Consistency of the speed of light

[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.

 

[clj4]

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

Interesting...I tried to do the numerical evaluation and I stumbled on the fact that the expression:

$$[\omega_2^2[1-\frac{v_x^2}{c^2}]-\omega^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}$$

happens to be imaginary.

 

[Aether]

Interesting...I tried to do the numerical evaluation and I stumbled on the fact that the expression:

$$[\omega_2^2[1-\frac{v_x^2}{c^2}]-\omega^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}$$

happens to be imaginary.

No wait, you have edited that equation. That isn't what I wrote in post #126.

 

[clj4]

No wait, you have edited that equation. That isn't what I wrote in post #126.

You are right.

 

[Aether]

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)...

You are right.

No wonder. :rofl:

 

[clj4]

No wonder. :rofl:

Not so fast again. Plugging in your omegas from post 144 in (9a) gives a much bigger number than the one you claim. Would you care to show how you get 0.0347?

 

[Aether]

Not so fast again. Plugging in your omegas in (9a) gives a much bigger number than the one you claim. Would you care to show how you get 0.0347?

OK.

This is how I got 0.0347 rad from the calculations in post #126. I'll re-do these calculations in a later post using $$L_1=L_2=2.4384 \meters$$, and plot Eq. (9) next to $$\Delta \phi(\theta)=k_2(\theta)L_2-k_1(\theta)L_1$$ over a 90-degree range of theta.

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

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 $$\Delta \phi$$.

In order to more plainly show how different measured output phase differentials are compared, let's label the various $$\Delta \phi$$ and $$k$$ functions with their absolute velocity components from now on:

For $$v_x=400 \ km/s$$ and $$v_z=0$$:
$$\Delta \phi (400000,0,0)=1753.498118-104.942309=1648.555808 \ rad$$

For $$v_x=0$$ and $$v_z=400 \ km/s$$:
$$\Delta \phi (0,0,400000)=1753.500365-104.979859=1648.520506 \ rad$$

$$(k_1(400000,0,0)+\frac{\omega}{c}\frac{v_z}{c})L_1=\frac{L_1}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_1^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}=104.942309 \ rad$$

$$(k_2(400000,0,0)+\frac{\omega}{c}\frac{v_z}{c})L_2=\frac{L_2}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_2^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}=1753.498118 \ rad$$

$$\Delta \phi(400000,0,0)=(k_2(400000,0,0)+\frac{\omega}{c}\frac{v_z}{c})L_2-(k_1(400000,0,0)+\frac{\omega}{c}\frac{v_z}{c})L_1=1648.555808 \ rad$$

$$\Delta \phi(0,0,400000)=(k_2(0,0,400000)+\frac{\omega}{c}\frac{v_z}{c})L_2-(k_1(0,0,400000)+\frac{\omega}{c}\frac{v_z}{c})L_1=1648.520506 \ rad$$

$$\Delta \phi(400000,0,0)-\Delta \phi(0,0,400000)=1648.555808-1648.520506=0.035302 \ rad$$

This is the change in $$\Delta \phi$$ predicted for a 90-degree rotation of the apparatus. This number will be slightly lower when re-calculated below using $$L_1=L_2=2.4384 \meters$$.

 

[clj4]

OK.$$(k_1+\frac{\omega}{c}\frac{v_z}{c})L_1=\frac{L_1}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_1^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}=104.942309 \rad$$

$$(k_2+\frac{\omega}{c}\frac{\v_z}{c})L_2=\frac{L_2}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_2^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}=1753.498118 \ rad$$

$$\Delta \phi=(k_2+\frac{v_z}{c}\frac{v_z}{c})L_2-(k_1+\frac{v_z}{c}\frac{v_z}{c})L_1=1648.555808 \ rad$$

Do you want me to show more detail for any of these terms?

You are missing the $$2\pi$$ factor again (see your other formula).
But the main point is how do you arrive from thousands of radians to 0.0347?

 

[clj4]

OK.Let's assume that $$L_1=L_2=L$$

$$(k_1+\frac{\omega}{c}\frac{v_z}{c})L=\frac{L}{c}[\omega^2[1-\frac{v_x^2}{c^2}]-\omega_1^2[1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2}]]^{1/2}=104.942309 \rad$$

Where is this coming from? You have never shown this before. I thought that we were talking about (9a)

 

[Aether]

You are missing the $$2\pi$$ factor again (see your other formula).

OK, give me a few minutes to finish constructing that post. The $$2\pi$$ factor was there originally because I was taking the omegas to be frequencies (as seemed to be implied throughout Gagnon), but I removed it when I changed the units on the omegas to radians.

