Acceleration of a circulatory particle

[ehasan]

a particle moves close to the speed of light, say u, around a circular path. what is acceleration according to particle's own reference frame?

 

[JesseM]

The acceleration in the particle's own instantaneous inertial rest frame (also called the 'comoving frame') is called the "proper acceleration", it also corresponds to the G-force that would be experienced by the particle (the reading on an accelerometer moving along with the particle). If I'm reading it correctly, equation 30.16 in this book says that if a particle is moving in a circle in some inertial frame, and its coordinate acceleration in that frame is a while its relativistic gamma-factor is \gamma_u = \frac{1}{\sqrt{1 - u^2/c^2}}, then its proper acceleration \alpha would be given by:

\alpha = \gamma_u^2 * a

Someone correct me if I'm misunderstanding the meaning of that equation. But assuming that's correct, then the coordinate acceleration a in the inertial frame where the particle is moving in a circle of radius r with tangential velocity u would just the standard centripetal acceleration, a = \frac{u^2}{r}, so putting that together with the above, the proper acceleration would be \alpha = \frac{u^2}{r * (1 - u^2/c^2)}

Of course this assumes we are talking about special relativity where some nongravitational force is causing the particle to move in a circle, in the curved spacetime of general relativity a particle can be moving in a circular orbit and experiencing no G-forces because it's in free fall (see the equivalence principle).

 

[ehasan]

thnaks