Derive a wavelength shift from vector momentum and total energy
A photon scatters from a free electron initially at rest. Let incoming photon energy be E, outgoing energy Eprime and deflection angle theta. Photon momentum magnitude is E/c. Resolve recoil momentum: px c=E−Eprime cos(theta), py c=−Eprime sin(theta). Vector subtraction is essential; subtracting momentum magnitudes would miss transverse recoil.
Learning outcome: Derive the shift before checking three angles, including zero and 180 degrees. Reconstruct both recoil components and compare relativistic kinetic energy with the photon loss. State what independent target-state evidence is needed before applying the free-stationary-electron model.
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Review definite integrals, complex amplitudes, probability moments, elementary differential equations and energy-momentum units. State boundary and normalization conditions, and distinguish calibration from unused validation.
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For E=100 keV and theta=90 degrees, Eprime=51100/611≈83.633387889 keV. Recoil components times c are approximately (16.366612111,−83.633387889) keV. Relativistic recoil kinetic energy equals the photon loss 16.366612111 keV. The wavelength increase is hc/(511 keV), approximately 2.42631 pm.
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Why is matching photon momentum magnitudes alone insufficient?
Experiment
Derive the shift before checking three angles, including zero and 180 degrees. Reconstruct both recoil components and compare relativistic kinetic energy with the photon loss. State what independent target-state evidence is needed before applying the free-stationary-electron model.
Close relativistic photon and electron accounts
Worked teaching model · synthetic values · Free electron initially at rest, single photon collision, rounded rest energy511 keV and hc1.239841984 keV nm. Bound-electron and material effects omitted. Recoil momentum has two signed components. No actual radiation source or detector operation.
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Energy and angle follow one conservation law
Graph axes rescale independently; read units and values, not screen angles. 1: free stationary-electron Compton model (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Energy and angle follow one conservation law
free stationary-electron Compton model
| photon angle (degrees) | scattered photon energy (keV) |
|---|---|
| 0 | 100 |
| 3 | 99.97319 |
| 6 | 99.89291 |
| 9 | 99.75965 |
| 12 | 99.57418 |
| 15 | 99.3376 |
| 18 | 99.05129 |
| 21 | 98.71688 |
| 24 | 98.33628 |
| 27 | 97.9116 |
| 30 | 97.44517 |
| 33 | 96.93949 |
| 36 | 96.39722 |
| 39 | 95.82111 |
| 42 | 95.21405 |
| 45 | 94.57896 |
| 48 | 93.91881 |
| 51 | 93.23661 |
| 54 | 92.53534 |
| 57 | 91.81794 |
| 60 | 91.08734 |
| 63 | 90.34638 |
| 66 | 89.59783 |
| 69 | 88.84435 |
| 72 | 88.08852 |
| 75 | 87.3328 |
| 78 | 86.57952 |
| 81 | 85.83092 |
| 84 | 85.08907 |
| 87 | 84.35595 |
| 90 | 83.63339 |
| 93 | 82.9231 |
| 96 | 82.22667 |
| 99 | 81.54557 |
| 102 | 80.88116 |
| 105 | 80.23466 |
| 108 | 79.6072 |
| 111 | 78.99983 |
| 114 | 78.41348 |
| 117 | 77.84899 |
| 120 | 77.30711 |
| 123 | 76.78854 |
| 126 | 76.29388 |
| 129 | 75.82367 |
| 132 | 75.3784 |
| 135 | 74.95849 |
| 138 | 74.56431 |
| 141 | 74.19619 |
| 144 | 73.85442 |
| 147 | 73.53925 |
| 150 | 73.25088 |
| 153 | 72.9895 |
| 156 | 72.75528 |
| 159 | 72.54833 |
| 162 | 72.36877 |
| 165 | 72.2167 |
| 168 | 72.09218 |
| 171 | 71.99527 |
| 174 | 71.92602 |
| 177 | 71.88446 |
| 180 | 71.8706 |
At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.
| Quantity | Value | Unit |
|---|---|---|
| Scattered photon energy | 83.63339 | keV |
| Electron recoil x momentum times c | 100 | keV |
| Electron recoil y momentum times c | -83.63339 | keV |
| Relativistic electron kinetic energy | 16.36661 | keV |
| Photon-energy loss | 16.36661 | keV |
| Energy closure residual | 3.552714e-15 | keV |
| Wavelength shift | 2.426305 | pm |
| Classical elastic no-shift photon prediction | 100 | keV |
Photon momentum is a vector; recoil energy must also obey the relativistic mass shell.
Energy gives E+mc²=Eprime+sqrt(m²c⁴+p²c²). Momentum gives p²c²=E²+Eprime²−2EEprime cos(theta). Eliminate p to obtain mc²(E−Eprime)=EEprime(1−cos(theta)), hence lambdaPrime−lambda=h(1−cos(theta))/(mc). Classical elastic scattering with no photon-energy change is a different comparison, not the recoil solution. The forward-angle boundary has exactly zero recoil in this ideal model.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
Photon momentum is a vector; recoil energy must also obey the relativistic mass shell.
Energy gives E+mc²=Eprime+sqrt(m²c⁴+p²c²). Momentum gives p²c²=E²+Eprime²−2EEprime cos(theta). Eliminate p to obtain mc²(E−Eprime)=EEprime(1−cos(theta)), hence lambdaPrime−lambda=h(1−cos(theta))/(mc). Classical elastic scattering with no photon-energy change is a different comparison, not the recoil solution. The forward-angle boundary has exactly zero recoil in this ideal model.
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