Consistency of the speed of light

[Hans de Vries]

At the very root of the issue lies a much bigger thing: the incorrect belief that the Mansouri-Sexl "aether" theory is indistiguishable from SR under ANY circumstances.

I agree.

This belief may look defendable from a pure theoretical point of view,
for practical purposes it becomes problematic. A good example is the
boundary condition of $$\bf{E}_\parallel=0$$.

If one uses a voltmeter with two probes to measure a static electric
field in the neighborhood of a conductor then one finds that the parallel
component is zero. However, it’s not supposed to be zero in a GGT
frame

To do the same measurement correctly for GGT one needs to measure
in the past with one probe and in the future with the other probe. The
problem is that one can't measure absolute voltages with a voltmeter
and subtract the results later. The voltmeter can't be used anymore.

One can theoretically proclaim that the synchronization between clocks
(or probes) is arbitrary, however, in the real world each reference frame
has its preferred way of synchronization between clocks. (and probes)

So the claim that one can not experimentally distinguish between SR and
MS is hard to keep. A simple voltmeter can do the trick.


Regards, Hans

 

[Aether]

-when Einstein wrote his 1905 paper he had a section dedicated on kinematics, one on dynamics, one on electromagnetism...

-MS stopped after the kinematics section (and they give the reason), they promised to write a dynamics section, it never happened (think about it)

So , no, the MS theory and SRT are not indistinguishable under ALL conditions. There are some very severe restrictions under which they produce the same predictions.

This is a good point.

It is precisely these restrictions that people who understand the theory (C.M.Will is a best example) decided to exploit in their experiments that expose the differences.

Where exactly?

MS theory is not equivalent to SRT, being an "aether" theory it requires additional ad-hoc assumptions to make the same predictions as SR (like all the other aether theories).
MS is a very valuable tool to test for SR violations but it is not going to supplant SR.

You have enough hints to start thinking.

What ad-hoc assumptions, that a locally preferred frame exists?

 

[Aether]

I agree.

This belief may look defendable from a pure theoretical point of view,
for practical purposes it becomes problematic. A good example is the
boundary condition of $$\bf{E}_\parallel=0$$.

If one uses a voltmeter with two probes to measure a static electric
field in the neighborhood of a conductor then one finds that the parallel
component is zero. However, it’s not supposed to be zero in a GGT
frame

To do the same measurement correctly for GGT one needs to measure
in the past with one probe and in the future with the other probe. The
problem is that one can't measure absolute voltages with a voltmeter
and subtract the results later. The voltmeter can't be used anymore.

One can theoretically proclaim that the synchronization between clocks
(or probes) is arbitrary, however, in the real world each reference frame
has its preferred way of synchronization between clocks. (and probes)

So the claim that one can not experimentally distinguish between SR and
MS is hard to keep. A simple voltmeter can do the trick.


Regards, Hans

Hans,
Please describe your experiment with the voltmeter mathematically.

 

[clj4]

This is a good point.

Where exactly?

What ad-hoc assumptions, that a locally preferred frame exists?

I will answer (I already answered quite a bit) after you answer the question that you signed up to answer weeks ago:

How does the alleged incorrect derivation of the boundary conditions in Gagnon affect the outcome of formula (9). If you cannot prove that mathematically, admit it and we can move on to my answering the two questions you just asked. We are under BAUT rules, you started this "Against the Mainstream" thread...Fair?

 

[clj4]

Where exactly?

OK, I'll answer one question : CMW sets the theoretical foundation of the the Krisher paper.

Now, your turn. Please defend your statements about the Gagnon paper.(see the previous post)

 

[clj4]

What ad-hoc assumptions, that a locally preferred frame exists?

Nope, this is part of the main theory. Remember : additional ad-hoc assumptions. You can even find them yourself. You can find them in both the MS papers and in the C.M.Will one.

 

[Aether]

I will answer (I already answered quite a bit) after you answer the question that you signed up to answer weeks ago:

How does the alleged incorrect derivation of the boundary conditions in Gagnon affect the outcome of formula (9). If you cannot prove that mathematically, admit it and we can move on to my answering the two questions you just asked.

I can't contribute anything in the near term to analyzing the PDE in Gagnon, and will learn what I can from the discussion. Nevertheless, that point is moot because the authors have recanted.

We are under BAUT rules, you started the "Against the Mainstream" thread...Fair?

You asked me to start that thread, and you are currently banned (for a week) from BAUT while I'm not, so what's your point?

 

[clj4]

I can't contribute anything in the near term to analyzing the PDE in Gagnon, and will learn what I can from the discussion.

