\- 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.
