Consistency of the speed of light

[clj4]

Lorentz transformation:
t_1=(1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2
x_1=(X_1-vT_1)/(1-v^2/c_0^2)^{1/2}

t_2=(1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2
x_2=(X_2-vT_2)/(1-v^2/c_0^2)^{1/2}

D/(t_2-t_1)=(x_2-x_1)/(t_2-t_1)=c_0

c(v,\theta)=c_0

M&S-I p.511 Eq. (6.16) gives the following result for first order effects when transport synchronization of clocks is used with the Lorentz transformation: c(\theta)=1-v(1+2\alpha)cos\theta where \alpha =-1/2 corresponds to perfect Lorentz symmetry.

"In discussing the experiments we need the inverse velocity of light to second order in v/c..." M&S-III p. 810 Eq. (2.1) - 1/c(\theta)=1+(\beta+\delta-1/2)v^2sin^2\theta+(\alpha-\beta+1)v^2

LET transformation:
t_1=(1-v^2/c_0^2)^{1/2}T_1
x_1=(X_1-vT_1)/(1-v^2/c_0^2)^{1/2}

t_2=(1-v^2/c_0^2)^{1/2}T_2
x_2=(X_2-vT_2)/(1-v^2/c_0^2)^{1/2}

D/(t_2-t_1)=(x_2-x_1)/(t_2-t_1)=c_0^2/(c_0+v)

c(v,\theta)=c_0^2/(c_0+v \cdot cos(\theta))

Hint: The speed of light c_0 is isotropic in the ether frame: (X_2-X_1)/(T_2-T_1)=c_0.

Assuming perfect Lorentz symmetry and using the LET transformation I get:
c_0/c(v,\theta)=1+(v/c_0) \cdot cos(\theta). This is a dimensionless ratio, and as such it is a measurable (e.g., physical) quantity. However, you must synchronize two clocks to make this measurement; and exactly how you choose to do that determines whether the Lorentz transformation or the LET transformation should be applied.


The speed of light is generally anisotropic in LET (e.g., except for within the ether frame), absolute simultaneity is maintained, and this is empirically equivalent to SR. The trick is that the -vx/c_0^2 term that is used to maintain a constant speed of light in the Lorentz transformation is used instead to maintain absolute simultaneity in the LET transformation. Both ways are equally valid.

Hmmm,

It is clear how you got the second formula but it is not clear at all how you got the first formula c(v,\theta)=c_0. Would you care to show the intermediate steps?

 

[clj4]

:
If the two arms of a round-trip interferometer are parallel, then these two terms cancel on subtraction because the two (\delta-\beta+1/2)*[(v/c)sin(\theta)]^2 terms are both proportional to sin^2(\theta). These two terms do not cancel in Michelson-Morley or Kennedy-Thorndike experiments because one of the terms is proportional to sin^2(\theta) while the other term is proportional to cos^2(\theta) (when the two arms of the interferometer are orthogonal... but they would cancel if the two arms of the interferometer were made parallel, and that is why they are not made parallel). Gagnon's interferometer is not round-trip so I'm not completely sure that this cancellation of terms applies, but he hasn't explained exactly how he got to Eq. (9)

According to (9) in Gagnon the terms do not cancel.
This has been your challenge from the beginning, to prove that expression (9) is zero. Your explanations have been all over the map, the previous one had to do with delta/beta being zero, the one before had to do with some impossible to correlate transformation thru the Lorentz transforms, most of them had to do with the confusion about the framework (which was incorrectly taken to be SR).

Let's try again:

1. How did you arrive to the expressions
\phi_1(t)=\phi_1(0)+\omega_1*(2L_1/c)(1/\eta_1+(\delta-\beta+1/2)*[(v/c)sin(\theta)]^2

\phi_2(t)=\phi_2(0)+\omega_2*(2L_1/c)(1/\eta_2+(\delta-\beta+1/2)*[(v/c)sin(\theta)]^2

for the two wave guides?

2. Are you aware that the parenses don't match, therefore the expressions are incomprehensible? You will need to correct that.

Look again at \phi_1(t)=\phi_1(0)+\omega_1*(2L_1/c)(1/\eta_1+(\delta-\beta+1/2)*[(v/c)sin(\theta)]^2

3. How does the term \1\/\eta_1 "disappear" from your considerations?

4. Where in your calculations do you factor in that the cutoff frequencies for the two waveguides are arranged to be very different? Gagnon seems to make a great deal of the fact that expression (9) is obtained by driving the two waveguides one just above the cutofff and the second one way above the cutoff frequency.

 

[Aether]

Hmmm,

It is clear how you got the second formula but it is not clear at all how you got the first formula c(v,\theta)=c_0. Would you care to show the intermediate steps?

This is just a statement that in SR one-way light speed is isotropic, but I will examine it anyway.

SR, RMS (Robertson-Mansouri-Sexl text theory), and Lorentz ether theory are all in agreement that the speed of light c_0 is defined to be isotropic with respect to at least one particular inertial frame (e.g., within the ether frame \frac{X_2-X_1}{T_2-T_1}=c_0 -- actually, all velocites are isotropic within this frame, so this equation really relates just to this example, I suppose that a more general equation like this \frac{X_2-X_1}{T_2-T_1}=v_0 is always valid). In SR however, the Lorentz transform relates this particular (arbitrary) inertial frame with every other inertial frame while explicitly preserving light speed, and therefore the speed of light is further defined (within SR) to be isotropic with respect to every other inertial frame as well.

For example, these are (two sets of) the Lorentz transforms for motion along the x-axis:

t_1=((1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2),
x_1=((X_1-vT_1)/(1-v^2/c_0^2)^{1/2}),
t_2=((1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2),
x_2=((X_2-vT_2)/(1-v^2/c_0^2)^{1/2}).

Rather than defining light-speed isotropy for all (inertial) frames, I will justify this equation c(v,\theta)=c_0 by defining light-speed isotropy only in the ether frame (e.g., \frac{X_2-X_1}{T_2-T_1}=c_0) and then combine this definition with the Lorentz transformations to show that light speed is therefore isotropic in all (inertial) frames \frac{x_2-x_1}{t_2-t_1}=c_0; such a showing is ultimately coordinate-system dependent, and it is definitely not something that could ever be proven by an experiment.

So, to verify that this equation \frac{x_2-x_1}{t_2-t_1}=c_0 is true (when \frac{X_2-X_1}{T_2-T_1}=c_0) then simply transform the four coordinates from any arbitrary inertial frame (e.g., use any v you like) into the ether frame using the Lorentz transform equations given above and verify that you always arrive back at this equation \frac{X_2-X_1}{T_2-T_1}=c_0 which we have defined above to be true.

Start with:
\frac{x_2-x_1}{t_2-t_1}=c_0

Transform the four space-time coordinates (by substitution) using the Lorentz transforms:
\frac{((X_2-vT_2)/(1-v^2/c_0^2)^{1/2})-((X_1-vT_1)/(1-v^2/c_0^2)^{1/2})}{((1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2)-((1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2)}=c_0




Now, simply reduce this equation to \frac{X_2-X_1}{T_2-T_1}=c_0 to show that c(v,\theta)=c_0 is true (this doesn't actually prove that it is true for all \theta because we're only looking at motion along the x-axis).

