NetNews Usenet Archive 1992 #18/NN_1992_18.iso/spool/sci/physics/12911

Text File | 1992-08-16 | 5.6 KB | 119 lines

Newsgroups: sci.physics
Path: sparky!uunet!sun-barr!ames!pacbell.com!well!sarfatti
From: [e-mail] (Jack Sarfatti)
Subject: New Time Arrow & Zero Point Engine?
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Organization: Whole Earth 'Lectronic Link
Date: Sun, 16 Aug 1992 22:02:13 GMT
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FINAL SOLUTION OF THE SUPERLUMINAL QUANTUM CONNECTION COMMUNICATION
PROBLEM
by Jack Sarfatti

ABSTRACT
It is shown that either I communication on the quantum connection is
impossible, or II if it is possible then: i. traditional "retarded"
causality (causes always before effects) is wrong; ii. parity is
violated; iii. time reversal symmetry is violated giving a new arrow of
time; iv: useful coherent energy can be drawn out of random zero point
quantum vacuum fluctuations by an optically induced first order vacuum
phase transition. The basic quantum mechanical equations are consistent
with either possibility.

The remark by the brilliant UCB physics graduate student Aephraim
Steinberg that the net reflection phase shift of a beam recombiner must
be pi/2 in order to have both time reversal symmetry and no cohering of
useful energy out of apparently random quantum vacuum fluctuations is
absolutely true in my opinion after having worked out several special
cases in detail.

Aephraim's basic argument is

"It can be shown from unitarity or from time-reversal
symmetry that this phase is always plus or minus pi/2.(For
any lossless beam-splitter, not just 50/50.) Thus cos(b) is
always 0, so p(e)=p(o), and there is no communication.

Born & Wolf treats it, Cohen-Tannoudji does it for general Schrodinger
barriers,and Ou & Mandel had a paper on it a few years back.

As for the Fresnel equations, think of the simplest beam-splitter,
a thin sheet of glass which reflects 4% at each surface (we'll half-
justifiably ignore multiple reflections for now).The amplitude for
transmission is just exp(ikd) to lowest order, where
k is the wavevector and d is the thickness. The amplitude for reflection
off the front surface is +0.2, no pi phase shift.The amplitude for
reflection off the back surface is -0.2exp(2ikd),where there is the pi
phase shift you noted, but there are two path lengths of d.If you add
the two phasors, you can see that the phase of the resultant is the
average of the two inital phases (2kd and pi),that is, kd+pi/2, that is,
phi(r)=pi/2 + phi(t) as claimed.

If you do it carefully for the infinite sum, you also get the right
answer,of course, that's just a plausibility argument in response to
your question about the Fresnel relations.

The general argument is easiest by time-reversal symmetry.

Suppose our input ports are I1 and I2, going into outputs O1 and O2.
If the amplitude incident on I1 is 1 and on I2 is 0, t comes out in O1
and r comes out in O2.Let us now time-reverse this, assuming a symmetric
beam-splitter (so r and t are the same from either side).
Now t* comes in O1 and r* comes in O2, so t*t+r*r comes out I1 and
t*r + r*t comes out I2. But we know nothing originally went into I2, so
t*r + r*t = 2Re (t*r) = 0.That is, if we write t*r as |t||r|exp(iphi(r)-
iphi(t)), cos(phi(r)-phi(t))=0, so the phases differ by plus or minus 90
degrees.There are other ways of showing this.If it's lossless but not
symmetric, then the AVERAGE of the two different reflection phase shifts
still differs from the transmission phase shift by 90 degrees.
If it's lossy, you can put an upper bound on that cosine, you can derive
it yourself or I can look for my old notes if you're curious.
I don't think it's reasonable to think that the losses would HELP you in
your scheme, though.(I admit I don't immediately understand how to
demonstrate this latter guess though.)

Hope this is clear,
aephraim."


Looking at Aephraim's micro-model "..but there are two path lengths of
d.." I suggested an optical inhomogeneity in one of those paths giving
an extra phase theta.

If the pi and zero phase shifts from the external and internal
reflections from local Fresnal matching conditions continues to hold
independent of the theta then we have a contradiction with time reversal
invariance and, more seriously, with conservation of energy.

The effective phase difference would then be b = (theta + pi)/2. There
is generally a "resonance" term in t*r + r*t (i.e., cos{kd + theta/2 -
pi/2} multiplying cos(theta+pi/2), in lowest order for small reflection
and ignoring multiple reflections) which allows it to vanish even when
theta is not zero. But that is only a special case when the thickness d
of the plate is tuned to the particular theta from the controllable
optical doping.

What happens in general when we are "off resonance"? Only two things can
happen either, I. the optical doping inside the glass plate changes the
boundary conditions at the front and back faces of the glass plate in
such a way that 0 and pi are no longer the phase shifts there so that
the net shift is still b = pi/2. In other words the internal theta shift
from the inhomogeneity is somehow screened out. This is interesting in
its own right.

Or, II the theta screening does not happen, the 0 and pi surface shifts
persist. If that happens, then time reversal symmetry is broken giving a
new arrow of time, and in order to preserve conservation of energy the
plate will cohere random quantum zero-point fluctuation energy into
additional real photons (depending on theta it can also absorb incident
photons into the vacuum).

The reason that II cannot be rejected as absurd out of hand is that the
quantum equations by themselves make perfect sense, conserving locally
observable detector probabilities for any value of b. Let experiment
decide!