Good, this is a major progress.

You asked me to start that thread, and you are currently banned (for a week) from BAUT and I'm not, so what's your point?

Right, what's your point?
Anyways , I gave you the answers to your questions.

 

[NotForYou]

The proponents of this idea "Aether" and "gregory=NotForYou" ...

Um, hello? How many times do I need to tell you that I am not gregory. Seriously, stick to the physics.


I must admit that I'm getting more curious about this topic. I worked out some of the math and disagree with gregory on where Gagnon's error is. When I worked it out, of course the Lorentz force law changes. And if I define the electromagnetic fields from the contravarient field tensor (to conserve the form of the 2 "source" Maxwell equations), the boundary conditions do not stay the same. However, if I define the electromagnetic fields from the covarient field tensor (to conserve the form of the 2 "non source" Maxwell equations), the boundary conditions do stay the same. Interestingly enough, the "wave equations" obtained from the standard method of evaluating$$\nabla \times (\nabla \times E)$$ and $$\nabla \times (\nabla \times B)$$, are the same regardless of whether you choose the contravarient or covarient field tensor to define the fields (should that be obvious for some reason?).

In case there is any confusion, clj4, in his https://www.physicsforums.com/showpost.php?p=950704&postcount=324" he attached ... he used the contravarient field tensor to define the electromagnetic fields (this is forced by his choice in his eq 3.6). Therefore the boundary conditions do not stay the same with that definition, ruining his arguement. However, Gagnon's paper uses the covarient field tensor to define the electromagntic fields, so if my calculations are correct (I think I see gregory's error, but I'll discuss this with him tomorrow), then the boundary conditions do stay the same using that particular definition (and therefore, clj4's error is moot since gregory is wrong as well).


Anyway, I read over the paper today and my calculation agrees with equation 5 (the wave equation in B is also of the same form). Equation 8 however is wrong, and unfortunately, they do not explain their calculation explicitly enough ... so while unsatisfying, we can not see the details of their mistake.

I found the form of physics in GGT frames not worth the effort. So instead I chose to do the calculations in a "lorentz frame", transform to some arbitrary "special frame" (where GGT and SR are defined to agree), then transform back to the "lab GGT frame". Because GGT and SR have identical metrics in the special frame, and have identical definitions of proper time (invarient interval $$ds^2=c^2 dt^2$$ is always true in the clock's rest/"proper" frame according to both SR and GGT), the frequency measured in a GGT frame agrees with the SR value (independent of the choice of "special frame"). They will however disagree on the value of k since that depends on the simultaneity convention.




To help this discussion along, I was wondering if anyone knew what the full form of E(x,y,z) is in equation 6. (ie. what is the result that they obtain for E(x,y) after applying their boundary conditions and the remaining maxwell's equations?). If they had included this in their paper, we could double check their intermediate results easier. So if you somehow figured out what they claim the full form of E(x,y,z) is in eq. 6, let us know.

 

[clj4]

Making progress in (re) establishing the truth

I see gregory's error, but I'll discuss this with him tomorrow), then the boundary conditions (for the Gagnon equations) do stay the same using that particular definition

Can you share your calculations with the rest of us? You know, like in an attachment?
Such that we can all see how you derive your conclusions.

 

[NotForYou]

Can you share your calculations with the rest of us? You know, like in an attachment?
Such that we can all see how you derive your conclusions.

I could try to find a scanner to scan in my work, but it is a bit messy. It really isn't worth my time to type it up. Besides, the important part is the proof that SR and GGT predict the same frequency (and the argument is so simple and straightforward that I've already included everything you need to see the result). Anyway, moving on to Gagnon details.



I think I am homing in on the location of Gagnon's error. As mentioned above, one huge issue here is that he didn't describe his calculations explicitly enough... so we first need to figure out what he did. I believe that I have "recreated" his calculations. It is imperitive that we agree on this recreation, otherwise no one can ever show the "very details" of any possible mistake Gagnon made. So everyone double check these, okay?


First, their appears to be a "typo" in the paper. Notice equation 7 and equation 8 treat $v_x$ and $v_y$ differently. Our problem is symmetric around the z axis, so there should be no distinction between the two. I believe what happenned is that their goal is to find the cutoff frequency of the lowest mode. All this means is that when they say mn like in $\omega_{mn}$ they really just mean to refer to the TE mode 10 cutoff (the lowest possible frequency of all modes in the waveguide). If one denies this is a "typo" then there is a serious error regarding the handling of $v_x$ and $v_y$. So, assuming no objections, I will continue with the assumption that we want to find the cutoff frequency of the lowest mode.