\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1-\frac{v(X_1-vT_1)}{c_0^2}}=c_0

\frac{(X_2-vT_2-X_1+vT_1)c_0^2}{(1-v^2/c_0^2)T_2c_0^2-v(X_2-vT_2)-(1-v^2/c_0^2)T_1c_0^2-v(X_1-vT_1)}=c_0

\frac{(X_2-vT_2-X_1+vT_1)c_0^2}{( c_0^2-v^2)(T_2-T_1)-v(X_2-vT_2-X_1+vT_1)}=c_0

(X_2-vT_2-X_1+vT_1)c_0^2=c_0(( c_0^2-v^2)(T_2-T_1)-v(X_2-vT_2-X_1+vT_1))

(X_2-vT_2-X_1+vT_1)c_0^2=c_0( c_0^2-v^2)(T_2-T_1)-v c_0(X_2-vT_2-X_1+vT_1)

(X_2-vT_2-X_1+vT_1)(c_0^2+v c_0)=c_0( c_0^2-v^2)(T_2-T_1)

(X_2-vT_2-X_1+vT_1)c_0(c_0+v)=c_0( c_0+v)( c_0-v)(T_2-T_1)

(X_2-vT_2-X_1+vT_1)c_0=c_0( c_0-v)(T_2-T_1)

X_2-vT_2-X_1+vT_1=c_0(T_2-T_1)-v(T_2-T_1)

X_2-vT_2-X_1+vT_1+vT_2-vT_1=c_0(T_2-T_1)

X_2-X_1=c_0(T_2-T_1)

\frac{X_2-X_1}{T_2-T_1}=c_0

 

[clj4]

This is just a statement that in SR one-way light speed is isotropic, but I will examine it anyway.

SR, RMS (Robertson-Mansouri-Sexl text theory), and Lorentz ether theory are all in agreement that the speed of light c_0 is defined to be isotropic with respect to at least one particular inertial frame (e.g., within the ether frame \frac{X_2-X_1}{T_2-T_1}=c_0 -- actually, all velocites are isotropic within this frame, so this equation really relates just to this example, I suppose that a more general equation like this \frac{X_2-X_1}{T_2-T_1}=v_0 is always valid). In SR however, the Lorentz transform relates this particular (arbitrary) inertial frame with every other inertial frame while explicitly preserving light speed, and therefore the speed of light is further defined (within SR) to be isotropic with respect to every other inertial frame as well.

For example, these are (two sets of) the Lorentz transforms for motion along the x-axis:

t_1=((1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2),
x_1=((X_1-vT_1)/(1-v^2/c_0^2)^{1/2}),
t_2=((1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2),
x_2=((X_2-vT_2)/(1-v^2/c_0^2)^{1/2}).

Rather than defining light-speed isotropy for all (inertial) frames, I will justify this equation c(v,\theta)=c_0 by defining light-speed isotropy only in the ether frame (e.g., \frac{X_2-X_1}{T_2-T_1}=c_0) and then combine this definition with the Lorentz transformations to show that light speed is therefore isotropic in all (inertial) frames \frac{x_2-x_1}{t_2-t_1}=c_0; such a showing is ultimately coordinate-system dependent, and it is definitely not something that could ever be proven by an experiment.

So, to verify that this equation \frac{x_2-x_1}{t_2-t_1}=c_0 is true (when \frac{X_2-X_1}{T_2-T_1}=c_0) then simply transform the four coordinates from any arbitrary inertial frame (e.g., use any v you like) into the ether frame using the Lorentz transform equations given above and verify that you always arrive back at this equation \frac{X_2-X_1}{T_2-T_1}=c_0 which we have defined above to be true.

Start with:
\frac{x_2-x_1}{t_2-t_1}=c_0

Transform the four space-time coordinates (by substitution) using the Lorentz transforms:
\frac{((X_2-vT_2)/(1-v^2/c_0^2)^{1/2})-((X_1-vT_1)/(1-v^2/c_0^2)^{1/2})}{((1-v^2/c_0^2)^{1/2}T_2-vx_2/c_0^2)-((1-v^2/c_0^2)^{1/2}T_1-vx_1/c_0^2)}=c_0

Now, simply reduce this equation to \frac{X_2-X_1}{T_2-T_1}=c_0 to show that c(v,\theta)=c_0 is true (this doesn't actually prove that it is true for all \theta because we're only looking at motion along the x-axis).

Well, not to be nitpicking but you wrote above
c(v,\theta)=c_0 for ANY \theta
The following derivation is true only for \theta=0

\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1-\frac{v(X_1-vT_1)}{c_0^2}}=c_0

There are some sign errors in the first expression. The correct thing is:

\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1+\frac{v(X_1-vT_1)}{c_0^2}}=c_0
You would then get:

\frac{X_2-vT_2-X_1+vT_1}{T_2-\frac{vX_2}{c_0^2}-T_1+\frac{vX_1}{c_0^2}}=c_0

If you divide both the numerator and the denominator by T_2-T_1 and you remember that \frac{X_2-X_1}{T_2-T_1}=c_0 you indeed get your result.
Very ugly...

 

[Aether]

Well, not to be nitpicking but you wrote above
c(v,\theta)=c_0 for ANY \theta
The following derivation is trur only for \theta=0

That's right. Do you think that it's important to show a more general proof for motion in 3D?

There are some sign errors in the first expression. The correct thing is:

\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1+\frac{v(X_1-vT_1)}{c_0^2}}=c_0
You would then get:

\frac{X_2-vT_2-X_1+vT_1}{T_2-\frac{vX_2}{c_0^2}-T_1+\frac{vX_1}{c_0^2}}=c_0

If you divide both the numerator and the denominator by T_2-T_1 and you remember that \frac{X_2-X_1}{T_2-T_1}=c_0 you indeed get your result.

There is a sign error in the first two equations that I gave, but the rest are OK. The correct equations are:

\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1+\frac{v(X_1-vT_1)}{c_0^2}}=c_0
\frac{(X_2-vT_2-X_1+vT_1)c_0^2}{(1-v^2/c_0^2)c_0^2T_2-v(X_2-vT_2)-(1-v^2/c_0^2)c_0^2T_1+v(X_1-vT_1)}=c_0

Very ugly...

What do you mean by this?

 

[clj4]

That's right. Do you think that it's important to show a more general proof for motion in 3D?

That the reasoning is only valid for theta=0.

There is a sign error in the first two equations that I gave, but the rest are OK. The correct equations are:

\frac{X_2-vT_2-X_1+vT_1}{(1-v^2/c_0^2)T_2-\frac{v(X_2-vT_2)}{c_0^2}-(1-v^2/c_0^2)T_1+\frac{v(X_1-vT_1)}{c_0^2}}=c_0
\frac{(X_2-vT_2-X_1+vT_1)c_0^2}{(1-v^2/c_0^2)c_0^2T_2-v(X_2-vT_2)-(1-v^2/c_0^2)c_0^2T_1+v(X_1-vT_1)}=c_0

Yes, I corrected it for you. I also cleaned up the proof a little.

What do you mean by this?

That the whole RMs theory is exceptionally ugly. But never mind this detour, let's return to my 4 questions on the Gagnon experiment.

 

[Aether]

But never mind this detour, let's return to my 4 questions on the Gagnon experiment.

clj4 and I are continuing a discussion on this topic that were were having here http://www.bautforum.com/showthread.php?t=38765 (see post #47) because Latex isn't supported by BAUT at this time.

