10.10: APPENDIX B- Radiation Damping as Functions of Angular Frequency, Frequency and Wavelength

It occurred to me while preparing this Chapter as well as the preceding and following ones, that sometimes I have been using angular frequency as argument, sometimes frequency, and sometimes wavelength. In this Appendix, I bring together the salient formulas for radiation damping in terms of $∆ω = ω - ω_0,\, ∆ν = ν - ν_0\text{ and }∆λ = λ - λ_0$. I reproduce equation 10.2.11 for the absorption coefficient for a set of forced, damped oscillators, except that I replace $n$, the number per unit volume of oscillators with $n_1f_{12}$, the effective number of atoms per unit volume in the lower level of a line, and I replace the classical damping constant $γ$ with $\Gamma$, which may include a pressure broadening component.

$$\tag{10.B.1}\alpha =\frac{n_1f_{12}\Gamma e^2 \omega^2}{mε_0c[(\omega^2-\omega_0^2)^2+\Gamma^2\omega^2]}\quad \text{m}^{-1}.$$

You should check that the dimensions of this expression are $L^{-1}$, which is appropriate for linear absorption coefficient. You may note that $[e^2 /ε_0] \equiv \text{ML}^3\text{T}^{-2}\text{ and }[\Gamma] \equiv \text{T}^{-1}$. Indeed check the dimensions of all expressions that follow, at each stage.

We can write $\omega^2-\omega_0^2=(\omega-\omega_0)(\omega+\omega_0)=∆ \omega (2\omega_0+∆ \omega)$, and the equation becomes

$$\alpha = \frac{n_1f_{12}\Gamma e^2 (\omega_0+∆ \omega)^2}{mε_0 c[(∆ \omega)^2(2\omega_0+∆ \omega)^2+\Gamma^2 (\omega_0 +∆ \omega)^2]}\quad \text{m}^{-1}.\tag{10.B.2}$$

Now I think it will be owned that the width of a spectrum line is very, very much smaller than its actual wavelength, except perhaps for extremely Stark-broadened hydrogen lines, so that, in the immediate vicinity of a line, $∆ω$ can be neglected compared with $ω_0$; and a very long way from the line, where this might not be so, the expression is close to zero anyway. (Note that you can neglect $∆ω$ only with respect to $ω$; you cannot just put $∆ω = 0$ where it lies alone in the denominator!) In any case, I have no compunction at all in making the approximation

$$\alpha (∆ \omega)=\frac{n_1f_{12}\Gamma e^2}{4mε_0 c[(∆ \omega)^2+(\frac{1}{2}\Gamma)^2]}\quad \text{m}^{-1}.\tag{10.B.3}$$

The maximum of the $\alpha (∆ \omega)$ curve is

$$\tag{10.B.4}\alpha(0)=\frac{e^2n_1f_{12}}{mε_0c\Gamma}\quad \text{m}^{-1}.$$

The optical thickness at the line centre (whether or not the line is optically thin) is

$$\tag{10.B.5}\tau (0)=\frac{e^2 \mathcal{N}_1f_{12}}{mε_0 c\Gamma}.$$

$\mathcal{N}_1$ is the number of atoms in level 1 per unit area in the line of sight, whereas $n_1$ is the number per unit volume.

The HWHM of $\alpha (∆ω)$ curve is

$$\tag{10.B.6}\text{HWHM}=\frac{1}{2}\Gamma \quad\text{rad s}^{-1}.$$

The area under the $\alpha (∆ω)$ curve is

$$\tag{10.B.7}\text{Area}=\frac{\pi e^2 n_1f_{12}}{2mε_0 c}\quad \text{m}^{-1}\text{rad s}^{-1}.$$

As expected, the area does not depend upon $\Gamma$.

To express the absorption coefficient as a function of frequency, we note that $ω = 2π\nu$, and we obtain

$$\tag{10.B.8}\alpha (∆\nu)=\frac{n_1f_{12}\Gamma e^2}{16\pi^2mε_0 c[(∆\nu)^2+\left (\frac{\Gamma}{4\pi}\right )^2]}\quad \text{m}^{-1}.$$

The maximum of this is (of course) the same as equation 10.B.4.

