Re: virtual particles preferred frame?

[John Denker <jsd@MONMOUTH.COM>]

At 10:59 AM 11/8/99 -0600, Glenn A. Carlson wrote:
>
> Can you provide a citation which describes the negative
> energy-imaginary momentum virtual particle theory?  I've never seen it
> described in any physics book I've read.

Alas, no.  I'm not sure I've ever seen a derivation anywhere.  That makes
it kinda hard for you to look up the details, but it doesn't make the
result any less true.

I am reminded of the oft-quoted words
> >Young man if I could remember the names of all of
> >these particles I would have become a botanist
> >                        Enrico Fermi

I, too, went into physics so I wouldn't have to remember details.  In this
case it is easier to derive the result from scratch than to find it in a
book.  See below.

> If momentum is imaginary, is the mass real and velocity imaginary, or
> is the mass imaginary and the velocity real, or both?

Huh?  It's got the plain old mass.  This is physics, not a freakshow.

>   Imaginary mass and real velocity sounds like tachyons to me.

If your "velocity" means group velocity, it's zero.  That's quite a bit
slower than most tachyons :-)

> I'm not skeptical because negative energy-imaginary momentum is just
> different than the explanation I described, and not even because I've
> never seen it in any physics book, but, rather, because it raises all
> kinds of questions for which I don't see an answer.  In fact, it seems
> to create a whole new area of physics (e.g., tachyons).

I can't imagine why you bring up tachyons.

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

Here's a simplified derivation:

Imagine we have a particle of mass M caught in a trap of depth M.  The trap
is centered at the origin and is very narrow spatially.

Imagine that everywhere outside the trap, our particle obeys the massive
scalar Klein-Gordon equation (aka the waveguide equation)

        (d/dt)^2 - (d/dx)^2 + M^2 = 0           (0)

You can treat this as your basic barrier-penetration problem.  The particle
will "leak" out of the trap for a certain distance which we shall calculate.

Notice that the escaped particle has exactly zero total energy -- it has
its usual, normal rest-mass energy M, but it had to borrow one M to climb
out of the trap.  This is an artificial zeroing of the energy, resulting
from my choice of the depth of the trap.  But there was a reason for
choosing this depth, because it corresponds exactly to the creation of a
virtual particle out of the previously-empty vacuum.

I assert that
        psi = exp(-|x|M)                        (1)
is a solution to the equation of motion for all nonzero x.  My assertion of
zero total energy is confirmed by the lack of any time dependence.  You
could write this as
        psi = exp(-|x|M) exp(i omega t)         (2)
but we see we my solution has omega=0.  Other solutions are possible, but
this is the one that propagates the farthest, so it best answers your
original question.

(Shorthand 1) Among the cognoscenti it is a common shorthand to pretend
that this nodeless decaying wave comes from the more-common nodeful
nondecaying wavefunction of the form
        psi = exp(i k |x|)                      (3)
by means of an imaginary value of k.  This is of course algebraically the
identical thing.

(Shorthand 2) Furthermore, you have the option of interpreting this
imaginary k as an imaginary momentum hbar k, but now we are two layers deep
in optional interpretation.

If imaginary momentum gives you indigestion, just go back to equation
(1).  It solves the equation of motion and that's really all that need be said.

==

The group velocity d(omega)/d(k) is zero.  This should be clear from
symmetry of the omega-versus-k graph if nothing else.