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.

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 spped 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).

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).

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.

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 spped 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 electic 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).

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.

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 spped 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).