Consistency of the speed of light

[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.