﻿ Solution: Compton's Formula | Primer on Geometric Algebra | David Hestenes

## Solution: Compton's Formula

Problem: Determine the shift in photon frequency due to the scattering:  γ + e  →  γ + e .

Compton scattering usually refers to high energy x-ray or gamma ray scattering off materials.
The photon scatters off a loosely bound, almost free, relatively low kinetic energy electron, which
we take to be "at rest" before the collision. As to why the photon would be scattered rather than
absorbed by the electron (as in the photoelectric effect),

Summarizing information known:

$\Large{\begin{array}{l}{P_{1\left( {{\rm{photon}}\,\,{\rm{before}}} \right)}}\,\, + \,\,{P_{2\left( {{\rm{electron}}\,\,{\rm{before,}}\,\,{\rm{at}}\,\,{\rm{rest}}} \right)}}\,\,\, = \,\,\,{P_{3\left( {{\rm{photon}}\,\,{\rm{after}}} \right)}}\,\, + \,\,{P_{4\left( {{\rm{electron}}\,\,{\rm{after}}} \right)}}\\\\\\\\{P_1}\,\widetilde {{P_1}}\,\,\, = \,\,\,0\,\,\, = \,\,\,{P_3}\,\widetilde {{P_3}}\,\,,\,\,\,\,~~~~~~~~~~~~~~\,\,\,\,\,\,\,\,{P_2}\,\widetilde {{P_2}}\,\,\, = \,\,\,{m^2}{c^2}\,\,\, = \,\,\,{P_4}\,\widetilde {{P_4}}\\\\\\{P_1}\,\,\, = \,\,\,\frac{h}{{{\lambda _1}}}\,\left( {1\,\, + \,\,{{{\bf{\hat p}}}_1}} \right)\,,\,\,\,\,\,\,~~~~~~~~~~~~~~\,\,\,\,\,\,\,\,{P_3}\,\,\, = \,\,\,\frac{h}{{{\lambda _3}}}\,\left( {1\,\, + \,\,{{{\bf{\hat p}}}_3}} \right)\,,\,\,\,\,~~~~~~~~~~~~~~\,\,\,\,\,\,\,\,{P_2}\,\,\, = \,\,\,mc\end{array}}$

Now, let's see if we can find a formula involving the scattering angle θ:

$\Large{\begin{array}{l}{m^2}{c^2}\,\,\, = \,\,\,{P_4}\,\widetilde {{P_4}}\,\,\, = \,\,\,\left[ {\left( {{P_1}\,\, - \,\,{P_3}} \right)\,\, + \,\,{P_2}} \right]\,\,\left[ {\left( {\widetilde {{P_1}}\,\, - \,\,\widetilde {{P_3}}} \right)\,\, + \,\,\widetilde {{P_2}}} \right]\\\\{m^2}{c^2}\,\,\, = \,\,\, - \,\,{P_1}\,\widetilde {{P_3}}\,\, - \,\,{P_3}\,\widetilde {{P_1}}\,\, + \,\,mc\,\left( {{P_1}\,\, + \,\,\widetilde {{P_1}}\,\, - \,\,{P_3}\,\, - \,\,\widetilde {{P_3}}} \right)\,\, + \,\,{m^2}{c^2}\\\\\frac{h}{{{\lambda _1}}}\,\left( {1\,\, + \,\,{{{\bf{\hat p}}}_1}} \right)\frac{h}{{{\lambda _3}}}\,\left( {1\,\, - \,\,{{{\bf{\hat p}}}_3}} \right)\,\, + \,\,\frac{h}{{{\lambda _3}}}\,\left( {1\,\, + \,\,{{{\bf{\hat p}}}_3}} \right)\frac{h}{{{\lambda _1}}}\,\left( {1\,\, - \,\,{{{\bf{\hat p}}}_1}} \right)\,\,\, = \,\,\,mc\,\left( {\frac{{2h}}{{{\lambda _1}}}\,\, - \,\,\frac{{2h}}{{{\lambda _3}}}} \right)\\\\\frac{{{h^2}}}{{{\lambda _1}{\lambda _3}}}\,\left( {2\,\, - \,\,2\,{{{\bf{\hat p}}}_1} \cdot {{{\bf{\hat p}}}_3}} \right)\,\,\, = \,\,\,2mch\,\,\left( {\frac{1}{{{\lambda _1}}}\,\, - \,\,\frac{1}{{{\lambda _3}}}} \right)\\\\\\{\bbox[8px,border:2px solid red]{\frac{h}{{mc}}\,\left( {1\,\, - \,\,\cos \theta } \right)\,\,\, = \,\,\,{\lambda _3}\,\, - \,\,{\lambda _1}\,\,\, = \,\,\,\Delta \lambda}} \end{array}}$

We have arrived at
Compton's formula. Arthur Compton's half of the 1927 Nobel prize was for
discovery of this scattering. The
Compton wavelength for the electron is h/(mc) = 0.00243 nm, and
its well-known rest energy is
mc2 = 511 keV. We can use these values to see one way to convert
quickly between a photon's wavelength and energy:

