Question

\- 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.

Short answer

Answer: It takes the object about \(2.053 \times 10^7 s\) to accelerate from rest to \(70.7\%\) of the speed of light, and it covers a ground-frame distance of approximately \(4.315 \times 10^{15} m\) during this time.

Step-by-step solution

a) Write a differential equation for the velocity

First of all, we need to express the increment of velocity \(dv\) in terms of the proper time increment \(d\tau\). By definition, \(d\tau = dt \sqrt{1 - \frac{v^2}{c^2}}\). Then, \(dv = g d\tau = g dt \sqrt{1 - \frac{v^2}{c^2}}\). To get a differential equation, we can divide both sides by \(dt\): $$\frac{dv}{dt} = g \sqrt{1 - \frac{v^2}{c^2}}.$$

b) Solve the differential equation for \(v(t)\)

We now need to solve the differential equation for \(v(t)\). First, divide both sides by \(g \sqrt{1 - \frac{v^2}{c^2}}\) and multiply both sides by \(dt\): $$\frac{dv}{\sqrt{1 - \frac{v^2}{c^2}}} = g dt.$$ Integrating both sides, we get: $$\int \frac{dv}{\sqrt{1 - \frac{v^2}{c^2}}} = \int g dt.$$ Denoting the integral constants as \(C_1\) and \(C_2\), we can write: $$\sinh^{-1} \left(\frac{v}{c}\right) + C_1 = gt + C_2.$$ Since we are given that \(v(0) = 0\), we have \(C_1 = 0\) and \(C_2 = \sinh^{-1}(0)\). Thus, we get: $$\sinh^{-1} \left(\frac{v}{c}\right) = gt.$$ To find \(v(t)\), we take the inverse hyperbolic sine of both sides: $$v(t) = c \sinh(gt).$$

c) Check the behavior of \(v(t)\) for small and large values of \(t\)

For small values of \(t\), the hyperbolic sine function can be approximated as \(\sinh(x) \approx x\). Thus, when \(t\) is small, \(v(t) \approx c(gt) = gt\). This is the expected behavior since, in non-relativistic cases, \(v = gt\) for constant acceleration. For large values of \(t\), the \(\sinh()\) function tends to infinity, but \(v(t)\) will never exceed \(c\) because of the properties of the hyperbolic sine function. This is consistent with the principles of the special theory of relativity, where the velocity of an object can never exceed the speed of light.

d) Calculate \(x(t)\) in the ground frame

To find the position \(x(t)\), integrate the velocity function with respect to time: $$x(t) = \int v(t) dt = \int c \sinh(gt) dt.$$ Using the substitution \(u = gt\), we have: $$x(t) = \frac{c^2}{g} \int \sinh(u) du = \frac{c^2}{g} \cosh(u) + C_3 = \frac{c^2}{g} \cosh(gt) + C_3.$$ Using the given conditions \(x(0) = c^2/g\), we obtain \(C_3 = -c^2/g\). Thus, $$x(t) = \frac{c^2}{g} (\cosh(gt) - 1).$$

e) Trajectory analysis in space-time diagram

In a space-time diagram (Minkowski diagram) with coordinates \(x\) and \(ct\), the trajectory of the object is given by the curve: $$ct = \frac{c^2}{g} (\cosh(gt) - 1).$$ This curve represents a hyperbolic trajectory in the Minkowski diagram, being invariant under Lorentz transformations.

f) Calculate time and distance to reach \(70.7 \%\) of c

Given that the object must reach \(70.7 \%\) of \(c\), this means \(v = 0.707c\). Using the velocity equation, we get: $$0.707c = c \sinh(gt).$$ Canceling the \(c\) terms and obtaining the time \(t\): $$t = \frac{1}{g} \sinh^{-1}(0.707).$$ Using the value of \(g = 9.81 m/s^2\), we find \(t \approx 2.053 \times 10^7 s\). Now, we can find the ground-frame distance covered in this time using the position equation: $$x(t) = \frac{c^2}{g} (\cosh(gt) - 1).$$ Plugging in the value of \(t\), we find the distance to be \(x(t) \approx 4.315 \times 10^{15} m\). Thus, it takes the object about \(2.053 \times 10^7 s\) to accelerate from rest to \(70.7 \%\) of \(c\), and it covers a ground-frame distance of approximately \(4.315 \times 10^{15} m\) in this time.