Derive fluorescent scattering cross section as an approximation when $\omega \sim \omega_0$

The scattering cross section for an bounded electron with resonant frequency $\nu_0 =\frac{\omega_0}{2 \pi}$ under the effect of an EM with frequency $\nu=\frac{\omega}{2\pi}$ is $$\sigma_d=\bigg( \frac{8}{3} \pi r_0^2 \bigg) \frac{\omega^4}{(\omega ^2 -\omega_0^2)^2+(\Gamma \omega)^2 }$$

Where $\Gamma$ is the dissipation factor and $r_0$ is electron classical radius.

On textbook, when fluorescent cross section is introduced, it is said that the previous expression is approximated to the following when $\omega \sim \omega_0$

$$\bigg( \frac{8}{3} \pi r_0^2 \bigg) \frac{\omega^4}{(\omega ^2 -\omega_0^2)^2+(\Gamma \omega)^2 }\to_{\omega \sim \omega_0} \bigg( \frac{3}{2} \pi \frac{\lambda_0^2}{ 4 \pi^2} \bigg) \frac{\gamma^2}{(\omega -\omega_0)^2+(\Gamma /2)^2 }$$

Where $\lambda_0 =\frac{c}{\nu_0}$ and $\gamma$ represents the part of $\Gamma$ related to dissipation due to irradiation. (In general $\Gamma= \gamma +\gamma'$ where, $\gamma'$ is dissipation due to other effects, such as the presence of atoms near the electron).

How to prove the previous approximation mathematically?

• Since you are adding $(\omega^2-\omega_0^2)^2$ and $(\Gamma\omega)^2$, then $[\Gamma] = [\gamma] = [\omega]$. Then, $[\bigg( \frac{3}{2 \cdot 4 \pi^2} \pi \lambda_0^2 \bigg) \frac{\gamma}{(\omega -\omega_0)^2+(\Gamma /2)^2 }] = \frac{[\sigma_d]}{[\omega]}$. Your formula can't be right. – Spirine Aug 19 '17 at 15:29
• @Spirine Thanks a lot, I apologize it is $\gamma^2$, I fixed it – Sørën Aug 19 '17 at 16:43

Maybe this helps.

$$\frac{\omega ^4}{(\omega ^2 - \omega_0^2)^2+(Γ\omega)^2} = \frac{\omega ^4}{(\omega - \omega_0)^2(\omega + \omega_0)^2+(Γ\omega)^2}$$

Since $\omega \approx\omega_0$, we can consider that $(\omega + \omega_0)^2\approx(\omega + \omega)^2 \approx (\omega_0 + \omega_0)^2\approx4\omega^2$ because good ol' physics and totally valid approximations. So, the relation becomes

$$\frac{\omega ^4}{4\omega^2(\omega - \omega_0)^2+(Γ\omega)^2} = \frac{1}{4} \frac{\omega ^2}{(\omega - \omega_0)^2+(Γ/2)^2}$$

This can also be considered as $$\frac{1}{4} \frac{\omega_0 ^2}{(\omega - \omega_0)^2+(Γ/2)^2}$$

Based on the same approximation made above.

Now you can see that the main fraction is similar to what you wanted. All that's left is to use the relations for the other parameters and it should get you to the result (as long as there are no typos there). I did not do it myself because it's not clear to me what $\gamma$ is.

In his answer, Victor Palea gave you the mathematical development that is needed for your derivation. Now, what you need is a answer from physics, that is what is $\gamma$ in your case?

From what I read, I understand non relativistic classical model of dipolar radiation with radiation dampening at the rate $\Gamma$. Something that classically correspond to an equation looking like the following:

$$\stackrel{..}{x} - \, \Gamma \stackrel{.}{x} + \, \omega_0^2 x = C$$

Short answer: $\gamma$ is the classical damping factor due to the radiation of moving charges (see Larmor's formula or Abraham–Lorentz force). For an electron it reads:

$\gamma = \dfrac{2}{3}\omega_0^2 \dfrac{r_0}{c}$ (check homogeneity)

So $\left(\dfrac{8}{3} \pi r_0^2 \right) \omega_0^2 = \left(\dfrac{8}{3} \pi\right) \dfrac{9}{4} \dfrac{\gamma^2 c^2}{\omega_0^2} = \left( 6 \pi\right)\dfrac{c^2}{4\pi^2\nu_0^2} \, \gamma^2 = 6 \pi \dfrac{\lambda_0^2}{4\pi^2} \, \gamma^2$

And you have a factor 1/4 to add from the Taylor expansion done by Victor Palea.

Finally, notice that I am not neglecting other effects represented through $\gamma'$, so we still have $\Gamma = \gamma + \gamma'$.