Fyneman exercise I ch31-4

Q.

It was deduced that the instantaneous energy flux of a wave is $S=\epsilon_0 c E^2$ watts per square meter:

1. Find the total rate at which energy is radiated by an electron which is oscillating with amplitude $x_0$ and angular frequency $\omega$.
2. Compare the energy radiated per cycle with the stored energy $\frac{1}{2}m\omega^2x_0^2$, and thus find the damping constant $\gamma_R$. This process is called radiation damping
3. An excited atom gives out radiation having a certain wavelength $\lambda$. Calculate theoretically the expected breadth $\Delta\lambda$ of the spectral line, if the breadth arises solely from radiation damping. (Think of the atom as being a tiny damped oscillator having a large Q.)

A.

1.

What is $E$ by an electron?

Let position $r$ and time $t$, referring FLP Volume I Chapter 28

$E(t, r)=-\frac{q}{4\pi\epsilon_0 c^2 r}a(t-r/c)\sin{\theta}$
, where $\theta$ is angle between sight line(i.e position vector r) and oscillation deriction.

Electron oscillates with amplitude $x_0$, angular frequency $\omega$, so retarded acceleration is

$a(t-r/c)=-\omega^2 x_0 e^{i\omega(t-r/c)}$

Then $E$ is

$E(t, r)=\frac{q}{4\pi\epsilon_0 c^2 r}\omega^2 x_0 e^{i\omega(t-r/c)}\sin{\theta}$

What is $E^2$?

At position $r$

$E^2=E\cdot E^\ast=(\frac{q}{4\pi\epsilon_0 c^2 r}\omega^2 x_0)^2\sin^2{\theta}$

Total rate at which energy is radiated by an electron

Total energy radiated in all directions is

$P=\int S dA=\int\epsilon_0 c E^2 dA$

$P= \epsilon_0 c \int_{0}^{\pi}{\int_{0}^{2\pi}{E^2 (r\sin{\theta}d\psi) (rd\theta})} = \epsilon_0 c\int_{0}^{\pi}{E^2 2\pi r^2 \sin{\theta}d\theta}$

Put $E^2$

$P= \epsilon_0 c \int_{0}^{\pi}{E(r)^2 2\pi r^2 \sin{\theta}d\theta}=\epsilon_0 c \int_{0}^{\pi}{(\frac{q}{4\pi\epsilon_0 c^2 r}\omega^2 x_0)^2\sin^2{\theta}\cdot 2\pi r^2 d\theta}$

Rearrange

$P_{total\ at \ same \ distance \ r}=\frac{q^2\omega^4 x_0^2}{8\pi \epsilon_0 c^3}\int_{0}^{\pi}{\sin^3{\theta}d\theta}$

$\int_{0}^{\pi}{\sin^3{\theta}d\theta}=\int_{0}^{\pi}{(1-\cos^2{\theta})\sin{\theta}d\theta}=\int_{0}^{\pi}{(\sin{\theta}-\cos^2{\theta}\sin{\theta})d\theta} = \frac{4}{3}$

$\therefore P_{total\ at \ same \ distance \ r}=\frac{q^2\omega^4 x_0^2}{6\pi \epsilon_0 c^3}$

FLP Volume I Chapter 32

remember that we have to be very careful when we square things that are written in complex notation—it really is the cosine, and the average of $\cos^2{\omega(t-r/c)}$ is one-half

So, time averaged $P$ is

$P=\frac{q^2\omega^4 x_0^2}{12\pi \epsilon_0 c^3}$

2.

Energy radiated per cycle = $time \ averaged \ P$

$\gamma_R = \frac{time \ averaged \ P}{stored\ energy}$


$\therefore \gamma_R=\frac{\frac{q^2\omega^4 x_0^2}{12\pi \epsilon_0 c^3}}{\frac{1}{2}m\omega^2x_0^2}=\frac{q^2 \omega^2}{6\pi \epsilon_0 c^3 m}$

3.

$\lambda = \frac{2\pi}{\omega} c$


$\Delta \lambda = 2\pi c\Delta(\frac{1}{\omega})=\frac{2\pi c}{\omega^2}\Delta \omega$

FLP Volume I Chapter 23

at one half the maximum height, the full width at half the maximum height of the curve is $\Delta \omega =\gamma$

As hint said, suppose the atom as being a tiny damped oscillator, we can replace $\Delta \omega = \gamma_R$

$\Delta \lambda = \frac{2\pi c}{\omega^2} \gamma_R = \frac{2\pi c}{\omega^2} \frac{q^2 \omega^2}{6\pi \epsilon_0 c^3 m} = \frac{q^2}{3\epsilon_0 c^2 m}$

$\Delta \lambda= \frac{(1.60 \times 10^{-19})^2}{3\times 8.85 \times 10^{-12}\times (3.00\times 10^8)^2 \times 9.11\times 10^{-31}}$

$\therefore \Delta \lambda = 1.18 \times 10^{-14}$