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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)
I wrote t' rather than the t component of S in a couple wrong
> It's convenient to make some substitutions for common
> 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.
- Re: C-ship
- From: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)
- Re: C-ship
- From: Steve VanDevender <email@example.com>