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[Phys-L] Re: relativistic doppler effect (was: The view from the Enterprise.)



Which makes clear the error in Rick's original analysis: changing reference frames *twice*. Adjusting the frequency already puts you in the observer's frame of reference. *Alternatively*, you could have calculated the wavelength in the frame of reference of the source, and then used length contraction to shift to the observer's frame. But you can't do both.

-----Original Message-----
From: Forum for Physics Educators [mailto:PHYS-L@list1.ucc.nau.edu]On
Behalf Of John Mallinckrodt
Sent: Wednesday, January 18, 2006 10:58 AM
To: PHYS-L@LISTS.NAU.EDU
Subject: Re: [PHYS-L] relativistic doppler effect (was: The view from
the Enterprise.)


Rick writes:

Which pretty much brings me back to my initial question. As an observer in
a ship moving at 87% the speed of light towards a star (but viewing the star
as moving towards me) and knowing special relativity, shouldn't I expect the
period of atomic emitters to have increased by a factor of two, therefore
the frequency to have decreased by a factor of two, but then the wavelengths
contracted by a factor of 2---washing out and leaving only a classical
Doppler shift? I know this is too simplistic, but where is the mistake?

Let's take this specific example and just run some numbers. Suppose
the rest frequency is 2 Hz so that, at v = (3/4)^1/2 c it becomes 1
Hz. The source emits a pulse and one second later (according to the
observer "at rest") it emits another pulse. In that one second the
source moves (3/4)^(1/2) of a light second while the first pulse
moves a full light second. The pulses are, thus, 1 - (3/4)^(1/2) of
a light second apart. This is the wavelength and it corresponds to
an observed frequency of

f_obs = [1 - (3/4)^(1/2)]^-1 Hz = 7.46 Hz

in agreement with the general formula.

--
John "Slo" Mallinckrodt

Professor of Physics, Cal Poly Pomona
<http://www.csupomona.edu/~ajm>

and

Lead Guitarist, Out-Laws of Physics
<http://www.csupomona.edu/~hsleff/OoPs.html>
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