The issue at hand is this paper [D.R. Gagnon et al., Guided-wave measurement of the one-way speed of light, Physical Review 38A(4), 1767 (1988); http://imaginary_nematode.home.comca...et_al_1988.pdf.] which seems to be claiming that RMS predicts a different outcome to their experiment than does SR (although even that isn't really made clear in the paper).

Now I've got to study some more about waveguides and try to reverse engineer Eq. (9) from (Gagnon et al., 1988) because it wasn't derived in the paper.

 

[clj4]

. No experiment has ever been able to distinguish between these two points of view, and if one ever does it could only favor the ether view.

What makes you think that? In light of the experimental proof that shows exactly the opposite...see our little discussion

 

[clj4]

Can you read Hans' version? If not, I'll post a pdf. I made a .pdf first, but it was very large so I wen't for a .doc; it wasn't any smaller though.

Suppose that LET turns out only to be useful as a waypoint (for some) on the philosophical journey to SR. Even so, what is the justification for tolerating false (however well intentioned) claims that the constancy of the speed of light is an empirically determined fact? Why not just state up front that partiality to SR is simply a matter of coordinate choice, and not an empirical necessity?

You know (by now) that the claims of one way light speed constancy are far from being false. You also know that there are several experiments proving it.

 

[clj4]

Lorentz transformations amount to an arbitrary choice of Einstein's clock synchronization convention. Lorentz symmetry is equally well represented by transformations in which absolute simultaneity is maintained.

Correct. The problem starts when the experiments come back and show that the Robertson-Mansoury-Sexl HYPOTHESIS of light speed anisotropy is wrong.

 

[clj4]

Their (M&S) general approach may also be useful in rotating coordinate systems, where the usual assumption of isotropy has issues. While one can always chose not to use rotating coordiantes, they are convenient enough that sometimes it's worth giving up the conveniences of isotropy for the convenience of using rotating coordinates.

It would be useful except that they derived it under very restrictive conditions (translation only).
The RMS (R stands for Robertson) is indeed a "test theory". Looks like there are quite a few one-way light speed experiments that refute the light speed anisotropy. A good thing because RMS is one of the ugliest theories ever.

 

[clj4]

This is misleading. The M&S paperer is referred to because of the
parameterization scheme for possible deviations of Special Relativity.
It's only you who uses it to promote your ether theory.

The math of M&S is correct in the preferred frame, not in any other.



Do you at all read my post? do you look at my examples. No you don't

SR is the ABC of physics. Something you have to understand pretty
well before you can start to learn some real physics. The examples I
gave are the simplest it gets in understanding the basic mechanisms in SR
and the simplest way to show that your Ether theory with absolute time
can never work.

Now try to do the math. Try to understand the physics. Don't just rely on
some statement you have found somewhere in a paper. It's now time for you
to prove your ether theory by actually showing how it can account for
these relativistic effects.

1) How can two observers both see the other in a Lorentz contracted state?

2) How can you rotate the deBroglie wavefronts of particles if you go from
one reference frame to another in order to keep them at right angles with
the direction of their speed? How can a single transformation rotate these
wavefront at all kinds of different angles depending on the speed of the
particles?

Let's see if you can do that without non simultaneity.

People here are willing to help others to get ahead. That's why it's called a
Physics Help and Math Help forum. But if there's no response and you just
keep repeating a statement from somewhere then things get pretty useless
after a while. I did the work, the math, the physics, showed you the images
from my simulations.

Now it's up to you.



Regards, Hans.

Hans,

I appreciate your passion. Have a little patience, RMS will go away. Soon.
It has hung around for nearly 30 years, it has evolved into the Standard Model (SM) which has, under Alan Kostelecky evolved into the Standard Model Extension. Test theories that look for "Lorentz symmetry violations" and find ...nothing. But they keep a lot of test theorists employed. No harm done, an interesting exercise.
RMS is very rarely taught in school (maybe because it is exceptionally ugly?, maybe because it assumes that light speed is anisotropic-something refuted by experiment?). Who knows? It is good to discuss such things but we also must tell the truth: there are quite a few OWLS experiments and they all concluded that the light speed is...isotropic. This is the theme of this thread, so we need to set the things right.

 

[clj4]

Equivalent? Only in the preferred frame.

Note that 3.6 makes both length and speed anisotropic in any other
reference frame besides the preferred frame. Round wheels would be
only really round in the preferred frame.

We at Earth would see al our wheels changing shape in a 24 hours cycle
corresponding with the rotation of the earth.


Regards, Hans

Hi Hans

Can you show this mathematically. I am just curious how one approaches this issue mathematically.

 

[Aether]

No experiment has ever been able to distinguish between these two points of view, and if one ever does it could only favor the ether view.

What makes you think that? In light of the experimental proof that shows exactly the opposite...see our little discussion

When going this far back, please give the post# that you are quoting from.

I no longer think that exactly, and I haven't said exactly that recently. The inherently coordinate-system dependent nature of speed measurements simply can't be overcome by any classical experiment, not even in principle. The eventual detection of any violation of local Lorentz invariance would allow a coordinate-system that maintains absolute simultaneity (e.g., the ether view) to be realized within any laboratory in a consistent way (e.g., a locally preferred frame would be measurable from within a windowless laboratory); such a coordinate system can be realized right now, but the choice of the locally preferred frame would be arbitrary since there isn't any known experiment to detect it. SR could still continue to be used for many purposes even if Lorentz symmetry is violated.

I agree that there is ample experimental evidence to constrain the possible violation of local Lorentz invariance to a small value (not counting gravity!) within the error-bars of the experiments. What you are claiming is that one-way speed of light measurements aren't coordinate-system dependent, and that is what I disagree with. Why aren't you proposing that the "postulates" of SR are actually proven facts, and therefore SR needs to be updated to reflect that? What about the quote from Albert Einstein in my signature below, what do you think that he meant by this?

 

[clj4]

What you are claiming is that one-way speed of light measurements aren't coordinate-system dependent, and that is what I disagree with. Why aren't you proposing that the "postulates" of SR are actually proven facts, and therefore SR needs to be updated to reflect that? What about the quote from Albert Einstein in my signature below, what do you think that he meant by this?

Sorry, I should have known better than to give you an opportunity for a diversion. We are still on the Gagnon experiment.

Let's not play games , I don't claim anything else than the fact that there are OWLS experiments that show light speed to be clearly isotropic. We are still on the Gagnon experiment, you have to provide a valid answer to it. We are waiting...

 

[Aether]

Suppose that LET turns out only to be useful as a waypoint (for some) on the philosophical journey to SR. Even so, what is the justification for tolerating false (however well intentioned) claims that the constancy of the speed of light is an empirically determined fact? Why not just state up front that partiality to SR is simply a matter of coordinate choice, and not an empirical necessity?

You know (by now) that the claims of one way light speed constancy are far from being false.

There are a handfull of published papers (listed here in sextion 3.2: http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html) that lend (meager) support to such claims, but still any claim that the constancy of the one-way speed of light is an empirically determined fact is false.

You also know that there are several experiments proving it.

I know that there are several old experiments claiming to prove it, I also know that Mansouri-Sexl and Zhang say that this isn't possible even in principle. Please cite a modern experiment to test for local Lorentz violations (or claiming to measure the one-way speed of light) that doesn't either cite Mansouri-Sexl directly, or at least indirectly; and then show where they claim (as you do) that Mansouri-Sexl is wrong. Or, cite one modern (or not so modern) reference that claims flatly (as you do) that Mansouri-Sexl is wrong.

