How can i derive the dynamic of a relativistic charged particle in a uniform magnetic field B = ( 0 , 0 , B )?

Question

How can i derive the dynamic of a relativistic charged particle in a uniform magnetic field   B  =  (  0  ,  0  ,  B  )?

Adeline Shah · Accepted Answer

Without relativity you would want to solve Newton's second law

F = m a

, where the force is supplied by the Lorentz force

F = q v \times B

, using standard notation for all the quantities involved. The simplest case is when the motion is in a plane and you want to solve for the trajectory

x (t), y (t)

, which ends up being a circle of radius related to the strength of the magnetic field. Do this exercise first since it is somewhat simpler and the real answer is not that different.
When the particle start moving fast you'd have to incorporate relativistic effects. The simplest way to proceed is to write Newton's second law as equation for 4-vectors,

f^{μ} = m \dot{p^{μ}}

, where both sides are 4-vectors and the nearly invisible dot is differentiation with respect to proper time. In a particular reference frame, these 4-vectors include factors of the relativistic

γ

-- look for any exposition of relativistic dynamics to learn about this.
When you do all that you can write the new equation of motion, which is only slightly modified compared to the non-relativisitic case. The motion is still in a circle (if we choose it to lie on a plane), whose radius now has a factor of

γ

relative to the Newtonian case.

How can i derive the dynamic of a relativistic charged

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