So we start with the wave equation:
$$[\nabla^2 + 2(\frac{\vec{v}}{c} \cdot \vec{\nabla}) \frac{1}{c} \frac{\partial}{\partial t} - (1-\frac{v^2}{c^2})\frac{1}{c^2} \frac{\partial^2}{\partial t^2}] \vec{E} = 0$$

Let's solve for the $$E_x$$ component.
Assume a seperable solution:
$$E_x(x,y,z,t) = X(x)Y(y)Z(z)T(t)$$

$$Z(z) = exp(+ikz)$$
$$T(t) = exp(-i\omega t)$$

Boundary condition $$E_\parallel = 0$$ only constrains Y(y) such that Y(0)=Y(a)=0 (where a=width of waveguide in y direction). We are looking for the lowest frequency mode, so we want $$\nabla^2 E_x$$ and $$\nabla E_x$$ to be a minimum. Since X(x) is unconstrained, the result is X(x) = constant for the lowest frequency mode.

Plugging into the wave equation we get:
$$[\frac{\partial^2}{\partial y^2} -k^2 + 2(\frac{v_y}{c} \frac{\partial}{\partial y} + ik \frac{v_z}{c})\frac{1}{c}(-i\omega) - (1-\frac{v^2}{c^2})\frac{1}{c^2} (-i\omega)^2] Y(y) = 0$$

Which can be rewritten as:
$$[\frac{d^2}{dy^2} +f \frac{d}{dy} + g] Y(y) = 0$$
where $$f=-2\frac{v_y}{c}\frac{i\omega}{c}$$
and $$g=-k^2 +2k\frac{v_z}{c}\frac{\omega}{c} +(1-\frac{v^2}{c^2})\frac{\omega^2}{c^2}$$

We know there should be two linearly independant solutions of this homogenous equation of the form $$e^{ry}$$.

Solving for the values of r:
$$r^2+fr+g=0$$
$$r_\pm = -f/2 \pm \sqrt{f^2/4 -g}$$

So this gives us $$Y(y) = A \exp(r_+y)+B \exp(r_-y)$$
Now applying the boundary condition $Y(0)=0$ gives $B=-A$. Thus:
$$Y(y) = A\exp(-y\ f/2)[\exp(y\sqrt{f^2/4 -g})-\exp(-y\sqrt{f^2/4 -g})]$$

Applying the boundary condition $Y(a)=0$ gives us the following condition for the lowest frequency mode.
$$\sqrt{f^2/4 -g} = i\frac{\pi}{a}$$
$$g-\frac{f^2}{4}=\frac{\pi^2}{a^2}$$
$$-k^2+2k\frac{v_z}{c}\frac{\omega}{c} +(1-\frac{v^2}{c^2})\frac{\omega^2}{c^2}+\frac{v_y^2}{c^2}\frac{\omega^2}{c^2} =\frac{\pi^2}{a^2}$$
$$k^2 - 2 \frac{v_z}{c}\frac{\omega}{c}k -(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega^2}{c^2}+ \frac{\pi^2}{a^2}}$$

$$k = \frac{v_z}{c}\frac{\omega}{c} \pm \sqrt{\frac{v_z^2}{c^2}\frac{\omega^2}{c^2} +(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega^2}{c^2} - \frac{\pi^2}{a^2}}$$

If I change the arbitrary sign choice in Z(z), we get:

$$k = - \frac{\omega}{c}\frac{v_z}{c} + \frac{1}{c}\sqrt{\omega^2(1-\frac{v_x^2}{c^2}) - \omega_{10}^2}$$
This is what Gagnon states in equation 7.
However, I chose the sign in Z(z) according to Gagnon's suggestion. Did I make a sign error? Or is this another typo?


Gagnon defined the cutoff frequency as that which made k=0.
$$k=0= -\frac{v_z}{c}\frac{\omega_c}{c}+\frac{1}{c}\sqrt{(1-\frac{v_x^2}{c^2})\omega_c^2 - \omega_{10}^2}$$
$$\frac{v_z^2}{c^2}\frac{\omega_c^2}{c^2}=(1-\frac{v_x^2}{c^2})\frac{\omega_c^2}{c^2} - \frac{\omega_{10}^2}{c^2}$$
$$\omega_c = \omega_{10} (1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})^{-1/2}$$
This is what Gagnon states in equation 8.



Please thoroughly check this. I want to agree on how Gagnon did the calculations before we continue further.

