Why the ether would be observable
z@z
"Why the Ether is Unobservable"
(http://www.deja.com/=dnc/getdoc.xp?AN=551470562):
| Lemma. In any theory of this class and for any assumed anisotropy
| in the 1-way speed of light, the time delay for a signal to make a
| round trip around any path of fixed shape and size is independent
| of the orientation of the path in an inertial frame.
| Two-Clock One-Way Experiments
| -----------------------------
|
| I consider experiments with two clocks fixed on a rotatable platform
| which look for variations in the 1-way propagation of a light signal
| between the clocks by comparing the phase of a local signal from one
| clock to the phase of a light signal from the distant clock. These
| (perfect) clocks emit a sinewave of fixed frequency. This discussion
| applies directly to the experiments of Krishner et al [27], Cialdea
| [24], Torr and Kolen [28], and similar experiments. It is simplest to
| build up to the final result via two gedanken experiments.
|
| Gedanken I
| ----------
| Consider the following physical setup:
|
| |----> X axis
| ----------- signal path in vacuum
| | clock A |>-----------------------------------------| mirror B
| ----------- |
| |
| v
| ----------------------
| | Phase Comparator 1 |
| ----------------------
|
| Here phase comparator 1 compares the phase of the local signal from
| clock A to the reflected vacuum signal.
|
| We will arbitrarily consider the positive peak of the signal's sinewave
| to be the "standard phase" (i.e. phase=0). We can draw the path of a
| particular phase=0 point of the signal in the X-T coordinates (T is
| elapsed time of clock A):
|
| T
| ^
| |
| |
| A3 |O
| | O
| | O
| | O
| | O
| | O
| | O
| | O
| | O
| | O
| | O
| | O
| | O |
| | o| B2
| | o |
| | o
| | o
| | o
| | o
| | o
| | o
| | o
| | o
| | o
| | o
| | o
| A1 |o
| |
| |--> X
| For variations in the one-way speed of light (due to orientation), B2
| will move up and down relative to A1 and A2, because the one-way speed
| is the slope of the two light rays in the plot. But the triangle remains
| a triangle because the mirror continues to reflect the light ray, and
| T(A3)-T(A1) remains constant as proven by the lemma.
That's SR reasoning which is incompatible with the assumption of
"variations in the one-way speed of light". The phase difference of
the signals between A and B does change even if we accept the above
lemma postulating a round-trip light-speed isotropy. The wave length
is different if the light speed is not the same in both directions,
so the number of phase=0 points between A1 and B2 cannot be the same
as between B2 and A3.
This fact totally undermines the following reasoning.
| Gedanken II
| -----------
| Add the following to the above physical setup:
|
| signal path in vacuum -----------
| (clock A) >-----------------------------------------<| clock B |
| | -----------
| |
| v
| ----------------------
| | Phase Comparator 2 |
| ----------------------
|
| Here phase comparator 2 compares the phase of the local signal from
| clock A to the incoming signal in vacuum from clock B. In one specific
| but arbitrary orientation we will initially adjust the output phase of
| clock B so that the two phase comparators (at clock A) give the same
| readout. In other words, we synchronize clock B to clock A using the
| outgoing signal from clock A. After this initial adjustment it remains
| fixed.
| For variations in the one-way speed of light (due to orientation), B2
| moves up and down as before. But the clock B output _must_ vary precisely
| as required to remain in phase with the incoming signal from clock A and
| the outgoing signal from the mirror. This is so because its output signal
| must exactly track the Gedanken I signal from B->A, and that signal
| exactly tracks the A->B signal at event B2. Equivalently, this is so
| because the A->B signal can be used to continuously synchronize clock B
| with clock A. Equivalently, in the X-T diagram the two events (reflection
| from mirror) and (clock B emission) cannot move apart: they cannot move in
| X because the distance is fixed, and they cannot move in T because the two
| clocks have the same frequency (the output of clock B is the same
| frequency as both the incoming signal from clock A and its outgoing
| reflection from the mirror).
|
| So for any theory in our class the phase difference in comparator 2 is
| constant, independent of the orientation of the X axis of the apparatus.
| This is clearly independent of the clocks' initial synchronization (or
| lack thereof). This can easily be extended to the case where the clock B
| -> clock A signal is carried in a cable or optical medium with refractive
| index > 1. And clearly this remains true even if clock A never emits a
| signal and both the mirror and comparator 1 are absent. Every theory of
| this class predicts a null result for all experiments of this type.
