Re: Does it move?

From: Old Man (nomail_at_nomail.net)
Date: 09/01/04 

Date: Tue, 31 Aug 2004 20:50:03 -0500

"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0408311508.39b108e8@posting.google.com...
> "Ian Macmillan" <iandmac@bigpond.com> wrote in message
news:<13kYc.12431$D7.8901@news-server.bigpond.net.au>...
>
> > Two masses together, one large, one small, are referred to a long
> > measuring rod. A force is applied between the masses to accellerate them
> > apart. The large mass accellerates at a modest rate, but the smaller
mass
> > reaches relativistic speed.
>
> <...>
>
> > However with the same forces applied for the same time (taken at the
> > large mass) the small mass will move a distance dependent on its
velocity
> > profile, so that when the system comes to rest, the position of it's
> > centre of mass relative to the reference rod can differ from that at the
> > start.
> >
> > Is this analysis correct?
>
> Regardless of the correctness of the details of your analysis, to
> which I harbor some uncertainty, your qualitative conclusion seems to
> be correct.
>
> Let me attempt a restatement of your set-up:
>
> Two point masses M,m interact in some way so that:
>
> (1) dP_M/dt = -dP_m/dt.
>
> (This bypasses the whole issue of the relativistic transformation of
> force).
>
> The masses are constrained to move along the x-axis, with coordinates
> X,x.
>
> Define a new coordinate:
>
> (2) xcom == [1/(M+m)][MX + mx]
>
> This is the non-relativistic definition of the position of the center
> of mass. But for our purposes, note that M,m are the rest masses, X,x
> the position coordinates, and that this algebraic combination is
> defined at all times, regardless.
>
> We know in Newtonian physics that, subject to the constraint of eq.
> (1), the velocity of the center of mass
>
> (3) x'com == [1/(M+m)][MX' + mx']
>
> always remains zero, given that the particles started at rest. So in
> this case the center of mass can never move, regardless of the
> acceleration profile.
>
> But the relation which enforces this immobility of xcom
>
> (4) x' = -(M/m)X'
>
> breaks down when the light particle reaches relativistic speeds, and
> in this case xcom acquires non-zero velocity. We can therefore arrange
> situations in which xcom moves between start and finish.*
>
> One objection to the argument is that condition (1) is unphysical,
> since it supposes instantaneous action at a distance (sans intervening
> fields). I don't know if this pulls the bacon out of the fire or not.
> However, despite Old Man's insinuation that SR does not violate
> conservation of momentum (but you have), we _haven't_ manifestly
> violated conservation of momentum! We put it in explicitly! What we
> have violated is "conservation of position of the center of mass",
> which may or may not be a law in SR.
>
> If you read my comments on the cat and the inchworm, and thought about
> how the cat pulls this thing off, it's because the cat is free to
> create and destroy "angular mass" (aka moment of inertia), and so pull
> itself along the angular coordinate, like an inchworm, all the time
> conserving angular momentum. It _appears_ that SR gives us the
> ability to create and destroy effective "linear mass" (aka the dreaded
> relativistic mass), and so pull ourselves along ordinary space, like
> an inchworm, all the time conserving linear momentum -- which must
> have been your original intuition, thereabout.
> ___
> *[At this point we might like to hope that even though xcom moves when
> the light particles reaches relativistic speeds, it reverses direction
> while the particles are deaccelerated, and obediently returns to its
> starting point by the time the particles stop. But this hope is
> dashed. For, let there be one acceleration profile with this
> property, and pick some intermediate epoch at which x'com = u, /= 0.
> Next, construct a new acceleration profile where we turn off the
> interaction at this epoch for a period T, then resume the former
> profile. During period T xcom has moved uT, and everything else being
> equal, this is the final diplacement of the center of mass under the
> modified profile.]

Since p_M = - p_m, then

d(p_M) / dt + d(p_m) /dt = 0

This defines the center-of-momentum which cannot accelerate, and, if
p_M = - p_m = 0 initially, then the center of momentum has zero velocity.

However, Ed's definition of center-of-mass is incorrect. In the CP frame,
the total rest mass is the relativistic mass, E = E_M + E_m wherein

E_m = m c^2 / sqrt[ 1 - (v_m / c)^2] = [p_m / v_m] c^2

E_M = M c^2 / sqrt[ 1 - (v_M / c)^2] = [p_M / v_M] c^2

The center-of-mass is now defined in terms of energy

X_cm = [X*E_M +x*E_m] / [E_M + E_m]

The coveted result is d(X_cm) / dt = 0.

[Old Man]

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