 

[clj4]

$$k = \frac{v_z}{c}\frac{\omega}{c} + \frac{1}{c}\sqrt{(1-\frac{v_x^2}{c^2})\omega^2 - \omega_{10}^2}$$
This is what Gagnon states in equation 7.

This doesn't appear to be what he's doing.

$$k = \frac{v_z}{c}\frac{\omega}{c} + \frac{1}{c}\sqrt{(1-\frac{v_x^2}{c^2})\omega^2 - \omega_{10}^2}$$ (1)
This is what Gagnon states in equation 7.Gagnon defined the cutoff frequency as that which made k=0.
$$k=0= -\frac{v_z}{c}\frac{\omega_c}{c}+\frac{1}{c}\sqrt{(1-\frac{v_x^2}{c^2})\omega_c^2 - \omega_{10}^2}$$
$$\frac{v_z^2}{c^2}\frac{\omega_c^2}{c^2}=(1-\frac{v_x^2}{c^2})\frac{\omega_c^2}{c^2} - \frac{\omega_{10}^2}{c^2}$$
$$\omega_c = \omega_{10} (1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})^{-1/2}$$
This is what Gagnon states in equation 8.
Please thoroughly check this. I want to agree on how Gagnon did the calculations before we continue further.

It is tough to comment since you don't number your equations.
Your general formula (1) for k seems wrong, contrary to what you claim, it is different from Gagnon (7).
In (7) Gagnon states:

$$k = -\frac{v_z}{c}\frac{\omega}{c} + \frac{1}{c}\sqrt{(1-\frac{v_x^2}{c^2})\omega^2 - \omega_{10}^2}$$

For k=0 the "minus" sign appears miraculously back in your calculations.

 

[NotForYou]

It is tough to comment since you don't number your equations.

Just quote them. It makes it easier for everyone to follow the discussion anyway.

...the "minus" sign appears miraculously back in your calculations.

If you can find my sign error, I would be much obliged. I had a friend do the calculations (independent, not checking my work) and he ran into the same sign problem on that term. It can be fixed by changing the Z(z) sign choice. I'm starting to think it is just a typo in the paper.

 

[clj4]

You just changed your posting by editing out a portion of your derivation.

 

[clj4]

$$k = \frac{v_z}{c}\frac{\omega}{c} \pm \sqrt{\frac{v_z^2}{c^2}\frac{\omega^2}{c^2} +(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega^2}{c^2} - \frac{\pi^2}{a^2}}$$

If I change the arbitrary sign choice in Z(z), we get:

$$k = - \frac{\omega}{c}\frac{v_z}{c} + \frac{1}{c}\sqrt{\omega^2(1-\frac{v_x^2}{c^2}) - \omega_{10}^2}$$
This is what Gagnon states in equation 7.
However, I chose the sign in Z(z) according to Gagnon's suggestion. Did I make a sign error? Or is this another typo?

What you were supposed to get is:$$k = -\frac{v_z}{c}\frac{\omega}{c} \pm \sqrt{\frac{v_z^2}{c^2}\frac{\omega^2}{c^2} +(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega^2}{c^2} - \frac{\pi^2}{a^2}}$$

Looks like you got your equation wrong , this is why you are missing the "minus" sign. The reason "plus-minus" is later taken as "plus" is that THIS is what produces k=minimum ( a little elementary algebra) . Actually , I take it back, you didn't recreate Gagnon (7) at all. There are two other issues with your stuff under the square root:

1. The term $$\frac{\pi^2}{a^2}}$$ has misteriously "disappeared"
2. The sign for the expression $$(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega^2}{c^2}$$ is wromg (it should be a minus)

 

[NotForYou]

You just changed your posting by editing out a portion of your derivation.

Yeah, I was still fixing up my own typos and stuff (there may be more in there still, let me know if you see anything else).

I don't think there is an error in the paper, but since you just changed your posting to match Gagnon perfectly, I have no objection.

Just to make sure, when you say "I don't think there is an error in the paper" ... you aren't referring to the minor/inconsequential things I claimed were typos, but instead the final "result" of the paper? (ie, do we agree on those typos so that we may continue on?)

 

[NotForYou]

Looks like you got your equation wrong , this is why you are missing the "minus" sign.

This is not helpful. Please be more specific ... in what step do I make a sign error.

Actually , I take it back, you didn't recreate Gagnon (7) at all. There are two other issues with your stuff under the square root:

1. The term $$\frac{\pi^2}{a^2}}$$ has misteriously "disappeared"
2. The sign for the expression $$(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega^2}{c^2}$$ is wromg (it should be a minus)

Huh?