 

[clj4]

There are a handfull of published papers (listed here in sextion 3.2: http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html) that lend (meager) support to such claims, but still any claim that the constancy of the one-way speed of light is an empirically determined fact is false.

I know that there are several old experiments claiming to prove it, I also know that Mansouri-Sexl and Zhang say that this isn't possible even in principle. Please cite a modern experiment to test for local Lorentz violations (or claiming to measure the one-way speed of light) that doesn't either cite Mansouri-Sexl directly, or at least indirectly; and then show where they claim (as you do) that Mansouri-Sexl is wrong. Or, cite one modern (or not so modern) reference that claims flatly (as you do) that Mansouri-Sexl is wrong.

We've been through this, let's not go around in circles. We agreed that you have an experiment to work on, you tried (and failed) to refute it, please stop regurgitating the same quotes from MS and/or Zhang.

You need to continue from here:

https://www.physicsforums.com/showpost.php?p=930823&postcount=107

 

[vladb]

Hi:

Einstein's second postulate states that the speed of light is constant as viewed from any frame of reference. <...>

Actually Einstein didn't postulate that. Just today I've read an article (which I'm going to attach. See part III there) about misconceptions about special relativity.
As far as I understood, Einstein postulated that the speed of light doesn't depend on the velocity of the source of light _only_ (in an inertial frame). And only from this he derived that the speed of light is also constant in all inertial frames of reference.

 

[Aether]

Lorentz transformations amount to an arbitrary choice of Einstein's clock synchronization convention. Lorentz symmetry is equally well represented by transformations in which absolute simultaneity is maintained.

Correct.

Good, I'm glad that we agree on this. Then our discussion on the coordinate-system dependent nature of one-way speed measurements is concluded in my favor?

The problem starts when the experiments come back and show that the Robertson-Mansoury-Sexl HYPOTHESIS of light speed anisotropy is wrong.

This is a different discussion entirely (about violations of Lorentz symmetry as opposed to coordinate-system dependency), but we can continue-on with this now as long as we're both on the same page with respect to the inherently coordinate-system dependent nature of one-way speed of light measurements.

We've been through this, let's not go around in circles. We agreed that you have an experiment to work on, you tried (and failed) to refute it, please stop regurgitating the same quotes from MS and/or Zhang.

You need to continue from here:

https://www.physicsforums.com/showpos...&postcount=107

OK.

Actually Einstein didn't postulate that. Just today I've read an article (which I'm going to attach. See part III there) about misconceptions about special relativity.
As far as I understood, Einstein postulated that the speed of light doesn't depend on the velocity of the source of light _only_ (in an inertial frame). And only from this he derived that the speed of light is also constant in all inertial frames of reference.

Thanks for the the article vladb. It looks like it's hot off the press. Here's Einstein's 1905 paper on special relativity: http://home.tiscali.nl/physis/HistoricPaper/Historic Papers.html.

 

[clj4]

Good, I'm glad that we agree on this. Then our discussion on the coordinate-system dependent nature of one-way speed measurements is concluded in my favor?

Not. we've been thru this before, please stop the diversions.

The only thing we agree on is that you need to finish the refutation of the Gagnon paper, pick up from here, please:

https://www.physicsforums.com/showpost.php?p=930823&postcount=107

 

[Aether]

Not. we've been thru this before, please stop the diversions.

The only thing we agree on is that you need to finish the refutation of the Gagnon paper, pick up from here, please:

https://www.physicsforums.com/showpos...&postcount=107

I am simply answering your questions (and/or responding to comments) that you addressed directly to me. If you don't want we to answer (or respond to) them at this time, then please withdraw them. They are diverting me from what I would rather be doing (e.g., studying Gagnon).

 

[clj4]

I am simply answering your questions (and/or responding to comments) that you addressed directly to me. If you don't want we to answer (or respond to) them at this time, then please withdraw them. They are diverting me from what I would rather be doing (e.g., studying Gagnon).

Study Gagnon .

 

[Hans de Vries]

There are a handfull of published papers (listed here in sextion 3.2: http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html)

I know that there are several old experiments claiming to prove it, I also know that Mansouri-Sexl and Zhang say that this isn't possible even in principle.

One-Way Test of Light Speed IsotropyOne needs at least two space dimensions to fully prove that
the speed of light is equal in all directions. (So MS&Z may indeed
claim that it's in principle impossible with one space dimension.)Definition of what we want to prove:

We experience the simultaneity of space-time in such a way
that the speed of light appears to be equal in all directions.
Yes one can do calculations equally well in any other reference
frame but what counts is the reference frame we experience.A visual prove without any math:

Look from the z-axis at a wheel which you experience as round,
that is, it's equal in the x and y directions. If you rotate the wheel
then its shape doesn't change.

Now start moving together with the wheel in arbitrary directions
at high speed (say 0.9 c) You will still observe the wheel as being
round. Length in x and y is isotropic with speed. You can rotate
the wheel but it doesn't change shape.

Now put some synchronized clocks at various locations at the edge
of the wheel, North, South, East, West, wherever you want. You
can rotate the wheel to any position but the clocks will always stay
synchronized.

Again start moving together with the wheel in an arbitrary direction
at high speed. the wheel stays round but the clocks have different
readings now!

Just synchronize the clocks so that they all have the same reading
again. Rotate the wheel to any position, the clocks will always show
equal times.

Now let each clock send a laser pulse to the clock which is at the
opposite site of the wheel from it, all at the same time. All clocks
will measure the arrival of the light pulses all at the same time.
There you are!

----------------------- end of proof --------------------
The only thing special what happened here was that the clocks
were not synchronized anymore when you changed your speed
while the wheel did stay round.

The wheel became Lorentz contracted in the original reference
frame, however, in your reference frame you appear to see the
front of the wheel (in the direction of speed) a little later together
with the end of the wheel a little earlier. The wheel looks round
again.

The things you experience as being simultaneously depend or
your reference frame. That is all there is to Special Relativity!Regards, Hans

 

[Aether]

Study Gagnon .

Eq. (6) of (Gagnon et al., 1988) is introduced like this: "For a waveguide lying along the z direction of the laboratory-coordinate system, the field in the waveguide is required to have the following form: E(x,y,z)=E(x,y)exp(ikz-i\omega t).", and Eqs. (7) & (9) flow from this assumption. However, the angle \theta apprearing in Eq. (9) "is the angle between the absolute velocity vector and the waveguide", and this experiment involves rotations of the waveguide on all three axes including passive rotation due to the Earth's spin, and active rotation of the apparatus within the horizontal plane. Eq. (6) is not valid for a waveguide that has been rotated away from "lying along the z direction of the laboratory-coordinate system", and therefore Eq. (9) is not generally valid because it is based on an assumption that the waveguides are "lying along the z direction of the laboratory-coordinate system."

If E(x,y,z)=E(x,y)exp(ikz-i\omega t) (Eq. (6) of Gagnon et al.) is valid for a waveguide lying along the z direction of the laboratory-coordinate system, then E(x,y,z)=E(y,z)exp(ikx-i\omega t) is valid for a waveguide lying along the x direction of the laboratory-coordinate system, and E(x,y,z)=E(x,z)exp(iky-i\omega t) is valid for a waveguide lying along the y direction of the laboratory-coordinate system.

