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Compton scattering occurs because of the particle-like nature of light, but does it occur between photons and charged particles only? Is it impossible for Compton scattering to occur between a photon and, e.g., a neutrino? A neutron? A hydrogen atom?

I think this questions splits into two questions:

  1. Can Compton scattering occur between a photon and a fundamental neutral particle?

  2. Is it correct to say that Compton scattering occurs between a photon and a composite neutral particle, it is only because of the interaction between the photon and a constituent sub-particle that is charged?

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    $\begingroup$ Remember that neutrons are composed of charged quarks, and photons do interact with quarks. So a photon can interact with neutrons. $\endgroup$
    – Triatticus
    Commented Apr 10, 2022 at 21:08
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    $\begingroup$ Neutrons have a magnetic moment (spin 1/2), so they can definitely couple to light that way too $\endgroup$
    – KF Gauss
    Commented Apr 10, 2022 at 22:09
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    $\begingroup$ see particles.uni-freiburg.de/dateien/vorlesungsdateien/… $\endgroup$
    – anna v
    Commented Apr 11, 2022 at 4:10

2 Answers 2

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The Compton scattering is an electromagnetic process and it will as we currently know happens between a (electromagnetic) charged particle (like a neutron does have a magnetic moment) and and photon. Given by the following Feynman diagram

Compton scattering

It is interpreted as an incoming electron absorbs a photon and travels some distance and releases a photon,

If the magnetic moment of neutrino is confirmed then in principle they should interact with electromagnetic field.

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$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mc}[1]{\mathcal {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\vlr}[1]{\left\vert#1\right\vert} \newcommand{\Vlr}[1]{\left\Vert#1\right\Vert} \newcommand{\lara}[1]{\left\langle#1\right\rangle} \newcommand{\lav}[1]{\left\langle#1\right|} \newcommand{\vra}[1]{\left|#1\right\rangle} \newcommand{\lavra}[2]{\left\langle#1\right|\left#2\right\rangle} \newcommand{\lavvra}[3]{\left\langle#1\right|#2\left|#3\right\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\Vp}[1]{\vphantom{#1}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\ox}{\bl\otimes} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad} \newcommand{\qqLraqq}{\qquad\boldsymbol{\e\!\e\!\e\!\e\!\Longrightarrow}\qquad} \newcommand{\tl}[1]{\tag{#1}\label{#1}} \newcommand{\hebl}{$\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$}$

Compton scattering $\:\gamma\p A\bl\longrightarrow\gamma\p A\:$ occurs between a photon and a massive particle at least, it's not necessary for the particle to be electrically charge.

Note that the Compton scattering equation for the wavelength of a photon that scatters off a particle, initially at rest, of mass $\:m\:$ as shown in Figure-01 is \begin{equation} \boxed{\:\: \lambda'\e \lambda\p\lambda_c\plr{1\m\cos\theta}\,,\qquad \lambda_c\e\dfrac{h}{mc}\e\texttt{Compton wavelength}\:\:\vp} \tl{01} \end{equation}

enter image description here

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    $\begingroup$ Interactions of photons have to be a quantum mechanical interaction modeled with Feynman diagrams Here is a text with the necessary Feynman diagrams researchgate.net/publication/… . Compton scattering has to happen either with an electron of a neutral atom or with a quark if one thinks of neutrons. $\endgroup$
    – anna v
    Commented Apr 11, 2022 at 4:22

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