# Compton scattering

In quantum mechanics, the Compton scattering or Compton effect, observed by Arthur Compton in 1923, is the increase in wavelength (decrease in energy) which occurs when X-ray (or gamma ray) photons with energies of around 0.5MeV to 3.5MeV interact with electrons in a material. Compton's experiment became the ultimate observation that convinced all physicists that light can behave as a stream of particles whose energy is proportional to the frequency.

Because the photons have such high energy, the interaction results in the electron being given enough energy to be completely ejected from its atom, and a photon containing the remaining energy being emitted in a different direction from the original, so that the overall momentum of the system is conserved. (If the photon still has enough energy, the process may be repeated.) Because of the overall reduction in energy of the photon, there is a corresponding increase in its wavelength. Thus overall there is a slight 'reddening' and scattering of the photons as they pass through the material. This scattering is known as Compton Scattering.

In a material where there are free electrons, this effect will occur at all photon energies and hence all wavelengths. In other materials, it is observed only with high-energy photons; photons of visible light, for example, do not have sufficient energy to eject the bound electrons.

The effect is important in scientific terms because it demonstrates that light cannot be explained purely as a wave phenomenon. Thomson scattering, the classical theory of charged particles scattered by an electromagnetic wave, cannot explain any shift in wavelength. Light must behave as if it consists of particles in order to explain the Compton scattering. It is also of prime importance to radiobiology, as it happens to be the most probable interaction of X rays with atomic nuclei in living beings and is applied in radiation therapy.

The Compton scattering has on occasion been proposed as an alternative explanation for the phenomenon of the Redshift by opponents of the Big Bang theory, although this is not generally accepted because the influence of the Compton scattering would be noticeable in the spectral lines of distant objects and this is not observed.

## The equations

File:Compton scattering diagram.png
Compton Scattering (in the rest frame of the target)

Compton used a combination of three fundamental formulas representing the various aspects of classical and modern physics, combining them to describe the quantum behaviour of light.

The final result gives us the Compton scattering equation:

$\displaystyle \Delta\lambda = \frac{h}{m_e c}(1-\cos{\theta})$
Here me is the electron mass and h/mec is known as the "Compton wavelength".

θ is the angle by which the photon's heading changes.

## Derivation

We use that:

$\displaystyle E_{\gamma} + E_{e} = E_{\gamma'} + E_{e'}\,$

(Conservation of energy, where $\displaystyle E_{\gamma}$ is the energy of a photon before the collision and $\displaystyle E_e$ is the energy of an electron before collision. (Its rest mass). The variables with a prime are used for those after the collision.
And:

$\displaystyle \vec p_{\gamma} + \vec{p_{e}} = \vec{p_{\gamma'}} + \vec{p_{e'}}\,$

(Conservation of impulse, with the $\displaystyle p_e=0$ because we assume that the electron is at rest.)
We then use $\displaystyle E = hf = pc$ :

$\displaystyle \vec{p_{e'}} = \vec{p_{\gamma}} - \vec{p_{\gamma'}}\,$
$\displaystyle {\vec{p_{e'}}}^2 = {(\vec{p_{\gamma}} - \vec{p_{\gamma'}})}^2$
$\displaystyle {\vec{p_{e'}}}^2 = \vec{p_{\gamma}}^2 - 2\cdot\vec{p_{\gamma}}\cdot\vec{p_{\gamma'}} + \vec{p_{\gamma'}}^2$
$\displaystyle \vec{p_{e'}} \cdot \vec{p_{e'}} = \vec{p_{\gamma}} \cdot \vec{p_{\gamma}}- 2\cdot\vec{p_{\gamma}}\cdot\vec{p_{\gamma'}} + \vec{p_{\gamma'}} \cdot \vec{p_{\gamma'}}$
$\displaystyle {p_{e'}}^2 \cdot \cos(0) = p_{\gamma}^2 \cdot \cos(0) - 2 \cdot p_{\gamma} \cdot p_{\gamma'} \cdot \cos(\theta) + p_{\gamma'}^2\cdot \cos(0)$

The $\displaystyle \cos(\theta)$ turns up because the impulses are actually vectors, and in 2D (just draw a collision, it's always 2D) the inner product (in this case, a dot product) of two vectors is the product of their norm times the cosine of the angle between them.
substituting $\displaystyle p_{\gamma}$ with $\displaystyle \frac{hf}{c}$ and $\displaystyle p_{\gamma'}$ with $\displaystyle \frac{hf'}{c}$ , we derive

$\displaystyle p_{e'}^2 = \frac{h^2 f^2}{c^2} + \frac{h^2 f'^2}{c^2} - \frac{2h^2 ff'\cos{\theta}}{c^2}$

Now we fill in for the energy part:

$\displaystyle E_{\gamma} + E_{e} = E_{\gamma'} + E_{e'}\,$
$\displaystyle hf + mc^2 = hf' + \sqrt{(p_{e'}c)^2 + (mc^2)^2}\,$

We solve this for pe':

$\displaystyle (hf + mc^2-hf')^2 = (p_{e'}c)^2 + (mc^2)^2\,$
$\displaystyle \frac{(hf + mc^2-hf')^2 -m^2c^4}{c^2}= p_{e'}^2\,$

Then we have two equations for $\displaystyle p_{e'}^2$ , which we equate:

$\displaystyle \frac{(hf + mc^2-hf')^2 -m^2c^4}{c^2} = \frac{h^2 f^2}{c^2} + \frac{h^2 f'^2}{c^2} - \frac{2h^2 ff'\cos{\theta}}{c^2}$

Now it's just a question of rewriting:

$\displaystyle h^2f^2+h^2f'^2-2h^2ff'+2h(f-f')mc^2 = h^2f^2+h^2f'^2-2h^2ff'\cos{\theta}\,$
$\displaystyle -2h^2ff'+2h(f-f')mc^2 = -2h^2ff'\cos{\theta}\,$
$\displaystyle hff'-(f-f')mc^2 = hff'\cos{\theta}\,$
$\displaystyle hff'(1-\cos{\theta}) = (f-f')mc^2\,$
$\displaystyle h\frac{c}{\lambda'}\frac{c}{\lambda}(1-\cos{\theta}) =\left(\frac{c}{\lambda}-\frac{c}{\lambda'}\right)mc^2$
$\displaystyle h\frac{c}{\lambda'}\frac{c}{\lambda}(1-\cos{\theta}) = \left(\frac{c\lambda'}{\lambda\lambda'}-\frac{c\lambda}{\lambda'\lambda}\right)mc^2$
$\displaystyle h(1-\cos{\theta}) = \frac{\lambda'}{c}\frac{\lambda}{c}\left(\frac{c\lambda'}{\lambda'\lambda}-\frac{c\lambda}{\lambda\lambda'}\right)mc^2$
$\displaystyle h(1-\cos{\theta}) = \left(\frac{\lambda'}{c}-\frac{\lambda}{c}\right)mc^2$
$\displaystyle \frac{h}{mc}(1-\cos{\theta}) =\lambda'-\lambda$