Newton's third breakage in Goldstein's classical mechanics

[BetterSense]

Newton's third breakage in Goldstein's "classical mechanics"

I was reading Goldstein's Classical Mechanics vol.2 to brush up, and didn't get far before I got stuck. The book warns that both the weak an strong forms of the action/reaction principle can be broken when forces predicted by the Biot-Savart law are involved. The offending passage is partially copied below (page 6).

If two charges are moving uniformly with parallel velocity vectors that are not perpendicular to the line joining the charges, then the mutual forces are equal and opposite but do not lie along the vector between the charges.[\quote]

So, I draw this system, with Particle 1 at the origin, and Particle 2 at x=y=1. They are both moving in the positive x-direction. Now, they will be repelled from each other due to Coulomb forces, but these forces point along the line between the charges so there is no problem. But since they are traveling in the x-direction, each will generate a magnetic field according to the right-hand rule, spiraling around the x-axis and the y=1 line. They will be influenced by each other's field and experience Lorentz forces so that P1 sees a force pointing upward, and P2 sees a force pointing downward. These forces are equal and opposite, but do not travel through the line between the charges, thus violating the weak principle of action and reaction. So, I see what he did there, but I have two issues.

Issue 1: The particles at the beginning of the problem are in uniform motion. However, since they are experiencing the Lorentz forces, they are accelerating, in particular, the angular momentum of the system is changing since the Lorentz forces form a torque. Angular momentum is conserved. If the angular momentum is changing, where is it coming from?

Issue 2: Suppose I transform to a coordinate system that is moving with the particles so that P1 is at the origin and P2 is not just sitting at x=y=1. The particles will be repelled from each other by the Coulomb force, but where is the magnetic force that was present in the first coordinate system? I should be able to transform coordinates systems and not have forces appear and disappear, they must manifest themselves as something else like a fictitious force or something, but I don't see it when I transform to moving coordinate system. The magnetic forces of the moving charges and thus the Lorentz forces of them on each other seems to just disappear.

 

[torquil]

I was reading Goldstein's Classical Mechanics vol.2 to brush up, and didn't get far before I got stuck. The book warns that both the weak an strong forms of the action/reaction principle can be broken when forces predicted by the Biot-Savart law are involved. The offending passage is partially copied below (page 6).

"If two charges are moving uniformly with parallel velocity vectors that are not perpendicular to the line joining the charges, then the mutual forces are equal and opposite but do not lie along the vector between the charges."

It is not strange that this can happened in the case of relativistic force fields (like electromagnetism).

So, I draw this system, with Particle 1 at the origin, and Particle 2 at x=y=1. They are both moving in the positive x-direction. Now, they will be repelled from each other due to Coulomb forces, but these forces point along the line between the charges so there is no problem. But since they are traveling in the x-direction, each will generate a magnetic field according to the right-hand rule, spiraling around the x-axis and the y=1 line. They will be influenced by each other's field and experience Lorentz forces so that P1 sees a force pointing upward, and P2 sees a force pointing downward. These forces are equal and opposite, but do not travel through the line between the charges, thus violating the weak principle of action and reaction. So, I see what he did there, but I have two issues.

Issue 1: The particles at the beginning of the problem are in uniform motion. However, since they are experiencing the Lorentz forces, they are accelerating, in particular, the angular momentum of the system is changing since the Lorentz forces form a torque. Angular momentum is conserved. If the angular momentum is changing, where is it coming from?

Have you calculated the resulting change of angular momentum. Does the magnetic field have an angular momentum that you have not thought of?

Issue 2: Suppose I transform to a coordinate system that is moving with the particles so that P1 is at the origin and P2 is not just sitting at x=y=1. The particles will be repelled from each other by the Coulomb force, but where is the magnetic force that was present in the first coordinate system? I should be able to transform coordinates systems and not have forces appear and disappear, they must manifest themselves as something else like a fictitious force or something, but I don't see it when I transform to moving coordinate system. The magnetic forces of the moving charges and thus the Lorentz forces of them on each other seems to just disappear.

If you transform to a coordinate system where P1 is always at the origin, then you have transformed into a non-inertial referance frame. E.g. centrifugal forces and coriolis forces might explain your quandry?

I haven't have time think about this, just throwing out some suggestions. Hope you sort it out!

Torquil

 

[Meir Achuz]

The process described there is the basis of the Trouton-Noble experiment.
This is explained in <arxiv.org/abs/physics/0603110>.

 

[mordechai9]

That's an interesting example which I hadn't thought about. Thanks for sharing that and your interpretation (which seems correct to me as well.)

In the way the problem has been stated, I assume these particles have been fired out with some initial velocity and now they are moving without dissipation through a vacuum. Consider we sit in the rest frame of one particle. This particle feels the instantaneous electric field from the other particle. This induces an acceleration to the reference frame. This will manifest itself as the other charge taking a relative motion, which looks like an instantaneous space current. This current creates a changing magnetic field in the original reference frame due to the Biot-Savart law. This field induces a new electric field due to the Faraday's law, and so on.

The presence of the magnetic field induces rotations due to the particle gyromotion -- charged particle trajectories rotating around magnetic field lines. (This is a basic, fundamental result of plasma physics.) The B field will point out of plane for the "left" particle (using the right hand rule for the BS law) but the B field will point into the plane for the "right" particle (using a similar rest frame analysis for that particle on the right.) Thus the particles will gyrate in opposite directions (assuming they are of the same charge) and this sense of angular momentum will be conserved. If the particles were differently charged, they would gyrate in different directions, but the original electric force would point in different directions, so you would still have oppositely oriented gyrations.

Hence the angular momentum should cancel out. Sorry if this was long-winded. Of course, this sounds correct, but I think you'd have to actually calculate it to make sure it works out right. It's probably not exactly solvable for more than two particles.

For more interesting info on relations between mechanical and e/m angular momentum, see the Einstein-de Haas effect http://en.wikipedia.org/wiki/Einstein-de_Haas_effect.

 

[BrunoIdeas]

Hello. I have just read the passage from Goldstein's text and have the same questions.
I however also question the validity of Biot-Savart Law for point charges.

If someone with a thorough understanding of the topic, who has gone through the same process of learning we are, could answer, it would be of great help.

Thanks.

 

[Meir Achuz]

The B-S law does not hold for point particles.
The E and B fields for a constant velocity are given by (in natural units)
$${\bf E}= \frac{q{\bf r}} {\gamma^2[{\bf r}^2-({\bf v\times r})^2]^{\frac{3}{2}}}$$
and
$${\bf B}={\bf v\times E} =\frac{q{\bf v\times r}} {\gamma^2[{\bf r}^2-({\bf v\times r})^2]^{\frac{3}{2}}}$$