u$

Where u is the four velocity of the particle.

So, whereas F is defined as

$F=c\partial\overline{A}$

We can define N as

$N=c\partial\left(\overline{A+\frac{m}{e}u}\right)$

So the point I am driving at here is this: on one hand we can derive the Hamiltonian of a particle in an electric field by starting with a free Hamiltonian, and extending the four momentum to the canonical momentum. So, what might happen if we were to start with the "free" Maxwell equations (free here means sourceless) and extend the four potential to the "canonical four potential"? My guess is:

Start with a sourceless field. Extend the potential. End up with source terms.

The free Maxwell equations are written as

$\partial\overline{F}=0$

Extending A to the canonical form converts F into N and the equation becomes

$\partial\overline{N}=0$

If the free vs. interaction Hamiltonian analogy holds true here, then this homogeneous equation may represent the inhomogeneous maxwell equations.

Splitting this up into field terms and source terms we have

$\partial\overline{F}+\frac{mc}{e}\partial\overline{\partial\overline{u}}=0$
$\partial\overline{F}=-\frac{mc}{e}\partial\overline{\partial\overline{u}}$

The empirical form of the inhomogeneous Maxwell equations with source terms is given in terms of currents as

$\partial\overline{\left[F\right]_V}\right=\frac{1}{\epsilon}\overline{j}$

We can make the correlation

$\overline{j}=-\frac{m}{e}\sqrt{\frac{\epsilon}{\mu}}\partial\left[\,\overline{\partial\overline{u}}\,\right]_V$

Of course, as stated previously, when the empirical current shows up as a source term, this represents approximating local disturbances in the potential wave as point disturbances, i.e. delta function sources. If we do not make this approximation, then j is always zero, and the local disturbances manifest themselves through the derivatives of the scalar EM field. Thus, if we do not make the approximation then we have

$\partial\overline{u}\partial=0$

Or, more importantly, taking into account the fact that the four vector potential also obeys this equation, we can make a homogenious wave from the four momentum.

$\partial\overline{p}\partial=0$