Talk:Compton scattering

Page contents not supported in other languages.
From Wikipedia, the free encyclopedia

Error in Derivation[edit]

Article claims The derivation which appears in Compton's paper is more terse, but follows the same logic in the same sequence as the following derivation

Compton's paper - PhysRev.21.483 - has relativistic momentum used in his derivation - see Compton's equation (1) from his paper which incidentally has a typo in the cosine rule formula - the cosine term should be negative.

Thus the vector momentum equation used in this article should have a gamma multiplying the electron momentum, which in turn means that equation (2) should have gamma-squared multiplying the left hand side.

Basically, you cannot have relativistic energy without relativistic momentum, so although the math of the article gives the correct formula for Compton Effect is does not represent the physics.

This would explain why Youtube videos are also incorrect, they seems to be sourcing the derivation from this Wiki JohnWki (talk) 13:17, 2 March 2022 (UTC)[reply]

Campton effect[edit]

English 117.248.148.230 (talk) 14:35, 9 July 2022 (UTC)[reply]

Compton Scattering is an inelasatic scattering[edit]

In the main text, there is a mistake in the introduction part. It states:

"Compton scattering is an example of elastic scattering of light by a free charged particle, where the wavelength of the scattered light is different from that of the incident radiation."


Of course, even that description describes that it is inelastic scattering


References:

Pratt, R. H., et al. "Compton scattering revisited." Radiation Physics and Chemistry 79.2 (2010): 124-131. Lupiloui (talk) 11:00, 4 January 2023 (UTC)[reply]

No sir. There is no energy transfer except for kinetic energy - just like, e.g., elastic electron-proton scattering. Kinetic energy ist transferred from the e to the p (in lab frame) but as long as the p does not go into an excited state (delta resonance = incease of mass), it is elastic. In Compton scattering neither the mass of the e changes (there is no such thing as an excited e) neither the mass of the photon (zero). — Wassermaus (talk) 07:31, 1 June 2023 (UTC)[reply]

Equation (4)[edit]

No derivation of equation (4) for the scattering angle of the electron is given. I have reason to doubt that this equation is correct. Specifically, see section 4.2 of tutorial on Compton Scattering written by Paul D'Alessandris to be found at https://phys.libretexts.org/Bookshelves/Modern_Physics/Book:_Spiral_Modern_Physics_(D'Alessandris)/4:_The_Photon/4.2:_Compton_Scattering TychoTor (talk) 23:45, 22 March 2023 (UTC)[reply]

Debye is under-represented here[edit]

Debye independently discovered this effect and actually published his theory in "Zerstreuung von Röntgenstrahlen und Quantentheorie" in Physikalische Zeitschrift in April 1923, before Compton's December 1923 paper. Debye apparently later said that Compton deserved the name of the effect because he had done more work but at the moment Debye also has a measly entry in the See also section. Any objections to giving Debye a bigger role in the article? Jason Quinn (talk) 11:24, 27 December 2023 (UTC)[reply]