Relativistic cyclotron frequency

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jdstokes
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Since the acceleration is transverse to the velocity, should we consider the transverse mass in the formula [itex]mv^2/r[/itex] ie

[itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m}[/itex]?
 
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How can you express the centripetal force in terms of momentum (with other terms)?
How is the relativistic momentum?
 
jdstokes said:
Since the acceleration is transverse to the velocity, should we consider the transverse mass in the formula [itex]mv^2/r[/itex] ie

[itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m}[/itex]?
Yes. But just to be clear, using your symbols, the transverse mass = gamma*m.

Pete
 
It's clearer in terms of momentum pv/r=qvB-->p=qBr.