# The damping of an oscillating electric dipole

I've read a lecture notes of solid state physics, specifically the subject of polarisation. A pair of electrons in a covalent bond in a dielectric material oscillates back and forth in the presence of an electric field $\vec E(t)$, and it forms an electric dipole.

• This harmonic oscillator suffers damping. Why is that? I don't understand the mechanism of this damping.

• In order to find the absorption coefficient of the material, we must introduce the damping. I know how to find the index of refraction, but without introducing the dumping. So why is it necessary to analyze the damping?

## 1 Answer

This harmonic oscillator suffers damping. Why is that? I don't understand the mechanism of this damping.

The electron pair is an oscillating electric dipole, so it radiates energy according to the Larmor formula ${dE \over dt}\sim-|\ddot P|^2$ ($P$ is the magnitude of the polarisation). The total energy of the oscillator (kinetic + potential) transforms into the energy of the produced EM waves, that's why the damping arises: the oscillator's energy decreases.

In order to find the absorption coefficient of the material, we must introduce the damping. I know how to find the index of refraction, but without introducing the dumping. So why analysing the dumping is necessary?

This can be shown by a rigorous mathematical analysis. In short, if you take the dumping into account, then the electric susceptibility $\chi_e$ is a complex quantity, its real part gives you the refractive index $n$, and its imaginary part gives you the absorption coefficient $\alpha$.

Edit. You start by developing the differential equation relating the polarisation $\vec P$ and the macroscopic electric field $\vec E$, this task is not quite obvious. This equation is similar to the forced harmonic oscillations's equation of motion:
$${d^2\vec P \over dt^2}+a{d\vec P \over dt}+\omega_0^2 \vec P=b\vec E$$

The damping is represented by the term ${d\vec P \over dt}$. You can interpret this equation as follows: the electric field forces (or induces) the polarisation.

The solution of this equation is trivial, is given by:
$$\vec P= \vec P_0 e^{i\omega t}$$ By plunging this solution into the equation we find the following relation:

$$\vec P=k(\omega)\vec E \tag 1\\$$ Where $k(\omega)$ is some complex parameter depends on the the electric field frequency $\omega$. $k(\omega)$ is complex due to the term ${d\vec P \over dt}$ in the equation. On the other hand, we have this relation:
$$\vec P=\chi_e \epsilon_0 \vec E \tag 2\\$$ $\chi_e$ is the electric susceptibility and $\epsilon_0$ is the permittivity of free space. By comparison, between $(1)$ and $(2)$:
$$\chi_e={k(\omega) \over \epsilon_0}$$ Since $k(\omega)$ is complex, so is $\chi_e$! The final step is writing $\chi_e$ in the algebraic form i.e $\chi_e=\chi_{e,r}+i\chi_{e,i}$, where the subscripts $r$ and $i$ refers to real and imaginary, respectively.

The math is straightforward. As I said before, the real part is related to the refraction index $n$ and the imaginary part is related the absorption coefficient $\alpha$.

• I got the first part. I'll accepet your response If you explain to me the mathematical analysis of the second part. Thanks. – Jonny Jul 19 '17 at 12:43
• @Jonny: I added some mathematical details and insights to my answer. I hope it's helpful and comprehensible. – Samà Jul 19 '17 at 13:33