action = reaction and lorenz force

0xDEADBEEF

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 B-Field of a moving point charge is:
[tex]\mathbf{B}=\frac{\mu_0}{4\pi}q \frac{\mathbf{v}\times\mathbf{r}}{\left|r\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.

Simon Bridge

Solve each force in the rest-frame of the other charge.

jtbell

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.

Andrew Mason

Interesting. If one interprets the third law as a statement about the reciprocity of changes in motion rather than reciprocity of forces there would be no third law problem. The reciprocity of forces holds only if the interacting bodies measure time the same.

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

DaleSpam

No, there would still be a problem since the change in momentum of one charge is not equal and opposite the change in momentum of the other. The key is to consider both the charges and the field. Then, to rescue Newton's third law you still need to express it in terms of forces since the field is interacting with both bodies.

Only after talking about equal and opposite forces twice and only with the caveat that while talking about changes in momentum he is specifically referring to a pair of isolated bodies with no other interaction forces. Your interpretation of Newtons third law is unsupported by any reference that I have seen.

stevendaryl

Well, Newton's laws really were not made to take into account momentum in fields. Interpreting his 3rd law in terms of forces, you need to consider a "force" on the field due to the particles. I guess you could do that, but it's a very different notion of "force'; it's certainly not equal to [itex]m a[/itex].

sophiecentaur

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?

stevendaryl

Momentum is certainly conserved in electrodynamics, but only if you take into account momentum in the field itself. It's not conserved if you only consider the momenta of particles.

sophiecentaur

But wouldn't that be a daft thing to do? It would be as unrealistic as to ignore the effect of (and on) a trampoline with two little boys fighting on it.

jtbell

With electric forces, there's no problem. With magnetic forces, on the other hand...

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

stevendaryl

Yes, it would be a daft thing to do. The question was whether we can interpret conservation of momentum in electrodynamics as consistent with Newton's laws of motion. I think that it's a mistake to even try. The modern view of conservation of momentum is more general than what Newton considered.

Dadface

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?

stevendaryl

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 ones--unlike Newton's action-at-a-distance, 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.

Dadface

But what about the points made previously regarding the deflector systems?

sophiecentaur

OK. I can see that you get a zero force with that model. That situation only applies at one instant, doesn't it (when the one electron is dead ahead of the other) and there will be the electric force, also, which needs to be considered and which is increasing to a maximum at some point during the near-miss. I can't say I have 'explained away' any apparent paradox but there is more going on than the initial description of the situation implies.
I get the strong message that "it can't be as simple as that" - situation normal then!

0xDEADBEEF

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?

DaleSpam

I agree.