# Relativistic cyclotron frequency

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

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

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

pmb_phy
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

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