# Relativistic cyclotron frequency

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

$\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m}$?

How can you express the centripetal force in terms of momentum (with other terms)?
How is the relativistic momentum?

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

$\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m}$?
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.