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Re: cross-product versus Clifford Algebra

Here is an outline of the Clifford-Algebra solution
to the problem Jack posed.

We start with the Maxwell equations in the form
del F = 4pi J (1)
It doesn't get much simpler than that.

If you need reminding what the symbols mean, you
can look at
http://www.monmouth.com/~jsd/physics/maxwell-ga.htm

We have a particle and an observer "Joe" initially comoving
with that test charge. We assume(*) this particle has
charge and mass but (unlike real electrons) no magnetic
moment.

Note: (*) marks assumptions that Jack made, and which
will be maintained here for consistency.

We assume(*) F represents a purely radiative wave, i.e.
no electrostatic or magnetostatic components.

The wave is plane-polarized. Joe chooses axes such that
the polarization is in the X-direction. This comes from
the statement of the problem.

In Joe's frame, F can be decomposed into an electrical
piece and a magnetic piece. The electrical piece will
be of the form (gamma0 gamma1). This comes from convention;
the piece of F involving gamma0 is, by definition, the
piece we call the electrical piece. And it involves
gamma1 because that is how people have been labelling
polarizers since the dawn of time.

The wave is propagating in Joe's Z-direction. If the
wave is monochromatic we can write it as
F = gamma0 gamma1 sin(k z - w t + p) + G (3)
where G is the aforementioned magnetic piece, which we
will determine shortly. G is purely spacelike, i.e. it
won't have any factors of gamma0.

We assume(*) we can work in the zeroth Born approximation,
i.e. we neglect any fields re-radiated by the particle.
Therefore we can replace equation (1) by
del F = 0 (4)
This is certainly valid for short times, but questionable
later.

Note that gamma2 does not appear in the first term in
equation (3). Also note that the Y-variable does not
appear there, so the del operator won't have any gamma2
component. And gamma2 definitely doesn't appear on the RHS
of equation (4). From this we conclude that G cannot involve
gamma2. So in fact the solution of equation (4) is
F = (gamma0 gamma1 + gamma1 gamma3)
sin(k z - w t + ph) (5)
and you can verify by direct evaluation of equation (4)
that this is a solution iff w equals k.

This is the entire solution if we assume(*) F is
monochromatic; more generally F will be a Fourier
superposition of such terms.

Next we will be needing the Lorentz force law
(d/d tau) p = F dot v (6)

In Joe's frame, the initial velocity is
v = [1 0 0 0]
i.e. proceeding toward the future at the rate of
60 minutes per hour.

Plugging this into equation (6) using the field (5)
we find that early times only the electric piece
contributes; the early force is
(d/d tau) p = gamma1 sin(k z - w t + ph) (8)

This tells us that at somewhat later times, the
velocity
v = p/m (9)
will contain gamma1 terms as well
as gamma0 terms. Plugging this new v into equation
(5) we find that now the magnetic term contributes
also, so gamma3 terms will eventually show up in the
momentum (and velocity).

This exhausts the possibilities. There is no way
for gamma2 terms to show up in the velocity. That
is, the particle cannot move in the Y-direction.
This is super-clear from the structure of the equations,
especially equation (5).

<digression>
Indeed, we haven't even assumed we are working
in D=1+3 spacetime. The equations do not even
require the existence of a Y direction. On the
other side of the same coin, there could be
eleventeen other directions, perpendicular to
the ones we care about (T, X, and Z) and the
equations would be the same.

There's no gamma2 in equation (5), and no way to
create any gamma2 contributions from scratch.

This illustrates the tremendous power of Clifford
Algebra to say things that are true regardless of
dimensionality. Contrast this with the old-fashioned
cross product, which doesn't exist in D=2 flatland
and is worse than useless in D=1+3 spacetime.
</digression>

The clear absence of any gamma2 components must be scored
as an advantage for the Clifford-Algebra formulation. Using
the cross-product formulation it is evidently possible to
get this wrong; see
http://lists.nau.edu/cgi-bin/wa?A2=ind0208&L=phys-l&P=R8914

Note that I have not used any cross products, nor have I
done anything involving the right-hand rule.

Jack Uretsky wrote:

1. So give me the non-quantum argument (without invoking the right-hand
rule) that the e-m wave carries momentum in the +z-direction.

I believe the paragraph following equation (8) above
answers this question.

2. Strictly speaking, the circles are the solutions in the limit of very
large eB/mw, where w is the angular frequency of the wave.

In this limit, the zeroth Born approximation is not valid
(except for super-short times). Re-radiation must be considered.

i've had to obtain the spatial solution numerically.

You can solve equation (6) by timestepping it, using
equation (5) and of course equation (9). This is
a) utterly straightforward, and
b) presumably identical to Jack's numerics (or nearly
identical, depending on how he handled the time
versus proper-time issues).

3. Incidentally, let's not forget that the B-pseudovector can be
visualized - as with patterns of iron filings.

a) It is not easy to decorate electric field lines in
the corresponding way. Does that mean that magnetic field
lines are more real than electric field lines? I don't
think so.

b) Note that the usual sort of iron filings are not sensitive
to the _direction_ of B. You could fix this using little
"compass needle" filings, with miniature markings (N and S)
on their poles. But compass needles are essentially chiral,
and you can't correctly mark them without invoking the right-
hand rule. Some kid named Pierre was intrigued by this
http://lists.nau.edu/cgi-bin/wa?A2=ind0104&L=phys-l&P=R2423

c) You can visualize B as a bundle of tubes (as opposed to
a bundle of lines) as emphasized and illustrated in Misner,
Thorne, Wheeler _Gravitation_.

if they're so conceptually friendly,

It's hard to imagine anything much more straightforward than
the calculation outlined above. The only real work involved
sorting out which assumptions that Jack wanted us to make.
After that, I just turned the crank on the Clifford-Algebra
formalism.