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Action = reaction and lorenz force 
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#1
Jan3113, 06:07 PM

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Newtons third law states that there is a counter force to every force. Unfortunately this doesn't seem to work for moving point charges. The Coulomb force cancels out but
the BField of a moving point charge is: [tex]\mathbf{B}=\frac{\mu_0}{4\pi}q \frac{\mathbf{v}\times\mathbf{r}}{\leftr\right^3}[/tex] And the Lorenz force is [tex]\mathbf{F}=q\, \mathbf{v}\times \mathbf{B}[/tex] Lets assume that the two charges have velocities [itex]\mathbf{v}_1,\mathbf{v}_1[/itex] Therefore the two Lorenz forces are [tex]\mathbf{F}_1=k(r)\, \mathbf{v}_1 \times (\mathbf{v}_2 \times\mathbf{r}) [/tex] and [tex]\mathbf{F}_2=k(r)\, \mathbf{v}_2 \times (\mathbf{v}_1 \times ( \mathbf{r})) [/tex] Due to the Jacobi identity the sum of the two forces is not zero [tex]\mathbf{F}_1+\mathbf{F}_2= k(r)\, \mathbf{r}\times(\mathbf{v}_1\times \mathbf{v}_2)[/tex] What is the solution here? The Pointing vector? Relativity? I think that the basic formulas must be correct for slowly moving charges. So it shouldn't be due to non linear trajectories, neglected acceleration or some such thing. 


#2
Jan3113, 09:58 PM

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Solve each force in the restframe of the other charge.



#3
Jan3113, 10:54 PM

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Indeed, Newton's Third Law does not apply to the Lorentz force on charges, in general.
Conservation of momentum (from which the Third Law can can be derived for mechanical forces) is more general. We can make conservation of momentum work in electrodynamics by associating a momentum density with the electromagnetic field: ##\vec g = \epsilon_0 \vec E \times \vec B##. The sum of the mechanical momenta of the particles, ##\vec p_i##, and the volume integral of ##\vec g## is conserved in a closed system. 


#4
Feb213, 10:58 PM

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Action = reaction and lorenz force
Physics texts often downplay Newton's "action/reaction" version of the third law and state it in terms of forces. Newton uses the term "force" (Latin: vis) in his first two laws but in the third he refers to "action" as in: "the mutual actions of two bodies upon each other are always equal, and directed to contrary parts." In his explanation of the law Newton talks about equal and opposed changes in motion: "If a body impinges upon another, and by its force change the motion of the other, that body also (because of the equality of, the mutual pressure) will undergo an equal change, in its own motion, towards the contrary part. The changes made by these actions are equal, not in the velocities but in the motions of bodies; that is to say, if the bodies are not hindered by any other impediments. For, because the motions are equally changed, the changes of the velocities made towards contrary parts are reciprocally proportional to the bodies."AM 


#5
Feb313, 06:39 AM

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#6
Feb313, 09:48 AM

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#7
Feb313, 10:33 AM

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Am I reading this correctly? Are we saying that momentum, here, is not conserved? Is it not true to say that the force on the deflector mechanism (magnetic or electric) of a crt is equal and opposite to the force that deflects the electron beam?



#8
Feb313, 10:42 AM

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#9
Feb313, 10:45 AM

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#10
Feb313, 10:50 AM

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Consider charged particle A traveling along the xaxis towards the origin (but not located at the origin), and charged particle B traveling along the yaxis and passing through the origin at a certain time. The magnetic field produced by A is zero along the xaxis, so there is no magnetic force on B at that instant. The magnetic field produced by B is nonzero at the location of A, so there is a magnetic force on A. 


#11
Feb313, 10:52 AM

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#12
Feb313, 10:58 AM

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B can be defined with reference to the force on a current carrying conductor in a magnetic field and it is standard practise in A level physics to measure this force by measuring the equal and opposite force on the magnet arrangement. In other words, in this example, it is the deflecting system that can be considered as the particle that the moving charges interact with. Is this not basically the same as the example given by sophiecentaur in post 7?



#13
Feb313, 11:08 AM

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On the other hand, Feynmann developed a theory of electrodynamics in which there were no extra degrees of freedom in the electromagnetic field; the theory could be interpreted as a pure particle theory with binary particle interactions (although strange onesunlike Newton's actionatadistance, the interactions aren't instantaneous).
His theory is equivalent to the usual classical electrodynamics if one assumes that all electromagnetic radiation is eventually absorbed by charged particles. If there is "free" electromagnetic radiation that spreads out forever never being absorbed, then Feynmann's theory doesn't work. 


#14
Feb313, 11:17 AM

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#15
Feb313, 11:42 AM

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I get the strong message that "it can't be as simple as that"  situation normal then! 


#16
Feb313, 12:24 PM

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So the solution lies in the Poynting vector. Does someone know if/where the momentum leaves the system? Where is the largest portion of the field's momentum?



#17
Feb313, 12:49 PM

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#18
Feb413, 06:52 AM

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Newton obviously had no idea about electromagnetic interactions and charge but his corpuscular model of light involved light carrying momentum. The difference is that in Newton's model, light had rest mass and one could imagine forces between light and matter. That really does not apply to interactions between photons and charges because photons lack rest mass and do not accelerate. It is not clear why Newton used the term "action" instead of "force" and it is not clear exactly what "action" means. He uses "action" to relate the concept of force to the movement of bodies. But what Newton's commentary shows is that he concluded from the Law III and Law II that "quantity of motion" (which is the term he uses for "momentum") is conserved in an isolated system. Take, for example, this Corollary to his Laws of Motion. The quantity of motion, which is collected by taking the sum of the motions directed towards the same parts, and the difference of those that are directed to contrary parts, suffers no change from the action of bodies among themselves.AM 


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