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*To*: DotarSojat@aol.com, stevev@efn.org*Subject*: Re: Doppler effect*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Sun, 06 Oct 1996 19:01:43 +0100

Hello Rex, On 10/1, 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... So we both say the same; The energy decrease is caused both by the red-shift and the photon density decrease. >>...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.") Yes, I heard of that name before. >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. Oh, you don't need to apologize, I only wonder why at 8/26 you mentioned and used both parts of the Doppler effect while at 9/11 (to SD) you used only one of both parts. Also I'm a bit confused now, about what the reduction of both parts of the Doppler effect are. If I understand correctly they are both the same and equal to: Pr = Pe * exp(-theta) or Pr = Pe * sqrt[(1 - beta)/(1 + beta)] So why do you admit that your 8/26 derivation was incorrect? (Did you mix up the date with the 9/11 letter?) I would think that the squaring of exp(-theta) in the 8/26 letter was correct. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - <Begin of your derivation removed> >Dividing the second differential relation by the first gives-- > dt'/dt = exp(theta) > = exp(a * t') . <End of your derivation removed> I can see that theta=a*t' gives the same result as Steve's. But what I really ment, was how you could prove that definition of theta=at'? By now I found the "proof" I was looking for myself: beta=Tanh[theta] -> v=Tanh[theta] v=Tanh[at'] with respect to x=Cosh[at']/a and t=Sinh[at']/a so v=Tanh[at']=Tanh[theta] -> theta=at' A confused Timothy

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