Similarly, if Eq. (7) of Gagnon et al. is valid for a waveguide lying along the z direction of the laboratory coordinate system, then at least five more equally valid equations may be generated for waveguides lying along the x, y, and z axes by an appropriate substitution of spatial coordinate indices. I'll write-out these equations later, and try to give a generalized expression for the guide wave number k(\theta) that is valid (or at least consistent with Gagnon's Eq. (6)) for any angle \theta between the absolute velocity vector and the waveguide.

The output signal phase difference between two parallel waveguides that are driven by a common input signal is given by \Delta \phi=2\pi(k_2L_2-k_1L_1) (where L_1 and L_2 are the lengths of the first and second waveguide respectively), and I expect to show that this equation is rotationally invariant (assuming Lorentz symmetry).

 

[clj4]

Eq. (6) of (Gagnon et al., 1988) is introduced like this: "For a waveguide lying along the z direction of the laboratory-coordinate system, the field in the waveguide is required to have the following form: E(x,y,z)=E(x,y)exp(ikz-i\omega t).", and Eqs. (7) & (9) flow from this assumption. However, the angle \theta apprearing in Eq. (9) "is the angle between the absolute velocity vector and the waveguide", and this experiment involves rotations of the waveguide on all three axes including passive rotation due to the Earth's spin, and active rotation of the apparatus within the horizontal plane. Eq. (6) is not valid for a waveguide that has been rotated away from "lying along the z direction of the laboratory-coordinate system", and therefore Eq. (9) is not generally valid because it is based on an assumption that the waveguides are "lying along the z direction of the laboratory-coordinate system."

You are going on a wrong path: the rotation in cause is the Earth rotation.
The Earth rotation will induce a change in \theta which in turn will induce a change in the component \sin^2(\theta) in (9). Further on in the paper you see that one would naturally expect a diurnal variation (very much like in the Kennedy-Thorndike experiment, see my earlier reference that you did not understand) and they found none.

So, sorry , (9) is generally valid, the other axis are not needed. The dependency on \theta is sufficient. Try deriving it, this is where we left off.

 

[Aether]

You are going on a wrong path: the rotation in cause is the Earth rotation.
The Earth rotation will induce a change in theta which in turn will induce a change in the component \sin^2(\theta) in (9).

On page 1770 they say that "Data was acquired as the arrangement was allowed to swing freely through just over 180-degrees of travel with a rotation period of about 30 seconds." This is what is being shown in Fig. 2. So, you agree that this part of the experiment (active rotations of the apparatus) is debunked?

Further on in the paper you see that one would naturally expect a diurnal variation (very much like in the Kennedy-Thorndike experiment, see my earlier reference that you did not understand) and they found none.

That would be a much smaller effect, and harder to detect. I thought that they were using the rotation of the Earth to change the orientation of the horizontal plane (within which they are actively rotating their apparatus) wrt the CMB. Where do they talk about a Kennedy-Thorndike type effect?

So, sorry , (9) is generally valid, the other axis are not needed. The dependency on is sufficient. Try deriving it, this is where we left off.

If they would leave the waveguides lying along the z direction of the laboratory-coordinate system, then maybe, but what about the active rotation of the waveguide within the laboratory frame?

Eq. (9) is an approximation for the phase difference between the two waveguides. This is the exact (albeit idealized, and assuming that Eq. (7) is right) expression for the theoretical phase difference between the two waveguides so long as they are lying along the z direction of the laboratory-coordinate system:

Here's a Kennedy-Thorndike type analysis (taking into account the Earth's rotation only):
\Delta \phi=k_2L_2-k_1L_1=\frac{L_2}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_2^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}-\frac{L_1}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_1^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}.

L_1=L_2=2.5 \ meters
\omega=2.52333\times 10^{11} \ rad/sec
\omega_1=2.52019\times 10^{11} \ rad/sec
\omega_2=1.39487\times 10^{11} \ rad/sec
c=2.99792458\times 10^8 \ m/s

For v_x=400 \ km/s and v_z=0:
\Delta \phi=1753.498118-104.942309=1648.555808 \ rad

For v_x=401 \ km/s and v_z=0:
\Delta \phi=1753.498110-104.942309=1648.555801 \ rad

This predicts a maximum daily phase shift of 0.401\times 10^{-3} degrees (when absolute motion is aligned with the x-axis). Compare this to this conclusion of (Gagnon et al., 1988): "...the phase shift attributable to reorientation of the apparatus by a 6-h rotation of the Earth does not exceed 8\times 10^{-3} degrees."

For v_x=0 and v_z=400 \ km/s:
\Delta \phi=1753.500365-104.979859=1648.520506 \ rad

For v_x=0 and v_z=401 \ km/s:
\Delta \phi=1753.500369-104.980047=1648.520322 \ rad

This predicts a maximum daily phase shift of 10.54\times 10^{-3} degrees (when absolute motion is aligned with the z-axis). Compare this to this conclusion of (Gagnon et al., 1988): "...the phase shift attributable to reorientation of the apparatus by a 6-h rotation of the Earth does not exceed 8\times 10^{-3} degrees."

These maximum daily phase shifts were not fully attainable by Gagnon et al. at 36-degrees N latitude, and using level waveguides (e.g., they should have been equatorially mounted, polar aligned, and made to track 11h RA +6-deg. dec).

Wow, maybe they did detect a diurnal Kennedy-Thorndike effect. I can't rule that out, but there isn't enough here to convince me that they did.

 

[clj4]

On page 1770 they say that "Data was acquired as the arrangement was allowed to swing freely through just over 180-degrees of travel with a rotation period of about 30 seconds." This is what is being shown in Fig. 2. So, you agree that this part of the experiment (active rotations of the apparatus) is debunked?

What do you mean? "debunked" implies some form of sleigh of hand from the authors, there is none of that.
As to the KT reference, read on your own explanation below and also what follows on page 1771 that discusses fig 2. (expected diurnal variation)

That would be a much smaller effect, and harder to detect. I thought that they were using the rotation of the Earth to change the orientation of the horizontal plane (within which they are actively rotating their apparatus) wrt the CMB. Where do they talk about a Kennedy-Thorndike type effect?

Again, look at fig 2.

Eq. (9) is an approximation for the phase difference between the two waveguides. This is the exact (albeit idealized, and assuming that Eq. (7) is right) expression for the phase difference between the two waveguides so long as they are lying along the z direction of the laboratory-coordinate system:

Here's a Kennedy-Thorndike type analysis (taking into account the Earth's rotation only):
\Delta \phi=2\pi(k_2L_2-k_1L_1)=2\pi\frac{L_2}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_2^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}-2\pi\frac{L_1}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_1^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}.

This looks reasonable but does not (yet) look like (9).
Either way, what follows below, in your own writing , results into NON-ZERO phase shift. And the authors measured...ZERO.

L_1=L_2=2.5 \ meters
\omega=40.16 \ GHz
\omega_1=40.11 \ GHz
\omega_2=33.50 \ GHz
c=2.99792458x10^8 \ m/s

For v_x=400 \ km/s and v_z=0:
\Delta \phi=1160.517675-104.968846=1055.548830 \ rad

For v_x=401 \ km/s and v_z=0:
\Delta \phi=1160.517670-104.968845=1055.548825 \ rad

This predicts a maximum daily phase shift of 0.287x10^{-3} degrees (when absolute motion is aligned with the x-axis). Compare this to this conclusion of (Gagnon et al., 1988): "...the phase shift attributable to reorientation of the apparatus by a 6-h rotation of the Earth does not exceed 8x10^{-3} degrees."

