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*To*: T.L.G.vanderLinden@student.utwente.nl*Subject*: Doppler effect*From*: DotarSojat@aol.com*Date*: Tue, 1 Oct 1996 03:13:25 -0400*cc*: stevev@efn.org

Hi Timothy On 9/30, you wrote-- >Steve's results are not related to the red-shift, they are based >on the signal travel time, which increases exponentially with >the trip-time. > >Am I right to say that during constant acceleration not only the >energy of the photons, but also the amount of photons decreases? >Steve's calculations pointed to the amount (density) of photons, >while yours seem to point only to their energy. I admit I wrote on 8/26-- >The energy of each photon is Doppler-shifted downward... The >arrival rate of the photons ... is also reduced by the Doppler >factor... I preceded the derivation that includes the above quote with the remark "I'm not totally comfortable with it...," in which I was referring to (without explicitly so stating) invoking the Doppler shift twice to cover two effects. In response to my 8/26 note, Steve wrote (also on 8/26)-- >...one can look at the relationship of received power to emitted >power just as well in terms of Doppler shifting as by using the >relationship between emitter time and receiver time (and in a >fundamental sense, they are the same thing expressed in differ- >ent terms; Doppler shifting is merely the result of different >rates of elapsed proper time between an emitter and a >receiver). The last part, "Doppler shifting is merely...," bothered me. I thought I had read something different. In going back to consult V.A. Ugarov, "Special Theory of Relativity," Mir, 1979, I found on page 86-- "...the Doppler effect is formed of two independent parts: (1) it is connected with the continuously changing distance between the observer and the source; (2) it is also connected with the transformation of time intervals between events on transition from one reference frame to another." (He went on to note that the second effect can be detected when there is no radial motion; he called it "the transverse Doppler effect.") I am willing to admit that my derivation of 8/26 was incorrect. I won't apologize too profusely for it because I qualified it by saying "...but it may be usable as a starting point for discussion," which it indeed has turned out to be. So much for background. Let me now address your quote of me-- >On 9/11 you wrote: >> >>... >> Pr = Pe * exp(-theta) >>... >> >>(I believe this relation, with theta = a * t' for constant a, >>is the source of the logarithmic time dependence introduced >>by Steve in his email of 8/20 to the Group.) (I had been hoping Steve would respond to this before anyone else would question it, so I wouldn't have to labor through the rigorous math.) As usual, if one thinks long and hard enough, one can come up with a "simple" derivation. So, here goes-- In the previous notation, the received power is Pr (constant for constant acceleration, a). If the sail/ship is receding with a velocity parameter, theta (which is a * t', where t' is the elapsed ship time from start), then the emitted power, Pe, must be greater than the received power by the Doppler factor-- Pe = Pr * exp(theta) . If there is no spillage, the total energy received, Er, must equal the total energy emitted, Ee, or-- Er = Ee = E . The differential reception of energy, dE, by the sail in differ- ential sail time, dt', is Pr, or-- dE/dt' = Pr , and the differential emission of energy, dE, from the source in differential Earth time, dt, is Pe, or-- dE/dt = Pe = Pr * exp(theta) . Dividing the second differential relation by the first gives-- dt'/dt = exp(theta) = exp(a * t') . Grouping terms, we get-- dt = exp(-a * t') * dt' . Integrating from t = 0 at t' = 0 to t and t', we get-- t = -(1/a) * [exp(-a * t') - 1] , or a * t = 1 - exp(-a * t') . Solving for t' gives-- t' = -[ln(1 - a * t)]/a . Q.E.D. So, we can get Steve's result solely from consideration of the Doppler shift. Regards, Rex

**Follow-Ups**:**Doppler effect***From:*Steve VanDevender <stevev@efn.org>

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