Motion of charged particles in an electric field > Motion in an alternating electric field > Problem 7.3.15

Condition

$7.3.15^*$. Determine the dielectric permittivity of a medium consisting of electrons elastically bound in molecules, in an electric field of strength $E = E_0 \sin \omega t$. The resonance frequency is $\omega_0$, the damping coefficient is $\gamma \ll \omega_0$, and the number of electrons per unit volume of the medium is $n_e$.

Solution

In an alternating field, a periodic force acts on the electron, and it begins to oscillate. The electrons are in molecules, so they can be considered in pairs with the protons in the nucleus. The nuclei are immobile, so these pairs become dipoles with time‑varying dipole moments as the electrons oscillate, and the medium becomes polarised.

It is known that the polarisation of the medium is related to the dielectric permittivity:
$$
\vec P = (\varepsilon - 1)\varepsilon_0 \vec E \tag{1}
$$
$$
n_e p(t) = e n_e \cdot x(t) = (\varepsilon - 1)\varepsilon_0 E(t) \tag{2}
$$
Taking into account the elastic bonds in the molecule and the force from the field, the equation of motion of the electron is:
$$
\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = \frac{E(t)e}{m_e} \tag{3}
$$
This is a typical equation of forced oscillations (the factor 2 in front of the damping coefficient may or may not be present, depending on the definition of $\gamma$). It is solved under the logical assumption that in the steady state the oscillation frequency of the electron coincides with the frequency of the driving force, but due to the dissipative force there may be a phase lag between them:
$$
x(t) = A\sin(\omega t + \varphi) \tag{4}
$$
This can be proved rigorously, but I will not spend time on it here: it can be found in many textbooks on oscillations.

Substituting (4) into (2):
$$
e n_e \cdot A = (\varepsilon - 1)\varepsilon_0 E_0 \quad\Rightarrow\quad \varepsilon = 1 + \frac{n_e e A}{\varepsilon_0 E_0} \tag{5}
$$
It remains to find the amplitude $A$.

Substituting (4) into (3):

$$
-A\omega^2 \sin(\omega t + \varphi) + 2\omega\gamma A \cos(\omega t + \varphi) + \omega_0^2 A \sin(\omega t + \varphi) = \frac{E_0 e}{m_e} \sin \omega t \tag{7}
$$
$$
(\omega_0^2 - \omega^2)\sin(\omega t + \varphi) + 2\omega\gamma \cos(\omega t + \varphi) = \frac{E_0 e}{m_e A} \sin \omega t \tag{6}
$$
Using trigonometry, expand the brackets, separate the coefficients of sine and cosine, and simplify:
$$
(\omega_0^2 - \omega^2)\cos\varphi - 2\omega\gamma \sin\varphi = \frac{E_0 e}{m_e A} \tag{7}
$$
$$
(\omega_0^2 - \omega^2)\sin\varphi + 2\omega\gamma \cos\varphi = 0 \tag{8}
$$
We are entitled to make such a separation because the equality must hold at any instant of time. Let

$$
P\cos(\omega t) + Q\sin(\omega t) = 0.
$$
Substitute $t = 0$:
$$
P\cdot 1 + Q\cdot 0 = 0 \quad\Rightarrow\quad P = 0.
$$
Substitute $t = \dfrac{\pi}{2\omega}$:
$$
P\cdot 0 + Q\cdot 1 = 0 \quad\Rightarrow\quad Q = 0.
$$
Thus $P = Q = 0$, so the coefficients of $\cos(\omega t)$ and $\sin(\omega t)$ can be equated to zero separately.

Returning to the main problem, square and add (7) and (8). Using the Pythagorean identity, we finally obtain:
$$
(\omega_0^2 - \omega^2)^2 + (2\omega\gamma)^2 = \left(\frac{E_0 e}{m_e A}\right)^2 \tag{9}
$$
$$
A = \frac{E_0 e}{m_e \sqrt{(\omega^2 - \omega_0^2)^2 + 4\gamma^2 \omega^2}} \tag{10}
$$
Substituting (10) into (5):
$$
\varepsilon = 1 + \frac{n_e e^2}{m_e \varepsilon_0 \sqrt{(\omega^2 - \omega_0^2)^2 + 4\gamma^2 \omega^2}} \tag{11}
$$

The author's answer (below) is also correct; I simply worked in SI, while Savchenko used CGS units.

Answer

$$
\boxed{\varepsilon = 1 + \frac{4\pi n_e e^2}{m_e \sqrt{(\omega^2 - \omega_0^2)^2 + 4\gamma^2 \omega^2}}}
$$