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# 14.4: Totally Elastic Collision - Compton Scattering

As a final example of a collision in special relativity, we consider the totally elastic case: a collision in which the total momentum, total kinetic energy, and the mass of all particles are conserved. An example of such a collision is Compton scattering: the collision between a photon and an electron, resulting in a transfer of energy from one to the other, visible in a change of wavelength of the photon. For our example, we’ll take the electron to be initially stationary, and the photon to be coming in along the $$x$$-axis; after the collision, both particles have nonzero velocities in both the $$x$$ and $$y$$ directions (see Figure 14.4.1).

Figure $$\PageIndex{1}$$: Compton scattering between a photon and an electron, resulting in a transfer of energy of the photon to the electron, measurable as a change in the photon’s wavelength.

The four-momenta of the electron and photon before and after the collision are given by:

$\overline{\boldsymbol{p}}_{\mathrm{e}, \mathrm{i}}=\left( \begin{array}{c}{m_{\mathrm{e}} c} \\ {0} \\ {0} \\ {0}\end{array}\right), \quad \overline{\boldsymbol{p}}_{\gamma, \mathrm{i}}=\frac{E_{\mathrm{i}}}{c} \left( \begin{array}{c}{1} \\ {1} \\ {0} \\ {0}\end{array}\right), \quad \overline{\boldsymbol{p}}_{\mathrm{e},\mathrm{f}}=\left( \begin{array}{c}{E_{\mathrm{e},\mathrm{f}} / c} \\ {p_{\mathrm{e}, \mathrm{f}} \cos \phi} \\ {p_{\mathrm{e}, \mathrm{f}} \sin \phi} \\ {0}\end{array}\right), \quad \overline{\boldsymbol{p}}_{\gamma, \mathrm{f}}=\frac{E_{\mathrm{f}}}{c} \left( \begin{array}{c}{1} \\ {\cos \theta} \\ {-\sin \theta} \\ {0}\end{array}\right) \label{14.4.1}$

We can now solve for the energy $$E_{\mathrm{f}}$$ of the outgoing photon (and thus its wavelength) in terms of that of the incoming photon ($$E_{\mathrm{i}}$$) and the scattering angle $$\theta$$. There are again (at least) two ways to do this. One is to compare the components of the initial and final energy-momentum four-vector term by term. The other is to again use the fact that we know about the length of the four-vector to immediately eliminate the scattering angle $$\phi$$ of the electron. To do so, we first rewrite the conservation of energy-momentum equation, $$\overline{\boldsymbol{p}}_{\mathrm{e}, \mathrm{i}}+\overline{\boldsymbol{p}}_{\gamma, \mathrm{i}}= \overline{\boldsymbol{p}}_{\mathrm{e}, \mathrm{f}}+\overline{\boldsymbol{p}}_{\gamma, \mathrm{f}}$$ to isolate the term of the outgoing electron, and then take the square, to get:

$\left(\overline{\boldsymbol{p}}_{\mathrm{e}, \mathrm{i}}+\overline{\boldsymbol{p}}_{\gamma, \mathrm{i}}-\overline{\boldsymbol{p}}_{\gamma, \mathrm{f}}\right)^{2}=\overline{\boldsymbol{p}}_{\mathrm{e}, \mathrm{f}}^{2} \label{14.4.2}$

$\overline{\boldsymbol{p}}_{\mathrm{e}, \mathrm{i}}^{2}+\overline{\boldsymbol{p}}_{\gamma, \mathrm{i}}^{2}+\overline{\boldsymbol{p}}_{\gamma, \mathrm{f}}^{2}+\overline{\boldsymbol{p}}_{\mathrm{e}, \mathrm{i}} \cdot \overline{\boldsymbol{p}}_{\gamma, \mathrm{i}}^{2}-2 \overline{\boldsymbol{p}}_{\mathrm{e}, \mathrm{i}} \cdot \overline{\boldsymbol{p}}_{\gamma, \mathrm{f}}-2 \overline{\boldsymbol{p}}_{\gamma, \mathrm{i}} \cdot \overline{\boldsymbol{p}}_{\gamma, \mathrm{f}}=\overline{\boldsymbol{p}}_{\mathrm{ef}}^{2}\label{14.4.3}$

$m_{e}^{2} c^{2}+0+0+2 m_{\mathrm{e}} E_{\mathrm{i}}-2 m_{\mathrm{e}} E_{\mathrm{f}}-2 \frac{E_{\mathrm{i}} E_{\mathrm{f}}}{c^{2}}(1-\cos \theta)=m_{e}^{2} c^{2}\label{14.4.4}$

from which we can solve for $$E_{\mathrm{f}}$$. Rewriting to wavelengths (through$$E=h f=h c / \lambda$$), we get

$\lambda_{\mathrm{f}}=\lambda_{\mathrm{i}}+\frac{h}{m_{\mathrm{e}} c}(1-\cos \theta)\label{14.4.5}$