Re: arbitrary choice of zero of potential

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

At 10:55 AM 10/18/01 -0700, John Mallinckrodt wrote:
> Consider a pair of electrons each with mass m (in the generally
> well understood sense of the "mass of an electron") that are
> released at rest from an initial separation along the x-axis that
> is specifically chosen so that each electron will approach an
> asymptotic kinetic energy equal to m (in units where c = 1) as
> they separate to large distances.  For the purpose of this
> problem, I would like to effectively "turn off" the EM radiation
> that would accompany this process so that we need not concern
> ourselves with accounting for it.
>
> 1. What is the initial momenergy four-vector, P = [E,px,py,pz],
> for either one of the two electrons?
        p = [m, 0, 0, 0]  in the obvious units.
> 2. How did you obtain the answer given in 1?
        p = m v
        where v = [1, 0, 0, 0] since the electron was
        specified to be at rest.
> 3. What is the initial momenergy four-vector, P = [E,px,py,pz],
> for the two electron system?
        I take this to be a question about the
        two-electrons-PLUS-FIELDS system, in which case
        the system 4-momentum is
        p = m [2 sqrt(2), 0, 0, 0]
> 4. How did you obtain the answer given in 3?
        Working backwards from answer 5.
        Neglecting radiation as requested.
> 5. What is the (asymptotically-approached) final momenergy
> four-vector for the single electron that is moving in the +x
> direction?
        p = m [sqrt(2), 1, 0, 0]
> 6. How did you obtain the answer given in 5?
        I interpreted the specification KE=m to mean p_x=m.
        Then calculate p_t from p_x to preserve |p|=m.

        Note: If the intended meaning was otherwise, then answer 5
        might become p=m[2, sqrt(3), 0, 0] or some such,
        in which case you have to calculate p_x from p_t,
        but it doesn't change the ideas much.
> 7. What is the (asymptotically-approached) final momenergy
> four-vector for the two electron system?
        p = m [2 sqrt(2), 0, 0, 0]
> 8. How did you obtain the answer given in 7?
        From item 5 by addition of 4-vectors.
> 9. How much of the system energy given in the answer to 3 is
> interaction energy?
        41%
> 10. How did you obtain the answer given in 9?
        2 sqrt(2) is 141% of 2

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Note that all the aforegoing questions and answers are gauge-independent.
The asymptotic potential could be 1,234,567 volts and it would make any
difference.