Relativistic charged particles in a constant uniform magnetic field

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

-
Your question and this one sound awfully similar. – user346 Mar 12 '11 at 12:09
Do you want us to derive the equation of motion or do you want us to solve it? In any case, the problem is basically the same as in the classical case, only difference being that you replace $t$ by $\tau$ and so the cyclotron frequency will get modified by a factor of $1 / \gamma$. – Marek Mar 12 '11 at 14:12
Marek is right, this is standard textbook stuff. Griffiths or Jackson cover that ground in great detail, as well as pretty much any other text on classical electrodynamics. – user566 Mar 12 '11 at 16:56

Since the other answer seem to lead to unnecessarily complicated territory, and conflates many different issues (classical versus quantum, particles versus fields, spin and fermionic statistics with charge), let me just sketch the answer whose details can be found in pretty much any text on classical electrodynamics.

Without relativity you would want to solve Newton's second law $F=ma$, 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, $f^\mu = m \dot{p^\mu}$, 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 $\gamma$ -- 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 $\gamma$ relative to the Newtonian case.

Again, this is just to direct you to the appropriate text, I am skipping all details because this is written very well and in complete detail in many many places. Look at the classical electrodynamics text of your choice, and you'll find it (I like the discussion in Griffiths book which includes all of the above including all details). If you have more specific questions, I'd be happy to answer. In any event, you don't need to discuss either classical or quantum fields, spin and the Dirac equation, or anything like that (which is why this things are not mentioned in those textbooks).

-
Just a minor quibble: the motion can be in a plane, but there are also more general helical trajectories. – Matt Reece Mar 12 '11 at 19:08
yep, but you can choose reference frame then where they are on a plane. I am also ignoring the issue of time dilation here, which is part of the more complete story. – user566 Mar 12 '11 at 19:16
I don't think you can map the helical solution into the circular one. Rotations preserve ${\bf v} \cdot {\bf B} \neq 0$, and boosts generate an ${\bf E}$ field, so they lead to a different problem. – Matt Reece Mar 12 '11 at 19:28
You are right, I'll correct the answer. I am really wondering what is the use of answering a question whose answer is written in much better form elsewhere. – user566 Mar 12 '11 at 19:36
@Moshe: most of the questions that people can come up with (and most of the question they can't come up with) have already been answered somewhere, so it pretty much renders this whole site invalid, right? :) Joking aside, if there is a precise answer to be found somewhere and you know the reference then just give it in the comment under question (as you did) and everyone should be happy. – Marek Mar 12 '11 at 20:16