If this SR reasoning were true, the measured one-way light speed
would be constant without slow clock transport. Let us assume A moves
inertially wrt the ether and B rotates around A at a "fixed" distance.
The phase difference in comparator 2 would remain constant if the
rates of both clocks were identical. Slow clock transport however
entails that they are not identical.
That's exactly the reason, why I had thought previously that
something like slow clock transport cannot be possible (within
SR or equivalent theories).
see: http://www.deja.com/=dnc/getdoc.xp?AN=548977376
| This result depends directly upon the way slow clock transport affects
| the clocks.
Strange logic!
Wolfgang Gottfied G. (0:017.77)
Tom Roberts
> Tom Roberts wrote in the interesting and transparent article
> "Why the Ether is Unobservable"
> (http://www.deja.com/=dnc/getdoc.xp?AN=551470562):
> | Lemma. In any theory of this class and for any assumed anisotropy
> | in the 1-way speed of light, the time delay for a signal to make a
> | round trip around any path of fixed shape and size is independent
> | of the orientation of the path in an inertial frame.
While I'm not sure what you mean by "transparent", I take it you do
not disagree with the lemma.
> | Two-Clock One-Way Experiments
> | -----------------------------
>
> [drawing not repeated]
>
> | [...]
> | T(A3)-T(A1) remains constant as proven by the lemma.
>
> That's SR reasoning which is incompatible with the assumption of
> "variations in the one-way speed of light".
This is not "SR reasoning", this is simple math. See below.
> The phase difference of
> the signals between A and B does change even if we accept the above
> lemma postulating a round-trip light-speed isotropy.
There is no way to measure or compare the phases at physically-
separated locations. You need to bring the signals together, and
the lemma ensures that the phase comparison between the signals
_when_brought_together_ will remain constant (independent of
orientation). [Strictly speaking one might need the corollary to
the lemma.]
Note please that the gedanken I presented makes no attempt to compare
the phase of signals at A and at B, it compares the phase of the
signal A->B->A with the signal from the clock at A. This is a
localized comparison which is valid, and my argument shows that this
phase difference remains constant independent of the orientation of
the path A->B.
> The wave length
> is different if the light speed is not the same in both directions,
> so the number of phase=0 points between A1 and B2 cannot be the same
> as between B2 and A3.
Sure. So what? The lemma ensures that THE PHASE WHICH IS MEASURED
remains constant. The lemma also ensures that the total number of
wavelengths along the path A1->B2->A3 remains constant, independent
of orientation (because the time delay must be constant, and because
a change in orientation cannot change the clock frequency):
The condition of this class of theories that the ROUND-TRIP speed of
light be isotropically c implies that the one-way speed in the
direction A->B (call it c1) is directly related to the one-way speed
in the direction B->A (call it c2):
2/c = 1/c1 + 1/c2
Using the fact that the frequency of the wave (call it f) cannot
change while it propagates, the number of wavelengths A1->B2->A3
is (with L = distance A to B = distance B to A):
(# wavelengths A->B) + (# wavelengths B->A) =
fL/c1 + fL/c2 = 2fL/c
This is manifestly constant, independent of orientation.
> [rest deleted - depends upon the above mistake]
Tom Roberts tjro...@lucent.com
Stephen
> Tom Roberts wrote in the interesting and transparent article
> "Why the Ether is Unobservable"
> (http://www.deja.com/=dnc/getdoc.xp?AN=551470562):
>
> | Lemma. In any theory of this class and for any assumed anisotropy
> | in the 1-way speed of light, the time delay for a signal to make a
> | round trip around any path of fixed shape and size is independent
> | of the orientation of the path in an inertial frame.
> [snip]
> | For variations in the one-way speed of light (due to orientation), B2
> | will move up and down relative to A1 and A2, because the one-way speed
> | is the slope of the two light rays in the plot. But the triangle remains
> | a triangle because the mirror continues to reflect the light ray, and
> | T(A3)-T(A1) remains constant as proven by the lemma.
