[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

*To*: stevev@efn.org, DotarSojat@aol.com*Subject*: The size of the problem (reply to Steve's 8/19 letter)*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Thu, 12 Sep 1996 13:27:25 +0100

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

- Prev by Date:
**Continuation of my 9/5 letter** - Next by Date:
**Mistake** - Prev by thread:
**Continuation of my 9/5 letter** - Next by thread:
**Mistake** - Index(es):