But the main point is how do you arrive from thousands of radians to 0.0347?

That is how Kennedy-Thordike experiments work. You record two different $$\Delta \phi$$ measurements (usually at different times of the year, but it can also be done at different times of the day). The 0.0347 radians is the difference between the two $$\Delta \phi$$ measurements, one with $$v_x=400 \ km/s$$ and the other with $$v_z=400 \ km/sec$$.

 

[clj4]

$$\Delta \phi=(k_2+\frac{\omega}{c}\frac{v_z}{c})L-(k_1+\frac{\omega}{c}\frac{v_z}{c})L_1=1648.555808 \ rad$$

This is new as well. How did you arrive to this new defiinition of

$$\Delta \phi$$

 

[Aether]

This is new as well. How did you arrive to this defiinition of

$$\Delta \phi$$

Let me finish constructing that post first. The "new" terms will cancel out, I'm just showing them here because you wanted to see my intermediate steps.

 

[clj4]

OK, give me a few minutes to finish constructing that post. The $$2\pi$$ factor was there originally because I was taking the omegas to be frequencies (as seemed to be implied throughout Gagnon), but I removed it when I changed the units on the omegas to radians.

That is how Kennedy-Thordike experiments work. You record two different $$\Delta \phi$$ measurements (usually at different times of the year, but it can also be done at different times of the day). The 0.0347 radians is the difference between the two $$\Delta \phi$$ measurements, one with $$v_x=400 \ km/s$$ and the other with $$v_z=400 \ km/sec$$.

Would you care to show the complete calculations? This is getting to look like a mess.

 

[Aether]

Would you care to show the complete calculations? This is getting to look like a mess.

Sure, I've got to run as soon as I repair a few things in post #156. I'll return later to fill-in the intermediate details.

 

[Aether]

Would you care to show the complete calculations? This is getting to look like a mess.

I have attached a text file containing the output (and source code) of a high-precision (e.g., 1000 digit computation(s) with output truncated at 9 decimal places) numerical comparison of the output of Eq. (9) of Gagnon et al. (described as an “approximation” by the authors) and my Eq. (9a) which applies Eq. (7) to the experiment through a 90-degree rotation in the x-z plane. Eq. (9) appears to be reasonably approximating the experimental hypothesis of Gagnon et al.. The textual claim appearing in (Gagnon et al., 1988) that a “peak-to-peak phase shift of at least 19-degrees is predicted as the apparatus turns in the laboratory” is apparently over-estimated by an order of magnitude (e.g., a peak-to-peak phase shift of 1.9-degrees seems to be more consistent with Eq. (9) of the paper).

The key functions implemented are:

Eq. (7):
$$k_1(v_x,v_y,v_z)=-\frac{\omega}{c_0}\frac{v_z}{c_0}+\frac{1}{c_0}[\omega^2(1-\frac{v_x^2}{c_0^2})-\omega_1^2(1-\frac{v_x^2}{c_0^2}-\frac{v_z^2}{c_0^2})]^{1/2}$$

$$k_2(v_x,v_y,v_z)=-\frac{\omega}{c_0}\frac{v_z}{c_0}+\frac{1}{c_0}[\omega^2(1-\frac{v_x^2}{c_0^2})-\omega_2^2(1-\frac{v_x^2}{c_0^2}-\frac{v_z^2}{c_0^2})]^{1/2}$$

Eq. (9):
$$\Delta \phi_9(v_x,v_y,v_z)=\phi_0+\frac{\omega_1L}{2c_0}[\frac{\omega_1}{2\delta}]^{1/2}\frac{v^2}{c_0^2}sin^2(\theta)$$

Eq. (9a):
$$\Delta \phi_{9a}(v_x,v_y,v_z)=k_2(v_x,v_y,v_z)L_2-k_1(v_x,v_y,v_z)L_1$$

Within this analysis I have arbitrarily defined $$\phi_0$$ as:
$$\phi_0=\Delta \phi_{9a}(v,0,0)-\Delta \phi_9(v,0,0)$$

This should allow us to focus on Eqs. (5) through (8) of Gagnon now. We may also return to Eq. (9a) later in order to apply length-contractions to $$L_1$$, and $$L_2$$.

 

[clj4]

So,

1. At 0 AM

$$\theta=0 degrees , v_x=0 , v_z=390km/sec$$

2. At 6 AM
$$\theta=90 degrees , v_z=0 , v_x=390km/sec$$

Now, if we work the peak to peak difference between the phase difference at 0AM and 6AM respectively we should get 2 degrees whether we use the author's formula (9) or the formulas that you derived (9a).

