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Consider a system of the two identical positive point charges situated in free space (isolated from the influence of any other external fields) as shown in the attached diagram. Particle $1$ is at $(a,a,0)$ and particle $2$ is at $(0,a,0)$. Their velocities, at the considered instant of time, are as shown in the diagram ($\mathbf{v}_1$ along the $+x$ axis, $\mathbf{v}_2$ along the $+z$ axis).

enter image description here

Now, by applying the Biot-Savart law, we find that the magnetic field due to particle $2$ at the position of particle $1$ is along $+y$ axis, which means that the force acting on particle $1$ is along $+z$ axis according to Fleming’s left-hand rule. A similar analysis shows that there is no magnetic force on particle $2$ as the magnetic field of particle $1$ should vanish at positions (relative to particle $1$) located along its velocity vector.

Now, if we observe the net torque on the system of two particles about the $y$ axis then it is non-zero and is directed along the $-y$ axis. Here, there are no external forces or torques acting on the isolated two-particle system, and yet the net torque, as well as the net force on the particles, are nonzero. Why? Also, Newton's third law of motion seems to be broken in this scenario. Why?

Edit $1$

I have come to know from the responses that the electromagnetic field itself takes away some momentum and angular momentum about the considered axis. However, I think that if I consider only two charged particles as my system then the Abraham-Lorentz force can be assumed to be acting upon the system, and that is sufficient to make sure that we have considered the momentum being carried away by the electromagnetic field itself.

Even after considering the action of the Abraham-Lorentz force for the two-particle system, the scenario breaks both Newton's third law and the conservation of linear and angular momentum. This is because the Abraham Lorentz forces do not exactly counterbalance the force and the torques, on the two-particle system under consideration, due to the magnetic field.

Edit $2$

The previous edit was a result of confusion and misunderstanding on my part. The force associated with the momentum carried away by the electromagnetic field as a result of the electromagnetic interaction described in the question is the Lorentz force itself which simply doesn't obey the third law of motion. The Abraham-Lorentz force is a different story. It is associated with the momentum carried away by the radiation emitted by the accelerated charged particles. This is an additional force apart from the Lorentz force and corresponds to an additional carriage of momentum by the electromagnetic field. The momentum carried away by the electromagnetic field in correspondence with the Lorentz force doesn't correspond to radiation.

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You are correct in your assertion that pairs of charged point particles can interact magnetically in ways that seemingly violate Newton's 3rd law, and therefore also seem to violate the conservation of both linear and angular momentum. This is a fundamental result and it is the decisive (thought) experiment which forces us to change our viewpoint on electrodynamics from something like

charged particles interact with each other

to a field-based one that says

charged particles interact with the electromagnetic field.

What this means, and the key point here, is that

  • the electromagnetic field should be considered as a dynamical entity of its own, on par with material particles, and it can hold energy, momentum, and angular momentum of its own.

The linear and angular momentum of the complete dynamical system, which includes the particles and the field, is indeed conserved. This means that in a situation like your diagram, where there is a net force and torque on the mechanical side of the system (i.e. the particles), there are corresponding and opposite net forces and torques on the electromagnetic field.

So, how much linear and angular momentum are there? This is a solid piece of classical electrodynamics: these momenta are 'stored' throughout space, with densities $$ \mathbf g =\epsilon_0 \mathbf E\times\mathbf B $$ and $$ \mathbf j =\epsilon_0\mathbf r\times\left( \mathbf E\times\mathbf B\right), $$ respectively. Once you account for these, it follows from Maxwell's equations and the Lorentz force that, for an isolated system, the total momenta are conserved. The details of the calculation are a bit messy, and so are the actual conservation laws; I gave a nice derivation of the linear momentum one in this answer.

