Editing Compton scattering (section)

=== Derivation of the scattering formula ===
[[File:ComptonEnergy-en.svg|thumb|Fig. 3: Energies of a photon at 500&nbsp;keV and an electron after Compton scattering.]]
A photon {{mvar|γ}} with wavelength {{mvar|λ}} collides with an electron {{mvar|e}} in an atom, which is treated as being at rest. The collision causes the electron to [[recoil]], and a new photon {{mvar|γ}}′ with wavelength {{mvar|λ}}′ emerges at angle {{mvar|θ}} from the photon's incoming path. Let {{mvar|e}}′ denote the electron after the collision. Compton allowed for the possibility that the interaction would sometimes accelerate the electron to speeds sufficiently close to the velocity of light as to require the application of Einstein's [[special relativity]] theory to properly describe its energy and momentum.

At the conclusion of Compton's 1923 paper, he reported results of experiments confirming the predictions of his scattering formula, thus supporting the assumption that photons carry momentum as well as quantized energy.  At the start of his derivation, he had postulated an expression for the momentum of a photon from equating Einstein's already established mass-energy relationship of <math>E=mc^2</math> to the quantized photon energies of <math>h f</math>, which Einstein had separately postulated. If <math>mc^2 = hf</math>, the equivalent photon mass must be <math>hf/c^2</math>. The photon's momentum is then simply this effective mass times the photon's frame-invariant velocity {{mvar|c}}. For a photon, its momentum <math>p=hf/c</math>, and thus {{math|''hf''}}  can be substituted for {{math|''pc''}} for all photon momentum terms which arise in course of the derivation below. The derivation which appears in Compton's paper is more terse, but follows the same logic in the same sequence as the following derivation.

The [[conservation of energy]] <math>E</math> merely equates the sum of energies before and after scattering.
: <math>E_\gamma + E_e = E_{\gamma'} + E_{e'}.</math>
Compton postulated that photons carry momentum;<ref name="taylor_136-9" /> thus from the [[conservation of momentum]], the momenta of the particles should be similarly related by
: <math>\mathbf{p}_\gamma = \mathbf{p}_{\gamma'} + \mathbf{p}_{e'},</math>
in which (<math>{p_e}</math>) is omitted on the assumption it is effectively zero.

The photon energies are related to the frequencies by
: <math>E_{\gamma} = hf</math>
: <math>E_{\gamma'} = hf'</math>
where ''h'' is the [[Planck constant]].

Before the scattering event, the electron is treated as sufficiently close to being at rest that its total energy consists entirely of the mass-energy equivalence of its (rest) mass <math> m_\text{e} </math>,
: <math>E_e = m_\text{e} c^2.</math>
After scattering, the possibility that the electron might be accelerated to a significant fraction of the speed of light, requires that its total energy be represented using the relativistic [[energy–momentum relation]]
: <math>E_{e'} = \sqrt{(p_{e'}c)^2 + (m_\text{e} c^2)^2} ~.</math>

Substituting these quantities into the expression for the conservation of energy gives
: <math>hf + m_\text{e} c^2 = hf' + \sqrt{(p_{e'}c)^2 + (m_\text{e} c^2)^2}.</math>
This expression can be used to find the magnitude of the momentum of the scattered electron,
{{NumBlk||<math display="block">p_{e'}^{\, 2}c^2 = (hf - hf' + m_{e}c^2)^2-m_{e}^2c^4. </math>|{{EquationRef|1}}}}
Note that this magnitude of the momentum gained by the electron (formerly zero) exceeds the energy/''c'' lost by the photon,
: <math>\frac{1}{c}\sqrt{(hf - hf' + m_{e}c^2)^2-m_{e}^2c^4} > \frac{hf - hf'}{c}~. </math>

Equation (1) relates the various energies associated with the collision. The electron's momentum change involves a relativistic change in the energy of the electron, so it is not simply related to the change in energy   occurring in classical physics.  The change of the magnitude of the momentum of the photon is not just related to the change of its energy; it also involves a change in direction.

Solving the conservation of momentum expression for the scattered electron's momentum gives
: <math>\mathbf{p}_{e'} = \mathbf{p}_\gamma - \mathbf{p}_{\gamma'}.</math>

Making use of the [[scalar product]] yields the square of its magnitude,
: <math>\begin{align}
p_{e'}^{\, 2} &= \mathbf{p}_{e'}\cdot\mathbf{p}_{e'} = (\mathbf{p}_\gamma - \mathbf{p}_{\gamma'}) \cdot (\mathbf{p}_\gamma - \mathbf{p}_{\gamma'}) \\
 &= p_{\gamma}^{\, 2} + p_{\gamma'}^{\, 2} - 2 p_{\gamma}\, p_{\gamma'} \cos\theta. \end{align}</math>
In anticipation of <math>p_{\gamma}c</math> being replaced  with <math>h f</math>, multiply both sides by <math>c^2</math>,
: <math>p_{e'}^{\, 2}c^2 = p_{\gamma}^{\, 2}c^2 + p_{\gamma'}^{\, 2}c^2 - 2c^2 p_{\gamma}\, p_{\gamma'} \cos\theta.</math>

After replacing the photon momentum terms with <math>h f/c</math>, we get a second expression for the magnitude of the momentum of the scattered electron,
{{NumBlk||<math display="block">p_{e'}^{\, 2}c^2 = (h f)^2 + (h f')^2 - 2(hf)(h f')\cos{\theta}~. </math>|{{EquationRef|2}}}}

Equating the alternate expressions for this momentum gives
: <math> (hf - hf' + m_\text{e} c^2)^2 - m_\text{e}^{\, 2}c^4 = \left(h f\right)^2 + \left(h f'\right)^2 - 2h^2 ff'\cos{\theta},</math>
which, after evaluating the square and canceling and rearranging terms, further  yields
: <math> 2 h f m_\text{e} c^2-2 h f' m_\text{e} c^2   = 2 h^2 f f' \left( 1 - \cos \theta \right) .</math>
Dividing both sides by {{nowrap|<math>2 h f f' m_\text{e} c</math>}} yields
: <math> \frac{c}{f'} - \frac{c}{f} = \frac{h}{m_\text{e} c}\left(1-\cos \theta \right) .</math>

Finally, since  {{math|''fλ''}} =  {{math|''f''{{&prime;}}''λ''&prime;}} = {{mvar|c}},
{{NumBlk||<math display="block">\lambda'-\lambda = \frac{h}{m_\text{e} c}(1-\cos{\theta})~ .</math>|{{EquationRef|3}}}}

It can further be seen that the angle {{mvar|φ}} of the outgoing electron with the direction of the incoming photon is specified by 
{{NumBlk||<math display="block"> \cot \varphi = \left ( 1+\frac{hf}{m_\text{e} c^2} \right ) \tan (\theta/2)~. </math>|{{EquationRef|4}}}}