 

[Aether]

So,

1. At 0 AM

$$\theta=0 degrees , v_x=0 , v_z=390km/sec$$

2. At 6 AM
$$\theta=90 degrees , v_z=0 , v_x=390km/sec$$

Now, if we work the peak to peak difference between the phase difference at 0AM and 6AM respectively we should get 2 degrees whether we use the author's formula (9) or the formulas that you derived (9a).

If you plug Eq. (7) into Eq. (9a), then that is correct. However, I will now argue that Eq. (6) is invariant over rotations in the x-y plane, but Eqs. (7) and (8) (which were derived using Eq. (6)) are not, and therefore Eqs. (7) and (8) are not generally valid (e.g., they are only valid, if at all, when $$v_y=0$$).

For a waveguide lying along the z direction of the laboratory-coordinate system:
Eq. (6):
$$E(x,y,z)=E(x,y)exp(ikz-i\omega t)$$

Eq. (7):
$$k_g(v_x,v_y,v_z)=-\frac{\omega}{c_0}\frac{v_z}{c_0}+\frac{1}{c_0}[\omega^2(1-\frac{v_x^2}{c_0^2})-\omega_c^2(1-\frac{v_x^2}{c_0^2}-\frac{v_z^2}{c_0^2})]^{1/2}$$

Eq. (8):
$$\omega_c(v_x,v_y,v_z)=\omega_{mn}[1-\frac{v_x^2}{c_0^2}-\frac{v_z^2}{c_0^2}]^{-1/2}$$

Therefore, I propose (provisionally) the following two new versions of these equations which are both invariant over rotations in the x-y plane:

Eq. (7a):
$$k_z(v_x,v_y,v_z)=-\frac{\omega}{c_0}\frac{v_z}{c_0}+\frac{1}{c_0}[\omega^2(1-\frac{v_x^2}{c_0^2}-\frac{v_y^2}{c_0^2})-\omega_z^2(1-\frac{v_x^2}{c_0^2}-\frac{v_y^2}{c_0^2}-\frac{v_z^2}{c_0^2})]^{1/2}$$

Eq. (8a):
$$\omega_i(v_x,v_y,v_z)=\omega_{mni}[1-\frac{v_x^2}{c_0^2}-\frac{v_y^2}{c_0^2}-\frac{v_z^2}{c_0^2}]^{-1/2}=\gamma (v_x,v_y,v_z)\omega_{mni}$$

If this is acceptable, then I will also propose (provisionally) four new equations (6a), (7b), (6b), and (7c) for waveguides lying along the x and y directions of the laboratory-coordinate system respectively (see post #124):

For a waveguide lying along the x direction of the laboratory-coordinate system:
Eq. (6a):
$$E(x,y,z)=E(y,z)exp(ikx-i\omega t)$$

Eq. (7b):
$$k_x(v_x,v_y,v_z)=-\frac{\omega}{c_0}\frac{v_x}{c_0}+\frac{1}{c_0}[\omega^2(1-\frac{v_y^2}{c_0^2}-\frac{v_z^2}{c_0^2})-\omega_z^2(1-\frac{v_x^2}{c_0^2}-\frac{v_y^2}{c_0^2}-\frac{v_z^2}{c_0^2})]^{1/2}$$

For a waveguide lying along the y direction of the laboratory-coordinate system:
Eq. (6b):
$$E(x,y,z)=E(x,z)exp(iky-i\omega t)$$

Eq. (7c):
$$k_y(v_x,v_y,v_z)=-\frac{\omega}{c_0}\frac{v_y}{c_0}+\frac{1}{c_0}[\omega^2(1-\frac{v_x^2}{c_0^2}-\frac{v_z^2}{c_0^2})-\omega_z^2(1-\frac{v_x^2}{c_0^2}-\frac{v_y^2}{c_0^2}-\frac{v_z^2}{c_0^2})]^{1/2}$$

If this is acceptable, then it is apparent that we could use either (7b), (7c), or (7a) to complete a $$k_i$$-invariant rotation of the apparatus about any of the three respective spatial axes. However, Eq. (7) leads us to conclude that we can't (in general) complete a $$k_z$$-invariant rotation of the apparatus about the y axis. Eq. (7c) generally contradicts both Eq. (7) and Eq. (7a) for rotations in the x-z plane.

 

[clj4]

Why would we complicate things this way? The lab (the Earth) is assumed to rotate in the x-z plane (y is assumed to be the axis going through poles, $$v_y=0$$).So the equations given in the paper (and your equivalent (9a)) are necessary and sufficient in describing the experiment.