Of course, this is an effect in (v/c)^2 , so it is very small but it is not zero, it is larger than what was measured in the experiment. Either if we take your prediction or the author's prediction, both are much bigger than the measurement.

For v_x=0 and v_z=400 \ km/s:
\Delta \phi=1160.521071-105.006386=1055.514685 \ rad

For v_x=0 and v_z=401 \ km/s:
\Delta \phi=1160.521083-105.006573=1055.514510 \ rad

This predicts a maximum daily phase shift of 10.0x10^{-3} degrees (when absolute motion is aligned with the z-axis). Compare this to this conclusion of (Gagnon et al., 1988): "...the phase shift attributable to reorientation of the apparatus by a 6-h rotation of the Earth does not exceed 8x10^{-3} degrees."

ok, so in this case you predict something very close to Gagnon. Perfect, so now, how do you explain the zero measurement (within the error bars). You have just proved Gagnon's point.

 

[Aether]

What do you mean? "debunked" implies some form of sleigh of hand from the authors, there is none of that.

No, I don't think there is any of that. I mean that their hypothesis isn't valid for rotations of the waveguides away from the z direction of the laboratory-coordinate system.

As to the KT reference, read on your own explanation below and also what follows on page 1771 that discusses fig 2. (expected diurnal variation)

Again, look at fig 2.

I do not see any hypothesized Kennedy-Thorndike effect, but you may have found a plausible one anyway! From page 1771 (emphasis is mine): "The 6 h separation between data sets was designed to provide a means of detecting direction-dependent effects by directly comparing data, averaged at pointing angles fixed in the laboratory, from each data set. Hypothesized effects, which vary with respect to direction fixed in the distant stars, could thereby be distinguished from purely systematic effects caused by rotations of the apparatus in the laboratory."

This looks reasonable but does not (yet) look like (9).
Either way, what follows below, in your own writing , results into NON-ZERO phase shift. And the authors measured...ZERO.

No, they measured "...does not exceed 8\times 10^{-3} degrees".

 

[clj4]

\Delta \phi=2\pi(k_2L_2-k_1L_1)=2\pi\frac{L_2}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_2^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}-2\pi\frac{L_1}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_1^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}.For v_x=0 and v_z=400 \ km/s:
\Delta \phi=1160.521071-105.006386=1055.514686 \ rad

Not so fast. You seem to have forgotten a term from expression (7). Where did

\frac{v_z}{c} term go? v_z is not zero,and it gets multiplied by

\frac{\omega_1}{c} and \frac{\omega_2}{c} respectively so they do not cancel out.

 

[Aether]

This looks reasonable but does not (yet) look like (9).

The authors claim on page 1771 that "Using these figures, a peak-to-peak phase shift of at least 19-degrees is predicted as the apparatus turns in the laboratory (at 36-degrees N latitude). Using Eq. (9) and these values for the variables, I only get 4.2933-degrees peak-to-peak. What do you get?

\omega_1=2.52019\times 10^{11} \ rad/sec
L=2.5 \ meters
v=\pm 400 \ km/s
\delta=3.14159\times 10^8 \ rad/sec

 

[Aether]

Not so fast. You seem to have forgotten a term from expression (7). Where did

\frac{v_z}{c} term go? v_z is not zero.

This term is the same for both waveguides and disappears on subtraction of the two output signal phases.

 

[clj4]

The authors claim on page 1771 that "Using these figures, a peak-to-peak phase shift of at least 19-degrees is predicted as the apparatus turns in the laboratory (at 36-degrees N latitude). Using Eq. (9) and these values for the variables, I only get 0.6833-degrees peak-to-peak. What do you get?

\omega_1=40.11 \ GHz
L=2.5 \ meters
v=\pm 400 \ km/s
\delta=50 \ MHz

In evaluating Gagnon (9) your values for \omega and \delta are off by a factor of 6.28 (see post #134). There may be other errors as well.

The conclusion of the paper gives a peak-to-peak of 19 DEGREES.
Considering that 1 radian corresponds to 57 degrees, it follows that 19 degrees means 0.33 radians.
I think that there may be a typo in the paper : I am evaluating (9) to 0.033 radians meaning 1.9 degrees.

Now, compare that to the measured data of 8x10^(-3) DEGREES inferred from fig 2.

 

[clj4]

This term is the same for both waveguides and disappears on subtraction of the two output signal phases.

Don't think so, there are missing terms and probably a swapping of the meaning of "omegas".

I think I may found the error, I think you may have switched the interpretations of "omegas" in (7)
Probably the correct formula is :\Delta \phi=2\pi(k_2L_2-k_1L_1)=2\pi\frac{L_2}{c}[\omega_2^2[1-\frac{v_x^2}{c^2}]-\omega^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}-2\pi\frac{L_1}{c}[\omega_1^2[1-\frac{v_x^2}{c^2}]-\omega^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}.You will also need to add the non-cancelable missing terms \frac{\omega_1}{c}*\frac{v_z}{c} and \frac{\omega_2}{c}*\frac{v_z}{c}

 

[clj4]

Another source of errors

\omega=40.16 \ GHz
\omega_1=40.11 \ GHz
\omega_2=33.50 \ GHz

Ahem,
f=40.16 \ GHz

\omega=2\pi*f

So your calculations are off by a factor of 6.28.

Out of curiosity, how did you get
f_2=33.50 \ GHz ?

 

[Aether]

In evaluating Gagnon (9) your values for \omega and \delta are off by a factor of 6.28 (see post #134). There may be other errors as well.

That's right. I'll go back and fix those.

Don't think so, there are missing terms and probably a swapping of the meaning of "omegas".

I think I may found the error, I think you may have switched the interpretations of "omegas" in (7)

What makes you think that? \omega isn't defined within the paper, but I'm assuming that it is the common driving (angular) frequency \omega=\omega_1+\delta.

 

[clj4]

What makes you think that? \omega isn't defined within the paper, but I'm assuming that it is the common driving (angular) frequency \omega=\omega_1+\delta.

Because you haven't yet rederived (9) successfully.
The reason (9) is necessary is that you are most likely subtracting two quantities that are very close, so your approach is susceptible to numerical errors. To make matters even worse, in order to convert the result (expressed in radians) to degrees you need to divide by \pi and this introduces even more errors. This is why (9) plays such a critical role in the paper.

By the way, how did you arrive to the value of f_2=33.50 \ GHz
? This is critical for your derivation, I asked this above and I got no answer.

 

[Aether]

Because you haven't yet rederived (9) successfully.

This doesn't answer my question.

The reason (9) is necessary is that you are most likely subtracting two quantities that are very close, so your approach is susceptible to numerical errors. To make matters even worse, in order to convert the result (expressed in radians) to degrees you need to divide by \pi and this introduces even more errors. This is why (9) plays such a critical role in the paper.

If you say so. I'm still working on Eq. (9). My current approach is to try and duplicate the output of Eq. (9) by rotating the "exact" equation from yesterday within the laboratory reference frame, and I have already shown that that's not allowed (e.g., that had better not be what Gagnon et al. did). I'm not saying that I'm sure that this approach will work out, just letting you know what my train of thought is at the moment.

