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Re: The dreaded derivation...
Thanks for the derivation in your email of 8/28. I'll save it as
a reference to help understand Lorentz dot-product manipulations
if I ever run across them again. Actually, the description of the
space-time diagram in your 8/26 email was more effective in con-
veying the relation to my intuition. Having gone through that
thought process, I can now rewrite the derivation more succinctly:
The distance, x, of the sail after accelerating at a level, a, for
a proper time, t', is
x = [cosh(a * t') - 1]/a .
The apparent (Earth) time, t, at which the sail reaches x is
t = [sinh(a * t')]/a .
The time, delta-t, that it takes light from Earth to reach the
sail at x is
delta-t = t - te
or = x for c = 1 lt-yr/yr,
where te is the Earth time at which the light was emitted.
Solving for te gives
te = t - x
= [1 + sinh(a * t') - cosh(a * t')]/a .
Replacing sinh and cosh with their exponential forms gives
te = [1 - exp(-a * t')]/a .
And solving for t' gives
t' = -[ln(1 - a * te)]/a . Q.E.D.
I'll be looking forward to the way you'll treat the "dreaded
derivation" for the SD Group. I'd feel honored if you used any
part of the above approach. You started a very interesting
learning process with your email of 8/20 to the Group.
"If mankind is ever to achieve the ultimate goal of traveling to
the stars, we will need an engineering understanding of relativ-
ity" (RGF, IDA Paper P-2361, Oct 1990).