Suppose there is an electromagnetic wave moving forward in the $\mathbf{\hat{k}}$ direction. Its electric/magnetic field components are given by: $$\mathbf{E} = E_0 \sin(kz - \omega t) \mathbf{\hat{i}}$$ $$\mathbf{B} = B_0 \sin(kz - \omega t) \mathbf{\hat{j}}$$ If a particle of charge $q$ was lying on the wave's trajectory, the Lorentz force law says that the force is given $\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$. However, is an electromagnetic wave a combination of both E and B fields, requiring both fields to be plugged into the equation, or does the force on the electron only depend on one of the fields, and is an EM wave only either an electric field or a magnetic field at one instant? Edit: changed x to z in expression for EM wave.


Note. As indicated by user23660, the EM wave must be transverse which means the $x$'s in your phases should instead be $z$'s.

At a given time $t$ and spatial point $\mathbf x = (x,y,z)$, the electromagnetic wave you consider is a combination of both of the fields; \begin{align} \mathbf E(t,\mathbf x) &= E_0\sin(kz-\omega t) \hat{\mathbf x} \\ \mathbf B(t,\mathbf x) &= B_0\sin(kz-\omega t) \hat{\mathbf y} \end{align} There are some special points at which both of the fields vanish though. In particular, any time the argument of the $\sin$ is an integer multiple of $\pi$; \begin{align} kz-\omega t = n\pi, \qquad n\in\mathbb Z \end{align} As a result, a particle sitting in the wave will experience both of the fields at once, and both of these fields will have to be plugged into the Lorentz force equation. Explicitly, Newton's Second Law along with the Lorentz force equation with both fields plugged in gives us the following equation of motion: \begin{align} \ddot{\mathbf x} =\frac{q}{m}(E_0\sin(kz-\omega t) \hat{\mathbf x} +B_0\sin(kz-\omega t)\dot{\mathbf x}\times\hat{\mathbf y}). \end{align} In components, this can be written as the following system of coupled differential equations: \begin{align} \ddot x &= \omega_0\sin(kz-\omega t)(c - \dot z) \\ \ddot y &= 0 \\ \ddot z &= \omega_0\sin(kz-\omega t) \dot x \end{align} where I've used the relationship $E_0 = cB_0$ and I have defined \begin{align} \omega_0 = \frac{qB_0}{m}. \end{align} As far as I can tell, this is a pretty nasty system, and I'm not sure if the general solution can be written in closed form (although admittedly I haven't really tried very hard to figure that out.) It's actually not so bad since $y$ is completely decoupled from $x$ and $z$, and it's differential equation simply implies constant velocity in $y$. This leaves a pair of coupled equations for $x$ and $z$.

  • 2
    $\begingroup$ It feels like the kind of system that should have some cute solution. Now I'm curious... $\endgroup$ – David Z Nov 2 '13 at 2:55
  • $\begingroup$ @DavidZ Yea I agree. I'm surprised I've never seen this problem tackled before in standard books, but perhaps I haven't really payed enough attention. I'd be curious to know if you find a clever method of solution. $\endgroup$ – joshphysics Nov 2 '13 at 3:06
  • $\begingroup$ Indeed, I'll let you know if I find anything. $\endgroup$ – David Z Nov 2 '13 at 3:08
  • $\begingroup$ If you consider the relativistic motion, ii will have analytical solution. deepblue.lib.umich.edu/bitstream/handle/2027.42/70253/… This paper equation 13 $\endgroup$ – user153663 Apr 26 '17 at 19:56
  • $\begingroup$ But indeed, recently my EM hw require us to solve the nonrelativisitc motion. That's your differential eqn, I don't know how to solve it. $\endgroup$ – user153663 Apr 26 '17 at 19:58

In linear polarization, the electron is accelerated by the electric field in x-direction, and therefore moves in the y-magnetic field of the wave. In the far field both fields are always present and reverse synchronously, so the electron performs an oscillation in the z-direction with doubled frequency. This oscillation vanishes with circularly polarized light. If you are interested in details, look here https://www.researchgate.net/publication/259232654_Inherent_Energy_Loss_of_the_Thomson_Scattering The formula for ω0 in the preceding answer makes no sense, because the magnetic component of the wave is not constant.


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