1. $$\frac{\pi^2}{a^2}} = \omega_{10}^2/c^2$$
2. I don't even understand here. I think you read something wrong. Please doublecheck.

 

[clj4]

This is not helpful. Please be more specific ... in what step do I make a sign error.Huh?

1. $$\frac{\pi^2}{a^2}} = \omega_{10}^2/c^2$$
2. I don't even understand here. I think you read something wrong. Please doublecheck.

In Gagnon (7), under the square root:

$$-(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega^2}{c^2}$$

In your solution:

$$+(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega^2}{c^2}$$

And, while you are at it, what about the discrepancy in the term $$\frac{v_z^2}{c^2}\frac{\omega^2}{c^2}$$. This is not what shows in Gagnon (7).

Looks like you have a lot of explaining to do. I am going skiing. Have fun!

 

[NotForYou]

In Gagnon (7), under the square root:

$$-(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega^2}{c^2}$$

Umm... no.

That is the $\omega_c$ term, ie $$-(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega_c^2}{c^2}$$. The calculation agrees with Gagnon here.

And, while you are at it, what about the discrepancy in the term $$\frac{v_z^2}{c^2}\frac{\omega^2}{c^2}$$. This is not what shows in Gagnon (7).

Looks like you have a lot of explaining to do.

Maybe it is just that you are in a rush, but you're not reading it correctly. I'm not sure what the problem is here.

Here, let me show you with much much more detail:

first, note: $$\frac{\pi^2}{a^2}=\frac{\omega_{10}^2}{c^2}$$

Also, note:$$\omega_{10}^2 = \omega_c^2 (1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})$$

So here we go:

start with equation you are complaining about -
$$k = \frac{v_z}{c}\frac{\omega}{c} \pm \sqrt{\frac{v_z^2}{c^2}\frac{\omega^2}{c^2} +(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})\frac{\omega^2}{c^2} - \frac{\pi^2}{a^2}}$$

$$k = \frac{v_z}{c}\frac{\omega}{c} \pm \sqrt{(1-\frac{v_x^2}{c^2})\frac{\omega^2}{c^2} - \frac{\pi^2}{a^2}}$$
See, that term that you thought was "extra" merely added with the other terms. Contrinuing on...

$$k = \frac{v_z}{c}\frac{\omega}{c} \pm \frac{1}{c}\sqrt{(1-\frac{v_x^2}{c^2})\omega^2 - c^2 \frac{\pi^2}{a^2}}$$

$$k = \frac{v_z}{c}\frac{\omega}{c} \pm \frac{1}{c}\sqrt{(1-\frac{v_x^2}{c^2})\omega^2 - \omega_c^2 (1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})}$$

$$k = \frac{\omega}{c}\frac{v_z}{c} \pm \frac{1}{c}\sqrt{\omega^2(1-\frac{v_x^2}{c^2}) - \omega_c^2 (1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})}$$

If I change the arbitrary sign choice in Z(z), we get:

$$k = - \frac{\omega}{c}\frac{v_z}{c} + \frac{1}{c}\sqrt{\omega^2(1-\frac{v_x^2}{c^2}) - \omega_c^2 (1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})}$$
This is what Gagnon states in equation 7.

As I stated before: I began the calcualtion with the sign in Z(z) according to Gagnon's suggestion. Did I make a sign error? Or is this another typo?

If you can find my sign error, I would be much obliged. I had a friend do the calculations (independent, not checking my work) and he ran into the same sign problem on that term. It can be fixed by changing the Z(z) sign choice. I'm starting to think it is just a typo in the paper.

 

[clj4]

OK, looks reasonable.

So we start with the wave equation:
$$[\nabla^2 + 2(\frac{\vec{v}}{c} \cdot \vec{\nabla}) \frac{1}{c} \frac{\partial}{\partial t} - (1-\frac{v^2}{c^2})\frac{1}{c^2} \frac{\partial^2}{\partial t^2}] \vec{E} = 0$$ [1]Plugging into the wave equation we get:
$$[\frac{\partial^2}{\partial y^2} -k^2 + 2(\frac{v_y}{c} \frac{\partial}{\partial y} + ik \frac{v_z}{c})\frac{1}{c}(-i\omega) - (1-\frac{v^2}{c^2})\frac{1}{c^2} (-i\omega)^2] Y(y) = 0$$ [2]

How do you get [2] from [1]? Can you show the intermediate steps, like how you calculated the divergence $$\nabla^2$$? How did you calculate the dot product $$(\frac{\vec{v}}{c} \cdot \vec{\nabla}) \frac{1}{c} \frac{\partial}{\partial t}$$?
It is not clear at all how you managed to separate the variables (you should still have terms in X(x)*Y(y) due to the second order partial derivative in z). How did you separate the variables?

