clj4 said:
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.
