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*To*: starship-design@lists.uoregon.edu*Subject*: starship-design: the beamed power problem, again*From*: Steve VanDevender <stevev@efn.org>*Date*: Wed, 4 Sep 1996 10:52:28 -0700*Reply-To*: Steve VanDevender <stevev@efn.org>*Sender*: owner-starship-design

Rex (DotarSojat@aol.com) and I have been having a private discussion on my assertion that if you are beaming power to a starship to give it a fixed ship-frame acceleration, you have only a finite amount of time during which you can send power to the ship, no matter how long the ship's trip time is. For example, if you are beaming power to accelerate a ship at 1 g, you have just short of a year after its departure to beam power to it, no matter how far away you are sending it, and you must beam increasing amounts of power towards the end of that time to sustain the 1 g ship-frame acceleration. I presented Rex with the derivation that led me to that conclusion. Unfortunately, my derivation was based on the solution to a substantially more complicated problem (where does an observer at an arbitrary spacetime location see an accelerated object travelling in an arbitrary direction?), which we agreed is not particularly accessible to the uninitiated. Rex came up with a simpler derivation specific to this problem that is much easier to understand, which I will quote below. DotarSojat@aol.com writes: > Actually, the description of the > space-time diagram in your 8/26 email was more effective in con- > veying the relation to my intuition. Having gone through that > thought process, I can now rewrite the derivation more succinctly: > > The distance, x, of the sail after accelerating at a level, a, for > a proper time, t', is > x = [cosh(a * t') - 1]/a . > > The apparent (Earth) time, t, at which the sail reaches x is > t = [sinh(a * t')]/a . > > The time, delta-t, that it takes light from Earth to reach the > sail at x is > delta-t = t - te > = x/c > or = x for c = 1 lt-yr/yr, > > where te is the Earth time at which the light was emitted. > > Solving for te gives > te = t - x > = [1 + sinh(a * t') - cosh(a * t')]/a . > > Replacing sinh and cosh with their exponential forms gives > te = [1 - exp(-a * t')]/a . > > And solving for t' gives > t' = -[ln(1 - a * te)]/a . Q.E.D. Note that as the quantity (a * te) approaches 1, the quantity -ln(1 - a * te) approaches infinity; in other words, as the Earth time te approaches 1/a, the ship time t' approaches infinity. If constant acceleration of the ship is maintained, communication from Earth to the ship after Earth time 1/a is impossible. (For solution in conventional units, replace a with aconv/c, where aconv is acceleration in conventional units.) My description of the spacetime diagram that inspired this, for those who are interested, was: Steve VanDevender writes: > The parametric equation [ t, x ] = 1/a * [ sinh(a * t'), (cosh(a * t')-1) ] > describes a hyperbola with asymptotes t = x + 1/a and t = x - 1/a > approached as t' goes to infinity. So draw this hyperbola and its > asymptotes on paper, putting t on the y-axis and x on the x-axis; the > hyperbola represents the worldine of the receiver. Also draw a heavy > line up the y-axis representing the worldline of the emitter. Now you > can draw lines with slope 1 (or parallel to the upper asymptote) between > the worldline of the emitter and the worldline of the receiver > representing light rays sent from the emitter in the direction of of the > receiver. Note that because of this asymptote rays that leave the > emitter after time 1/a can never reach the receiver (they all travel > above the asymptote), and that as the emitter time approaches 1/a > photons emitted quite close together in emitter proper time are received > with a much larger difference in receiver proper time.

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