Re: [Phys-l] Equations (causal relationship)

[John Denker <jsd@av8n.com>]

Michael Edmiston wrote:
> The picture is great.  Now I understand the previous description.

:-)


> Thinking out loud here... If your drawing is used for a case in which 
> F(net) is not zero, then your red, green, blue stages cannot have 
> arbitrary accelerations, but must have the F = ma relationship if the 
> red stage and particle are to have the same acceleration.

I think it is more underdetermined than that.

I think the full solution-set is:
  a1 = F1/m + q1
  a2 = F2/m + q2
  a3 = F3/m + q3
for any arbitrary q1, q2, and q3 subject to the constraint that
(q1 + q2 + q3) = 0.  As always, the roles of force and acceleration are
completely symmetric;  one can equally well write:
  F1 = m a1 + g1
  F2 = m a2 + g2
  F3 = m a3 + g3
for arbitrary g1 g2 g3 such that (g1 + g2 + g3) = 0.

My rationale for saying this is that the net acceleration (a1 + a2 + a3)
is the only acceleration that matters to the particle ... just as the net
force (F1 + F2 + F3) is the only force that matters to the particle.  This
can be quantified using the notion of _sufficient statistic_.

This is true no matter whether the net force and net acceleration are zero
or nonzero.

We know there is plenty of arbitrariness in how we decompose the net force,
and plenty of arbitrariness in now we decompose the net acceleration;  the
fudge factors q1 q2 q3 tell us that these two decompositions are _separately_
arbitrary.

The particular solution
  F1 = m a1
  F2 = m a2
  F3 = m a3
is preferable in terms of elegance, but is not required by the physics.

As a real-world application, typical automotive traction-control systems
work by applying the brakes as needed while you are trying to accelerate.
(*Some* systems also try to cut back on engine torque, to minimize the
amount of opposing forces within the system ... but AFAIK most of them
don't bother.  It's easier to rely on the brakes alone, and live with the
opposing forces.  As Mr. Spock would say:  it's crude, but effective.)