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The size of the problem (reply to Steve's 8/19 letter)

Hi Steve,

It took quite some time, to respond, but here it is. In the beginning I had
some trouble getting the same formula you did. Somehow I got to a formula
that gave the same answers, while I was not able to rewrite it so that it
looked like yours.
In the meantime I have seen how you and Rex got to the answer, so that is
clear now.

However I am not certain about your final conclusion. While the calculations
are right, the interpretation may be wrong.

>Note that this is an asymptotic relationship -- as the frame time t of
>the beamer approaches 1/a, the object proper time t1 approaches
>infinity.  This consequently means that the beamer must send energy for
>any possible trip within a time 1/a, no matter how far the acclerated
>object goes, and that the rate at which power is sent increases
>asymptotically to infinity as t approaches 1/a.  The relative rate of
>time passage between the beamer and the accelerated object at frame time
>t has the relationship
>dt1 / dt = 1/(1 - a*t)
>In the case where a = 9.8 m/s^2 (or in geometrized units, 3.267e-8 c/s),
>the asymptote is reached within about year of beamer time (3.06e8 s).

Should have been 3.06e7

>The good news is that to boost an object at 1 g up to its turnaround
>point and then provide deceleration power to its destination, you beam
>power for no more than two years, no matter how far away you send the

Why two years? Once the ship is at turnaround, the beaming station is not
limited by the above calculated time of 3.06e7 seconds.

>The bad news is that at the turnaround point you are beaming
>some large multiple of the power needed to keep the object accelerating
>at 1 g at the beginning and end of the trip, because of the relative
>rate of time lapse between the beamer and the accelerated object.
>In fact, given the relationship between t1 and t, we can characterize
>just what this multiple is based on halfway trip time of the object.
>Solving t1 = -(ln(1 - a * t))/a for t, we get:
>t = (1 - e^(-a*t1))/a
>Substituting into 1/(1 - a*t), we get:
>dt1 / dt = e^(a*t1)
>In other words, the maximum power output at turnaround is exponentially
>related to trip time for the object.

I assume you mean not the trip time, but the acceleration time limit.

Anyway I think your conclusion is not right. While in Earth's frame the
signal-receive-interval increases exponentially, to the ship the interval
stays the same, since its time slows down exponentially:

t = 1/a * sinh(a * t1) = 1/a (Exp[a t1]-Exp[-a t1])/2
When t1 becomes bigger, the term Exp[-a t1] is goes to 1 and is neglectable.
So then t is approximately equal to Exp[a t1]/2a

So the people in the ship would not notice any difference if Earth keeps
sending in regular intervals. Well of course they would notice the
red-shift, but I believe that's all.