Consistency of the speed of light

[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.

 

[clj4]

No wonder.

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

 

[clj4]

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 $$\omega_c=0$$):

Thank you, this is much better.

But $$\omega_c$$ is clearly not 0 in the experiment, so I cannot accept (7d). I thought that we were done with the ad-hoc invention of formulas, that you were going to rederive things from base principles. You come back with the same thing. can you solve partial differential equations?
K is the solution of the partiial differential equation (5), you cannot keep coming with cooked up formulas. If you want to refute the paper you need to solve (5) from base principles. I gave you a tool, the conversion to polar coordinates. On a different issue, k and $$\omega$$ are variables that get tied together by (5), so this should be your starting point, not all the different speculations as to how to connect the two, They are connected by equation (5).

 

[Aether]

Thank you, this is much better.

But $$\omega_c$$ is clearly not 0 in the experiment, so I cannot accept (7d). I thought that we were done with the ad-hoc invention of formulas, that you were going to rederive things from base principles. You come back with the same thing. can you solve partial differential equations?
K is the solution of the partiial differential equation (5), you cannot keep coming with cooked up formulas. If you want to refute the paper you need to solve (5) from base principles.

OK, I'm working up to that.

I gave you a tool, the conversion to polar coordinates.

You said that $$k$$ was a "unit vector", but it is defined on p. 212 of ref (10) as: $$k^2=(\frac{\pi p}{a})^2+(\frac{\pi q}{b})^2$$ where p=1, q=0, (in this experiment) and (a,b) are the interior cross-sectional dimensions of a waveguide. Apparently, how the waveguide coordinates $$(x,y,z)=(a,b,L)$$ transform over a rotation of the apparatus (combined with the one-way speed of light) is what ultimately determines the outcome of this experiment/analysis. We can expect that any anisotropy in the one-way speed of light due to RMS is going to be exactly offset by how these spatial coordinates transform; not so in Galileian relativity however.

On a different issue, k and $$\omega$$ are variables that get tied together by (5), so this should be your starting point, not all the different speculations as to how to connect the two, They are connected by equation (5).

This is how $$k$$ (in addition to the equation that I just gave) and $$\omega$$ are defined:

...bearing in mind that $$k$$ is independent of $$\omega$$ and $$\beta$$--it is a constant depending only on the mode concerned and the geometry of the waveguide cross-section.

The "mode concerned" is/are an integer(s), so $$k$$ varies with the geometry of the waveguide cross-section only.

Since it is phase locked to a reference oscillator which is at rest in the laboratory, the output frequency of the klystron [$$f=\frac{\omega}{2\pi}$$] is unaffected by rotation of the apparatus.

 

[777tiM777]

prior to einsteins special theory of relativity maxwell proved (theoretically) that the propagation of electromagnetic waves is 3.00*10^8 m/s (c = 1/sqrt(e0*u0)), all einstein did essentially in his special theory of relativity was expand galilean relativity(which stated something along the lines that the velocity viewed from different frames of reference will not be the same, but the laws such as the law of conservation of eneregy will still hold true in any inertial frame of reference) to say that all laws of physics hold true in all inertial frames of reference, and since maxwell's equations showed that the speed of light is constant it fell under the "all laws of physics"...im really sorry if someone else posted this argument before me, i didnt look through all the posts... hope this still helps answer your question

 

[Aether]

You come back with the same thing. can you solve partial differential equations? K is the solution of the partiial differential equation (5), you cannot keep coming with cooked up formulas. If you want to refute the paper you need to solve (5) from base principles.

I ordered a couple of advanced calculus textbooks to study partial differential equations in more depth; it will be at least a few weeks before I solve (5) for k. I think that I'll probably also need that to derive an anisotropic RMS transformation tensor for use with Eq. (6) from ref (9). I'll use that to transform both k and Eq. (8b) independent of Eq. (5).

In ref (10) the mode of the waveguide is identified by (p,q) rather than (m,n) as is used in Gagnon et al. (I will use (p,q) to identify the mode and reserve (m,n) for a different purpose), and on pp. 212-213 the cutoff angular frequency is defined as:

$$\omega_{pq}=[p^2\pi^2\epsilon \mu \frac{c_0^2}{a_0^2}+q^2\pi^2\epsilon \mu \frac{c_0^2}{b_0^2}]^{1/2}$$ Eq. (8b),​

where $$\epsilon \mu=1$$ for a vacuum “filled” waveguide.

