From Wikipedia, I'm copying two things:

It is the low-energy limit of Compton scattering: the particle's kinetic energy and photon frequency do not change as a result of the scattering.

In the low-energy limit, the electric field of the incident wave (photon) accelerates the charged particle, causing it, in turn, to emit radiation at the same frequency as the incident wave, and thus the wave is scattered.

These two seem to contradict each other. One says the kinetic energy of the free electron doesn't change, the other one says the free electron is accelerated.

Question 1: how is it possible that electron is accelerated, but its kinetic energy didn't change? Doesn't the electron scatter?

Question 2: If the photon hit the electron exactly (not passing by, but hit it in the spot), would there be a case where neither Compton nor Thomson scattering takes place? How is this possible? Then what will happen exactly to the photon and electron?

  • $\begingroup$ are you aware that photons and electron interactions need quantum mechanical models? $\endgroup$
    – anna v
    Commented May 3, 2023 at 13:20

3 Answers 3


You're mixing classical EM theory description with photons, that's why the contradiction.

In classical EM theory, there are no photons. The incoming wave field makes the electron oscillate. It does give electron energy in one quarter of period, then the electron gives energy back to EM field in the following quarter of period, then the field gives energy to the electron again, and so on. After many periods of oscillation, the electron has about the same energy (depending on the current phase of oscillation); it does not absorb energy systematically.

In the naive billiard photon model of the Thomson scattering, the incoming photon disappears and new photon of almost the same frequency and possibly different direction appears.

  • $\begingroup$ So in photon model, when photon hits an electron, electron does not move by even a tiny bit ? $\endgroup$
    – Matt
    Commented May 3, 2023 at 13:54
  • $\begingroup$ In the simplest (and very deficient) model, the collision is instantaneous, so it happens at single point of space. If the photon is scattered into different direction, the electron velocity changes, so after some time, the initially non-moving electron will move. $\endgroup$ Commented May 3, 2023 at 13:59
  • $\begingroup$ Thats what’s bugging me. Why then wikipedia at one point says that particle kinetic energy does not change ? If it moved, then it definately changed kinetic energy $\endgroup$
    – Matt
    Commented May 3, 2023 at 14:14
  • 1
    $\begingroup$ Wikipedia is unreliable on physics, prefer books and papers. The statement "the particle's kinetic energy does not change" is not really true. In scattering off an electron, there is almost always non-zero change of energy of the photon (except if the scattering produces photon in the same direction as the incoming photon has), and due to energy conservation, equal change of energy but with opposite sign happens to the electron. $\endgroup$ Commented May 3, 2023 at 14:47
  • 1
    $\begingroup$ Calling the scattering elastic, or energy conserving, is like saying ordinary motions do not manifest the Doppler effect on light. If we do not detect change in frequency of the light that is coming off a light source, we might tolerate statements such as "velocity of the emitting object does not change in time". While a more correct statement would be "velocity of the emitting object can be changing in time only by a small amount, because we are not detecting variations in frequency". $\endgroup$ Commented May 3, 2023 at 15:03

As I explained in the comments to the linked question. In the classical picture of Thomson scattering, the electric field of the light accelerates the electron but you nevertheless assume that this kinetic energy is negligible compared with both any photon energy and the rest-mass energy of the electron.

Thomson scattering is a classical model, so there are no photons to directly hit the electron. Compton scattering is the quantum-mechanical treatment of the process (which also works for low energy photons) and in quantum mechanics you cannot say, "let the photon exactly hit the (point-like) electron". Those are Newtonian mechanics ideas that aren't valid here. The scattering has a cross-section (of $\leq 6.6 \times 10^{29}$ m$^2$), which is the effective area presented by an electron to the radiation.

Using Newton's second law for an incoming wave polarised along the x-axis. $$ m\ddot{x} = -eE_0 \cos (\omega t)\ , $$ where we ignore the Lorentz force due the magnetic field, which is valid so long as the electron does not move at anywhere near the speed of light.

You can now integrate to get the velocity and speed. $$ \dot{x} = -\frac{eE_0}{m \omega} \sin(\omega t)\ ,$$ where a boundary condition has been set to make sure the initial velocity was zero when $t=0$.

The kinetic energy (again assuming non-relativistic motion) $$ K = \frac{e^2 E_0^2}{2m \omega^2} \sin^2 (\omega t)\ ,$$ with a time-average of $$<K> = \frac{e^2 E_0^2}{4m \omega^2} = 4.4\times 10^{-23} E_0^2 \left(\frac{\omega}{10^{15}\ {\rm rad/s}}\right)^{-2}\ {\rm keV}\ . $$

Thus for visible light, with a photon energy of $\sim$ eV, then the kinetic energy of the electron will be much less than the photon energy and much, much less than the electron rest-mass energy (511 keV) unless the electric field strength is something massive like $E_0\geq 10^{13}$ V/m (equivalent to a power per unit area of about $10^{21}$ W/m$^2$ - about $10^{18}$ times more powerful than sunlight at the Earth). In this broad range of applicability, the classical Thomson scattering approach, which assumes the electron picks up negligible energy is clearly valid.

The small kinetic energy that is picked up by the electron in this regime is due to work done by the electric field. However the power transfer is transient and the kinetic energy of the electron gets no bigger because the electric field and electric current are $\pi/2$ out of phase and thus the work done has a time-average of zero in the steady-state.

The Thomson scattering assumption can break down in TWO ways. First, there is the commonly understood case that $\hbar \omega \sim m_e c$ - in which case the photons will transfer significant momentum to the electrons and a quantum mechanical treatment leading to Compton scattering is required.

The second case, but less well-known, is that the electric field amplitude $E_0$ becomes large enough that the kinetic energy of the accelerated electrons becomes comparable with their rest-mass energy. The magnetic field component of the Lorentz force then becomes non-negligible and the electrons suffer a recoil even though the photon energy is still much lower than the electron rest-mass energy. This is known as high-intensity Compton scattering (e.g., Kibble 1965; Moore 1995).


You're thinking mathematically, where you assume perfection. This is a physical model. All physical models are approximate. In this approximation, the electron accelerates, but the acceleration is too small to transfer a significant amount of energy to it. So, assume the energy transfer is zero and calculate.

Note that classical models of point charges have mathematical problems. Still, if you ignore the problems, you can often coax excellent, accurate predictions from the models.


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