Question 13.2: The Compton effect depicted in Fig. 13.1 is a collision even...
The Compton effect depicted in Fig. 13.1 is a collision event in which a photon scatters off an electron that is initially at rest
γ + e → γ + e.
Here the photon is denoted by γ . Find the frequency of the scattered photon as a function of the scattering angle
We shall denote the momentum four-vectors of the initial and final photons by P^{\mu }_{γ } and P^{\prime \mu }_{γ }, and momentum four-vectors of the initial and final electrons by P^{\mu } and P^{\prime \mu }. The condition that the energy and momentum be conserved is equivalent to the requirement that the sum of the momentum four-vectors before and after the collision be equal
P^{\mu }_{γ }+P^{ \mu }=P^{\prime \mu }_{γ }+P^{\prime \mu }.Taking the electron four-vector P^{\prime \mu } over to the left-hand side of the equation and the photon four-vector P^{\mu }_{γ } over to the right-hand side and forming the square of the sum of four-vectors on the left and the right, we obtain
\left|P^{\mu }+P^{\prime \mu }\right|^{2}=\left|P^{\prime \mu }_{γ }- P^{ \mu }_{γ }\right| ^{2}. (13.16)
The electron four-momenta are now on the left and the photon four-momenta are on the right.
If the electron is initially at rest, the initial energy of the electron, which we denote by E, is equal to the rest energy mc^{2} and the spatial part of the initial momentum of the electron p is zero. We denote the energy and the spatial part of the momentum of the outgoing electron by E^\prime and P^\prime . The left-hand side of Eq. (13.16) can be evaluated by taking the difference of the square of the zeroth and spatial components of the four-vector P^{\mu} −P^{\prime \mu } . We get
\left|P^{\mu }+P^{\prime \mu }\right|^{2}= \frac{1}{c^{2}}(mc^{2}-E^{\prime })^{2}-P^{\prime 2} .Using the energy-momentum relation (13.14), the right-hand side of this last equation can be simplified giving
E^{2} = p^{2}c^{2} + m^{2}c^{4}. (13.14)
\left|P^{\mu }+P^{\prime \mu }\right|^{2}=2m^{2}c^{2}-2mE^{\prime } . (13.17)
We shall denote the initial and final energies of the photon by hf and hf^\prime , respectively. According to Eq. (13.15), the magnitude of the momentum of a photon is its energy divided by c. We shall thus denote the initial and final momenta of the photon by (hf/c)\hat{p} and (hf^\prime /c)\hat{p}^\prime , where \hat{p} and \hat{p}^\prime are unit vectors pointing in the incoming and outgoing directions of the photon. With this notation, the right-hand side of Eq. (13.16) can be written
E = |p|c. (13.15)
\left| P^{\prime \mu }_{γ }-P^{\mu }_{γ }\right|^{2}=\frac{h^{2}}{c^{2}}(f^\prime-f)^{2}-\frac{h^{2}}{c^{2}}(f^\prime \hat{p}^\prime – f \hat{p}^\prime)^{2}We note that the terms on the right-hand side of this last equation which involve the squares of the frequencies cancel out, and we get
\left| P^{\prime \mu }_{γ }-P^{\mu }_{γ }\right|^{2}=-\frac{2h^{2}ff^{\prime }}{c^{2}}(1-\cos \theta ). (3.18)
where θ is the angle between the outgoing and incoming directions of the photon. Recall that Eqs. (13.17) and (13.18) give the left- and right-hand sides of Eq. (13.16). Equating these two results, we obtain
2m^{2}c^{2}-2mE^\prime =-\frac{2h^{2}ff^{\prime }}{c^{2}}(1-\cos \theta ). (13.19)
The energy E^\prime of the outgoing electron, which appears on the left-hand side of this equation, can be removed by using the conservation of the energy. The condition that energy isconserved in the process is
h f +m c^{2}= h f^\prime+ E^\prime.Solving this equation for E^\prime and substituting this value into Eq. (13.19), we get
-2mh(f-f^\prime )=-\frac{2h^{2 }f f^\prime }{c^{2}}(1-\cos \theta ) .Finally, multiplying this last equation through by −c^{2}/2h, we obtain
mc^{2}(f-f^\prime )=h f f^\prime (1-\cos \theta ) . (13.20)
In 1923, Arthur Compton performed an experiment in which light scattered off electrons in graphite obtaining results consistent with the above formula. His experiment played a very important role in the development of physics because it showed that photons could scatter like ordinary particles.