Re: [Phys-l] basic laws of motion +- vectors +- angular momentum

[John Denker <jsd@av8n.com>]

On 12/09/2008 02:08 PM, LaMontagne, Bob wrote:
> I'm still unclear how this resolves the issue. 

I'm not sure either.  Partly I'm confused because there
are divergent opinions as to what "the issue" is.  So
let's soldier on and see what we can figure out.

> The torque bivector still
> requires a sum over the /\ for the individual forces and moment arms.

OK, that's not a problem, is it?

> 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. 

Let's be careful.  If we're summing over force vectors 
strictly speaking, force vectors don't have attachment 
points.  So, as Henny Youngman said, don't do that.
Instead let's sum over "physical interactions" or some
such, where each physical interaction has two vectors,
i.e. a force vector and a point-of-attachment vector.

> It appears that the operations on the two sides of N2
> are not equivalent in any obvious way. 

Lost me there.  The LHS is _equal_ to the RHS.  Is that
trying to draw a distinction between equal and "equivalent"?
I'm not following that.

Or is that trying to say that the LHS is not not _compatible_
with the RHS?  I'm not following that, either.  I hope nobody
thought I was suggesting using angular momentum as a direct
plug-in replacement for linear momentum.  You can derive 
conservation of linear momentum starting from conservation 
of angular momentum, but they are not equivalent.  They don't
even have the same dimensions.

> 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.

We agree that things are simple for pairwise interactions
and central forces.

> 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.

Please let's not confuse "conserved" with "constant".
  http://www.av8n.com/physics/conservative-flow.htm#sec-conservation+-constancy

Angular momentum is _constant_ when the torque is zero.

Angular momentum is _conserved_ always.  Always.  If
there is a net torque, that shows up as a boundary term
in the conservation law:

   change(XX inside boundary) = − flow(XX, outward across boundary)