I Don't Why its necessary for a photon to have a frequency more than the plasma frequency to be able to pass through.On what factors does plasma frequency depends on?.


If we consider a dielectric with N molecules per volume and Z electrons per molecule, then the dielectric constant is given by

$$ \frac{\epsilon(\omega)}{\epsilon_0} = 1 + \frac{N e^2}{\epsilon_0 m} \sum_j \frac{f_j}{\omega_j^2 - \omega^2 - i \omega \gamma_j} $$ where $\omega_j$ is the binding frequency of the molecules, $\gamma_j$ is the damping constant, and $f_j$ are the oscillator strengths, satisfying $\sum_j f_j = Z$.

This comes from simply solving the equation of motion for a damped, driven harmonic oscillator. According to Jackson, this is generalizable to QM with suitable definitions for $f_j, \gamma_j, \omega_j$.

If we define the plasma frequency as $\omega_p = \frac{NZe^2}{\epsilon_0 m}$, and take free electrons and damping to be $0$, we can see that the above takes the form

$$ \frac{\epsilon(\omega)}{\epsilon_0} = 1 - \frac{\omega_p ^2}{\omega^2} $$

With this, the wavenumber is given as $ck = \sqrt{\omega^2 - \omega_p^2}$, and so we can see that for any $\omega < \omega_p$, we are left with an imaginary wavenumber, indicating exponential decay in the material and complete reflection.

So, for any ideal plasma or metal with free electrons and no damping, we must have a frequency $\omega > \omega_p$ to have transmission.

Source: Classical Electrodynamics, Third Edition by John David Jackson, Chapter 7.

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