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  • $\begingroup$ You defined the EM field as having it's own momentum. So does it mean that it has some mass and is always moving with some velocity? That's what I always associate the word momentum with. $\endgroup$ – harshit54 Mar 19 at 14:10
  • $\begingroup$ Momentum is best thought of as the conserved quantity associated with translation invariance. In terms electromagnetic fields, this is completely unrelated to the $p=mv$ form. (In fact, relating the momentum to the electromagnetic energy gives the massless relativistic version of the energy-momentum relation $E^2=m^2c^4+p^2c^2$.) $\endgroup$ – Bob Knighton Mar 19 at 23:21
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    $\begingroup$ @harshit54 "I've never seen it, so therefore it doesn't exist"? The EM field can have momentum without having mass - momentum is a much more general quantity than just $mv$ for a massive particle. For more details see If photons have no mass, how can they have momentum? and links therein. $\endgroup$ – Emilio Pisanty Mar 21 at 10:16
  • $\begingroup$ Thanks for the reply. But a photon is still not a field and it moves which should give it some momentum. $\endgroup$ – harshit54 Mar 21 at 14:15
  • $\begingroup$ @harshit54 If this is still unclear, ask separately. $\endgroup$ – Emilio Pisanty Mar 21 at 14:40
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I want to give an answer for the case, that the mentioned particles are electrons. Let us consider that the magnetic dipole moments of this two electrons are aligned in a straight line through the points (0,a,0) and (a,a,0). Since both electrons are moving their magnetic dipole moments begin to turn when the electrons leave the mentioned points.

Perhaps it is not necessary and maybe not allowed to apply laws for macroscopic objects to particles? The phenomenon which undergoes the electrons is obvious. Any change of the direction of the magnetic dipole moment changes the direction of the electrons intrinsic spin too. As known this led to a deflection in the trajectory of the electron. This is one result of the interaction between these electrons.

A second result of the interaction between these electrons is that any deflection of an electron is an acceleration and this led to the emission of photons. The electrons velocity decreases.

A third result of their interaction is based on the repelling each over due to negative electric charges. This led to deflection and photon emission too. And there is a difference in the magnetic induced deflection and the electric deflection. The electric induced force acts in the plane spanned from the two velocities, the magnetic induced force acts perpendicular to this plane.

This is not a full answer but it shows why

electromagnetic field itself takes away some momentum and moment of momentum.

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  • $\begingroup$ An electron could accelerate and be at rest, in which case the velocity can't decrease because it is already zero. What happens is in the frame where the electron is instantaneous at rest the electron increases its energy less than it otherwise would without radiation, that's where the energy comes from. Its like friction, things can speed up even with friction acting, just not as much as without friction acting. $\endgroup$ – Timaeus Jun 16 '15 at 15:18
  • $\begingroup$ @Timaeus Sorry. I'm not sure what you want to say. $\endgroup$ – HolgerFiedler Jun 16 '15 at 15:29
  • $\begingroup$ You wrote "A second result of the interaction between these electrons is that any deflection of an electron is an acceleration and this led to the emission of photons. The electrons velocity decreases." But velocity is zero in the frame where the electron is instantaneously at rest, in that frame it increases its speed less than it otherwise would, so in all frames the electron just accelerates less than it otherwise would. The speed can't can't decrease when it is already at zero speed. $\endgroup$ – Timaeus Jun 16 '15 at 15:33
  • $\begingroup$ @Timaeus If you move a frame with a deflected electron, this frame could not be at rest because in this frame one feel an acceleration. For an deflected electron it is not possible to get a frame at rest. $\endgroup$ – HolgerFiedler Jun 16 '15 at 17:55
  • $\begingroup$ The electron has a velocity at that moment, hence there is a frame moving at that same velocity. That frame keeps moving at that speed so the electron is only at rest in that frame for that instant, but it is at rest for that instant. And thus in that frame at that moment it has no kinetic energy to give so the energy comes from it not gaining as much kinetic energy as it would have. Thus in all frames the kinetic energy just changes differently than it otherwise would have. $\endgroup$ – Timaeus Jun 16 '15 at 19:09
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The Lorentz force law adequately describes the acceleration of the point charges. Therefore we are dealing here with a paradox and it is misguided to hypothesise that the field carries off or brings in momentum.

The solution is in the interpretation of the Lorentz force law. Since the electric force does conserve momentum, as the forces on the charges are opposite and equal in size, I will concentrate on the magnetic part $f_B = q \vec v \times \vec B = q \vec v \times \vec \nabla \times \vec A$. Writing in components is easier, so $f_{Bi} = d(mv_i)/dt = q v_j \partial_j A_i - q v_j \partial_i A_j$. This can be written as $f_{i} = d \left(mv_i - qA_i \right)/dt = - q v_j \partial_i A_j$. It can easily be checked that $m \vec v - q \vec A$ is conserved.

The conclusion is that total momentum consists of a kinetic contribution, $m v_i$, and a potential contribution, $-q\vec A$, in the presence of electromagnetism.

