How to obtain the high frequency dispersion relation for transverse waves in the Drude model?

Question

Context

I am studying electrodynamics using Zangwill [1]. I have a narrow question regarding a proposition found there in. In the section on the Drude model of conducting matter that applies to particles with mean travel time $\tau$, a characteristic plasma frequency $\omega_p$, the Fourier transform of the permittivity, $\hat{\epsilon}(\omega)$, the Fourier transform of the permiability, $\hat{\mu}(\omega)$, and frequency parameter $\omega$. Adjusting from what Zangwill writes in [1], the proposition is that

In the high frequency limit when $$\frac{\hat{\epsilon}(\omega)}{\epsilon_o}\approx 1 - \frac{\omega_p^2}{\omega^2} \quad (\omega\,\tau \gg 1)$$ is valid, the dispersion relation for transverse waves $$k(\omega) = \omega\sqrt{\hat{\mu}(\omega)\,\hat{\epsilon}(\omega)} = \frac{\omega}{c}\,\hat{n}(\omega)$$ reads $$ \boxed{\omega^2 = \omega^2_p + c^2\,k^2 .}\tag{1} $$ I have made several attempts to maniupulate these and other equations to obtain (1). Yet, I come up short.

Question

The narrow question here is, how does one derive the high-frequency (or collisionless) dispersion relation, i.e., Eq.(1), for transverse waves using the Drude model?

Bibliography

Zangwill, Modern Electrodynaics, 2013 pp. 630-632 .

Answer 1

The dispersion relation follows from a simple substitution of your first equation into the second. The square of the 2nd equation gives $$ k^2 = {\omega^2 \over c^2}{\epsilon \over \epsilon_0} $$ Substituting the 1st equation $$ k^2 = {\omega^2 \over c^2}{(1-{\omega_p^2 \over \omega^2})} $$

$$ \omega^2 = {\omega_p^2+k^2 c^2} $$ Edit based on comments about refractive index:

I am not sure what is troubling you regarding refractive index. Look at your 2nd equation. If you eliminate $\omega$, you have the definition of refractive index. $$n(\omega) = c \sqrt{\epsilon(\omega) \mu(\omega)}$$ $$n(\omega) = \sqrt{{\epsilon(\omega) \mu(\omega)} \over {\epsilon_0 \mu_0}}$$ With the usual assumption of non-magnetic media, $\mu(\omega)=\mu_0$ , then $$n(\omega) = \sqrt{{\epsilon(\omega)} \over {\epsilon_0}}$$ Also note that the ratio, $\epsilon(\omega) / \epsilon_0$ is called the relative permittivity or dielectric constant.