0

A moving charged particle moves in a magnetic field, the field exerts a force to the particle $F=q v \times B$. Now I have two questions:

  1. What is the reference frame for the velocity $v$? In other words, how can observers with different velocities agree on the calculated force and measured force (each observer thinks he/she is stationary and the other observer is moving)?

  2. Suppose an electron moving in a magnetic field senses a force say pushing the electron to the left ($F=q v \times B$). When we think of symmetry, what breaks the symmetry and makes the force point to the left rather than the right?

Qmechanic
  • 201,751
Ebi
  • 1,106

2 Answers2

4

Force (as a 3-vector) is not an invariant in special relativity. Whilst different observers agree that the charge is being accelerated, they will disagree on what that acceleration is and disagree on what electromagnetic fields are responsible for the acceleration.

To directly answer your first question, the velocity is measured in the frame of an observer who says that the magnetic field is ${\bf B}$. Observers in other moving frames will measure a different magnetic field and measure an electric field. They will say that the force on the charged particle is different, but is still given by $q({\bf E}' + {\bf v}'\times {\bf B}')$, where the primes indicate quantities measured in an alternate inertial reference frame.

The magnetic part of the Lorentz force acts in a particular direction because that is the definition of the B-field. It is that field which is deemed responsible for a force $q{\bf v}\times {\bf B}$ in the frame of reference where a charge moves with velocity ${\bf v}$.

Why is there a force at all? Well if there is a frame of reference where there is just a B-field, then if you move to the stationary frame of the charged particle there will also be an E-field. This will apply a force on the charge in a particular direction, and whilst different observers will not agree on the force (see above), they will agree that the charge is being accelerated according to the transformation rules of special relativity - which requires the magnetic component of the Lorentz force to act in the direction given by ${\bf v}\times {\bf B}$.

The underlying symmetry of the situation is broken by the direction of relative motion between the two observers.

ProfRob
  • 130,455
2

The force on charge particle in electromagnetic field given by $$\mathbf{F}=q\mathbf{E}+q\mathbf{v}\times\mathbf{B}$$ The transformation of field from one frame to another frame $S'$ which is moving with velocity $\mathbf{v}$ with respect to $S$ given by $$\mathbf{E}'_\parallel=\mathbf{E}_\parallel\ \ \ \ \ \ \mathbf{E}'_\perp=\gamma(\mathbf{E}_\perp+\mathbf{v}\times \mathbf{B}_\perp)$$ $$\mathbf{B}'_\parallel=\mathbf{B}_\parallel\ \ \ \ \ \ \mathbf{B}'_\perp=\gamma(\mathbf{B}_\perp-(\mathbf{v}/c^2)\times \mathbf{E}_\perp)$$ You can verify that two observer conclude the same trajectory of the particle not necessary the force.

If you reverse the direction of the magnetic field then force will also be reversed and the same for the velocity of the particle. Now if you fix the direction of field and velocity of the particle, then force acts in one direction namely which is perpendicular to both. If a charge is positive then it's anti-clockwise and if negative then clockwise.

The law of force specifies a unique direction of force, Hence the left and right are different. If the law specifies two-direction then particles can go either left or right.

Young Kindaichi
  • 11,746