 

[NotForYou]

How do you get [2] from [1]?

Huh? You just plug $E_x=X(x)Y(y)Z(z)T(t)$ into the wave equation, as I said.

Can you show the intermediate steps, like how you calculated the divergence $$\nabla^2$$? How did you calculate the dot product $$(\frac{\vec{v}}{c} \cdot \vec{\nabla}) \frac{1}{c} \frac{\partial}{\partial t}$$?

Are you kidding me? I know you know how to solve differential equations, let alone just take derivatives of an exponential. So I don't understand what is going on. Are we having some kind of communication problem or something?

It is not clear at all how you managed to separate the variables (you should still have terms in X(x)*Y(y) due to the second order partial derivative in z). How did you separate the variables?

What? Can you show your work? You are not making sense at all.

------------------------------------------
Again, adding in more detail...

$$E_x(x,y,z,t) = X(x)Y(y)Z(z)T(t)$$

At that point in the calculation, we don't know Y(y), but we have (up to a constant factor and a phase):
$$X(x) = 1$$
$$Z(z) = \exp(+ikz)$$
$$T(t) = \exp(-i\omega t)$$

So we have:
$$E_x = Y(y)\exp(+ikz-i\omega t)$$

$$\nabla^2 E_x = (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})Y(y)\exp(+ikz-i\omega t)$$
$$= (0 + \frac{\partial^2}{\partial y^2}+(ik)^2)Y(y)\exp(+ikz-i\omega t)$$

Now for the other derivative you asked about:
$$(\frac{\vec{v}}{c} \cdot \vec{\nabla}) \frac{1}{c} \frac{\partial}{\partial t}E_x = (\frac{v_x}{c} \frac{\partial}{\partial x} + \frac{v_y}{c} \frac{\partial}{\partial y} + \frac{v_z}{c} \frac{\partial}{\partial z})\frac{1}{c} \frac{\partial}{\partial t}[Y(y)\exp(+ikz-i\omega t)]$$
$$= (0 + \frac{v_y}{c} \frac{\partial}{\partial y} + \frac{v_z}{c} (ik))\frac{1}{c} (-i\omega)[Y(y)\exp(+ikz-i\omega t)]$$

 

[clj4]

You don't have to continue to be insolent.
This is precisely why I asked, what allows you to make the assumption X(x)=1?
With X(x) a non constant function of x you will have terms in X(x)* Y(y) and you would have a much more difficult time separating the variables.
You introduce X(x)=1 late, in post 371, why wasn't introduced upfront?

 

[NotForYou]

You don't have to continue to be insolent.

I'm not trying to be insolent. You keep asking questions about incredibly basic things. For instance several of your complaints above are you mistaking $$\omega$$ and $$\omega_c$$, not noticing a simple addition, and complaining about some easy derivatives. Since I know you are capable of doing all these things, it appears to me that either a) you aren't taking the time to read through / think through things before complaining or b) we are having a communication problem. In no way am I questioning your intelligence. It is because I know you can do all this math that your questions come off as so bizarre.

This is precisely why I asked, what allows you to make the assumption X(x)=1?
With X(x) a non constant function of x you will have terms in X(x)* Y(y) and you would have a much more difficult time separating the variables.
You introduce X(x)=1 late, in post 371, why wasn't introduced upfront?

See, this is what I am talking about. I didn't randomly call X(x) a constant. And I didn't "introduce it late in post 371". Go back and read the original post at the top of this page.

We seem to be having some serious communication problems. If there is something I can do to help rectify this, please let me know as this is frustrating for both of us.

 

[clj4]

Yes, I found it:

Since X(x) is unconstrained, the result is X(x) = constant for the lowest frequency mode.

I missed it in the first pass. OK, so you have rederived Gagnon (7). The sign issue is due to the fact that, if you want minimum k , you need to take "minus" in front of the square root. When you do this, k becomes negative. In oreder to rectify this, Gagnon inverts the whole expression. So now, we have Gagnon (7) exactly.
Gagnon (8) is also correct and it is inconsequential (it is just a notation).

So, to recap, you convinced yourself, contrary to your earlier claims that :

1. The initial conditions transform correctly.
2. Gagnon (7,8) are both correct.

 

[clj4]

BTW, the miscommunication and the miscues go in both directions:

In case there is any confusion, clj4, in his https://www.physicsforums.com/showpost.php?p=950704&postcount=324" he attached ... he used the contravarient field tensor to define the electromagnetic fields (this is forced by his choice in his eq 3.6). Therefore the boundary conditions do not stay the same with that definition, ruining his arguement.

