- 35.87 Although it deals with inertial reference frames, the special theory of relativity describes accelerating objects without difficulty. Of course, uniform acceleration no longer means $d v / d t=g,$ where $g$ is a constant, since that would have $v$ exceeding $c$ in a finite time. Rather, it means that the acceleration experienced by the moving body is constant:
In each increment of the body's own proper time, $d \tau,$ the body experiences a velocity increment $d v=g d \tau$ as measured in the inertial frame in which the body is momentarily at rest. (As it accelerates, the body encounters a sequence of such frames, each moving with respect to the others.) Given this interpretation:
a) Write a differential equation for the velocity $v$ of the body, moving in one spatial dimension, as measured in the inertial frame in which the body was initially at rest (the "ground frame"). You can simplify your equation by remembering that squares and higher powers of differentials can be neglected.
b) Solve the equation from part (a) for $v(t),$ where both $v$ and $t$ are measured in the ground frame.
c) Verify that $v(t)$ behaves appropriately for small and large values of $t$.
d) Calculate the position of the body, $x(t),$ as measured in the ground frame. For convenience, assume that the body is at rest at ground-frame time $t=0$ and at ground-frame position $x=c^{2} / g$.
e) Identify the trajectory of the body on a space-time diagram (a Minkowski diagram, for Hermann Minkowski) with coordinates $x$ and $c t,$ as measured in the ground frame.
f) For $g=9.81 \mathrm{~m} / \mathrm{s}^{2},$ calculate how much time it takes the body to accelerate from rest to $70.7 \%$ of $c$, measured in the ground frame, and how much ground-frame distance the body covers in this time.