Re: 4/3 problem resolution/Action-reaction paradox in pdf format

[Bob Sciamanda <trebor@VELOCITY.NET>]

----- Original Message -----
From: "David Rutherford" <drutherford@SOFTCOM.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, June 19, 2001 01:02 PM
Subject: Re: 4/3 problem resolution/Action-reaction paradox in pdf format


I said:
> > 3.) Therefore, an observer travelling with this CM is viewing the world
> > from a non-inertial frame.
You objected:
> No, he's viewing the world from an inertial frame. Maybe it would be
easier
> if you explain to me why you think his frame is non-inertial.

In the Newtonian world, there is a simple criterion which applies here:
any frame which is accelerating (as viewed from an inertial frame) is a
non-inertial frame (also true in the world of special relativity).  The CM
of your electron system is accelerating as viewed in the lab . . . Ergo .
. . -->

I said:
> > He will see objects accelerate with no forces
> > acting on them;
You replied:
> All sorts of weird things will happen because the Lorentz force
equations
> give conflicting results, in this situation. That's my point.

I don't know what this means.  (The Lorentz transformations connect only
inertial frames.)

I said:
> > he will of course see objects at rest relative to him NOT
> > accelerate even though there is a net force on them
You asked:
> > From where?
If they are at rest in an accelerating frame, there must be a net force on
them.  When I somersault, the coins in my pocket would be an example of
such objects.

I added:
> > (as there must be on him).
You replied:
> The observers are assumed not to interact with the particles.
He must be experiencing a net force from something (probably not the
particles), because he must accelerate in order to stay with the
accelerating CM.  How else does he become a "CM observer"?

I said:
> > He cannot apply Newton's laws to his observations without adding
> > corrections which take into account his own acceleration.
You articulated:
> As I said, he's inertial according to himself.
I haven't a clue as to what this means!

We went on - I speculated:
> > 4.)  I think that by your phrase "then the center of mass is inertial"
> > you are asserting something other than the (false)statement that an
> > observer travelling with the CM is at rest in an inertial frame.  What
is
> > it that you really mean?
 You repied:
> By definition, if two particles start out with equal and opposite (or
zero)
> velocities, as in this case, and the net forces on the particles sum to
zero
> (which they do in the CM frame), then the velocity of the center of mass
of
> the particles is constant (or zero). That means that the reference frame
of
> the center of mass (CM frame) is inertial from the point of view of an
> observer at rest in the CM frame.
The accelerations sum to zero in the CM frame; the forces do not.  In fact
in the (v<<c) Newtonian scheme forces are invariant  - the same to all
observers - and acceleration is only a valid measure of force in inertial
(non-accelerating) frames.  The last sentence is inscrutable.  To be an
inertial frame it is not sufficient for the CM to be at rest in its own
frame (all frames are!) - it must be at rest in some inertial frame - and
have a constant velocity in all other inertial frames (eg: your lab
frame).

> Imagine two identical pool balls travelling with equal and opposite
> velocities (CM frame view). They collide elastically then move away from
> each other with equal and opposite velocities. The center of mass of the
two
> balls is stationary throughout the interaction, therefore it's inertial.
If
> there was a very short observer standing at the center of mass, he would
say
> that he is inertial, too.
>
> Now just insert pool balls=electrons and you've got the situation in the
CM
> frame.

If the pool balls constitute an isolated system and their forces of
interaction obey the third law, then their CM will not accelerate and will
be at rest in an inertial frame.  The Maxwellian forces between two
electrons do not obey Newton's third law.  In order to preserve our very
useful Conservation of Momentum calculational tool, we assert that neither
are they isolated - there are other momentum/energy sharing entities
involved (the fields).  If you have a better (or even equivalent) model,
it is welcome - but you should still be bound by logic, consistency and
lucid terminolgy.

Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor