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



> 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.

But now adding the Lorentz contraction lowers this to 3.75 Hz according to
the relativistic Doppler effect. I'm not sure to what frame I apply the
contraction here to get that result rather than twice John's answer.
Someone want to explain?

There is no Lorentz contraction to add.

In that one second (according to the observer) the first pulse moves
1 light second toward the observer (according to the observer) and
the source moves (3/4)^(1/2) of a light second toward the observer
(according to the observer). As a result, the wavelength (according
to the observer) is [1 - (3/4)^(1/2)] light seconds.

--
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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