From: sarfatti@well.sf.ca.us (Jack Sarfatti)
Subject: Math Appendix Final Solution FTL
Organization: Whole Earth 'Lectronic Link
Date: Mon, 17 Aug 1992 04:54:32 GMT


Mathematical Appendix to Final Solution of FTL Communication
by Jack Sarfatti
#1 "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."
aephraim

#2 "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."
aephraim

Let front to back face optical path length be d1 and the return path be
d2. A localized optical inhomogeneity in the d2 segment makes d2
different from d1. Assume that there is no "screening" so that the
surface shifts are unaffected by the inhomogeneity. The externally
reflected phasor (TM wave) from the front surface is 0.2, the internally
reflected phasor from the back face is (-0.2)(0.98) exp[ik(d1 + d2)] = -
0.196 exp[ik(d1 + d2)]. Add these to reflection phasors to get the total
reflection amplitude r(1) ("1" signifies to first order approximation in
the multiple reflection perturbation series):

r(1) = {0.2 - 0.196exp[ik(d1 + d2)]}

= exp[ik(d1 + d2)/2]{0.2 exp[-ik(d1 + d2)/2] - 0.196 exp[ik(d1 + d2)/2]}

= exp[ik(d1 + d2)/2]{-i0.392 sin[k(d1 + d2)/2]

+ 0.004 exp[-ik(d1 + d2)/2]} (1a)

Similarly for the first order transmission amplitude t(1):

t(1) = (0.98)(0.98)exp[ikd1]

+ (0.98)exp[ikd1](0.2)exp[ikd2](0.2)exp[ikd1](0.98)

= 0.96exp[ikd1] + 0.038exp[ik(2d1 + d2)]

= 0.96exp[ikd1]{1 + 0.04exp[ik(d1 + d2)]}   (2a)

t(1)*r(1) = -i(0.96)(0.392)exp[ik(d2 - d1)/2]sin[k(d1 + d2)/2]

{1 + i(0.004/0.392)exp[-ik(d1 + d2)/2]}{1 + 0.04exp[ik(d1 + d2)]}

= -i0.376exp[ik(d2 - d1)/2]sin[k(d1 + d2)/2]

{1 + i.01exp[-ik(d1 + d2)/2]}{1 + 0.04exp[ik(d1 + d2)]}   (3)

We can see that the leading term has the phasor

iexp[ik(d2 - d1)/2] = exp[ik(d2 - d1)/2 + ipi/2]     (4)

If the calculation is done exactly then there will be some correction to
this phase delta. That is in general

b = k(d2 - d1)/2 + pi/2 + delta     (5)

There is no reason to think that this b will always add up to pi/2 if
the surface reflection shifts are fixed at 0 or pi as this crude model
assumes.

The only way to get b = pi/2 for all different values of reflectance
magnitude is for some kind of a screening effect to happen. If the
screening effect does not happen then there is a violation of
conservation of energy unless the zero point fluctuations contribute
either as a source or a sink.

Remember t*r + tr* = 2Re{t*r} is excess energy either positive or
negative compared to the total input energy t*t + r*r.

The one other special option is the "resonance" factor, for example,
sin[k(d1 + d2)/2] in equation (3). At a plate thickness resonance tuned
to the optical doping shift theta = k(d2 - d1) we have k(d1 + d2)/2 =
pi. Therefore, t*r + tr* = 0 even when b is not equal to pi/2. That is,
at these special resonances that can be engineered in the plate, the
phase screening to b = pi/2 need not occur and energy can still be
conserved without coupling to the zero point fluctuations. And in this
resonant case, the quantum mechanical equations predict faster than
light communication.

Since anomalous production and loss of light energy is not normally
noticed in everyday optical beam recombiners, the screening must be
happening. This would mean that b = pi/2 in ordinary situations which
are off-resonance. The question is, is it possible to trick nature into
letting b not equal pi/2 at a resonance in presence of optical doping?
Perhaps we can get a specially doped glass that can bifurcate or make a
phase transition that supresses the phase screening thereby allowing
superluminal and backwards in time communications without transferring
energy out of or into the vacuum.

Such resonant quantum connection communication would not perturb the
vacuum. Off-resonant quantum connection communication, if it were
possible, would perturb the vacuum. Indeed it would suggest a new kind
of advanced propulsion system.

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