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When the photon is incident on free electron, we say that compton happens. Though, we require that photon is x-ray or gamma ray. I wonder why at least x-ray is required.

I have my own observation and correct me if I'm wrong. I just want to make sure. Due to the formula of $$ λ_f - λ_i = \frac{h*(1-\cos a)}{mc}$$ We can see that if compton happens, the maximum shift the original photon might shift is by $2ℎ/𝑚𝑐 = 0.005\mathrm{nm}$. Due to this, if visible light is incident(let's say 500nm) on free electron, the new photon would have 500.005nm and we can't see this difference in our labs, whereas if x-ray is incident(0.05nm), it would result in 0.05 + 0.005 = 0.055 and it's easier for us to see the actual difference between original and reflected/emitted wavelength, hence we assume that if we see the wavelength difference, it must be compton.

If everything is correct above, could we assume that even visible light photon could cause the compton, it's just we don't care if it does as the difference in wavelength is so small, we neglect it ?

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2 Answers 2

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The "low energy limit" of Compton scattering is called Thomson scattering, which occurs when the energy of the photon $E_\gamma \ll m_ec^2$. Since $m_e c^2 = 511$ keV, then this condition is true for photons with wavelengths longer than X-rays.

In Thomson scattering, negligible momentum is imparted to the electron, the problem can be treated with classical electromagnetism, and scattering occurs with a fixed cross-section of $6.6\times 10^{-29}$ m$^2$ leaving the energy and wavelength of the scattered photons the same. The distribution of scattered light follows that of a classical oscillating electric dipole if the incoming light is linearly polarised, with no scattered light in the direction of the original polarisation.

Here's my favourite application - measuring the density and structure of the solar corona using white light from the visible photosphere scattered towards us from free electrons in the hot coronal plasma (image from the SOHO satellite, with the visible Sun shown inside the coronagraphic occulting disc for comparison).

Scattering in the solar corona.

The Thomson scattering cross-section represents the maximum of the scattering cross-section for free electrons as a function of energy. Inelastic Comption scattering has a cross-section that decreases with increasing energy. See the plot below from Cerutti et al. (2007). $x$ on this plot is the ratio of the photon energy to $m_e c^2$.

Scattering cross-section

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  • $\begingroup$ Hi Rob. Does the electron accelerate or photon energy is not enough for that ? I believe it should accelerate a little bit. You said negligible momentum, but it is still more than zero. $\endgroup$
    – Matt
    May 3 at 9:25
  • $\begingroup$ But if electron accelerates a little bit, then it means it received energy from photon, which means photon must have lost small bit of energy and it should have an increased wavelength as reflected one. How is this not true ? $\endgroup$
    – Matt
    May 3 at 9:29
  • $\begingroup$ @Matt, negligible compared with $mc$. Thomson scattering assumes no energy is transferred. The time-averaged work done by the electric field on the electron is zero because the electric field and current density are out of phase. The electron "borrows" some energy and then gives it back. Quantum electrodynamics is a more accurate theory and suggests a tiny amount of energy is transferred to the electron. Thomson scattering is the low energy limit of Compton scattering, when $E_\gamma \rightarrow 0$. $\endgroup$
    – ProfRob
    May 3 at 9:59
  • $\begingroup$ It is really hard to grasp though. Then we should say that electron stays at the same as it was before photon hit it. Otherwise if it moved by a hugely tiny bit, then it must have gotten energy from photon $\endgroup$
    – Matt
    May 3 at 10:09
  • $\begingroup$ @Matt and then when the light leaves the electron it has no kinetic energy. It's a model (like all Physics). $\endgroup$
    – ProfRob
    May 3 at 10:37
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If everything is correct above, could we assume that even visible light photon could cause the compton, it's just we don't care if it does as the difference in wavelength is so small, we neglect it?

You pretty much have the right idea. Compton scattering can happen with any energy photon, it's just that when its a low energy photon there will be essentially no measurable wavelength shift so if the wavelength shift is the thing you're interested in then low energy photons won't do much for you. That being said, even with low wavelength photons you still have the effect that they will be scattered in different directions even if the wavelength doesn't change, which is an important effect in its own right. People often call this Thompson scattering and its studied in a wide variety of contexts (especially plasma physics as a diagnostic tool), but it's really nothing more than the low photon energy limit of Compton scattering.

TL;DR-- Compton scattering is still present at low photon energies, it just becomes nearly elastic and so people call the low energy limit Thompson scattering to distinguish it from the more clearly inelastic behavior at higher energies. But they're really the same phenomenon.

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  • $\begingroup$ Compton scattering in QFT calculations can be found here sites.ualberta.ca/~gingrich/courses/phys512/node102.html and agrees with your argument. $\endgroup$
    – anna v
    May 3 at 3:34
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    $\begingroup$ Amazing answer. Can we say that when photon hits a free electron, there will ALWAYS be thompson scattering or compton scattering or is there a chance none of it happens and if so why ? $\endgroup$
    – Matt
    May 3 at 3:38
  • $\begingroup$ @Matt most light will not scatter from an electron at all and will pass through. You can look up the Compton cross section for estimates of the scattering rate. $\endgroup$
    – KF Gauss
    May 3 at 3:54
  • $\begingroup$ @Matt KF Gauss is right that most light does not interact with the electron-- to answer your question of why, it is because thompson/compton scattering has a finite cross section, so each photon passing by an electron does have a chance to not interact. Not all interactions between particles are like this-- in particular, Rutherford scattering has an infinite cross section. The thing that helps you with photons is that they're electrically neutral so you don't have any "long range" interactions between the photons and electrons. $\endgroup$
    – el duderino
    May 3 at 3:59
  • $\begingroup$ So even with free electron and photon interaction, sometimes neither thompson nor compton will happen right ? Note that I emphasided free electron. $\endgroup$
    – Matt
    May 3 at 8:55

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