From wikipedia (wikilink[/PLAIN] ):

The covariant version of the field strength tensor $$F_{ab}$$ is related to to contravariant version $$F^{ab}$$ by the Minkowski metric tensor η

$$F_{ab} = F^{ab}$$

 

[NotForYou]

The sign issue is due to the fact that, if you want minimum k , you need to take "minus" in front of the square root. When you do this, k becomes negative. In oreder to rectify this, Gagnon inverts the whole expression.

Hmm... I like that as it sounds reasonable. But I don't see what allows us to do that. For instance we have:

$$k = \frac{\omega}{c}\frac{v_z}{c} \pm \frac{1}{c}\sqrt{\omega^2(1-\frac{v_x^2}{c^2}) - \omega_{10}^2}$$

Your argument works in general only if,
$$\frac{\omega}{c}\frac{v_z}{c} < \frac{1}{c}\sqrt{\omega^2(1-\frac{v_x^2}{c^2}) - \omega_{10}^2}$$

Can you help me see how that works out?

Anyway, if we accept that explanation, then the sign issue is gone. And we agree on the other "typo" (the fact that they don't actually solve for the general $$\omega_{mn}$$ case but just the specific $$\omega_{10}$$ case), right?

So, to recap:

1. The initial conditions transform correctly.
2. Gagnon (7,8) are both correct.

1. I never discussed initial conditions (and they are irrelevant). However, the boundary conditions of the electromagnetic fields on the waveguide walls do stay the same using their choice of defining the electromagnetic fields as the components of the covarient field tensor. I assume that is what you meant (if not, please tell me what you'd like to discuss about the initial conditions).

2. I am not agreeing that they are correct. I am merely trying to
a) figure out what their calculations were
b) get us to agree on what their calcualtions were
c) since I already proved that $$\omega$$ does not depend on v, I have already shown their "solution" is wrong ... but I want us to agree on what their calculations were so that I may proceed forward to point out where their error comes in.

That is why it is important that we agree on what Gagnon's calculations were here.

 

[clj4]

1. I never discussed initial conditions (and they are irrelevant). However, the boundary conditions of the electromagnetic fields on the waveguide walls do stay the same using their choice of defining the electromagnetic fields as the components of the covarient field tensor. I assume that is what you meant (if not, please tell me what you'd like to discuss about the initial conditions).

yes, right...this was not the first argument as "gregory"

2. I am not agreeing that they are correct.

Then you'll have to prove it. You are down to one equation, Gagnon (9)

I am merely trying to
a) figure out what their calculations were
b) get us to agree on what their calcualtions were

So far so good, you rederived their formulas ad-literam in a very restrictive situation (unidimensional)

c) since I already proved that $$\omega$$ does not depend on v, I have already shown their "solution" is wrong ... but I want us to agree on what their calculations were so that I may proceed forward to point out where their error comes in.

The authors don't claim that $$\omega$$ depends on v so I fail to see any relevance.
Can you explain what does this have to do with what the experiment is all about, equation (9)?

 

[clj4]

Hmm... I like that as it sounds reasonable. But I don't see what allows us to do that. For instance we have:

$$k = \frac{\omega}{c}\frac{v_z}{c} \pm \frac{1}{c}\sqrt{\omega^2(1-\frac{v_x^2}{c^2}) - \omega_{10}^2}$$

Your argument works in general only if,
$$\frac{\omega}{c}\frac{v_z}{c} < \frac{1}{c}\sqrt{\omega^2(1-\frac{v_x^2}{c^2}) - \omega_{10}^2}$$

Can you help me see how that works out?

Sure, it works for $$\omega\sqrt{(1-\frac{v_x^2}{c^2})}> \omega_c$$

which is the case when driving the waveguide well above the cutoff.

 

[NotForYou]

BTW, the miscommunication and the miscues go in both directions:

...

The covariant version of the field strength tensor $$F_{ab}$$ is related to to contravariant version $$F^{ab}$$ by the Minkowski metric tensor η

$$F_{ab} = F^{ab}$$

I assume you meant:
$$F_{ab} = g_{a\mu}g_{b\nu}F^{\mu \nu}$$
(where g is the metric = your "η")

What do you feel I misunderstood?
(Maxwell's equations and the Lorentz force law look different in GGT depending on whether you define the electromagnetic fields from the components of the covarient field tensor or the contravarient field tensor. You implicitly chose the contravarient field tensor, which is the opposite of what Gagnon chose, and also with that choice the boundary conditions do NOT stay the same.)

