## Saturday, June 26, 2010

### The "N" Field

I have recently been exposed to a terrific idea, proposed by my friend Bill Polson, while enjoying a nice breakfast overlooking the coast.

He expressed a few ideas about Lagrangians that I think are rather revolutionary.

We usually think of an action integral is the integral of a scalar Lagrangian function along a parametrized path, with time as the parameter. The path that minimizes the result of this integral is the path that a particle takes.

$S=\int_{t_0}^{t_f}L\,dt$

Bill Polson introduced the idea of a vector valued Lagrangian. This is something I've done as well, but Bill does something rather special with his vector Lagrangian. He forms the action integral out of a line integral on a closed loop. The first branch of the loop represents the path the particle might take, while the returning branch contains a slight variation. He is then able to make a correlation, via Stokes Law, between the abstract idea of minimizing action, and the geometric properties of the vector Lagrangian field.

$S_{BP}=\oint_C\vec{L}\cdot\vec{n}\,dt=\int_A\nabla\times\vec{L}\,d\sigma=\int_A N\,d\sigma$

Bill introduces a vector field he calls N, which should contain vector and pseudo-vector portions. The action is stationary when integrated over the "area" contained inside of the closed path. He defined N as the "curl" of the vector Lagrangian field, but he meant the generalization of the curl to 4 dimensions, of course.

Now, I think this idea is rockin' as is. However, Bill went further and figured out what N needed to be in order to reproduce the Lagrangian of a charged particle in an EM field.

$N=F+\nabla\times u$

Again, the curl here is a generalization of the 3D curl of a vector field into 4 dimensions. Also, there are several factors such as c, charge and mass that Bill neglected for convenience.

Using the Clifford Algebra notation, this is how I would define N

$\partial\overline{L} = N = F + \frac{mc}{e}\partial\overline{u}$

Here, I have inserted the mass, charge, and speed of light constants where they are appropriate. If I replace F with its definition in terms of the four-vector potential A, this becomes

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

The quantity under the conjugation bar is the canonical momentum p of a particle in an EM field. Using this we can build an analogy. As p is to A, N is to F.

Now we have

$N=\frac{c}{e}\partial\overline{p}=\partial\overline{L}$

Or, just

$L=\frac{c}{e}p$

The Action integral SBP expressed in terms of Clifford Algebra notation is now

$S_{BP}=\oint_C\frac{c}{e}\left_S\,dt$

Now, let's compare this with my version of the action integral, which is an integral of a scalar Lagrangian density function over a volume.

$S_{EB}=\int_V\left_S$

Thus, according to this definition of action, my vector valued Lagrangian field is

$L=p$

I don't define an N field, but my vector Lagrangian is the same as Bill's apart from a constant factor.