A force acts on a particle moving at a relativistic speed in a direction perpendiculal' to its path. How will the particle move? Express the force in terms of the speed and of the radius of curvature of the path.

Answer

The particle moves in a circle. The expression for the force will be formally the same as in Newtonian mechanics.

Explanation

Since the force is perpendicular to the path, the magnitude of the momentum does not change (Fig. 6.13). `abs(p_(1))= abs(p_(2))=p` .Tne maghudae of the momentum.s variation is ) `Deltap=pabs(Deltaphi)`. The normal force is
`F_(n)=(abs(Deltap))/(Deltat)=p(abs(Deltaphi))/(Deltat)=pomega=`
`=momega^(2)r=(mv^(2))/r`

Formally, we have obtained the Fig. 6.13 same result as in the case of the nonrelativistic motion of a particle in a circular path (see 7.2). However, in the relativistic case the expression for the force contains the relativistic mass. Thus, in the particular case the force is perpendicular to the path) the relativistic approach produces formally ihe same result as the classical approach, although in general the second law of Newton in the form F=ma is not valid for relativistic motion.