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An electromagnetic plane wave moving in the $z$-direction

$E(x,t) = E_x \cos(k z - \omega t), \hspace{1cm} B(x,t) = B_y \sin(kz-\omega t)$

has field momentum in $z$-direction. But according to this paper, it induces motion in $x$ and $y$ direction in a charged particle. How is this possible? Does conservation of momentum not apply? If not, are field and particle momentum completely separated concepts?

Remark: My question is less focused on the precise movement described in the paper, but generally how it is possible for a wave with momentum in $z$ direction to induce motion in $x$ and $y$ direction.

Qmechanic
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spastikatenpraedikat
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2 Answers2

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Does conservation of momentum not apply?

Of course it applies. The conservation of momentum is built into Maxwell’s equations. The EM Lagrangian is independent of spatial translations, so there is a corresponding conserved momentum, per Noether’s theorem.

It is easiest to think of this in terms of scattering. A photon, with momentum in $z $ collides elastically with a particle at rest. After the collision the particle has momentum along $x$ so the photon is scattered and has equal momentum along $-x$ in addition to its momentum along $z$

Dale
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In the simple classical picture, the electromagnetic field does no work on the charged particle (because the E-field and particle velocity are $\pi/2$ out of phase) in a time-averaged sense. Similarly, there is no time-averaged momentum transferred to the particle because it oscillates up and down along the polarisation vector of the E-field. In a bit more detail we would include the Lorentz force due to the magnetic part of the field and find that there was some momentum transferred to the particle along the z-axis.

I'm not sure it is that helpful to examine the classical picture of wave scattering from a charged particle too closely. In detail it doesn't work unless the photon energy is much less than the rest mass energy of the charged particle and even then is just an approximation for what goes on.

ProfRob
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