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  • $\begingroup$ The potential contribution precisely captures the momentum carried away/brought in by the fields. Correct me if I misunderstood your post but it seems to me that it is not misguided to say that the field carries away or brings in momentum. $\endgroup$ – Dvij Mankad Mar 19 at 22:55
  • $\begingroup$ @Dvij Mankad It is up to you to show that. I maintain that it is misguided. The potential l contribution does not "carry away" momentum. The potential is momentum. $\endgroup$ – my2cts Mar 19 at 23:23
  • $\begingroup$ To clarify, by "carry away", I don't mean that it carries away momentum in the form of radiation. The momentum doesn't run off to spatial infinities but is stored in the electromagnetic field which can later take the form of the kinetic momentum of a particle via some interaction. The relevant detailed calculation can be found here: physics.stackexchange.com/questions/83743/… $\endgroup$ – Dvij Mankad Mar 20 at 0:03
  • $\begingroup$ "as the forces on the charges are opposite and equal in size" - OP has clearly and concisely shown that for the configuration at hand this is not the case. $\endgroup$ – Emilio Pisanty Mar 21 at 10:27
  • $\begingroup$ @Emilio Pisanty as specified in my text, this statement refers to the electrical contribution to the Lorentz force. $\endgroup$ – my2cts Mar 21 at 15:39
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I have been searching for a satisfying answer for this question for long time. The momentum explanation provided by Pisanty is excellent to prove that the Netwon’s third law holds but it does not explain how. it tells us that for sure Netwon’s third law is correct because the momentum equations indicate that the opposite force exists and should be there in the system, the momentum is conserved, but it does not explain how this opposite force exists, its nature and why it does not appear in some cases for the magnetic force which I believe it was the reason behind this question. Momentum explanation usually gives general statements about it such as “there are corresponding and opposite net forces and torques on the electromagnetic field” and this is not satisfying for me.

Few months ago I came across a work that explains how these opposite forces exist and why they do not appear in some cases and it was satisfying to me. Shadid in his work “Two new theories for the current charge relativity and the electric origin of the magnetic force” provided a successful and proved explanation to the magnetic force as purely electrostatic through careful analysis for the spreading electric field pattern in the space for moving positive and negative charges in current elements. The magnetic force is explained as a result of electric interaction between current charges and the charges at discontinuity locations where the electric field changes from positive to negative and negative to positive due to the switching of places between the moving positive and negative charges in a current element. These discontinuity charges surround current elements and are produced when charges move to carry on electric field changes in the space. The existence of these discontinuity charges is proved using Gauss law and explained by the photons that travel to indicate the changes in the electric field, these photons are assumed to be charged as assumed in Altschul’s work "Bound on the photon charge from the phase coherence of extragalactic radiation".

Moving electric charges of current elements interact with each other through discontinuity charges, since current elements are electrically neutral. The electric force between a current charge and a discontinuity charge obeys Newton's third law as in Coulomb’s law. The forces exerted on current charges allow the charges to produce either a non-zero or zero net force on the containing infinitesimal current element. This net force on the current element is the observed magnetic force. The produced net force is non-zero on the current element when the positive and negative charges push the current element in the same direction. The push occurs when the exerted forces are perpendicular to the motion direction of the charges and they are not allowed to move outside the containing filamentary current element, while they are free to move along that element. Notice that the push interaction between the current charges and the containing current element obeys Newton’s third law as in particles interaction. However, the net force is zero when these charges push the current element in opposite directions thereby canceling each other or when the exerted forces on the current charges are completely along the direction of movement for the charges so no push force is produced on the containing current element. This explanation was proved by deriving the exact magnetic force law and Biot-Savart law using the basis of electric forces as specified in the electromagnetic theory.

The details of the proof and calculation are a bit long; I gave a brief overview of it. Details can be found at http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=7546893

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The momentum of massless radiation is not enough to consider Lorentz' force law in agreement with momentum conservation. Radiation momentum is negligible for magnetostatic situations, and the Lorentz force law violates Newton's third principle of motion also for magnetostatic currents that are not closed-on-itself. The previous answer given is therefore wrong.

All textbooks define magnetostatics as the physics of "closed loop" electric currents (such that the Lorentz force does not violate Newton's third law), which is not based on experiments. This definition is deceptive, and obscures the inconsistency of Lorentz' force law with classical mechanics.

It is often suggested that "fields have momentum". That is not true. Only a Poynting energy flow represents momentum, according to Classical Electro-Dynamics theory (see the usual power theorem of CED). It is highly unlikely that the ohmic 'heat radiation' losses in stationary current circuits compensate for Lorentz force violation of Newton's third principle of motion, such that momentum is conserved.

