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# [Phys-L] Re: Area vector

• From: John Denker <jsd@AV8N.COM>
• Date: Wed, 30 Mar 2005 06:09:19 -0500

Tony Wayne wrote:
Does any one know why the "area vector" for a surface is defined as
being normal to the plane of the surface instead of parallel to the
plane of the surface?

As others have mentioned, a vector in the plane of the surface
doesn't tell you what you need to know. There isn't a one-to-one
mapping between vectors and areas. Dimensional analysis suffices
to demonstrate the point: vectors have length; areas have area.

What you want is a *bivector*. If P and Q are vectors, the best
representation of the parallelogram spanned by P and Q is the
wedge product P /\ Q. In particular, the amount of paint
required to cover such a region is given by the norm thereof,
i.e. ||P /\ Q||. This is so elegant, so simple, and so powerful
that I shudder at the thought of representing areas any other
way.

For details, see
http://www.av8n.com/physics/area-volume.htm
and references therein.

Bivectors have all the properties you want when representing
areas.
-- For starters, the antisymmetry of the wedge product
guarantees that the area P /\ Q is manifestly zero if P happens
to be parallel to Q.
-- A patch of area might be part of the boundary of a volume.
The volume V is best represented as V = P /\ Q /\ R. This is
manifestly zero if R happens to lie in the plane of P /\ Q.

In a three-dimensional space where you have basis vectors
(unit vectors) X, Y, and Z, the natural basis for computing
bivectors is Y/\Z, Z/\X, and X/\Y. These critters are isomorphic
to quaternions.
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