What is the force direction of the radiation of the electromagnetic field

Question

We know the radiation waves are transversal wave. That means the electric field $\boldsymbol{E}$ is perpendicular to the radiation direction $\boldsymbol{r=x-x'}$. Here $\boldsymbol{x}$ is the position of the field point. $\boldsymbol{x}'$ is the position of the sources, $\boldsymbol{E}\bot\boldsymbol{r}$. Assume we put an charge of an electron at the place $\boldsymbol{x}$ What is the force on this electron? The force should at the electric field direction or at the direction of the radition ($\boldsymbol{r}$ direction)?

Consider the Poynting vector $\boldsymbol{S}=\boldsymbol{E}\times\boldsymbol{H}$ is at the direction of radiation. $\boldsymbol{S}\Vert\boldsymbol{r}$. The momentum of the photon should at the direction of Poynting vector $\boldsymbol{S}$. If the momentum of photon at the direction of $\boldsymbol{S}$, when the electron receive the photon should get a force alos at the radiation direction simlar to $\boldsymbol{S}$. However according to the electric field $\boldsymbol{E}$, the electron should get a force at the direction of $\boldsymbol{E}$ which is at the direction perpendicular to the radiation direction. I am confused.

Answer 1

When an electron interacts with a classical electromagnetic wave it does not absorb the electromagnetic wave, it scatters it. Since it does not absorb the light, and because the scattering is symmetric around a line parallel to the original electric field of the EM wave, then the change of momentum that you have in mind associated with the Poynting vector, does not occur.

The electron feels the Lorentz force due to the electromagnetic fields of the wave. Since we can usually assume that $E = cB$ (SI units), then one can often ignore the magnetic part of the Lorentz force as small (if the electron moves non-relatvistically). As a result, the electron is accelerated in the direction of the electric field and then radiates as a classical oscillating dipole. This radiation is symmetrically emitted about the axis of oscillation and so there is no net change of momentum in the original direction of the wave.

If one does consider the magnetic field then there is also a small, oscillatory (at twice the wave frequency) force component along the direction of the original wave, caused by the $-e\vec{v} \times \vec{B}$ term. But of course magnetic fields don't do any work on a charged particle, so there is no net acceleration in that direction. See for example Kruger & Bovyn (1976) and https://physics.stackexchange.com/a/313743/43351 .

Things change when the waves have more energy (when the photons have energies that are non-negligible) with respect to the rest mass energy of the electron. Then you have Compton scattering which does result in momentum transfer to the electron, but this is not treated classically.

Answer 2

What you are saying is maybe true when light encounters electrically neutral objects.

The force imparted by light is not necessarily in the direction of its Poynting vector when you look at charged particles. As an example, consider an e.m. field in vacuum described by $$\mathbf{E} = E_0 \, \cos{\omega t}\, \hat{y} \\ \mathbf{B} = B_0 \, \cos{\omega t} \, \hat{x}.$$ The Poynting vector $$\mathbf{S} = \frac{1}{\mu_0} (\mathbf{E} \times \mathbf{B})$$ is clearly in the $-\hat{z}$ direction, as you can verify.

Now, let an electron be moving around with a velocity $$\mathbf{v}= v_x \, \hat{x} + v_y \, \hat{y} + v_z \, \hat{z} $$ in space. The electron, being a charged particle, experiences the Lorentz-force $$\mathbf{F} = -e \left(\mathbf{E} + \mathbf{v} \times \mathbf{B}\right) = -e \, [(E_0 + B_0 v_z) \, \hat{y} - B_0 v_y \, \hat{z}],$$

which, unless you are considering the special case $v_z = -E_0/B_0$, lies in the $y-z\,$ plane and is not strictly restricted to the $\hat{z}$ direction in general.