Re: electron circles, or not

["John S. Denker" <jsd@MONMOUTH.COM>]

Jack Uretsky posed the challenge:

> Without invoking the
> cross-product, answer the following:  right-handed coordinates (x,y,z), an
> e-m wave in the +z-direction is polarized with the E-vector along the
> x-axis.  An electron, at rest at a very early time before the e-m wave is
> turned on, will move which way?

Jack asserts that the answer involves circles.  I am still
unable to confirm this answer.

In particular, my calculations show that neither strictly
circular circles nor the curlicues depicted at
  http://www.hep.anl.gov/jlu/polarelec.jpg
can be solutions to the challenge as posed.

To simplify my calculation, let us recall that Jack said
> Strictly speaking, the circles are the solutions in the limit of very
> large eB/mw, where w is the angular frequency of the wave.

So in particular let's consider a really, really small angular
frequency, but not quite zero.   (BTW, this means we can
neglect the particle's magnetic moment, since it is a _dipole_
moment and the B-field _gradient_ is small.)

I continue to assume
  1) this EM wave is a plain old propagating wave, as opposed to an
electrostatic or magnetostatic field.
  2) we are supposed to neglect the magnetic moment of the particle.
  3) we are supposed to neglect radiation emitted from the particle,
i.e. we are working in the zeroth Born approximation.

Because this is still a real-live radiating wave, we know that
        E^2-B^2 = 0.
That's true for any EM wave.  (Hint:  for any EM field, radiating
or not, E^2-B^2 is a relativistic invariant scalar.  Use this fact
to convince yourself that the value of this scalar is zero for
EM waves.)

Let's consider the X-component of momentum of the charged particle.
It will pick up some momentum due to the E-field piece of the wave.
This will depend on the timelike component of the particle velocity;
the longer the particle sits in the field, the more momentum it picks
up.  There will be another contribution to the X-component of momentum
from the B-field piece of the wave.  This will depend on the spacelike
Y component of the particle velocity.

Now the timelike component of velocity goes like cosh(rho), while the
spacelike component goes like sinh(rho) and cosh is _always_ bigger
than sinh.  So the derivative of X-momentum cannot change sign (over
timescales where the wave field doesn't change sign).  (If you
don't like the cosh/sinh argument, you can make a stricter argument
using the invariant norm of the 4-velocity.)

I surmise that the curlicues shown in Jack's .jpg were produced using
a field that is not a solution to the EM wave equation, but rather has
been augmented with a magnetostatic piece, so that E^2-B^2 is
significantly
negative.

Psychological / pedagogical remark:  We all know about cyclotron orbits.
But we shouldn't let that unduly color our intuition about this
challenge,
because that's a different set-up.  The cyclotron is dominated by a
magnetostatic field.  Waves are not.  E^2-B^2=0 for waves.

===============

In the figure
  http://www.hep.anl.gov/jlu/polarelec.jpg
it appears the particle is drifting in the -Z direction.  That
seems backwards.

Speaking of backwards, I previously wrote
> >   F = (gamma0 gamma1 + gamma1 gamma3)
> >                sin(k z - w t + ph)              (5)
This is a typo.  If we want it to be a solution to the
correct wave equation, we need a minus sign.
  F = (-gamma0 gamma1 + gamma1 gamma3)
                  sin(k z - w t + ph)              (5')
Sorry.  Note this just fixes a typo; there is no
ad-hockery here;  if F solves the wave equation at all,
it propagates in the correct direction and only that
direction.

Jack most recently wrote:
> Nor have you made an argument that I can see that the
> electron moves in the Positive z-direction.

In addition to my previous qualitative explanations of
why it must do so, numerical timestepping of equation (5')
gives the desired result, as shown in this figure:
  http://www.monmouth.com/~jsd/physics/em-ga.gif

> How do I set up my Netscape so it shows your Greek letters.
Netscape version 6 is a loser.  Use an earlier version.  Or
use IE.  For other options (mostly for unix) see:
  http://www.monmouth.com/~jsd/fly/how/htm/font-fixing.html
  http://hutchinson.belmont.ma.us/tth/Xfonts.html
  http://pauillac.inria.fr/~maranget/hevea/doc/browser.html

I am not planning to convert everything to html-4 entity codes,
because that would make things worse for the 90% of my readers
who are not running netscape-6.  And I don't feel like supporting
multiple versions of all my pages.

================

I'm more convinced than ever that Clifford Algebra is the Right
Way to do physics, as opposed to cross products and suchlike.  It's
like eating soup with a spoon -- it's less work and gives a better
result than eating soup with a fork.