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I'm just a beginner so bear with me. Consider two frames at rest wrt to each other separated by distance enough for light to take a minute or so. At a given instant we create two large dipoles by some method. Electric field will start to travel at the time of creation at speed of light. But before electric field reaches frame 1, we reunite the dipole in frame 1. so when electric field reaches from frame 2 to frame 1, there'll be no effect on it, since net charge is zero. But electric field from frame 1 when reaches, creates force on dipole of frame 2. So in reality what happens? Is conservation of momentum violated? Or law of equal and opposite forces???

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I think this is due to the fact that fields can carry away momentum. Not sure :/ –  Manishearth Apr 11 '12 at 15:22
+1 to John McVirgo. Momentum conservation is always true, but law of equal and opposite forces is concept of Newtonian physics where it is assumed that light travels at infinite speed. –  anuragsn7 Apr 11 '12 at 15:38
Related: physics.stackexchange.com/q/7218/2451 –  Qmechanic Nov 2 '12 at 17:32
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2 Answers 2

Although the total momentum of any electromagnetic system is conserved, once you take into account the momentum stored in the field, as John McVirgo points out, the action-reaction law only really holds in electrostatics.

In fact, you don't even need the finite propagation speed of the field as in your example. Consider two equal positive charges moving towards the origin along the positive $x$ and $y$ axes. Then they experience magnetic fields in the positive and negative $z$ directions, respectively, and therefore magnetic forces in the positive $y$ and positive $x$ directions, resp., and these cannot cancel each other out. In a full description, this motion induces crossed electric and magnetic fields throughout space, which store the momentum not accounted for by the non-cancelling forces on the charges.

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Thanks a lot. So how exactly does the momentum is carried by emw? through some quantum particles? And what about conservation of energy in the above problem? –  Aman Rusia Apr 18 '12 at 14:51
How exactly the EM field carries momentum depends on how exactly you describe the field. If you insist on the field being made up of photons, then each photon of well-defined direction $\hat{n}$ and wavelength $\lambda=:2\pi/k$ carries momentum $\mathbf{p}=\hbar k \hat{n}$. If you're within a classical description, then the field has a momentum density $\mathbf{g}=\frac{1}{4\pi c}\mathbf{E}\times\mathbf{B}$. –  Emilio Pisanty Apr 19 '12 at 10:59
If we consider photons, they have energy. does that means a charged particle continuously emits energy? Its energy decreases continuously? –  Aman Rusia Apr 19 '12 at 11:05
It is only accelerated charges that can emit photons/radiate (classical) EM energy. This applies in principle to electrons circling an atom, so they would spiral into the nucleus within a picosecond (which doesn't happen!). Thus for bound electrons Bohr and later Schrödinger and Dirac developed quantum mechanics, which predicts that bound electrons only emit photons when changing between different energy levels in an atom. These photons then have energy equal to the difference between the two levels (and of course the associated momentum, which imparts a recoil on the atom). –  Emilio Pisanty Apr 19 '12 at 11:53
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Action equals reaction isn't generally true in electrodynamics such as the simple case of magnetic forces acting upon moving charges, whereas the conservation of momentum holds in all areas of physics. There is momentum associated with the electromagnetic field which must be added to that of the momentum of the charges it acts upon to get the total momentum of the electrodynamic system.

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@aman according to quantum theory electromagnetic radiation is made up of massless particles called photons which possess momentum $p = E/c$ where $E$ is energy, and $c$ is the speed of light. An object initially at rest will recoil upon emitting radiation to conserve momentum. The linear momentum carried by electromagnetic radiation is given by $\epsilon_0(E\times B)$ –  John McVirgo Apr 18 '12 at 22:29
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