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Re: Doppler effect



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