Question 6.1: Study the one-dimensional motion of a relativistic particle ......

This example might describe a charged particle being accelerated by a uniform electric field. In the one-dimensional case, Eq (6.99) takes the form

\frac{dp}{dt} \equiv \frac{d}{dt}\left(\frac{mv}{\sqrt{{1-v^{2}}/{c^{2}}} } \right)=F.   (6.99)

\frac{d}{dt} \left(\frac{m\dot{x} }{\sqrt{{1-\dot{x}^{2}}/{c^{2}}} } \right)=F.  (6.114)

An immediate integration yields

\frac{\dot{x} }{\sqrt{{1-\dot{x}^{2}}/{c^{2}}}}= at+b , a= \frac{F}{m},  (6.115)

where b is an arbitrary constant. Solving this equation for \dot{x} we are led to

\dot{x}= \frac{at+b }{\sqrt{{1+\left( at+b\right)^{2}}/{c^{2}}}} ,   (6.116)

which shows that, differently from the Newtonian problem, the particle’s acceleration is not constant.^{12} After another elementary integration, we find

x − d = \frac{c^{2}}{a} \sqrt{1+ \left(\frac{at+b}{c}\right)^{2}}.   (6.117)

Let us consider the particularly simple case in which the particle starts from the origin at rest – that is, x(0) = 0 and \dot{x}(0) = 0. In such circumstances b = 0 and d = −c²/a, whence

x= \frac{c^{2}}{a} \left[-1 + \sqrt{1+ \left(\frac{at}{c}\right)^{2}}\right] .   (6.118)

This last equation can be cast in the more elegant form

\left(x+\frac{c^{2}}{a}\right)^{2} – c^{2}t^{2}= \frac{c^{4}}{a^{2}},  (6.119)

which is the equation of a hyperbola in the x^{0}x plane (see Fig. 6.5).

^{12}It is the proper acceleration of the particle that is constant and equal to F/m (Problem 6.13).