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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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    When the particle acquires x momentum, the field acquires negative x momentum. (Don’t forget that charged particles make their own fields.) – knzhou Nov 02 '22 at 20:48
  • Can this be true? The particles momentum is dependent on the mass, but the charged particles field is indipendent of the mass. So how can they cancel out reliably? Or did I missunderstand you? – spastikatenpraedikat Nov 02 '22 at 22:07
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    The field the charged particle makes depends on how fast it's moving, which in turn depends on its mass, because $F = ma$. – knzhou Nov 02 '22 at 22:10
  • see https://physics.stackexchange.com/q/275930/ . – anna v Nov 03 '22 at 06:01
  • The paper you link in your question is too involved for me to check, In the linked answer to a similar question it is seen that the oscilations happen in the direction of the propagation of the wave, – anna v Nov 03 '22 at 06:43
  • @annav I am sorry if I misunderstand your linked question, but doesn't it say that in the non-relativistic case the movement happens along E, therefore transversal? – spastikatenpraedikat Nov 03 '22 at 09:59
  • @knzhou Let's imagine two particles of same charge, but unequal mass. Since they have the same charge, they will gain the same momentum in an electric field, namely \Delta p = F = qE. But since they have unequal masses, their velocities due to that momentum gain will not be equal. And since the field momentum depends on charge and velocity, but not math, the field momenta will not be euqal either, meaning at least one of these two field momenta will not compensate the linear momenta of its particle. Right? – spastikatenpraedikat Nov 03 '22 at 10:18
  • @spastikatenpraedikat The formula has the x direction of the beam in its variations. – anna v Nov 03 '22 at 10:26

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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