Editing Compton scattering (section)

== Description of the phenomenon ==
[[Image:Compton-scattering.svg|thumb|200px|right|Fig. 2: A photon of wavelength <math>\lambda</math> comes in from the left, collides with a target at rest, and a new photon of wavelength <math>\lambda'</math> emerges at an angle <math>\theta</math>. The target recoils, carrying away an angle-dependent amount of the incident energy.]]
By the early 20th century, research into the interaction of [[X-ray]]s with matter was well under way. It was observed that when X-rays of a known wavelength interact with atoms, the X-rays are scattered through an angle <math>\theta</math> and emerge at a different wavelength related to <math>\theta</math>. Although [[classical electromagnetism]] predicted that the wavelength of scattered rays should be equal to the initial wavelength,<ref name="taylor_136-9">{{cite book|last1=Taylor|first1=J.R.| last2=Zafiratos|first2=C.D.|last3=Dubson|first3=M.A.|year=2004|title=Modern Physics for Scientists and Engineers| edition=2nd| publisher=[[Prentice Hall]]|isbn=0-13-805715-X|pages=136–9}}</ref> multiple experiments had found that the wavelength of the scattered rays was longer (corresponding to lower energy) than the initial wavelength.<ref name="taylor_136-9" />

In 1923, Compton published a paper in the ''[[Physical Review]]'' that explained the X-ray shift by attributing particle-like momentum to light quanta ([[Albert Einstein]] had proposed light quanta in 1905 in explaining the photo-electric effect, but Compton did not build on Einstein's work). The energy of light quanta depends only on the frequency of the light. In his paper, Compton derived the mathematical relationship between the shift in wavelength and the scattering angle of the X-rays by assuming that each scattered X-ray photon interacted with only one electron.  His paper concludes by reporting on experiments which verified his derived relation:
<math display="block">\lambda' - \lambda = \frac{h}{m_\text{e} c}(1 - \cos{\theta}),</math>
where
* <math>\lambda</math> is the initial wavelength,
* <math>\lambda'</math> is the wavelength after scattering,
* <math>h</math> is the [[Planck constant]],
* <math>m_\text{e}</math> is the [[electron rest mass]],
* <math>c</math> is the [[speed of light]], and
* <math>\theta</math> is the scattering angle.

The quantity {{sfrac|''h''|''m''<sub>e</sub>''c''}} is known as the [[Compton wavelength]] of the electron; it is equal to {{val|2.43|e=-12|u=m}}. The wavelength shift {{nowrap|''λ′'' − ''λ''}} is at least zero (for {{nowrap|1=''θ'' = 0°}}) and at most twice the Compton wavelength of the electron (for {{nowrap|1=''θ'' = 180°}}).

Compton found that some X-rays experienced no wavelength shift despite being scattered through large angles; in each of these cases the photon failed to eject an electron.<ref name="taylor_136-9" /> Thus the magnitude of the shift is related not to the Compton wavelength of the electron, but to the Compton wavelength of the entire atom, which can be upwards of 10000 times smaller.  This is known as "coherent" scattering off the entire atom since the atom remains intact, gaining no internal excitation.

In Compton's original experiments the wavelength shift given above was the directly measurable observable.  In modern experiments it is conventional to measure the energies, not the wavelengths, of the scattered photons.  For a given incident energy <math>E_\gamma = hc/\lambda</math>, the outgoing final-state photon energy, <math>E_{\gamma^\prime}</math>, is given by
<math display="block">E_{\gamma^\prime} = \frac{E_\gamma}{1 + (E_\gamma/m_\text{e} c^2)(1-\cos\theta)}.</math>

=== 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}}}}