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



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?

Rick

----- Original Message -----
From: "Roger Haar" <haar@PHYSICS.ARIZONA.EDU>


Hi,

Nicholas, I think you are somewhat mistaken. The
relativistic breakdown in simultaneity implies
that one cannot talk about "the distance between
two points at a single time, in two different
reference frames." It is even more complex, if
you look at the effect of Lorentz transforms on x
coordinates, the values become larger not shorter
with increased speed. Length contraction occurs
because the two observers disagree about when each
end of the rod is measured.

Relativistic Doppler shift is the combined effect
of several things. First an atom emitting light
is a clock (Some of our best clock are just
this.), so the frequency emitted changes from rest
frame to rest frame. There is a fundamental
relationship

the speed of light = wavelength * frequency.

Since the speed of light is constant, the
lengthing of the wavelength this is another way
derive length contraction.
There is also the part of the Doppler shift which
depends on the motion toward or away from the
source like the case of sound.

*******
In general it is at best dangerous to try to
separate what happens to time from what happens
to space at relativistic speeds. I think all of
the claimed contradictions of special relativity
are based on applying the relativity to either
only time or only space. Just be carefull.



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