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The dreaded derivation, Re: your starship-design email of 8/20

Now that I have a little time to kill, let me derive in more concrete
mathematical terms the equations that show the relationship between
proper time of emission vs. proper time of reception for a system
consisting of an inertial transmitter and a uniformly-accelerating

First, a few preliminaries.  Let the dot product operator 'dot' be
defined by:

[ u0 u1 u2 u3 ] dot [ v0 v1 v2 v3 ] = u0*v0 - u1*v1 - u2*v2 - u3*v3

Then the spacetime interval for a displacement U = [ u0 u1 u2 u3 ]
is sqrt(U dot U), if U describes a timelike or lightlike interval (U dot
U >= 0).

We can describe the worldline of an object undergoing uniform
acceleration with the parametric equation in t':

S(t') = 1/a^2 * [ a * sinh(a * t')
                  ax * cosh(a * t')
		  ay * cosh(a * t')
		  az * cosh(a * t') ]

where A = [ 0 ax ay az ] describes the acceleration, and a = abs(A) =
sqrt(-(A dot A)).

Note that at time t' = 0, S(t') = 1/a^2 * [ 0 ax ay az ] and the object
is instantaneously at rest in the frame of description.

We have an observer P whose spacetime location is [ t x y z ].  We want
to find possible paths that correspond to light rays passing from S(t')
to P, or from P to S(t').  We can characterize this relationship with
the equation:

(P - S(t'))^2 = 0

That is, values of S(t') for which the displacement between P and S(t')
has the spacetime interval 0, or a lightlike interval.  If we solve for
t', we can get the frame locations of the accelerated object relative to
P for which light can pass between P and S(t').

Expanding, we get:

   (t - (a/a^2) * sinh(a * t'))^2
 - (x - (ax/a^2) * cosh(a * t'))^2
 - (y - (ay/a^2) * cosh(a * t'))^2
 - (z - (az/a^2) * cosh(a * t'))^2
 = 0

The math to solve this, is, as you might guess, exceptionally tedious.
If you're really interested, I can expand on the steps to the solution
in detail, but since you both seem to be competent algebraists, I will
only point out the particularly helpful steps to the solution that might
not be immediately obvious in the middle of all the rote work.

Once you expand the above, you can collect these terms together:

(a^2/a^4) * sinh^2(a * t') - ((ax^2 + ay^2 + az^2)/a^4) * cosh^2(a * t')

Note that a^2 = (ax^2 + ay^2 + az^2), and that sinh^2(u) -
cosh^2(u) = -1, which means those terms simplify to merely:


At this point, you have an equation with several terms in sinh(a * t')
and cosh(a * t').  It is then useful to do a couple of things.  Multiply
through by 1/a^2 to get rid of those denominators (this is OK because
you should still have an equation of the form {blah} = 0).  Then rewrite
the sinh and cosh terms in terms of exp:

sinh(a * t') = 1/2 * (exp(a * t') - exp(-a * t'))
cosh(a * t') = 1/2 * (exp(a * t') + exp(-a * t'))

Once you've done this, expand the terms and multiply through by exp(a *
t') to get all the terms as multiples of exp(2 * a * t'), multiples of
exp(a * t'), or constants.  If you did the same things I did, then at
this point you should have:

   exp(2 * a * t') * (ax * x + ay * y + az * z - a * t)
 + exp(a * t') * (a^2 * (t^2 - x^2 - y^2 - z^2) - 1)
 + (ax * x + ay * y + az * z + a * t)

This can be written in terms of the original vector quantities A and P

   exp(2 * a * t') * (-(A dot P) - a * t)
 + exp(a * t') * (-(A dot A) * (P dot P) - 1)
 + (-(A dot P) + a * t)

This is a quadratic equation in terms of exp(a * t'), so a solution for
t' can be obtained through the quadratic formula.


f = -(A dot P) - a * t
g = -(A dot A) * (P dot P) - 1
h = -(A dot P) + a * t


exp(a * t') = (-g +/- sqrt(g^2 - 4 * f * h)) / (2 * f)

t' = 1/a * ln((-g +/- sqrt(g^2 - 4 * f * h)) / (2 * f))

I still probably need to do a bunch of work on rearranging this into a
simplified form that would also be numerically accurate in the extreme
cases, but I've found it adequate for my purposes so far.

The reason that I want this general of a solution is that I want to be
able to model accelerating objects in a general way in my planned
spaceflight simulator; I want to be able to accurately calculate the
apparent visual location and appearance of an accelerated object, or
calculate the appearance of other objects in the universe as seen by the
accelerated object.

There are a couple of interesting properties of this solution.  You
should note that there are regions in which only one solution produces a
real value for t', and regions in which there is no real solution for
t'; in other words, there are places where an observer cannot see the
accelerated object, places where the accelerated object cannot see the
observer, and even places where neither can see each other.  For a
diagram illustrating these properties, see chapter 6 of _Gravitation_.

Now, if you got through all that, we can obtain substantially simpler
solutions for the case of interest, a "stationary" emitter that is
sending light signals to a uniformly-accelerating receiver that was
initially at rest and coincident with the emitter.  First, we can reduce
the problem into only one space dimension to reduce the complexity of
the math.  Then:

A = [ 0 a ]

The frame location P of the emitter at the emitter's proper time t is:

P = [ t 1/a ]

The frame location of the accelerated object at its proper time t' is

1/a * [ sinh(a * t') cosh(a * t') ]

We can use the two-dimensional version of the Lorentz dot product to
calculate the terms f, g, and h from t, a, A and P to obtain a
much-simplified solution for t' in terms of t, or the proper time of
reception for the acclerating object in terms of the proper time of

t' = -ln(1 - a * t) / a

Note what this solution implies.  Unlike many relativity problems, we
are not concerned about times or positions purely in terms of abstract
frame coordinates -- these describe what the observers _see_, not what
they measure in their reference frames.

More importantly, the solution for time of reception vs. time of
emission is asymptotic -- as t approaches 1/a, t' approaches infinity.
This is the purely mathematical statement of what I was trying to
describe more intuitively before: in order to maintain constant
acceleration, the amount of power beamed to the starship rises rapidly
as the time approches 1/a in the emitter's frame, and that no matter how
far out the starship goes, power is never beamed for longer than 1/a in
the emitter's frame, because power beamed after that time would never
reach the receiver.  (In practical terms, of course, indefinitely
sustained continuous acceleration is impossible, so this statement isn't
quite as dire as it sounds).

The other solution (because of the +/- in the general solution)
describes the proper time of emission of a light signal from the
accelerating object in terms of the time of reception for the
"stationary" observer:

t' = ln(1 + a * t) / a

This other, seemingly paradoxical result, is that while the starship
sees the emitter slowly approach the time 1/a as the starship's proper
time goes to infinity, the emitter can see signals beamed back from the
starship forever, and the signals from the starship come from starship
proper times that don't approach some finite limit.