Consistency of the speed of light

[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?

 

[clj4]

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.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?

Ah, now the "gregory" personality re-emerges (and the same prose, with no math).
Which reminds me , how did you convince "gregory" that he was wrong about the boundary conditions? You never produced the mathematical proof. Knowing "gregory", this wasn't such an easy task. Can you show us?

 

[NotForYou]

Ah, now the "gregory" personality re-emerges (and the same prose, with no math).
Which reminds me , how did you convince "gregory" that he was wrong about the boundary conditions? You never produced the mathematical proof. Knowing "gregory", this wasn't such an easy task. Can you show us?

I AM NOT GREGORY! You have been warned repeatedly to stick to the physics, and stop making accusations. Please do so.

He's my roommate. We can sit down and discuss the tensor equations. Two other friends have worked through the equations as well. Since the result was that I don't believe Gagnon made an error by assuming the boundary conditions were the same using his definition, there is no point in showing all the calculations (since everyone agrees with that conclusion).



As for the proof, why prose and no math? Because the result is basically by definition, so there is nothing really to show. Here, let me show you step by step. Tell me which step you disagree with.

1] DEFINITION: GGT and SR agree on the physical laws in one "special frame".

2] DEFINITION: let w = cutoff frequency of waveguide according to SR in the lab frame. This is measured by two events on a stationary clock, the time between two peaks T = 2 pi / w.

3] Transform into the "special frame". The proper time of the clock is invarient. So if we did ALL the calculations according to the physical laws in this moving frame ... while more complicated ... we know that the result is that SR will predict that the clock will still measure T.

4] GGT and SR agree on the laws of physics in this special frame. So doing ALL the calculations according to SR in this "special" moving frame is equivalent to doing the calculations for GGT in this frame.

5] Thus GGT also predicts that the clock will measure T.

6] Thus GGT and SR agree on the cutoff frequency of the waveguide, independent of the choice of the "special frame".


Which step do you deny?

 

[clj4]

I AM NOT GREGORY! You have been warned repeatedly to stick to the physics, and stop making accusations. Please do so.

He's my roommate. We can sit down and discuss the tensor equations. Two other friends have worked through the equations as well. Since the result was that I don't believe Gagnon made an error by assuming the boundary conditions were the same using his definition, there is no point in showing all the calculations (since everyone agrees with that conclusion).
As for the proof, why prose and no math? Because the result is basically by definition, so there is nothing really to show. Here, let me show you step by step. Tell me which step you disagree with.

1] DEFINITION: GGT and SR agree on the physical laws in one "special frame".

2] DEFINITION: let w = cutoff frequency of waveguide according to SR in the lab frame. This is measured by two events on a stationary clock, the time between two peaks T = 2 pi / w.

3] Transform into the "special frame". The proper time of the clock is invarient. So if we did ALL the calculations according to SR in this moving frame ... while more complicated ... we know that the result is that SR will predict that the clock will still measure T.

4] GGT and SR agree on the laws of physics in this special frame. So doing ALL the calculations according to SR in this "special" moving frame is equivalent to doing the calculations for GGT in this frame.

5] Thus GGT also predicts that the clock will measure T.

6] Thus GGT and SR agree on the cutoff frequency of the waveguide.Which step to you deny?

It doesn't work this way: you do the math.
Besides, you miss the point: Gagnon measured a difference in phase
\phi. Your task is to disprove (9). It has always been , but you keep trying to expedite it with prose or to do it by disproving other things. Remember, Gagnon sets to measure \phi.
\omega_c is not relevant to the discussion. It is just a byproduct of the derivation of (7). You can view it as a value very close to \omega_1_0.
Think about it: what would you need to write to Phys Rev to refute the Gagnon paper? For sure, the prose above wouldn't wash.

Let's see the equations that disprove Gagnon (9).
While you are at it, I still want to see what you wrote to convince "gregory" of his mistake.

 

[clj4]

6] Thus GGT and SR agree on the cutoff frequency of the waveguide.

Ok, and the impact on (9) is? Can we have the mathematical disproof of (9)?

 

[NotForYou]

\omega_c is not relevant to the discussion.

It is. Because I can show that GGT predicts the same \omega_c as SR. Therefore Gagnon is wrong (eq. 8).

If you disagree with me, go back to my previous post and tell me which step you disagree with and why.