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Re: cross-product versus Clifford Algebra

Hi all-
In response to John D.:
But that is the wrong answer. The electron moves in the direction of the
Poynting vector (as would be evident from a quantum answer). The E-Field
accelerates the electron in the +E-direction, the B-field supplies a force
in the vxB direction (same direction as ExB drift). So the electron
executes circles in the x-y plane as it "drifts" in the z-direction.

That curvy-little arrow in the x-z plane (my problem requires the
B-pseudovector to lie along the y-axis) has an orientation that
is given by the right hand (or pseudopod?) rule. John accepted that point
when he agreed that a pseudovector describes and "oriented" area. We need
the RHR (or its equivalent) to describe the orientation).
Regards,
Jack

On Sun, 11 Aug 2002, John S. Denker wrote:

Jack Uretsky wrote:
I like the cross-product because it tells me quickly how to apply
the right hand rule,

But if you don't use the cross product you don't need
the right hand rule. You can't give the cross product
credit for solving a problem of its own making.

do not forget that the cross-product
(pseudovector) describes an (oriented) area.

Yes! I 100% agree with that.

Consider the following concrete application of the idea:
Draw a picture of a charged, spinning object and its
magnetic field at a nearby point. Suppose the field
is in the +Z direction.
a) If you draw the field as a little straight arrow
(representing a PSEUDOvector) in the +Z direction,
the mirror-image of this picture is wrong.
b) If instead you draw the field as a little curvy
arrow in the XY plane, the mirror image of this picture
is correct. No right-hand rule, no problems.

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?

Why does anybody need a cross product to answer that?
To first order, the electron will move in the direction
of the E vector, directly specified as the x-direction.
To higher order, the e-m wave will interact with the
electron's magnetic moment, in a direction that cannot
be determined because the orientation of the electron
was not specified. To yet higher order, there will
Compton scattering with a predictable component in the
+z direction, plus unpredictable components in the other
directions.

The cross product doesn't help with the answerable parts.
The cross product doesn't help with the unanswerable parts.
Perhaps I'm overlooking some additional parts, but I doubt
the cross product is necessary for them, either.

Related question? What language do I us to describe what us older
guys called E-cross-B drift?

You could call it E-wedge-B drift.

More generally:
There is a mechanical procedure for rewriting cross
products in terms of wedge products.
1) Sometimes that's the whole story.
2) On other occasions you should look around a little
bit and see if the wedge product is part of a more
general expression. This is what happens in electromagnetism.
The general expression is the geometric product
del F
which can (like any geometric product) be expanded as
del F = del dot F + del wedge F
and both terms on the RHS are significant. Del dot F
gives the vector terms in
http://www.monmouth.com/~jsd/physics/maxwell-ga.htm#eq-max-exp
while del wedge F gives the trivector terms.


--
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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
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<The 5-MINUTE ILIAD and Other Classics>