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Re: cross-product versus Clifford Algebra



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.
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. The motion
superimposed on the drift velocity in the general case can include
circles. I'll try posting a graph tonight on my web-site tonight (and let
you know when it's posted). The solution for the velocity imposed on the
drift velocity can be obtained in straightforward fashion (hint: its
magnitude is constant), but i've had to obtain the spatial solution
numerically. I'm sure that this is all old-hat to beam physicists.
3. Incidentally, let's not forget that the B-pseudovector can be
visualized - as with patterns of iron filings. The filings line up to
define smooth curves between the poles of a magnet, for example, and the
tangent to any curve at any point is the direction of the pseudovector!
4. An aside to Leigh, including a hearty "Welcome Back!". A number of
years ago, as I was returning to the practice of physics, I worked all of
the wedge-product problems in MSW's Gravitation text, so I'm not
unfamiliar with the stuff. But if they're so conceptually friendly, why
don't you help John out by showing him how to argue the direction of
motion of an electron in "crossed E- and B-fields" without invoking the
right-hand rule.
Regards,
Jack


On Mon, 12 Aug 2002, John S. Denker wrote:

Jack Uretsky wrote:

the circles....are in the x-z plane.

1) I have tried, but I cannot reproduce this version
of the circles, either. Please elaborate.

The
cross-product predicts the direction of drift, so that the drift is
imposed on the circular motion.

Earlier he wrote:

it "drifts" in the z-direction.

2) The incoming wave had momentum in the z-direction and
only in the z-direction. Why do we need anything like a
cross-product to tell us that the preferred direction for
drift imparted to the electron will be the z-direction?
Why do we even need to know the polarization state of
the incoming wave?

3) Even if we did need something like a cross product, why
would not a wedge product do the job at _least_ as nicely?


--
"But as much as I love and respect you, I will beat you and I will kill
you, because that is what I must do. Tonight it is only you and me, fish.
It is your strength against my intelligence. It is a veritable potpourri
of metaphor, every nuance of which is fraught with meaning."
Greg Nagan from "The Old Man and the Sea" in
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