Hugo Stokes

Hugo Stokes

Answered

2022-10-21

Why the relationship v p v g = c 2 (phase velocity multiply group velocity equals lightspeed squared) exists in waveguide structures?

Do you have a similar question?

Recalculate according to your conditions!

Answer & Explanation

silenthunter440

silenthunter440

Expert

2022-10-22Added 19 answers

This fact exists in classical wave theory. The phase velocity is v p = ω / β, whereas the group velocity is v g = d ω / d β (i.e. velocity of a narrow band signal, calculated from the dispersion relation ω = ω ( β ) for your waveguide), where ω is the angular frequency of the propagating wave and β is its (guided) propagation constant (same as the wavenumber, k, in lossless guides). Now, in a typical waveguide, you have β derived (say as eigenvalue of the Helmholtz wave equation and given boundary conditions) to be β = k 0 2 k c 2 , where k 0 = ω / c is the free-space wavenumber (if waveguide weren't there) and k c 2 is a geomery-based cutoff wavenumber (constant), calculated based on the particular boundary conditions of the structure. Therefore, you can see that
d β d ω = 2 ω / c 2 2 ( ω / c ) 2 k 0 2 = ω c 2 β v p v g = ω β c 2 β ω = c 2 .
As energy propagates down the waveguide at the group velocity, which may be slower than light in such structures, you can now see easily why the phase velocity becomes higher than the speed of light ( v g c v p , with c as their geometric mean).

Still Have Questions?

Ask Your Question

Free Math Solver

Help you to address certain mathematical problems

Try Free Math SolverMath Solver Robot

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?