John Kennaugh skrev:
Paul B. Andersen wrote:
John Kennaugh skrev:
This is my version of a paradox described by Waldron.
S |---L--| A B--v
Source S. Observer A stationary w.r.t S Observer B moving at v away
from S. A has an interferometer and shows that exactly n wavelengths
occupy a distance L
------------------------------------------------------------------------
OK so let us devise another experiment. Before Setting out B,
together with A, sets up the interferometer to have exactly n
wavelengths in length L. Let us assume that the output is by way of
a dark fringe and this is fed into a light detector and comparator.
The threshold of the comparator is set such that a small increase in
brightness will give an output and the output is fed to a detonator
attached to a pile of explosives.
Sorry I have not been ignoring your post. All my spare time has been
used putting a new roof on my garage and I have only just got back to
less important matters.
A 'one way interferometer'? There is no such thing.
I never suggested there was.
An interferometer is when to rays interfere.
I was aware of that.
So there must be two rays going along separate paths.
Let's make a simple one, as equal to yours as possible.
***** |--------|
-*-*-*|--------|
| |
half sivered mirror
mirror
HSM M
***** incident ray, half reflected, half transmitted
----- transmitted ray, reflected at right mirror
-*-*- interference ray
OK I'm with you so far.
We can determine the phase difference between the reflected ray
and the incident ray at the half silvered mirror by counting
the _instant_ number of wavelengths in the ray.
instant?
See:
http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf
http://home.c2i.net/pb_andersen/pdf/...ror_sagnac.pdf
http://home.c2i.net/pb_andersen/pdf/...optic_gyro.pdf
This can be done in any frame of reference, so let's choose B's rest
frame.
That is the frame of the observer moving away from both the source and
the interferometer at v.
The interferometer is moving at the speed v to the left in this frame.
All primed entities are referred to B's rest frame.
N' = Nf' + Nb'
N' number of wavelengths in the ray from HSM-M-HSM
Nf' number of wavelengths in the ray from HSM-M
Nb' number of wavelengths in the ray from M-HSM
I'm still with you.
Nf' = L'/l_f'
Nb' = L'/l_b'
L' length of interferometer
l_f' wavelength of right going ray
l_b' wavelength of left going ray
L' = L.sqrt(1-v^2/c^2)
l_f' = l.sqrt((1+v/c)/(1-v/c))
l_b' = l.sqrt((1-v/c)/(1+v/c))
L proper length of interferometer
l wavelength of ray in interferometer-frame
Nf' = L.sqrt(1-v^2/c^2)/(l.sqrt((1+v/c)/(1-v/c))) = (L/l)(1-v/c)
Nb' = L.sqrt(1-v^2/c^2)/(l.sqrt((1-v/c)/(1+v/c))) = (L/l)(1+v/c)
N' = (L/l)(1-v/c)+(L/l)(1+v/c) = 2L/l
So the phase difference at the half silvered mirror doesn't depend
on the speed of the interferometer in the (arbitrary) chosen frame.
This will be the case for all interferometers. Obviously.
The phase difference between the HSM and M will however
depend on the speed. But this phase difference depend
on the definition of simultaneity at the two mirrors,
and it has no physical significance.
(Different observers will have different ideas of what is simultaneous
at the mirrors, and thus of what the phase difference is.)
There is no way an interferometer can compare these phases.
I see something of a problem. Transmission through a half silvered
mirror changes neither the speed nor the wavelength of the light
therefore l_f' is the wavelength of the both the forward going ray and
the incident ray. l_b' is the wavelength of the left going ray and your
maths requires that l_b and l_f differ.
In B's FoR both travel at c therefore the incident ray and the left
going ray have different frequencies. The interference ray consists of
a mixture of two different frequencies, and two different frequencies do
not result in an interference pattern.
Read the description of the interferometer again, please.
The interference wave is between the light reflected from the left
half silvered mirror and the light reflected from the right mirror,
both going left. This wave would not be easy to observe, of obvious
reasons, so this interferometer isn't very practical.
But I think you get the principle.
--
Paul
http://home.c2i.net/pb_andersen/