>
> That's SR reasoning which is incompatible with the assumption of
> "variations in the one-way speed of light". The phase difference of
> the signals between A and B does change even if we accept the above
> lemma postulating a round-trip light-speed isotropy. The wave length
> is different if the light speed is not the same in both directions,
> so the number of phase=0 points between A1 and B2 cannot be the same
> as between B2 and A3.
That would only be true if the distance between A1 and B2 were the same
as between B2 and A3.
--
Felix qui potuit rerum cognoscere causas - Virgil.
Tom Roberts
> "z@z" <z...@z.lol.li> wrote:
> > [about my article "Why the Ether is Unobservable"
> > (http://www.deja.com/=dnc/getdoc.xp?AN=551470562)]
> > The wave length
> > is different if the light speed is not the same in both directions,
> > so the number of phase=0 points between A1 and B2 cannot be the same
> > as between B2 and A3.
> as between B2 and A3.
The notation is from my above-mentioned original article, and A and B
are located at definite positions in the inertial frame; A1 and A3
are events occuring at the location of A; B2 is an event occuring after
A1 and before A3 at the location of B. So the distance between A1 and B2
is indeed the same as the distance between B2 and A3.
Note that all three events are connected by light rays, so the
time-ordering I mentiond is valid in all inertial frames, as
is the equality of distances.
Z@Z's claim is indeed true if the one-way speed of light is anisotropic.
But as I explained in another response directly to his article, this
does not invalidate my argument, and the ether is indeed unobservable.
Tom Roberts tjro...@lucent.com
z@z
:: = Wolfgang G. in http://www.deja.com/=dnc/getdoc.xp?AN=574329583
If I have understood correctly the part of your article we are dealing
with, then you postulate round-trip light-speed isotropy and derive
from this postulate that one-way experiments must yield null results.
For a one-way experiment two clocks A and B at different locations
are required. If a signal is emitted from A at A-time A1 and received
at B at B-time B2, then a null result implies that B2-A1 does not
change when the orientation of the path (wrt the ether frame) is
changed.
Quotes from your gedanken II:
"In other words, we synchronize clock B to clock A using the outgoing
signal from clock A. After this initial adjustment it remains fixed.
For variations in the one-way speed of light (due to orientation), B2
moves up and down as before. But the clock B output _must_ vary precisely
as required to remain IN PHASE with the incoming signal from clock A and
the outgoing signal from the mirror. THIS IS SO BECAUSE its output signal
must exactly track the Gedanken I signal from B->A, and that signal
exactly tracks the A->B signal at event B2. Equivalently, this is so
because the A->B signal can be used to continuously synchronize clock B
with clock A." [emphasis mine]
:: The phase difference of
:: the signals between A and B does change even if we accept the above
:: lemma postulating a round-trip light-speed isotropy.
:
: There is no way to measure or compare the phases at physically-
: separated locations. You need to bring the signals together, and
: the lemma ensures that the phase comparison between the signals
: _when_brought_together_ will remain constant (independent of
: orientation).
Your reasoning requires that the signal sent from A to B remains IN
PHASE with the signal B independently of the one-way speed of light.
If we replace the signals by short light pulses emitted at a
certain frequency we can recognize more easily that it is possible
to determine wavelength and phase of a signal, because we can
measure the location of the pulses at a given time. The distance
between two neighbouring pulses does obviously depend on the one-way
light speed.
A | > > > > > > > > |
|< < < < < < < < | B
Fig. 1. Pulse senders A and B at rest in the ether frame
A | > > > > > > > > > > > | v
|< < < < < | B -->
Fig. 2. Pulse senders moving at v wrt ether frame
:: The wave length
:: is different if the light speed is not the same in both directions,
:: so the number of phase=0 points between A1 and B2 cannot be the same
:: as between B2 and A3.
:
: Sure. So what? The lemma ensures that THE PHASE WHICH IS MEASURED
: remains constant. The lemma also ensures that the total number of
: wavelengths along the path A1->B2->A3 remains constant, independent
: of orientation (because the time delay must be constant, and because
: a change in orientation cannot change the clock frequency).
I agree. But if you want to derive conclusions concerning one-way
experiments then you need synchronization between two clocks. And
for that it does not help that phase and total number of wavelengths
remain constant along round-trip paths.
