# Imagine a solid, non-conducting cylinder at rest in IRF K, with a

• GRDixon
In summary, two charges on opposite ends of a non-conducting cylinder in IRF K experience an electric force towards each other. When viewed from frame K', the charges experience a net electromagnetic force that does not coincide with the cylinder's axis, causing a non-zero torque around the cylinder's midpoint. However, the cylinder does not rotate in K' due to a counter torque, possibly caused by the compressed cylinder exerting a reaction torque on the charges. This is consistent with the Lorentz transformation result that the cylinder's ends are not perpendicular to its walls in K'. The situation is similar to the "lever paradox" and the Trouton-Noble experiment, where the internal forces do not contribute to angular momentum and the energy and angular
GRDixon
Imagine a solid, non-conducting cylinder at rest in IRF K, with a positive spherical charge centered on one end and a negative spherical charge on the other end. Each charge experiences an electric force toward the other. The cylinder is compressed, but there is no torque.
Let us say that the cylinder’s axis makes an angle, A, with the x-axis, such that 0<A<90 degrees. Viewed from frame K’, each charge experiences a net electromagnetic (Lorentz) force that does not point along the cylinder’s axis. Together the charges exert a nonzero torque about the cylinder’s midpoint. Yet in K’ (as in K) the cylinder does not rotate. What counteracts the Lorentz torque in K’?
Bonus question: why don’t the charges slide/roll off the cylinder ends in K’?

It doesn't seem like you defined K'. A diagram would really help here.

Assuming that K' is moving relative to K along the x-axis, then K' will see a magnetic field that cancels the torque. I haven't got time to do the calculation, but that's the usual answer.

bcrowell said:
It doesn't seem like you defined K'. A diagram would really help here.

K and K' are both rectangular coordinate systems, with their x/x', y/y' and z/z' axes overlaid at t=t'=0. Let K move in the negative x direction at constant speed v. In this case the cylinder and charges move in the positive x' direction of K' at constant speed v. I don't know how to do a diagram in one of these replies, but hope you get the idea.

Mentz114 said:
Assuming that K' is moving relative to K along the x-axis, then K' will see a magnetic field that cancels the torque. I haven't got time to do the calculation, but that's the usual answer.

The idea is that, relative to K', the electric force still points along the chord between the 2 charges, but the magnetic force is parallel to the y' axis. Thus the sum of the 2 forces (i.e., the Lorentz force) does not generally point along the chord between the 2 charges. Together, the 2 Lorentz forces create a force couple (and torque) around the chord midpoint.

GRDixon said:
The idea is that, relative to K', the electric force still points along the chord between the 2 charges, but the magnetic force is parallel to the y' axis. Thus the sum of the 2 forces (i.e., the Lorentz force) does not generally point along the chord between the 2 charges. Together, the 2 Lorentz forces create a force couple (and torque) around the chord midpoint.
Could you show your calculation ?

Mentz114 said:
Could you show your calculation ?

I must confess that I haven't considered any particular case. I simply used the right hand rule to convince myself that the electric force and the magnetic force on a given charge don't point in the same direction, when things are viewed from K'. The net Lorentz forces on the two charges produce a torque on the cylinder. Yet the cylinder doesn't rotate. (All parts of it move with the same, common velocity in K'.) Since the cylinder doesn't rotate, there must be a counter torque. The only thing I can think of is that the compressed cylinder itself must exert a reaction torque on the charges. This is consistent with the Lorentz transformation result that the cylinder's ends are not at right angles to its walls, when things are viewed from K'. (That is why the charges don't roll off the ends in K'.) Thanks for your interest. I've never seen this problem discussed in an electromagnetism text.

I haven't read this closely, and I can't quite envision the situation, but the general description sounds a lot like the classic "lever paradox", first proposed by Lewis and Tollman,and related to the Trouton-Noble experiment with the rotation of a charged capacitor.

There's a paper at http://arxiv.org/abs/0805.1196 with a bibliography, I'm not terribly sure of this paper's approach yet (the paper does appear to be peer-reviewed). However, it's different from the papers I've read in the past. So as usual, let the reader beware.

Thanks, Pervect. I have checked out the link and will spend more time thinking about the whole thing. I was aware of the Trouton-Noble experiment.

The standard derivation that internal forces don't contribute to angular momentum requires that the force between any pair of particles lies along a line connecting them, if you go back and read the fine print in your standard physics textbook (i.e. Goldstein, for example).

This assumption becomes a bit problematical in the context of SR. I've seen various discussions of the issue but I'm not really super-happy with any of them.

pervect said:
The standard derivation that internal forces don't contribute to angular momentum requires that the force between any pair of particles lies along a line connecting them

And that assumption can fail for electromagnetic forces. The interpretation is that the angular momentum is carried away by electromagnetic waves.

As an example, if you start a dipole spinning, its spin will slow down as it dissipates its energy and angular momentum into the energy and angular momentum of the electromagnetic waves it radiates. I don't know if this is quite what the OP had in mind.

Mentz114 said:
Could you show your calculation ?

You might try Googling "Trouton Noble Revisited".

GRDixon said:
You might try Googling "Trouton Noble Revisited".

Thanks. I will.

## 1. What is an IRF K?

An IRF K stands for an Inertial Reference Frame K. It is a frame of reference in which Newton's first law of motion holds true, meaning that an object at rest will remain at rest unless acted upon by an external force.

## 2. What is a solid, non-conducting cylinder?

A solid, non-conducting cylinder refers to a physical object that has a cylindrical shape and is made of a material that does not allow electricity or heat to pass through it easily. This can include materials such as wood, plastic, or rubber.

## 3. How does a solid, non-conducting cylinder behave in IRF K?

In IRF K, a solid, non-conducting cylinder at rest will remain at rest unless an external force is applied to it. This is because the cylinder does not conduct electricity or heat, so there are no internal forces acting on it. It will only move if an external force, such as a push or pull, is applied.

## 4. What happens if a force is applied to the solid, non-conducting cylinder in IRF K?

If a force is applied to the cylinder, it will accelerate in the direction of the force until it reaches a new state of rest. This is in accordance with Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

## 5. Can the solid, non-conducting cylinder ever reach a state of constant motion in IRF K?

Yes, the cylinder can reach a state of constant motion in IRF K if the forces acting on it are balanced. This means that there is no net force causing the cylinder to accelerate or decelerate. It will continue to move at a constant velocity until an external force is applied to it.

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