Eq. (7c) generally contradicts both Eq. (7) and Eq. (7a) for rotations in the x-z plane.

Not so.

(7a),(8a) are nothing but (7) and (8) for $$v_y=0$$ which is the case of the experiment. The experiment clearly assumes $$v_y=0$$. There is no need for the newly introduced (7a),(7b),(7c).

 

[Aether]

Why would we complicate things this way? The lab (the Earth) is assumed to rotate in the x-z plane (y is assumed to be the axis going through poles, $$v_y=0$$).So the equations given in the paper (and your equivalent (9a)) are necessary and sufficient in describing the experiment.

Eq. (7) does not appear (to me) to be generally valid over a rotation in the x-z plane. If this is true, then the equations given are not necessary and sufficient for correctly describing the experiment.

Not so.

(7a),(8a) are nothing but (7) and (8) for $$v_y=0$$ which is the case of the experiment. The experiment clearly assumes $$v_y=0$$. There is no need for the newly introduced (7a),(7b),(7c).

I have introduced them to show that Eq. (7) does not appear (to me) to be generally valid over a rotation in the x-z plane.

 

[clj4]

Eq. (7) does not appear (to me) to be generally valid over a rotation in the x-z plane. If this is true, then the equations given are not necessary and sufficient for correctly describing the experiment.

I have introduced them to show that Eq. (7) does not appear (to me) to be generally valid over a rotation in the x-z plane.

I disagree. Your equation is nothing but (7) with $$v_y=0$$ which also happens to be the case in the actual experiment. Looks like a last ditch diversion.

 

[Aether]

I disagree. Your equation is nothing but (7) with $$v_y=0$$ which also happens to be the case in the actual experiment.

Eq. (7a) helps establish the pattern that I used to generate Eq. (7c), and Eq. (7c) generally contradicts Eq. (7). I am still working toward this, to "try to give a generalized expression for the guide wave number that is valid (or at least consistent with Gagnon's Eq. (6)) for any angle between the absolute velocity vector and the waveguide." I don't think that Eq. (7) does that, not even for rotations in the x-z plane.

Looks like a last ditch diversion.

I called this shot in post #124.

 

[clj4]

Eq. (7a) helps establish the pattern that I used to generate Eq. (7c), and Eq. (7c) generally contradicts Eq. (7). I am still working toward this, to "try to give a generalized expression for the guide wave number that is valid (or at least consistent with Gagnon's Eq. (6)) for any angle between the absolute velocity vector and the waveguide." I don't think that Eq. (7) does that, not even for rotations in the x-z plane.

I called this shot in post #124.

Make $$v_y=0$$ , (7) and (7a) are identical .
Now, if you want something really general, you shouldn't simply permute the indexes but you should derive (7) under the circumstance that the two waveguides make an angle $$\theta$$ with the vector v. I think that this is exactly what (9) is. I don't think that your simple permutation of indexes produces what you are after.

If you wanted to do this, I think that you would need a better definition of the "field of the waveguide" (6):

$$\exp(i<k,\frac{r}{|r|}>-i\omega*t)$$ instead of

$$\exp(ikz-i\omega*t)$$

where <*,*> is the dot product between the unit vector k and the unit positional vector $$\frac{r}{|r|}$$

Transform Gagnon equation (5) into polar coordinates in the x-z plane and work things from base principles
For example:
Set
$$z=r*cos(\theta)$$
$$x=r*sin(\theta)$$
with $$\theta$$ varying from 0 to $$\pi/2$$
I am willing to bet that the final outcome will be a much more ellegant derivation of Gagnon (9)

I am willing to work with you on this.

 

[Aether]

Make $$v_y=0$$ , (7) and (7a) are identical .
Now, if you want something really general, you shouldn't simply permute the indexes but you should derive (7) under the circumstance that the two waveguides make an angle $$\theta$$ with the vector v. I think that this is exactly what (9) is. I don't think that your simple permutation of indexes produces what you are after.

I only showed that to prove that Eq. (7) is not generally valid. It still remains to be seen exactly what is generally valid.

If you wanted to do this, I think that you would need a better definition of the "field of the waveguide" (6):

$$\exp(i<k,\frac{r}{|r|}>-i\omega*t)$$ instead of

$$\exp(ikz-i\omega*t)$$

where <*,*> is the dot product between the unit vector k and the unit positional vector $$\frac{r}{|r|}$$

Something like that, yes.