According to Gagnon et al., Eq. (8) is supposed to transform a cutoff angular frequency $$\omega_{mn}$$ in the absolute frame into a cutoff angular frequency $$\omega_c$$ in a moving frame:

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

however, this seems to be inconsistent with the RMS transforms (their Eq. (1), also see post #92):

$$x=\gamma(x_0-vt_0),\ t=\gamma^{-1},\ y=y_0,\ z=z_0$$ Eq. (1).​

For example, with $$(p,q)=(1,0)$$, $$\epsilon \mu =1$$, where $$a_0$$ is a distance along the x-axis and absolute motion is along the x-axis, then Eq. (1) transforms Eq. (8b) into Eq. (8c) (note that transverse wave motion is round-trip averaged):

$$\omega_c=\frac{2\pi c_0}{(a_0-vt_0)+(a_0+vt_0)}[1-\frac{v_x^2}{c_0^2}]^{1/2}=\frac{\pi c_0}{a_0}[1-\frac{v_x^2}{c_0^2}]^{1/2}$$ Eq. (8c);​

but Eq. (8) transforms Eq. (8b) into Eq. (8d) (the exponent in (8d) is of the opposite sign as in (8c)):

$$\omega_c=\pi \frac{c_0}{a_0}[1-\frac{v_x^2}{c_0^2}]^{-1/2}$$ Eq. (8d);​

 

[clj4]

I ordered a couple of advanced calculus textbooks to study partial differential equations in more depth; it will be at least a few weeks before I solve (5) for k. I think that I'll probably also need that to derive an anisotropic RMS transformation tensor for use with Eq. (6) from ref (9). I'll use that to transform both k and Eq. (8b) independent of Eq. (5).

In ref (10) the mode of the waveguide is identified by (p,q) rather than (m,n) as is used in Gagnon et al. (I will use (p,q) to identify the mode and reserve (m,n) for a different purpose), and on pp. 212-213 the cutoff angular frequency is defined as:

$$\omega_{pq}=[p^2\pi^2\epsilon \mu \frac{c_0^2}{a_0^2}+q^2\pi^2\epsilon \mu \frac{c_0^2}{b_0^2}]^{1/2}$$ Eq. (8b),​

where $$\epsilon \mu=1$$ for a vacuum “filled” waveguide.

According to Gagnon et al., Eq. (8) is supposed to transform a cutoff angular frequency $$\omega_{mn}$$ in the absolute frame into a cutoff angular frequency $$\omega_c$$ in a moving frame:

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

however, this seems to be inconsistent with their Eq. (1):

$$x=\gamma(x_0-vt_0),\ t=\gamma^{-1},\ y=y_0,\ z=z_0$$ Eq. (1).​

For example, with $$(p,q)=(1,0)$$, $$\epsilon \mu =1$$, where $$a_0$$ is a distance along the x-axis and absolute motion is along the x-axis, then Eq. (1) transforms Eq. (8b) into Eq. (8c) (note that transverse wave motion is round-trip averaged):

$$\omega_c=\frac{2\pi c_0}{(a_0-vt_0)+(a_0+vt_0)}[1-\frac{v_x^2}{c_0^2}]^{1/2}=\frac{\pi c_0}{a_0}[1-\frac{v_x^2}{c_0^2}]^{1/2}$$ Eq. (8c);​

but Eq. (8) transforms Eq. (8b) into Eq. (8d) (the exponent in (8d) is of the opposite sign as in (8c)):

$$\omega_c=\pi \frac{c_0}{a_0}[1-\frac{v_x^2}{c_0^2}]^{-1/2}$$ Eq. (8d);​

You are inverting (8b), right? I would expect to see a term in $$\omega_{pq}$$ in (8c). I am not seeing it. Could you show the steps, at a superficial view, if I look at (8b) and at (1), i would expect to see the exponent $${-1/2}$$ in (8c)
Who are a0 and b0?