 

[NotForYou]

Then you'll have to prove it. You are down to one equation, Gagnon (9)

I already proved that they are incorrect. We are merely trying to find where their mistake is now.

So far so good, you rederived their formulas ad-literam in a very restrictive situation (unidimensional)

As long as you don't believe I was more restrictive than they were, then we are fine.

The authors don't claim that $$\omega$$ depends on v so I fail to see any relevance.
Can you explain what does this have to do with what the experiment is all about, equation (9)?

They do claim that $$\omega$$ depends on v. They operate one waveguide "very close to the cutoff frequency of its fundamental mode". This minimum allowed frequency depends on v (equation 8). This why they believe they can experimentally distinguish between the coordinate systems.

Sure, it works for $$\omega\sqrt{(1-\frac{v_x^2}{c^2})}> \omega_c$$

which is the case when driving the waveguide well above the cutoff.

But they are not driving the waveguide well above the cutoff.

 

[NotForYou]

Oh, I'm an idiot. It looks like we both made algebra errors.

The condition required by your argument is:
$$\frac{\omega}{c}\frac{v_z}{c} < \frac{1}{c}\sqrt{\omega^2(1-\frac{v_x^2}{c^2}) - \omega_{10}^2}$$

which is:
$$\omega^2\frac{v_z^2}{c^2} < \omega^2(1-\frac{v_x^2}{c^2}) - \omega_{10}^2$$

$$\omega_{10}^2 < \omega^2(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})$$

$$\omega > \omega_c$$

Yeh! You solved my sign error problem.

 

[clj4]

I assume you meant:
$$F_{ab} = g_{a\mu}g_{b\nu}F^{\mu \nu}$$
(where g is the metric = your "η")

What do you feel I misunderstood?
(Maxwell's equations and the Lorentz force law look different in GGT depending on whether you define the electromagnetic fields from the components of the covarient field tensor or the contravarient field tensor. You implicitly chose the contravarient field tensor, which is the opposite of what Gagnon chose, and also with that choice the boundary conditions do NOT stay the same.)

Read the wiki page.

 

[clj4]

Oh, I'm an idiot. It looks like we both made algebra errors.

The condition required by your argument is:
$$\frac{\omega}{c}\frac{v_z}{c} < \frac{1}{c}\sqrt{\omega^2(1-\frac{v_x^2}{c^2}) - \omega_{10}^2}$$

which is:
$$\omega^2\frac{v_z^2}{c^2} < \omega^2(1-\frac{v_x^2}{c^2}) - \omega_{10}^2$$

$$\omega_{10}^2 < \omega^2(1-\frac{v_x^2}{c^2}-\frac{v_z^2}{c^2})$$

$$\omega > \omega_c$$

Yeh! You solved my sign error problem.

Thank you :!)

 

[clj4]

I already proved that they are incorrect. We are merely trying to find where their mistake is now.

Huh? You haven't proven anything.

They do claim that $$\omega$$ depends on v. They operate one waveguide "very close to the cutoff frequency of its fundamental mode". This minimum allowed frequency depends on v (equation 8). This why they believe they can experimentally distinguish between the coordinate systems.


But they are not driving the waveguide well above the cutoff.

What they believe is encapsulated in (9). So far , you have reproduced the whole paper exactly without any ability to refute it. You are simply repeating your earlier claim that (8) is wrong with no mathematical proof whatsoever and in the context of them measuring a phase difference shown in (9).

 

[NotForYou]

Read the wiki page.

Again. What do you feel I have said that is incorrect regarding the field tensor? What do you feel I have misunderstood?

Are you denying that the form of Maxwell's equations and the Lorentz force law change form in GGT depending on whether you choose the contravarient or the covarient field tensor to define the electromagnetic fields in the GGT frame? The form of maxwell's equations requires this. And also, you'd be contradicting Gagnon's ref 9.

Huh? You haven't proven anything.

Sure I have. I'll post it again. Follow through the proof yourself:

I found the form of physics in GGT frames not worth the effort. So instead I chose to do the calculations in a "lorentz frame", transform to some arbitrary "special frame" (where GGT and SR are defined to agree), then transform back to the "lab GGT frame". Because GGT and SR have identical metrics in the special frame, and have identical definitions of proper time (invarient interval $$ds^2=c^2 dt^2$$ is always true in the clock's rest/"proper" frame according to both SR and GGT), the frequency measured in a GGT frame agrees with the SR value (independent of the choice of "special frame").

Which statement do you deny?