Secondly, the 'non-radiative' Poynting flow in stationary current circuits does not change in time (it is stationary as well), such that it does not contribute to momentum change.

My conclusion is: we need Whittaker's force law to replace Grassmann's force law (which is a special case of the Lorentz force law, for stationary currents in circuits) in order to define an electrodynamics theory that is consistent with classical mechanics, since Whittaker's force law obeys Newton's third law of motion. This means that the Maxwell-Lorentz theory is inconsistent with classical mechanics, and should be replaced by a consistent theory, see http://philpapers.org/archive/VANGCE-2.pdf

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    $\begingroup$ It is well known that Maxwell's electrodynamics is at odds with Newtonian mechanics because Newtonian mechanics is just an approximation. It should be superseded by special relativity (and GR if gravity must be taken into account, in which case electrodynamics must also be modified). We don't need to change electrodynamics for disagreeing with Newtonian mechanics. In fact, that's a virtue! That aside, I think your analysis is wrong, too, so the discussion doesn't even apply to this particular point. $\endgroup$ – Danu Dec 20 '15 at 16:04
  • $\begingroup$ The fact that the Lorentz force law does not conserve momentum isn't fixed by relativistic electrodynamics, nor by the extra momentum of massless radiation. These unproven suggestions do not "fix" the inconsistencies of the Maxwell-Lorentz theory. My conclusion is that the Lorentz force law must be replaced by the Whittaker force law (which agrees with Newton's third law) in order to formulate a consistent classical electrodynamics, see academia.edu/20508645/General_Classical_Electrodynamics $\endgroup$ – ProfRaccoon Jan 27 '16 at 13:00
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I don't understand why all answerers immediately jump to the topic of momentum, carrying by the field. Nowadays nobody doubt field can carry momentum.

The questions like this and especially mine one: How to observe Newton's third law violation? have different aspect on how this is possible, that third law is violated. Third law is the main workhorse of nature, when it is avoiding perpetuate motors and UFO's reactionless engines.

If the explanation would "momentum is transmitted to the field" then the next idea would be "okay, let's build engine to gain endless energy from the field".

But this is impossible. And the field does not matter.

Below is an explanation why.

Regard two charges moving at the same speed but in perpendicular diections:

enter image description here

The point is: yes, it is true, that according to laws of electrodynamics, the third law of Newton is violated here.

But this happens only temporary, at this very moment.

So, let's regard the situation at some moments after:

enter image description here

You see, that one charge have passed the crossroad, while another one have reached it.

And you can notice, that the charges have flipped they roles.

So, any momentum, which was owed by one charge, will be returned to it at this moment.

In other words, the reason of momentum conservation is not the field, but the symmetry. Field is just a credit holder here.

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    $\begingroup$ I don't think the free charges in the first configuration would ever reach the second configuration illustrated by you. And there certainly exist electrodynamic systems which irreversibly transfer momentum and energy to the fields so that even if we consider as long time as pleases us we can't get the momentum conserved just within the particles' momentum. $\endgroup$ – Dvij Mankad Apr 12 '16 at 12:37
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    $\begingroup$ Devices which transfer momentum (and energy) to the field are called cellular phones and are definitely exist :) $\endgroup$ – Dims Apr 12 '16 at 12:50
  • $\begingroup$ Also, charges can definitely keep moving on straight trajectories, if they were mounted on rails (i.e. not free). Violation of 3rd law is still violation of 3rd law, even if bodies are mounted somehow. $\endgroup$ – Dims Apr 12 '16 at 12:51
  • $\begingroup$ There are points of emphasis where it comes down to opinion, but blaming 'symmetry' seems misguided to me, since there is nothing forcing the situation to be symmetric; and indeed there is no requirement that the particles in your first picture reach the second configuration. (For one, there could be additional forces at play, without affecting the conclusions about momentum balance.) $\endgroup$ – Emilio Pisanty Apr 12 '16 at 13:53
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    $\begingroup$ Similarly, saying "If the explanation would "momentum is transmitted to the field" then the next idea would be "okay, let's build engine to gain endless energy from the field"" is pretty naive - the field can store momentum, much like a flywheel stores angular momentum, but that does not mean it is an inexhaustible source of it. $\endgroup$ – Emilio Pisanty Apr 12 '16 at 13:53

protected by Qmechanic♦ Sep 20 '17 at 23:08

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