Aether
Gold Member
- 714
- 2
No.clj4 said:You are inverting (8b), right?
I would expect to see a term in \omega_{pq} in (8c). I am not seeing it.
\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),
with (p,q)=(1,0), and \epsilon \mu =1:
\omega_{10}=\pi \frac{c_0}{a_0}.
\omega_{10} is a cutoff angular frequency in the absolute frame, and \omega_c is that cutoff angular frequency in a moving frame:
\omega_c=\frac{\pi c_0}{a}.
To transform \omega_{10} to \omega_c using Eq. (1) we need to suppose that a_0 is a distance along the x-axis, and absolute motion is along the x-axis (this is necessary for now because Eq. (1) can only be used for motion and distance along the x-axis):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)
\omega_{c+}=\frac{\pi c_0}{\gamma (a_0-vt_0)}.
However, a_0 lays along the x-axis while the wave propagates along the z-axis, and E reciprocates in the \pm x-direction. So, this last equation only applies while the transverse wave motion is in the +x direction; while the transverse wave motion is in the -x direction this equation applies:
\omega_{c-}=\frac{\pi c_0}{\gamma (a_0+vt_0)}.
Eq. (8c) is an attempt to compute the average of these two equations, and may change slightly if this average turns out not to be done quite right.
\omega_c=\frac{2\pi c_0}{\gamma((a_0-vt_0)+(a_0+vt_0))}=\frac{\pi c_0}{\gamma a_0};
\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);
\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);
a_0 and b_0 are the dimensions (in the absolute frame) of the interior cross-section of the waveguide. a_0 is taken to be along the x-axis, b_0 is taken to be along the y-axis. The electromagnetic wave propagates longitudinally along the z-axis. In a moving frame, these two dimensions transform to a and b respectively.Who are a0 and b0?
It is possible to compute \omega_c and k for a waveguide in a moving frame simply by transforming these coordinates (a,b).
Last edited: