Motion of charged particle in em wave

Question

What is the direction of oscillation of a charged particle when an electromagnetic wave hits it? I think it would in a circle whose plane is along the direction of em wave and perpendicular to magnetic field of wave. Am I right?

Answer 1

You can divide the problem up into two regimes: (Non-relativistic) where the charged particle velocity is $\ll c$ and the solution is simple, shown below. (Relativistic} where the non-relativistic assumption cannot be made and the solution is quite difficult.

Non-relativistic. We say that $$ m \frac{d^2 \vec{r}}{dt^2} = q(\vec{E} + \vec{v} \times \vec{B}).$$ But for a plane wave in vacuum we know that $B = E/c$ and so the second term on the right hand side can be ignored.

So for example, if you have a plane wave travelling along the x-axis with amplitude $E_0$ and frequency $\omega$: $$m \frac{d^2 \vec{r}}{dt^2} = qE_0 \sin (\omega t - kx) \vec{j},$$ where I have chosen to polarise the wave along the y-axis and the B-field is along the z-axis.

If we now assume the charged particle starts at rest and at the origin, then by integrating twice, we have $$ \vec{r} = \frac{q E_0}{m \omega} \left( t - \frac{\sin \omega t}{\omega} \right) \vec{j}.$$

There is some second-order oscillation in the $x$ direction, at twice the frequency, but with a much smaller amplitude.

Relativistic. Much harder.

The four components of acceleration are: $$m \ddot{x} = -q\dot{y} B_0 \sin \omega t $$ $$m \ddot{y} = q( \dot{x}B_0 \sin \omega t + \dot{t} E_0 \sin \omega t)$$ $$m \ddot{z} = 0$$ $$m \ddot{t} = \frac{q \dot{y}}{c}E_0 \sin \omega t $$ where the derivatives here are with respect to proper time $\tau$ and the relationship between proper time and coordinate time $t$ is $$ d\tau = dt \left[ 1 - \frac{1}{c^2}\left[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right) \right]^2 \right]^{1/2}$$

As you might expect, the results of integrating these are messy. The whole thing is gone through in Kruger & Bovyn (1976) and the results are expressed as functions of $\tau$ and the cyclotron frequency $\Omega = eB_0/m$ (equations 12a-d in that reference give general expressions for a plane wave excitation). However the results can be expressed in terms of a series expansion of harmonics of frequency $\omega^*$, where $\omega^*$ is related to $\omega$ by a doppler shift (see p. 1844 of the paper) - that is the oscillations can no longer be considered to be at a single frequency and take place in both the $x$ and $y$ directions.

Answer 2

The Lorentz force acting on a free charged particle in an em field ${\bf E},{\bf B}$ is ${\bf F} = q({\bf E} + {\bf v}\times{\bf B})$ and if you are using non-relativistic mechanics then you have to solve the equation $$m\frac{d {\bf v}}{dt} = q({\bf E} + {\bf v}\times{\bf B})$$ with (from your question I presume you mean a plane wave travelling in the $z$ direction)$${\bf E}= E_0{\bf i} sin(\omega t + kz), {{\bf B}= B_0{\bf j} sin(\omega t + kz)}$$

(it may be easier to keep the complex form for the em field, i.e replace $\sin$ by $e^{i(\omega t + k z)}$?). For a plane wave there will be a relationship between the amplitudes $E_0, B_0$ but usually the magnetic force term will be much smaller than the electric force term (certainly if you are assuming non-relativistic mechanics then this will be true).

To solve the problem for ${\bf v}$ you will need the initial conditions. In the simplest case that the charge at $t = 0$ is initially at rest at the origin I suspect that it may be easier looking for a numerical solution: When the charge is at rest it feels the force of the electric field which moves it in the direction ${\bf i}$ and then the magnetic force pushes it in the direction ${\bf i}\times{\bf j} = {\bf k}$ so that you have to take into account the variation with distance of the plane wave.

PS: Note that besides being non-relativistic it also ignores the charge (since it is accelerating) also radiates energy.