Cheers, Wolfgang
Joe Fischer
: : = Tom Roberts in http://www.deja.com/=dnc/getdoc.xp?AN=574678842
: :: = Wolfgang G. in http://www.deja.com/=dnc/getdoc.xp?AN=574329583
:
: If I have understood correctly the part of your article we are dealing
: with, then you postulate round-trip light-speed isotropy and derive
: from this postulate that one-way experiments must yield null results.
:
: For a one-way experiment two clocks A and B at different locations
: are required. If a signal is emitted from A at A-time A1 and received
: at B at B-time B2, then a null result implies that B2-A1 does not
: changed.
This is all so silly, and even moreso with modern
radar having the ability to send coded pulses.
All OWL lovers should set up an experiment, with
one radar transmitter-receiver unit on Mount Wilson Calif.,
and another just like it on Mount Palomar Calif.
These peaks are at least 100 kilometers apart,
and not moving relative to each other very fast, just
a few millimeters per year due to plate tectonic motion.
The pulse coding controlled by computer can tell
the other unit what time the pulse left, and the two
computers can be made to measure the relationship
between OWL and TWL.
Then etherists can let their brain rest.
Joe Fischer
Wayne Throop
: The notation is from my above-mentioned original article, and A and B
: are events occuring at the location of A; B2 is an event occuring after
: A1 and before A3 at the location of B. So the distance between A1 and B2
: is indeed the same as the distance between B2 and A3.
:
: Note that all three events are connected by light rays, so the
: time-ordering I mentiond is valid in all inertial frames, as
: is the equality of distances.
( ie, we are talking about http://sheol.org/throopw/androcles-diagram-03.gif,
except that A is "k-origin", and B is "mirror" )
Huh? The equality of distances between the EVENTS (that is,
delta-x between A1->B2 vs B2->A3) is NOT true in all inertial frames.
Of course, the spacetime interval is the same; they are both zero.
I must be missing something. Can you be more specific?
Wayne Throop thr...@sheol.org http://sheol.org/throopw
z@z
________________
If I'm not completely confused by all possible changes in space,
time, simultaneity and even velocity, resulting from the Lorentz
transformation, then I've found a very simple way to determine
our velocity wrt the ether in LET.
I suppose that this reasoning also entails that Special Relativity
is empirically refuted. Anyway, I'm sure that SR is fundamentally
wrong (despite of being a very elegant and astonishingly well
working theory).
Let as assume that the earth moves wrt the ether in such a way
that point C moves at v = 0.001 c (300 km/s) wrt the ether in
direction of point R. The distances from C to L and to R are both
10'000 nano light seconds (3 km).
L C R v wrt ether
---+--------------+--------------+--- --->
surface of the earth
The gamma factor of 1.000'000'5 can be ignored, because the
following reasonings are based on a first order effect proportional
to v.
From C short laser pulses are sent to L and R at a frequency of
10^8 (10 ns delay between two successive emissions). By comparing
the number of pulses on the two paths (from C to L and from C to R)
it becomes possible to determine the velocity wrt the ether.
At the same moving-frame time, there are only 999 pulses between L
and C, but 1001 between C and L.
This becomes obvious if we take into consideration that the same
pulses can be used to synchronize the clocks at L and R from C. As
observed from the ether frame, it needs 9'990 ns to synchronize
L and 10'010 ns to synchronize R from C.
During these 10'010 ns 1001 pulses have been sent in both
directions. When the first of these pulses reaches R, then the
first two pulses in the other direction have already surpassed L
and it is the third pulse at L which is declared to be moving-frame
simultanous with the first pulse at R. Therefore we have 999
pulses in the one and 1001 in the other direction.
________________
Just after having written the last sentence, I recognized the big
CONTRADICTION I have committed. It is of course in both cases the
first pulse which is declared moving-frame simultanous. If the
first pulse is emitted in both directions from the center at ether
time t = 0 and at moving-frame time t' = 0 then at t' = 10'000 ns
we have this situation:
Locus t' t Pulse
----------------------------------------------
L 10'000 ns 9'990 ns 1
C 10'000 ns 10'000 ns 1000
R 10'000 ns 10'010 ns 1
The moving-frame time t' = 10'000 ns as observed from the ether frame
appears at first at L (exactly when pulse 1 arrives), then moves
continuously to C where it arrives 10 ns later (exactly after pulse
1000 has been emitted) and arrives another 10 ns later at R (again
when pulse 1 arrives). So we have complete symmetry between the
pulses in the moving frame. Really astonishing!
