Transformation of Acceleration

Consider a particle that moves with constant acceleration ${\bf a}'=a_0\,{\bf e}_x$ in its instantaneous rest frame, $S'$. Let the particle be located at the origin of $S'$ at time $t'=0$. It follows that a very short time, $\delta t$, later, the particle's spacetime coordinates in $S'$ are

$$x' =\frac{1}{2}\,a_0\,\delta t^{\,2},$$ (3.139)

$$y' =0,$$ (3.140)

$$z' =0,$$ (3.141)

$$t' =\delta t.$$ (3.142)

Consider a second frame of reference, $S$, that moves with velocity - $v\,{\bf e}_x$ with respect to $S'$, and is also in a standard configuration with respect to $S'$. Thus, the particle's instantaneous velocity in $S$ is ${\bf v} = v\,{\bf e}_x$. In $S$, the particle moves from the origin at time $t=0$, to a point whose spacetime coordinates are

$$x = \gamma\,(x'+v\,t')=\gamma\left(\frac{1}{2}\,a_0\,\delta t^{\,2} + v\,\delta t\right),$$ (3.143)

$$y =0,$$ (3.144)

$$z =0,$$ (3.145)

$$t = \gamma\left(t' + \frac{v\,x'}{c^2}\right)=\gamma\left(\delta t + \frac{v\,a_0\,\delta t^{\,2}}{2\,c^2}\right),$$ (3.146)

where $\gamma=(1-v^2/c^2)^{-1/2}$, a very short time later. Here, use has been made of Equations (3.104)–(3.107), as well as Equations (3.139)–(3.142). If ${\bf a}= a\,{\bf e}_x$ is the particle's instantaneous acceleration in $S'$ then we expect the relation

$$x = v\,t+ \frac{1}{2}\,a\,t^{\,2}$$ (3.147)

to hold for a short time interval (i.e., in the limit $\delta t\rightarrow 0$). It follows from Equations (3.143) and (3.146) that

$$\gamma\left(\frac{1}{2}\,a_0\,\delta t^{\,2} + v\,\delta t\right)... ...2}\,a\,\gamma^2\left(\delta t + \frac{v\,a_0\,\delta t^{\,2}}{2\,c^2}\right)^2.$$ (3.148)

Note that the terms in the previous equation that are first order in $\delta t$ cancel one another. Equating the terms that are second order in $\delta t$, we obtain

$$\gamma\,a_0 = \gamma\,\frac{v^2}{c^2}\,a_0 + \gamma^2\,a,$$ (3.149)

or

$$a = \frac{a_0\,(1-v^2/c^2)}{\gamma},$$ (3.150)

which reduces to

$$a = \frac{a_0}{\gamma^3}.$$ (3.151)

Thus, we conclude that the particle's instantaneous acceleration in a frame of reference in which it has a finite speed is always less that that in its instantaneous rest frame.

Given that, by definition,

$$\frac{dv}{dt} =a,$$ (3.152)

the particle's equation of motion in $S$, which is assumed to be an inertial frame, is

$$\frac{dv}{dt} = \frac{a_0}{\gamma^3} = a_0\left(1-\frac{v^2}{c^2}\right)^{3/2}.$$ (3.153)

If $v=0$ at $t=0$ then the previous equation can be integrated to give

$$\frac{v}{(1-v^2/c^2)^{1/2}} = a_0\,t,$$ (3.154)

or

$$v = \frac{a_0\,t}{(1+a_0^{\,2}\,t^2/c^2)^{1/2}}.$$ (3.155)

Thus, as seen by an observer at rest in frame $S$, our particle initially (i.e., for $t\ll c/a_0$) accelerates such that

$$v\simeq a_0\,t,$$ (3.156)

in accordance with Newtonian dynamics. However, when the speed of the particle becomes comparable with the speed of light in vacuum, the linear increase in speed with time specified in the previous equation breaks down, and, instead,

$$v \rightarrow c~~ ~~t\rightarrow\infty.$$ (3.157)

Thus, despite the particle's constant acceleration, $a_0$, in its instantaneous rest frame, the particle never appears to move faster than the speed of light to a stationary observer.