By the way, how did you arrive to the value of f_2=33.50 \ GHz ? This is critical for your derivation, I asked this above and I got no answer.

On page 1769 Gagnon et al. say that "The WR-28 waveguide is driven well above cutoff, giving a phase velocity of about 1.2 times the speed of light". So, \frac{k_0}{k_2}=1.2=\frac{\omega}{(\omega^2-\omega_2^2)^{1/2}}. I was doing that calculation on the fly and gave you (\omega^2-\omega_2^2)^{1/2}=33.50 \GHz. Solving for \omega_2 it is f_2=22.20 \ GHz so \omega_2=1.395\times 10^{11} \ rad/sec.

 

[Aether]

I think that there may be a typo in the paper : I am evaluating (9) to 0.033 radians meaning 1.9 degrees.

Now I get 0.0375 radians (2.149 degrees). This is the amplitude of the predicted signal, so we should double it to get a peak-to-peak value of 4.297 degrees. So, this doesn't look like it's just a typo anymore.

 

[clj4]

If you say so. I'm still working on Eq. (9). My current approach is to try and duplicate the output of Eq. (9) by rotating the "exact" equation from yesterday within the laboratory reference frame, and I have already shown that that's not allowed.

You haven't shown anything except some formulas that look very shaky.

I'm not saying that I'm sure that this approach will work out, just letting you know what my train of thought is at the moment.

OK, carry on.

On page 1769 Gagnon et al. say that "The WR-28 waveguide is driven well above cutoff, giving a phase velocity of about 1.2 times the speed of light". So, \frac{k_0}{k_2}=1.2=\frac{\omega}{(\omega^2-\omega_2^2)^{1/2}}. I was doing that calculation on the fly and gave you (\omega^2-\omega_2^2)^{1/2}=33.50 \GHz. Solving for \omega_2 it is f_2=22.20 \ GHz so \omega_2=1.395\times 10^{11} \ rad/sec.

Thank you. all of the above contradicts your earlier statement that \omega_2=33.50 \ GHz compounded by the error of using \omega_2 as f_2.

Looking over you calculations above I still doubt that you got the values correctly. You have a rather annoying habit of making a lot of mistakes in elementary calculations. (see also below, at post 140)

 

[clj4]

Now I get 0.0375 radians (2.149 degrees). This is the amplitude of the predicted signal, so we should double it to get a peak-to-peak value of 4.297 degrees. So, this doesn't look like it's just a typo anymore.

No, you shouldn't double it. In the author's parlay, peak to peak is taken for the extreme values of \sin^2(\theta) and these are obviously 0 and 1.

 

[Aether]

No, you shouldn't double it. In the author's parlay, peak to peak is taken for the extreme values of \sin^2(\theta) and these are obviously 0 and 1.

OK, so it's 2.149 degrees. How did the author's get 19 degrees then? Do you agree with my new value for \omega_2?

 

[clj4]

OK, so it's 2.149 degrees. How did the author's get 19 degrees then?

I told you originally about this a few posts above. (post 132)

Do you agree with my new value for \omega_2?

No, I don't, see the above post. You'll need to do the elementary calculations over.Look at post 139.

 

[clj4]

On page 1769 Gagnon et al. say that "The WR-28 waveguide is driven well above cutoff, giving a phase velocity of about 1.2 times the speed of light". So, \frac{k_0}{k_2}=1.2=\frac{\omega}{(\omega^2-\omega_2^2)^{1/2}}. I was doing that calculation on the fly and gave you (\omega^2-\omega_2^2)^{1/2}=33.50 \GHz. Solving for \omega_2 it is f_2=22.20 \ GHz so \omega_2=1.395\times 10^{11} \ rad/sec.

In addition to all the problems with your reasoning, the equation above is also patently wrong. The paper text, taken llterally means:

1.2c=\frac{(\omega^2-\omega_2^2)^{1/2}}{k_2}

I doubt that this is what they really mean since later on, they say something that translates into:

20c=\frac{(\omega^2-\omega_1^2)^{1/2}}{k_1}

 

[Aether]

In addition to all the problems with your reasoning, the equation above is also patently wrong. The paper text, taken llterally means:

1.2c=\frac{(\omega^2-\omega_2^2)^{1/2}}{k_2}

I doubt that this is what they really mean since later on, they say something that translates into:

20c=\frac{(\omega^2-\omega_1^2)^{1/2}}{k_1}

On page 1768 they say "Notice that for v=0, the conventional results for waveguide propagation are obtained." Therefore, Eq. (7) reduces to: k_g=\frac{1}{c}[\omega^2-\omega_c^2]^{1/2}.

k_0=\frac{\omega}{c}.
k_1=\frac{1}{c}[\omega^2-\omega_1^2]^{1/2}.
k_2=\frac{1}{c}[\omega^2-\omega_2^2]^{1/2}.

\frac{k_0}{k_1}=\frac{841.6923}{41.9770}=20.0513
\frac{k_0}{k_2}=\frac{841.6923}{701.399}=1.2000

\omega=2.52333\times 10^{11} \ rad/sec
\omega_1=2.52019\times 10^{11} \ rad/sec
\omega_2=1.39487\times 10^{11} \ rad/sec

I rounded L to 2.5 meters before, but I would like to be more precise from now on and use L=2.4384 \ meters.

Using L=2.4384 \ meters I now get 0.0365 rad from Eq. (9) if I use v=400 \ km/s as is suggested on page 1771. However, when I use v=390 \ km/s as is suggested on page 1767 I get 0.0347 rad. Compare this to 0.0343 rad (see post #126) predicted by a 90-degree spatial rotation of the apparatus in the laboratory frame (e.g., the difference between v_x=400 \ km/s and v_z=400 \ km/s).

These values of v are somewhat higher than Gangnon et al. should have used because the CMB rest frame is seldom (if ever) on the horizon. We need to be using a projection of that frame onto the horizontal plane because that is presumably the plane in which Gagnon rotated his apparatus. I'll look into finding the correct projection angle. Also, the velocity (relative to the Earth) of the CMB rest frame varies by about \pm 30 \ km/s depending on the time of year due to the orbit of the Earth around the Sun. I'll also look into that.

 

[clj4]

On page 1768 they say "Notice that for v=0, the conventional results for waveguide propagation are obtained." Therefore, Eq. (7) reduces to: k_g=\frac{1}{c}[\omega^2-\omega_c^2]^{1/2}.

k_0=\frac{\omega}{c}.
k_1=\frac{1}{c}[\omega^2-\omega_1^2]^{1/2}.
k_2=\frac{1}{c}[\omega^2-\omega_2^2]^{1/2}.

\frac{k_0}{k_1}=\frac{841.6923}{41.9770}=20.0513
\frac{k_0}{k_2}=\frac{841.6923}{701.399}=1.2000

OK, so you read between the lines better than I do :-). It was not about phase velocity proper since phase velocity is defined as I showed above.

\omega=2.52333\times 10^{11} \ rad/sec
\omega_1=2.52019\times 10^{11} \ rad/sec
\omega_2=1.39487\times 10^{11} \ rad/sec

I rounded L to 2.5 meters before, but I would like to be more precise from now on and use L=2.4384 \ meters.

