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, the length of the interferometer. He can
calculate the wavelength as L/n so
frequency Fo = cn/L ---------------- [1]
He signals the value n to B
Now in B's FoR length L becomes L' = L x sqr(1-vv/cc)
He calculates the wavelength as L'/n

Then 'he' doesn't know how to use the LT.
Wavelength doesn't transform like the length of a rod!

The length of the interferometer *does* transform 'like the length of a
rod'.

Indeed.
See my other posting.

--
Paul

http://home.c2i.net/pb_andersen/

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.

--
John Kennaugh

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/

Paul B. Andersen wrote:
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.


OK I understand the concept. The maths work I think. Of course as a
relativist you won't worry about the implication that the same light
inside the same apparatus has different wavelengths depending on who is
looking at it.

S **************** X ************** Y

Suppose light hits detectors X and Y such that 'light' (radio
frequency?) is converted into electrical signals, passed via equal
lengths of co-axial cable to a mixer. The distance between X and Y can
then be adjusted to get a null. What does relativity say about how the
choice of FoR affects the motion of an electrical signal in co-axial
cable?

--
John Kennaugh

John Kennaugh wrote:
OK I understand the concept. The maths work I think. Of course as a
relativist you won't worry about the implication that the same light
inside the same apparatus has different wavelengths depending on who is
looking at it.

OF COURSE there's no problem in that -- since simultaneity is different
for different frames, it immediately follows that the wavelength of a
given light beam is not the same when measured in different frames, with
the measurement made SIMULTANEOUSLY in each frame.


S **************** X ************** Y

Suppose light hits detectors X and Y such that 'light' (radio
frequency?) is converted into electrical signals, passed via equal
lengths of co-axial cable to a mixer. The distance between X and Y can
then be adjusted to get a null. What does relativity say about how the
choice of FoR affects the motion of an electrical signal in co-axial cable?

The thing you keep forgetting, over and over, is to discuss
MEASUREMENTS. Regardless of how many wavelengths a given observer counts
(simultaneously in her frame) between X and Y, and regardless of how the
observer observes the signal to behave in the cable, the MEASUREMENT
will be independent of frame -- here that is the null in the mixer
(after adjustment).

To compute the speed of a signal in a moving cable, note that in the
cable's rest frame the signal has speed c/n, where n is the effective
"index of refraction" of the cable (n 1). In a frame relative to which
the cable is moving, use the Lorentz composition of velocities. Note you
will also get different values for the wavelength and frequency of the
signal....

Just because you PERSONALLY do not understand SR does not mean it is
wrong, or inconsistent. Ditto for Waldron. The real measure in science
is how well the theory predicts the results of experiments, and by that
measure SR is unrefuted within its domain of applicability.


Tom Roberts

Tom Roberts wrote:
John Kennaugh wrote:
OK I understand the concept. The maths work I think. Of course as a
relativist you won't worry about the implication that the same light
inside the same apparatus has different wavelengths depending on who
is looking at it.

OF COURSE there's no problem in that -- since simultaneity is different
for different frames, it immediately follows that the wavelength of a
given light beam is not the same when measured in different frames,
with the measurement made SIMULTANEOUSLY in each frame.


S **************** X ************** Y
Suppose light hits detectors X and Y such that 'light' (radio
frequency?) is converted into electrical signals, passed via equal
lengths of co-axial cable to a mixer. The distance between X and Y can
then be adjusted to get a null. What does relativity say about how the
choice of FoR affects the motion of an electrical signal in co-axial cable?

The thing you keep forgetting, over and over, is to discuss
MEASUREMENTS. Regardless of how many wavelengths a given observer
counts (simultaneously in her frame) between X and Y, and regardless of
how the observer observes the signal to behave in the cable, the
MEASUREMENT will be independent of frame -- here that is the null in
the mixer (after adjustment).

To compute the speed of a signal in a moving cable, note that in the
cable's rest frame the signal has speed c/n, where n is the effective
"index of refraction" of the cable (n 1). In a frame relative to
which the cable is moving, use the Lorentz composition of velocities.
Note you will also get different values for the wavelength and
frequency of the signal....

Just because you PERSONALLY do not understand SR does not mean it is
wrong, or inconsistent. Ditto for Waldron. The real measure in science
is how well the theory predicts the results of experiments, and by that
measure SR is unrefuted within its domain of applicability.


So was the geocentric theory of the solar system.


Tom Roberts

--
John Kennaugh

John Kennaugh wrote:
Tom Roberts wrote:
Just because you PERSONALLY do not understand SR does not mean it is
wrong, or inconsistent. Ditto for Waldron. The real measure in science
is how well the theory predicts the results of experiments, and by
that measure SR is unrefuted within its domain of applicability.


So was the geocentric theory of the solar system.

That is not a useful comparison at all. In particular, the geocentric
theory of the solar system WORKS -- it's much more complicated than the
heliocentric model, but with the correct formulas it accommodates all
observations of the solar system.

Within SR's domain, no other theory works, unless it happens to be one
of the theories that are experimentally indistinguishable from SR.

This is science, and a theory "works" iff it agrees with all of the
experiments and observations within the relevant domain.


Tom Roberts