# 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
>object.

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.

Timothy

```