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Re: Doppler effect
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
>>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
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
>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
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
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