Re: arbitrary choice of zero of potential

["John S. Denker" <jsd@MONMOUTH.COM>]

At 05:20 PM 10/17/01 -0700, John Mallinckrodt wrote:

> ... isn't it trivial to see that Sherwood and Chabay are correct?

Au contraire, it is trivial to show that it is incorrect.

>   Here's a simple example:
>
> Take a system of two electrons each of mass m that are at rest and
> separated by a large distance like, say, a lightyear.  "Let" the
> potential energy be zero.  The components of the momenergy four
> vector (E,px,py,pz) with c = 1 are
>
>        [2m,0,0,0]

So far so good.  That's the total 4-momentum of the two-particle system.

> and the invariant mass, (E^2-p^2)^1/2, is 2m.

OK.

> Now use the Lorentz transformation to transform to a reference
> frame moving at v = 4/5 in the +x direction. The new components of
> the momenergy four vector are
>
>       [(10/3)m,-(8/3)m,0,0]
>
> We find that the invariant mass is still 2m and, therefore, indeed
> invariant.  No surprise here, this is what the Lorentz
> transformation *does*.

So far so good.

> Now suppose that we "let" the potential energy be nonzero, say, m.
> The components of the momenergy four vector in the initial rest
> frame are
>
>        [3m,0,0,0]

Oooops, AFAIK there is no such 4-vector.

 1) If you want an "authority", take a look at section 2.5 of Misner /
Thorne / Wheeler, wherein the 4-momentum is defined as the invariant mass
times the 4-velocity, with no contribution from potential energy.  If you
don't have MTW handy you can read the online reference cited in my previous
note in this thread.
 2) If you want an explanation without appeal to authority, there a couple
of ways to proceed:
   2a) Fancy general argument: The gauge adds a constant to the
Lagrangian.  Notice that things we care about, such as the momentum and/or
the equations of motion, depend on _derivatives_ of the Lagrangian, so they
cannot be affected by the choice of gauge.
   2b) Simple argument:  Drop a brick.  It moves in the gravitational
field with constant total energy;  what it loses in potential energy it
gains in kinetic energy.  But its momentum is changing.  So we have
constant total E, constant mass, and changing momentum.  This trivially
disproves the hypothesis that the total E is the zeroth component of the
momentum 4-vector.

The rest of JM's argument is predicated on this impossible 4-vector, and
does not require further comment.

> As Sherwood and Chabay point out, "If an arbitrary constant is
> added, energy and momentum will not transform correctly between
> different reference frames."

That statement implies that we are going to add an arbitrary constant to
the zeroth component of the 4-momentum.  This implication is totally false.

> Recognizing the absolute nature of
> potential energy has far-reaching pedagogical consequences,

The laws of physics do not recognize any absolute potential energy.  Making
false assumptions has all sorts of consequences.

=========================

By the way:  Although S+C try to "prove" a certain gauge must be used, you
don't even need to know how to spell gauge invariance in order to refute
the "proof".  Argument (2b) is independent of gauge;  pick any gauge you
want, or no gauge at all, and you can easily see that we are dealing with a
purely imaginary bogey-man.  It is suitable for scaring young
trick-or-treaters, but not much else.