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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.

ShRenZhao
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    You aren’t considering the magnetic force. – G. Smith Dec 19 '20 at 03:27
  • The real answer: https://www.feynmanlectures.caltech.edu/I_34.html#Ch34-S9 – R. Emery Dec 19 '20 at 06:45
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    @R.Emery Feynman is always right, except when he's trying to stretch some analogy or simplification. That analysis you refer to just doesn't work for electrons. The electric field and velocity are out of phase by $\pi/2$ so the time average $=0$. Neither the electric field or magnetic field do any work on the charge when considered classically. The conclusion that the force is the energy absorbed divided by $c$ is ok, but free point-like charges can't absorb light. – ProfRob Dec 19 '20 at 10:39
  • Which is why he says: Therefore the force, the “pushing momentum,” that is delivered per second by the light, is equal to 1/c times the energy absorbed from the light per second! That is a general rule, since we did not say how strong the oscillator was, or whether some of the charges cancel out. In any circumstance where light is being absorbed, there is a pressure. The momentum that the light delivers is always equal to the energy that is absorbed, divided by c: – R. Emery Dec 19 '20 at 10:50
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    @R.Emery and I say again, a free point-like charged particle, which is exactly the setup considered by Feynman, cannot absorb light. If the latter point is what you mean by a "real answer" to this question, then I agree, but it hardly assists the OP. – ProfRob Dec 19 '20 at 11:11
  • He never called it a free charge. – R. Emery Dec 19 '20 at 11:13
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    R.Emery so what other forces on the charge are considered if it is not a free charge? – ProfRob Dec 19 '20 at 11:17
  • Thought experiments are not limited by whats practical or even by whats possible. BTW I found this: https://physics.stackexchange.com/questions/132515/does-a-reflection-still-transfer-momentum-to-an-mirror It looks interesting. – R. Emery Dec 20 '20 at 11:19
  • @R.Emery In general, light scattered from a particle imparts a force too (energy does not need to be absorbed). – Gilbert Dec 20 '20 at 16:52
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    What has coherent reflection from a mirror got to do with scattering from a free electron? Of course there is an onward force in the former case because back scattering is not axially symmetric. – ProfRob Dec 21 '20 at 09:38
  • @R.Emery like ProfRob said, if we are talking about a free charge, half of the time the electric field points to the opposite direction of the velocity, so magnetic force is also oscillating. Situation changes if the motion of the charge is sufficiently damped, then most of the time E and v point to the same direction, and then you get to see the charge getting pushed in the direction of the radiation. – Michael Mitsopoulos Dec 13 '22 at 08:07

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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.

ProfRob
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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.

Yejus
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