The HWHM of the $\alpha (∆\nu)$ curve is

$$\tag{10.B.9}\text{HWHM}=\Gamma /(4\pi)\quad \text{s}^{-1}.$$

The area under the $\alpha (∆\nu)$ curve is

$$\tag{10.B.10}\text{Area}=\frac{e^2n_1f_{12}}{4mε_0c}\quad \text{m}^{-1}\text{s}^{-1}.$$

$$\tag{10.B.11}\alpha=\frac{n_1f_{12}\Gamma e^2}{mε_0 c}\cdot \left ( \frac{λ^2λ_0^4}{4\pi^2c^2\left (λ_0^2-λ^2 \right )^2+λ^2 λ_0^4\Gamma^2}\right )\quad \text{m}^{-1}.$$

In a manner similar to our procedure following equation 10.B.12, we write $λ_0^2-λ^2=(λ_0 -λ)(λ_0 +λ)$, and , $λ=λ_0+∆ λ$, and neglect $∆ λ$ with respect to $λ_0$, and we obtain:

$$\tag{10.B.12}\alpha (∆ λ)=\frac{n_1f_{12}\Gamma e^2}{16\pi^2mε_0c^3}\cdot \left (\frac{λ_0^4}{(∆ λ)^2+\frac{λ_0^4 \Gamma^2}{16\pi^2c^2}}\right )\quad \text{m}^{-1}.$$

The maximum of this is (of course) the same as equation 10.B.4. (Verifying this will serve as a check on the algebra.)

The HWHM of the $\alpha (∆λ)$ curve is

$$\tag{10.B.13}\text{HWHM}=\frac{λ_0^2\Gamma}{4\pi c}\quad \text{m}.$$

The area under the $\alpha (∆λ)$ curve is

$$\tag{10.B.14}\text{Area}=\frac{λ_0^2e^2n_1f_{12}}{4mε_0 c^2}.$$

Did I forget to write down the units after this equation?

These results for $\alpha$ might be useful in tabular form. For $\tau$, replace $n_1\text{ by }N_1$.

$$\begin{array}{c c c c}&∆ \omega &∆\nu & ∆ λ \\ &\frac{\Gamma e^2 n_1 f_{12}}{4mε_0 c[(∆ \omega)^2+(\frac{1}{2}\Gamma)^2]} & \frac{\Gamma e^2 n_1 f_{12}}{16\pi^2mε_0 c[(∆\nu)^2+\left (\frac{\Gamma}{4\pi}\right )^2]} & \frac{\Gamma e^2λ_0^4 n_1f_{12}}{16\pi^2 mε_0 c^3 \left [(∆ λ)^2+\frac{λ_0^4\Gamma^2}{16\pi^2c^2}\right ]} \\ \text{Height} & \frac{e^2n_1f_{12}}{mε_0 c\Gamma} & \frac{e^2 n_1 f_{12}}{mε_0c\Gamma}&\frac{e^2 n_1 f_{12}}{mε_0c\Gamma}\\ \text{Area} & \frac{\pi e^2 n_1 f_{12}}{2mε_0 c} & \frac{e^2n_1 f_{12}}{4mε_0 c} & \frac{λ_0^2 e^2 n_1 f_{12}}{4mε_0 c^2} \\ \text{HWMH} & \frac{1}{2}\Gamma & \Gamma/(4\pi) & \frac{λ_0^2 \Gamma}{4\pi c} \\ \end{array}$$

It is to be noted that if the radiation damping profile is thermally broadened, the height of the absorption coefficient curve diminishes, while the area is unaltered provided that the line is optically thin. The optically thick situation is dealt with in the following chapter. It might also be useful to note that a gaussian profile of the form

$$\tag{10.B.15}\alpha (∆ λ) = \alpha (0) \text{exp}\left (-\frac{c^2 (∆ λ)}{V_\text{m}^2 λ_0^2}\right )$$

has an area of $\frac{λ_0^2 e^2 n_1 f_{12}}{4mε_0 c^2}$ if

$$\tag{10.B.16}\alpha (0) =\frac{λ_0 e^2n_1 f_{12}}{4\sqrt{\pi}mε_0 cV_\text{m}}.$$