- #1

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## Summary:

- I want to understand how the experimentalist deals with Compton Scattering theory and see some examples.

## Main Question or Discussion Point

I've been studying Compton Scattering (Mandl & Shaw, pages 142-147) and the derivation of the following formulas:

$$\Big( \frac{d \sigma}{d \Omega}\Big)_{LAB} = \frac{1}{(4 \pi)^2} \Big( \frac{\omega'}{\omega}\Big)^2 |\mathscr{M}|^2 \tag 1$$

$$\Big( \frac{d\sigma}{d \Omega} \Big)_{LAB} = \frac{\alpha^2}{2m^2} \Big(\frac{\omega'}{\omega}\Big)^2 \Big\{ \frac{\omega}{\omega'}+\frac{\omega'}{\omega}-\sin^2 \theta \Big\} \tag 2$$

Which in the low-energy limit (i.e. ##\omega' \sim \omega##) reduces to the differential cross-section for Thomson scattering

$$\Big( \frac{d\sigma}{d \Omega} \Big)_{LAB, \omega' \sim \omega} = \frac{\alpha^2}{2m^2} \Big\{ 2-\sin^2 \theta \Big\}=\frac{\alpha^2}{2m^2} \Big\{ 1+\cos^2 \theta \Big\} \tag 3$$

$$\Big( \frac{d\sigma}{d \Omega} \Big)_{LAB, pol} = \frac{\alpha^2}{4m^2} \Big(\frac{\omega'}{\omega}\Big)^2 \Big\{\frac{\omega}{\omega'}+\frac{\omega'}{\omega}+4(\epsilon \epsilon')^2 -2 \Big\} \tag 4$$

The unpolarized cross-section ##(2)## can be obtained due to

$$\frac 1 2 \sum_{pol} (\epsilon \epsilon')^2 = \frac 1 2 (1 + \cos^2 \theta) \tag 5$$

So my questions are:

Is ##(2)## the only important equation for the experimentalist? Based on what I've read most experiments involve unpolarized beams. Thus, the polarization of the particles produced in the collision are not observed. Are lasers an exception?

Why is Thomson approximation useful?

Thank you

**1)**Differential cross section in the LAB frame$$\Big( \frac{d \sigma}{d \Omega}\Big)_{LAB} = \frac{1}{(4 \pi)^2} \Big( \frac{\omega'}{\omega}\Big)^2 |\mathscr{M}|^2 \tag 1$$

**2)**The unpolarized differential cross-section in the LAB frame$$\Big( \frac{d\sigma}{d \Omega} \Big)_{LAB} = \frac{\alpha^2}{2m^2} \Big(\frac{\omega'}{\omega}\Big)^2 \Big\{ \frac{\omega}{\omega'}+\frac{\omega'}{\omega}-\sin^2 \theta \Big\} \tag 2$$

Which in the low-energy limit (i.e. ##\omega' \sim \omega##) reduces to the differential cross-section for Thomson scattering

$$\Big( \frac{d\sigma}{d \Omega} \Big)_{LAB, \omega' \sim \omega} = \frac{\alpha^2}{2m^2} \Big\{ 2-\sin^2 \theta \Big\}=\frac{\alpha^2}{2m^2} \Big\{ 1+\cos^2 \theta \Big\} \tag 3$$

**3)**The polarized cross-section in the LAB frame (known as Klein-Nishina formula)$$\Big( \frac{d\sigma}{d \Omega} \Big)_{LAB, pol} = \frac{\alpha^2}{4m^2} \Big(\frac{\omega'}{\omega}\Big)^2 \Big\{\frac{\omega}{\omega'}+\frac{\omega'}{\omega}+4(\epsilon \epsilon')^2 -2 \Big\} \tag 4$$

The unpolarized cross-section ##(2)## can be obtained due to

$$\frac 1 2 \sum_{pol} (\epsilon \epsilon')^2 = \frac 1 2 (1 + \cos^2 \theta) \tag 5$$

So my questions are:

Is ##(2)## the only important equation for the experimentalist? Based on what I've read most experiments involve unpolarized beams. Thus, the polarization of the particles produced in the collision are not observed. Are lasers an exception?

Why is Thomson approximation useful?

Thank you