$\Large{\begin{array}{l}hc\,\,\, = \,\,\,\frac{h}{{mc}}\,m{c^2}\,\,\, = \,\,\,\left( {0.00243\,{\rm{nm}}} \right)\,\left( {511\,\,{\rm{keV}}} \right)\,\,\, = \,\,\,1.24\,\,{\rm{keV}} \cdot {\rm{nm}}\\\\\\{E_{{\rm{photon}}}}\,\,\, = \,\,\,\frac{{hc}}{\lambda }\,\,\, = \,\,\,\frac{{1.24\,\,{\rm{keV}} \cdot {\rm{nm}}}}{\lambda }\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{{1.24\,\,{\rm{keV}} \cdot {\rm{nm}}}}{{59.54\,\,{\rm{keV}}}}\,\,\, = \,\,\,0.0208\,\,{\rm{nm}}\end{array}}$

For our numerical example, we've used a gamma ray emitted from the nucleus of radioactive 241Am.
From its wavelength, it might be classified as a hard (high energy) x-ray, but it's usually called a
low energy gamma ray because its source was a nucleus. We'll use it in the interactive diagram
below. Meanwhile, we might also determine the
kinetic energy of the electron which was kicked
from rest:

$\Large{\begin{array}{l}{P_4}\,\,\, = \,\,\,\frac{{{E_4}}}{c}\,\, + \,\,{{\bf{p}}_4}\,\,\, = \,\,\,{P_1}\,\, - \,\,{P_3}\,\, + \,\,{P_2}\,\,\, = \,\,\,\frac{h}{{{\lambda _1}}}\,\left( {1\,\, + \,\,{{{\bf{\hat p}}}_1}} \right)\,\, - \,\,\frac{h}{{{\lambda _3}}}\,\left( {1\,\, + \,\,{{{\bf{\hat p}}}_3}} \right)\,\, + \,\,mc\\\\{K_4}\,\,\, = \,\,\,{E_4}\,\, - \,\,m{c^2}\,\,\, = \,\,\,\frac{{hc}}{{{\lambda _1}}}\,\, - \,\,\frac{{hc}}{{{\lambda _3}}}\,\,\, = \,\,\,{\rm{energy}}\,\,{\rm{lost}}\,\,{\rm{by}}\,\,{\rm{photon}}\\\\{\lambda _{3\max }}\,\,\, = \,\,\,{\lambda _1}\,\, + \,\,\frac{h}{{mc}}\,\left( {1\,\, - \,\,\cos \pi } \right)\,\,\, = \,\,\,0.0208\,{\rm{nm}}\,\, + \,\,2\,\left( {0.00243\,{\rm{nm}}} \right)\\\\\\{K_{4\max }}\,\,\, = \,\,\,\frac{{1.24\,\,{\rm{keV}} \cdot {\rm{nm}}}}{{0.0208\,{\rm{nm}}}}\,\, - \,\,\frac{{1.24\,\,{\rm{keV}} \cdot {\rm{nm}}}}{{0.02566\,{\rm{nm}}}}\,\,\, = \,\,\,59.6\,{\rm{keV}}\,\, - \,\,48.3\,{\rm{keV}}\,\,\, = \,\,\,11.3\,{\rm{keV}}\end{array}}$

We see the maximum kinetic energy given to the electron is only about 2% of its rest energy.
The photon's original energy was only about 12% of the electron's rest energy, and with its max
kick, it boosts the electron energy 2% and loses 19% of its own energy. The above expression for P4
also allows us to calculate
the electron's recoil momentum (and therefore recoil angle, ϕ), which
is useful for creating the interactive diagram:

$\Large{\begin{array}{l}{{\bf{p}}_4}\,\,\, = \,\,\,\frac{h}{{{\lambda _1}}}\,{{{\bf{\hat p}}}_1}\,\, - \,\,\frac{h}{{{\lambda _3}}}\,{{{\bf{\hat p}}}_3}\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{{\bf{\hat p}}}_4}\,\,\, = \,\,\,\frac{{{\lambda _3}\,{{{\bf{\hat p}}}_1}\,\, - \,\,{\lambda _1}\,{{{\bf{\hat p}}}_3}}}{{\left| {{\lambda _3}\,{{{\bf{\hat p}}}_1}\,\, - \,\,{\lambda _1}\,{{{\bf{\hat p}}}_3}} \right|}}\\\\\\{{{\bf{\hat p}}}_4}\,\,\, = \,\,\,\frac{{\left\langle {{\lambda _3}\,\, - \,\,{\lambda _1}\,\cos \theta \,,\,\,\, - \,{\lambda _1}\,\sin \theta } \right\rangle }}{{\sqrt {{\lambda _1}^2\,\, + \,\,{\lambda _3}^2\,\, - \,\,2\,{\lambda _1}\,{\lambda _3}\,\cos \theta } }}\,\,\, = \,\,\,\left\langle {\cos \phi \,,\,\,\,\sin \phi } \right\rangle \end{array}}$

