Re: C-ship

[Steve VanDevender <stevev@efn.org>]

Steve VanDevender writes:
 > Things are substantially more complicated when dealing with
 > accelerated worldlines.  I've got a preliminary solution stated
 > in similar terms as the above discussion; perhaps you'd like to
 > check my math :-).
 > 
 > You are at the point S = [ t x y z ] attempting to view an object
 > whose coordinates are P(t') =
 > 
 > 1/a^2 * [ a * sinh(a * t')
 >           ax * (cosh(a * t') - 1)
 > 	  ay * (cosh(a * t') - 1)
 > 	  az * (cosh(a * t') - 1) ]
 > 
 > where the acceleration is represented by A = [ 0 ax ay az ] (in
 > the object local frame) and a = sqrt(ax^2 + ay^2 + az^2) (the
 > magnitude of the acceleration).
 > 
 > So again, we want to solve the equation (S - P(t'))^2 = 0.  Of
 > course, the components of P(t') are much more complicated.  I
 > won't bore you with the full derivation, other than to note that
 > it becomes easier to isolate t' by writing the sinh and cosh
 > terms in terms of their definitions using exp (e^x).
 > 
 > Eventually, you get:
 > 
 > exp(a * t')^2 * (1 - A|S - a * t') +
 > exp(a * t') *   (a^2 * S^2 - 2 * (1 - A|S)) +
 >                 (1 - A|S + a * t')
 > = 0

Sigh.  That should actually be:

exp(a * t')^2 * (1 - A|S - a * t) +
exp(a * t') *   (a^2 * S^2 - 2 * (1 - A|S)) +
                (1 - A|S + a * t)
= 0

I wrote t' rather than the t component of S in a couple wrong
places.

 > It's convenient to make some substitutions for common
 > subexpressions:
 > 
 > k = 1 - A|S
 > p = k - a * t'
 > q = a^2 * S^2 - 2 * k
 > r = k + a * t'

And these should be

p = k - a * t
q = a^2 * S^2 - 2 * k
r = k + a * t

 > So then applying the quadratic formula and isolating t' gives:
 > 
 > t' = 1/a * ln((-q - sqrt(q^2 - 4 * p * r)) / (2 * p))

Fortunately I still copied that correctly.