Dependence of Phase Velocity on Wavelength

AI Thread Summary
The discussion centers on a homework problem regarding the phase velocity of electromagnetic waves in a polymer with a dielectric constant inversely proportional to the square root of frequency. The phase velocity is expressed as v_p = Ck^(1/3), while the group velocity is given by v_g = (4/3)Ck^(1/3), leading to a ratio of v_p/v_g = 2. Participants express frustration over the reliance on wavenumber instead of wavelength, prompting a clarification on the relationship between wavenumber and wavelength. The correct substitution of wavenumber into the phase velocity equation is necessary to derive a relationship involving wavelength. The discussion concludes with a reminder to verify the ratio of phase to group velocity.
keV.92
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Homework Statement



This is Problem 7.6 from Electronic Properties of Engineering Materials by Livingston.

"Over a wide range of frequencies, the dielectric constant of a polymer is found to be proportional to the inverse square root of frequency. (a) How does the phase velocity of EM-waves vary with wavelength in this polymer? (b) What is the ratio between phase and group velocities in this material?"

This is frustrating for many reasons, not the least of which is that the answers we've been given are in terms of wavenumber, not wavelength. The (presumed) answer follows.

Problem 7-6: (a) ##v_p = Ck^\frac{1}{3} ##, (b) ##v_g = \frac{4}{3}Ck^\frac{1}{3}## so ##\frac{v_p}{v_g} = 2##

Homework Equations



##\epsilon_r \propto \frac{1}{\sqrt{\omega}}##

Phase Velocity: ##v_p = \frac{\omega}{k}##

Group Velocity: ##v_g = \frac{\partial \omega}{\partial k}##

The Attempt at a Solution



For an EM-Wave in a non-magnetic dielectric polymer, ##\mu = \mu_0## and ##\epsilon = \epsilon_r \epsilon_0##. From the solutions to Maxwell's Equations, we produce ##\frac{\omega}{k} = (\mu \epsilon)^{-1/2}##. Thus,

##v_p = \omega / k = (\mu_0 \epsilon_0 \epsilon_r)^{-1/2}##

By adding a constant of proportionality to the given relation, I've been able to insert the equations into each other. This produced the following:

##\frac{\omega}{k} = \frac{Ac_0^2}{\sqrt{\omega}} k##

That seems to show a linear dependence on k, and the wavelength never enters into it. This problem has me running in mathematical circles; this was the only time I got an answer that didn't just prove ##v_p = c_{polymer}##.
 
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keV.92 said:
That seems to show a linear dependence on k, and the wavelength never enters into it.
Do you know the relation between wavenumber and wavelength?
 
Shouldn't that be ##\lambda = \frac{2\pi}{k}## ?
 
Yes that's correct. So if you rearrange that for ##k## and then substitute into your expression for phase velocity you'll get a relationship between phase velocity and wavelength. Is that sufficient for the question you're attempting?

Also check again your ratio between phase velocity and group velocity.
 
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