Transform Gagnon equation (5) into polar coordinates in the x-z plane and work things from base principles
For example:
Set
$$z=r*cos(\theta)$$
$$x=r*sin(\theta)$$
with $$\theta$$ varying from 0 to $$\pi/2$$
I am willing to bet that the final outcome will be a much more ellegant derivation of Gagnon (9)

If so, then we look at Eq. (5). If not, and Gagnon falls, then we may still want to look at Eq. (5) and ref (9) anyway. We'll need to see full 3D rotations before moving on in any case.

I am willing to work with you on this.

Thank-you. I am studying ref (10) at the moment.

 

[clj4]

I only showed that to prove that Eq. (7) is not generally valid. It still remains to be seen exactly what is generally valid.

You haven't proved that "(7) is not generally valid". As you can see I rejected all the "made up" formulas (7b),(7c) as being baseless. This is also why I suggested that you worked from base principles. Actually you haven't proved anything yet. Until you do, (7) stands.

 

[Aether]

Immediately following Eq. (5), Gagnon et al. state the following: “Substituting the usual solution for a plane wave traveling in the z direction, i.e., $$E(z)=Eexp(ikz-i\omega t)}$$ and solving for k, we find that the wave number corresponds to a wave with phase velocity c+v, to the first order of approximation. The classical velocity addition is thus obtained for electromagnetic waves moving in a reference frame.” Eq. (7) is given soon thereafter as:

$$k_g=-\frac{\omega}{c_0}\frac{v_z}{c_0}+\frac{1}{c_0}[\omega^2(1-\frac{v_x^2}{c_0^2})-\omega_c^2(1-\frac{v_x^2}{c_0^2}-\frac{v_z^2}{c_0^2})]^{1/2}$$ (Eq. (7)).​

This term in Eq. (7):

$$\omega_c^2(1-\frac{v_x^2}{c_0^2}-\frac{v_z^2}{c_0^2})$$​

is invariant over rotations in the x-z plane, so we may simplify our analysis by applying Eq. (7) to the case of an unguided electromagnetic wave traveling in a vacuum along the z-direction of the laboratory-coordinate system (e.g., where $$\omega_c=0$$):

$$k_0(v_x,0,v_z)= \omega \frac{(1-\frac{v_x^2}{c_0^2})^{1/2}-\frac{v_z}{c_0}}{c_0}$$ (Eq. (7d)).​

The vacuum wave number for an unguided electromagnetic wave in an isotropic coordinate system (such as SR, or the $$\Sigma$$ frame of RMS) is:

$$k_0=\frac{\omega}{c_0}$$ (Eq. (7e)),​

and I will now argue that the vacuum wave number for an unguided electromagnetic wave in an anisotropic coordinate system such as RMS is:

$$k_0(v_z)=\frac{\omega}{c_z}$$ (Eq. (7f)​

where $$c_z$$ is the one-way speed of light in the longitudinal z-direction of propagation…Note: the round-trip speed of light is isotropic in RMS as well as SR, so the average speed of light in the transverse direction is always $$c_0$$...see post #92).

Now, let’s consider two special cases of Eq. (7d): 1) Absolute motion is in the longitudinal direction of wave propagation:

$$k_0(0,0,v_z)= \omega \frac{(1-\frac{v_z}{c_0})}{c_0}$$ (Eq. (7g));​

and 2) Absolute motion is transverse to the direction of wave propagation:

$$k_0(v_x,0,0)= \omega \frac{(1-\frac{v_x^2}{c_0^2})^{1/2}}{c_0}$$ (Eq. (7h)).​

Eq. (7g) closely approximates Eq. (7f) if:

$$\frac{1}{c_z}=\frac{1}{c_0+v_z}$$,​

and Eq. (7h) closely approximates (to second order) Eq. (7f) if:

$$\frac{1}{c_z}=\frac{\frac{1}{c_0+v_x}+\frac{1}{c_0-v_x}}{2}$$​

(round trip speed of light is always $$c_0$$ in SR and RMS, but not in Galileian relativity). However, in the RMS coordinate system Eq. (7f) evaluates to:

$$k_0(0,0,v_z)=\frac{\omega}{c_z}=\omega \frac{c_0+v_z}{c_0^2}$$ ,​

and for absolute motion normal to the z-direction of wave propagation:

$$k_0(v_x,0,0)=\omega (\frac{\frac{c_0+v_x}{c_0^2}+\frac{c_0-v_x}{c_0^2}}{2})=\frac{\omega}{c_0}$$.​

Gagnon et al. appear to have based Eq. (7) on a first order approximation to a wave’s phase velocity (e.g., the Galilean transform) rather than the GGT/RMS transform, and their experimental hypothesis appears to be directed ((un)intentionally?) toward a refutation of Galilean relativity as opposed to RMS.

 

[clj4]

The above is unreadable, could you reformat it a little? Thank you