But I cannot believe that the basic asymmetry in the ether has no
effect at all. Isn't there a problem with the intensity of light?
Intensity is inversely proportional to the distance square. If the
ether is the medium of light, then light intensity decreases with
ether distance, doesn't it?
In the above case (300 km/s wrt ether) the result is a difference
in light intensity corresponding to a factor of 1.004, certainly
enough to be measurable.
Wolfgang Gottfried G. (0:022.65)
My previous post of this thread:
http://www.deja.com/=dnc/getdoc.xp?AN=574910921
Tom Roberts
> But I cannot believe that the basic asymmetry in the ether has no
> effect at all.
This is not really a question for beliefs, it is a question for
_COMPUTATIONS_. If one generalizes the clear experimental result
that the round-trip speed of light is isotropic in all inertial
frames occupied by the earth to all inertial frames whatsoever,
then computations show that any ether theory consistent with that
is experimentally indistinguishable from SR. Need I reference
my trio of articles which discuss this again?
> Isn't there a problem with the intensity of light?
Not for any theory in the equivalence class of theories I discuss.
> Intensity is inversely proportional to the distance square. If the
> ether is the medium of light, then light intensity decreases with
> ether distance, doesn't it?
Yes, when measured in the ether frame. But intensity is also
proportional to the solid angle subtended by the detector, and
_that_ is not invariant. Do the computation and you will find
that competing effects cancel out exactly.
Tom Roberts tjro...@lucent.com
z@z
:: = Wolfgang G.
:: But I cannot believe that the basic asymmetry in the ether has no
:: effect at all. Isn't there a problem with the intensity of light?
:: Intensity is inversely proportional to the distance square. If the
:: ether is the medium of light, then light intensity decreases with
:: ether distance, doesn't it?
:
: Yes, when measured in the ether frame. But intensity is also
: proportional to the solid angle subtended by the detector, and
: _that_ is not invariant. Do the computation and you will find
: that competing effects cancel out exactly.
| - - - - - - - o - - - - - - - |
detector source of detector
sound
The sound intensity registered by the two detectors, unlike the
measured frequency, is not independent of the speed of the medium
(the air) wrt the system.
A spherical light wave is Lorentz invariant:
x^2 + y^2 + z^2 = (ct)^2 --> x'^2 + y^2 + z^2 = (ct')^2
But I don't think that all segments corresponding to given solid
angles are Lorentz invariant in the same way. We can for instance
divide the original spherical wave into two parts separated by
the circularly propagating wave
x = 0, y^2 + z^2 = (ct)^2
Thinking about stellar aberration is enough to recognize that
this circular wave cannot be Lorentz-transformed to
x' = 0, y^2 + z^2 = (ct')^2
This means that if the number of photons per solid angle is
uniform in the system at rest then it cannot be uniform in the
moving system and vice versa.
In SR it is reasonable to assume that it is always the frame
of the source where the number of photons per solid angle is
uniform. I doubt however that this is a reasonable assumption
in an ether theory.
| - - - - - - - o - - - - - - - | -->
detector spherical detector v
light source
Let us assume that this system moves at v = sqrt(0.99)c wrt the
ether and that the distance between the source and each dectector
is compressed from 10 m (restlength) to 1 m in the ether.
Whereas light to the right detector has to travel over 199.5 m in
the ether, the light path to the opposite detector is only 0.5 m.
1 m / (1 - sqrt(0.99) = 199.5 m
1 m / (1 + sqrt(0.99) = 0.5 m
The sound analogy would result in an intensity difference
factor of (199.5 m / 0.5 m)^2. So the intensity registered by
the left dectector would be (at least) 159 thousand times
stronger than by the right detector.
Wolfgang Gottfried G. (0:032.6)
Wayne Throop
: This means that if the number of photons per solid angle is
: uniform in the system at rest then it cannot be uniform in the
: moving system and vice versa.
I think you are correct.
And the experimental confirmation of this agrees with SR's prediction.
( I'm refering to the "searchlight" effect in emissions from
moving sources, mentioned fairly often by Tom Roberts, IIRC. )
What you are still missing is any way of determining an ether rest frame.