Using L=2.4384 \ meters I now get 0.0365 rad from Eq. (9) if I use v=400 \ km/s as is suggested on page 1771. However, when I use v=390 \ km/s as is suggested on page 1767 I get 0.0347 rad. Compare this to 0.0343 rad (see post #126) predicted by a 90-degree spatial rotation of the apparatus in the laboratory frame (e.g., the difference between v_x=400 \ km/s and v_z=400 \ km/s).

In #126 you were telling us (textually :

"This predicts a maximum daily phase shift of 8x10^(-3) DEGREES (when absolute motion is aligned with the z-axis). "

So, it looks like you changed your mind. Are you saying that now your formula shows 0.0347 RADIANS? That is 1.9 DEGREES (or so) ? Then we are done. You know why.

 

[Aether]

In #126 you were telling us (textually :

"This predicts a maximum daily phase shift of 8x10^(-3) DEGREES (when absolute motion is aligned with the z-axis). "

So, it looks like you changed your mind.

No, not at all. I told you that "My current approach is to try and duplicate the output of Eq. (9) by rotating the "exact" equation from yesterday within the laboratory reference frame, and I have already shown that that's not allowed." I have now done that. My next step will be to plot the output from Eqs. (7) and (9) side-by-side over a 90-degree rotation in the x-z plane to show that they are substantially similar.

Are you saying that now your formula shows 0.0347 RADIANS? That is 1.9 DEGREES (or so) ?

This isn't my formula, it is Eqs. (7) & (9). Yes, it shows 0.0343 radians and it closely approximated this yesterday too. However, this isn't what you (seem to) think (or maybe it is, we'll see). The rotation that gives 0.0343 radians is for the CMB rest frame rotating in the x-z plane, not the waveguides. Eq. (7) is only valid while the waveguides are "lying along the z direction of the laboratory coordinate system". If the rotation of the Earth can accomplish this rotation of the CMB rest frame in the x-z plane while leaving the waveguides lying along the z direction of the laboratory coordinate system, then I'll want to know if Mansouri-Sexl leads to Eqs. (7) & (9).

Then we are done. You know why.

Not until you admit that Gagnon et al. is toast we aren't.

Even if this turns out to be a case of "experiment contradicts theory", we're not done until this traces back to Mansouri-Sexl. I'm not sure that GGT is the same as RMS. Note, Gagnon et al. do not even reference Mansouri-Sexl directly.

 

[clj4]

No, not at all. I told you that "My current approach is to try and duplicate the output of Eq. (9) by rotating the "exact" equation from yesterday within the laboratory reference frame, and I have already shown that that's not allowed." I have now done that.

BS. You haven't done anything but to do the first steps that lead to (9). That's all.

My next step will be to plot the output from Eqs. (7) and (9) side-by-side over a 90-degree rotation in the x-z plane to show that they are substantially similar.

How could you? (7) represents a wave number and (9) represents a phase differential. What sort of physics is this?

This isn't my formula, it is Eq. (7).

Hmmm...would you care to show us how you did your calculations?
How could it be, the output of (7) is a wave number. Wave number expressed in radians?
How do all the THREE "omegas" intervene in (7)? You have fed us so much BS in the past that I am sorry to say but I don't believe that you are giving us the output of (7).

Then keep working on your formula, the one that gave us 8*10(-3) degrees in post 126. The one that has all the omegas and both k's. The one that is supposed to be equivalent to (9).

Yes, it shows 0.0347 radians and it closely approximated this yesterday too. However, this isn't what you (seem to) think. The rotation that gives 0.0347 radians is for the CMB rest frame rotating in the x-z plane, not the waveguides. Eq. (7) is only valid while the waveguides are "lying along the z direction of the laboratory coordinate system". If the rotation of the Earth can accomplish this rotation of the CMB rest frame in the x-z plane while leaving the waveguides lying along the z direction of the laboratory coordinate system, then I'll want to know if Mansouri-Sexl leads to Eq. (9).

Not until you admit that Gagnon et al. is toast we aren't.

Even if this turns out to be a case of "experiment contradicts theory", we're not done until this traces back to Mansouri-Sexl. I'm not sure that GGT is the same as RMS. Note, Gagnon et al. do not even reference Mansouri-Sexl directly.

This is just a secondary point. Spare us the diversions. The primary point is that you set to disprove Gagnon. Stick to the subject. If you manage to prove that "Gagnon is toast" you get the next paper in the series.

 

[Aether]

Hmmm...would you care to show us how you did your calculations? How do all the THREE "omegas" intervene in (7)? You have fed us so much BS in the past that I am sorry to say but I don't believe that you are giving us the output of (7).

Apply Eq. (7) to a WR-28 waveguide to get a first guide wave number k_1, and multiply that by the length of the first waveguide L_1 to get a first phase; then apply Eq. (7) to a near-cutoff waveguide to get a second guide wave number k_2, and multiply that by the length of the second waveguide L_2 to get a second phase; subtract the first phase from the second phase to get \Delta \phi.

 

[clj4]

Apply Eq. (7) to a WR-28 waveguide to get a first guide wave number k_1, and multiply that by the length of the first waveguide L_1 to get a first phase; then apply Eq. (7) to a near-cutoff waveguide to get a second guide wave number k_2, and multiply that by the length of the second waveguide L_2 to get a second phase; subtract the first phase from the second phase to get \Delta \phi.

You mean this:

\Delta \phi=2\pi(k_2L_2-k_1L_1)=2\pi\frac{L_2}{c}[\omega_2^2[1-\frac{v_x^2}{c^2}]-\omega^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}-2\pi\frac{L_1}{c}[\omega_1^2[1-\frac{v_x^2}{c^2}]-\omega^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}.

(9a)
This is not (7), this is where you left off in your quest to get to (9). And this is exactly what I referred to in post 145. Let's call it (9a). So , let me ask you one more time :
1. is the output of (9a) now equal to 0.0347 radians or (2 degrees)? The one that you claimed to be 8*10^(-3) DEGREES in post 126? Yes or no?

2. is (9a) what you plan to "rotate"? Yes or No?

 

[Aether]

You mean this:

\Delta \phi=2\pi(k_2L_2-k_1L_1)=2\pi\frac{L_2}{c}[\omega_2^2[1-\frac{v_x^2}{c^2}]-\omega^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}-2\pi\frac{L_1}{c}[\omega_1^2[1-\frac{v_x^2}{c^2}]-\omega^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}.

(9a)
This is not (7), this is where you left off in your quest to get to (9). And this is exactly what I referred to in post 145. Let's call it (9a).

Yes, that's what I mean, that's Eq. (7) applied to the experiment.

So , let me ask you one more time : is the output of (9a) now equal to 0.347 radians? The one that you claimed to be 8*10(-3) DEGREES in post 126? Yes or no?

The difference between (9a) when the absolute velocity is in the x-direction vs. when the absolute velocity is in the z-direction is 0.0347 radians (using L-2.5 meters, I think that it may be closer to Eq. (9) using the new value of L).

With the absolute velocity staying fixed in the z-direction, the difference between (9a) due to 1/2 rotation of the Earth (1 km/s change in absolute velocity in the z-direction) is approximately 10\times 10^{-3} degrees.

Note that in post #126 I said "Here's a Kennedy-Thorndike type analysis (taking into account the Earth's rotation only)". I gave separate K-T analyses for absolute motion aligned with the x-axis, and absolute motion aligned with the z-axis. The difference in \Delta \phi for these two alignments of the absolute motion is 0.0347 rad.