I'm still unclear how this resolves the issue. The torque bivector still
requires a sum over the /\ for the individual forces and moment arms.
However, the other side of Fdt=dp does not have an equivalent set of
attachment points and lever arms if we are dealing with a rigid
continuous object. It appears that the operations on the two sides of N2
are not equivalent in any obvious way. This is the same difficulty as
when traditional vectors are used. The problem only seems to disappear
for a point object where the moment arm is simple to identify on the
right hand side of the equation.
Conservation of angular momentum supposedly comes about when the net
torque is zero - but I don't see the entity we call angular momentum
appearing in either formulation.
Bob at PC
-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of John
Denker
Sent: Tuesday, December 09, 2008 1:52 PM
To: Forum for Physics Educators
Subject: [Phys-l] basic laws of motion +- vectors +- angular momentum
The point is that in terms of vectors, strictly speaking,
there is no difference between diagram (A) and diagram (B).
Vectors have magnitude and direction, period.
If you want to talk about something having magnitude,
direction, and point of attachment, you need a bivector:
torque = force /\ lever_arm
So ... as far as I can see, so long as the third law is
formalized in terms of forces i.e. vectors, the formal
law does not capture the full physical reality.