What is the relativistic kinetic energy for an object in a circular motion?

Question

Actually I was listening to a about cyclotrons and the lecturer used $$KE=(\gamma -1)mc^2$$

For calculating radius of the particle's path. But can we use that equation as kinetic energy equation. Because in a circular motion the force is independent of velocity and thus $\gamma$ (as said by https://physics.stackexchange.com/a/20922/314854).

So while calculating the kinetic energy for this one we should use: $$F=\gamma ma$$

Or shouldn't we?

(Edit:

If we use the $F=\gamma ma$ for calculating kinetic energy we take $\int \gamma mvdv$ ($adx = vdv$) instead of $\int \gamma^3 mvdv$. My question is whether I should use first method or the second method for calculating kinetic energy?

)

(Edit2:

If we have to use other one then does that mean kinetic energy is not $(\gamma -1)mc²$

)

I am really confused. Please help me in this and tell me whether my assumption is right or not. If not please rectify me.

Answer 1

The derivative of the momentum with respect to time is the force acting on the particle. If the velocity of the particle varies only in deraction, which means that the force is perpendicular to the velocity, we have:$$\frac{d\vec{p}}{dt}=\gamma m\frac{d\vec{v}}{dt} $$

If it is the velocity modulus that is variable, and the direction of the force coincides with the velocity, we have: $$\frac{d\vec{p}}{dt}=\frac{m}{\left(1-\frac{v^{2}}{c^{2}}\right)^{3/2} }\frac{d\vec{v}}{dt} $$

So this is the 1st formula to use.

ps: extracted and translated from: Theoretical Physics,volume II, field theory, by L.Landau, E.Lifchitz