Just below is the
interactive diagram for the scattering of this particular photon (you can drag the

scattering angle slider and watch energies and wavelengths update). The diagram also allows comparison
of this
Compton scattering with the scattering from classical billiard balls in a 2D elastic collision
(
see below the diagram for solution to the classical case). We can see from the numbers above that the ratio of

total electron energy to photon energy before the collision is 8.6, while that ratio after is 10.8. We've added
a
time slider to watch the billiards collision and a slider to vary the ratio of target ball mass to incident
cue ball mass
. We'd expect some differences between photon scattering obeying our relativistic conservation
laws and slow speed billiard balls obeying classical conservation laws, but when the ratio of masses is set
to, for example, 9.8,
the scattering and recoil angles in the two cases track each other very closely.
You can turn on the billiards traces to see their scattering angles more clearly.

Compare to billiards and notice that to sink the electron in the side pocket at almost a right angle, the
photon has to just barely kiss the electron, giving the electron hardly any speed and altering the photon's
path and energy only very slightly. On the other hand, with a direct shot, the electron gets maximum
speed. Actually, this is more like a defective billiards set, where the cue ball weighs quite a bit less than
the other balls.

(Click here for a derivation of classical 2D elastic scattering formulas used in the interactive diagram above.)

Photon scattering versus absorption:

We are familiar with the idea of a photon getting absorbed by an electron in an atom. If the photon has
just the right energy, the electron can jump to a higher energy state. If the photon energy is more than the
binding energy of the electron, the electron will be freed from the atom (the photoelectric effect). It may seem
a bit odd, but an incoming photon can have "too much" energy, in which case it can only be scattered, as
investigated above.

To see such possibilities in an elementary lab setting, consider the setup in the diagram below. These are
carts on a "frictionless" track. The side magnets bind the large cart (electron). We can change the strength
of the electromagnets to change the amount of binding.

The small leftmost cart represents the incoming photon. If we give it a small push (low energy photon), it will get
captured by the electron (largest cart) and the coupled carts will vibrate a little (
excited state). If the push is large
enough, the coupled carts (photon absorbed by electron) will have enough energy to break loose from the side binding
magnets (
the photoelectric effect) and move on down the track. Finally, if we fire the small cart with high speed,
it will bounce off the large cart (
scattering). {Well, at least this sounded right in Rowley's imagination. He could be
wrong, but it sounds like a fun lab experiment to try.}

It is interesting to note that
a photon cannot be absorbed by a free electron, because this would cause the
energy-momentum conservation law to be violated:

$\Large{\begin{array}{l}{P_{1{\rm{(photon)}}}}\,\, + \,\,{P_{2{\rm{(electron}}\,\,{\rm{before)}}}}\,\,\, = \,\,\,{P_{{\rm{(electron}}\,\,{\rm{after)}}}}\,\\\\\\{m^2}{c^2}\,\,\, = \,\,\,P\widetilde P\,\,\, = \,\,\,{P_1}\widetilde {{P_1}}\,\, + \,\,{P_2}\widetilde {{P_2}}\,\, + \,\,{P_1}\widetilde {{P_2}}\,\, + \,\,\widetilde {{P_1}\widetilde {{P_2}}}\,\,\, = \,\,\,0\,\, + \,\,{m^2}{c^2}\,\, + \,\,2{\left\langle {{P_1}\widetilde {{P_2}}} \right\rangle _0}\\\\\\0\,\,\, = \,\,\,{\left\langle {{P_1}\widetilde {{P_2}}} \right\rangle _0}\,\,\, = \,\,\,{\left\langle {\frac{h}{{{\lambda _1}}}\,\left( {1\,\, + \,\,{{{\bf{\hat p}}}_1}} \right)\,{\gamma _2}m\left( {c\,\, - \,\,{{\bf{v}}_2}} \right)} \right\rangle _0}\\\\\\0\,\,\, = \,\,\,1\,\, - \,\,{{{\bf{\hat p}}}_1} \cdot \left( {\frac{{{{\bf{v}}_2}}}{c}} \right)\end{array}}$

We've run into a problem. If the last term were positive (as it always would be), it would mean the mass of
the electron had increased (in other words, it's
not an electron anymore). We're not aware of any such event
predicted theoretically or detected experimentally. And, that term could only be zero if the photon were
traveling initially in the same direction as the electron, which would also have to be traveling at the speed
of light. Therefore,
a photon can only scatter off a free electron. The electron can't simply absorb the
photon with increased energy and altered momentum.
Other outcomes of a photon --> free electron

collision are known to be possible if the photon energy is high enough, such as the creation of an extra
electron <--> positron pair, but that would leave more particles to help conserve total energy and momentum.

On the other hand, if the electron is bound to an atom, there are more particle momenta involved in both
sides of the starting equation here, so absorption can occur without violating the conservation laws. As
incident photon energy increases, loosely bound electrons appear relatively more free, so scattering becomes
the more likely event.