Irodov 1.367: The Crew Ages Only Logarithmically – Amit & Deepali Sinha Foundation

Problem Irodov 1.367

A rocket leaves the Earth from rest and travels in a straight line. Its motor is run so that at every moment of the flight the rocket’s acceleration, measured in the inertial frame that momentarily travels with the rocket and in which the rocket is momentarily at rest, has the same value $w' = 10g$ , with $g = 9.8$ m/s $^2$ . The boost lasts $\tau = 1.0$ year as measured by clocks of the geocentric frame, which is treated as inertial throughout, and $v(t)$ denotes the rocket’s speed in that frame at the geocentric time $t$ . The time elapsed on a clock carried by the rocket itself over the same boost is

\tau_0 = \int_0^{\tau}\sqrt{1-\left(\frac{v}{c}\right)^{2}}\;dt .

Taking $c = 3.00\times10^{8}$ m/s, evaluate $\tau_0$ .

The moment the boost ends. The clocks of the geocentric frame K, synchronized and spread along the path, all read τ; the rocket's single clock, passing the middle one, reads the smaller τ_0. Its speed is v. — The moment the boost ends. The clocks of the geocentric frame $K$ , synchronized and spread along the path, all read $\tau$ ; the rocket’s single clock, passing the middle one, reads the smaller $\tau_0$ . Its speed is $\vect v$ .

Given

$w' = 10g$
$g = 9.8$ m/s $^2$
$\tau = 1.0$ year $=3.156\times10^{7}$ s
$c = 3.00\times10^{8}$ m/s

Find

the boost time

\tau_0

on the rocket’s own clock

Solution

Step 1The rocket’s speed as a function of geocentric time

The integral to be done has $v(t)$ inside it, so the first job is to get $v(t)$ out of the statement about the co-moving frame. Take the moment when the geocentric frame $K$ measures the speed $v$ , and let $K'$ be the frame moving at exactly that speed, in which the rocket is momentarily at rest. In a short interval the rocket picks up the speed $du'$ in $K'$ ; velocity composition converts it into the speed $K$ sees, and the expansion to first order in $du'$ strips it down.

\begin{aligned} v+dv &= \frac{v+du'}{1+\dfrac{v\,du'}{c^2}} = \left(v+du'\right)\left(1-\frac{v\,du'}{c^2}\right) \\[3pt] &= v+du'\left(1-\frac{v^2}{c^2}\right) \end{aligned}

The two moments are events at one place in $K'$ , so $K$ records the dilated interval between them. Dividing the velocity increment by the time increment turns the stated proper acceleration into an equation of motion in $K$ .

dt = \frac{dt'}{\sqrt{1-\dfrac{v^2}{c^2}}}, \qquad \frac{dv}{dt} = \frac{du'}{dt'}\left(1-\frac{v^2}{c^2}\right)\sqrt{1-\frac{v^2}{c^2}}

The ratio $du'/dt'$ is the acceleration measured in the co-moving frame, which is the stated constant $w'$ , and the two roots combine into one power.

\frac{dv}{dt} = w'\left(1-\frac{v^2}{c^2}\right)^{3/2}

That equation separates. Integrating it from the launch, where the rocket is at rest, uses the fact that $v/\sqrt{1-v^2/c^2}$ differentiates to exactly the integrand on the left.

\begin{aligned} \int_0^{v}\left(1-\frac{v_1^2}{c^2}\right)^{-3/2}dv_1 &= \int_0^{t}w'\,dt_1 \\[3pt] \frac{v}{\sqrt{1-\dfrac{v^2}{c^2}}} &= w't \end{aligned} \tag{1}

Squaring and collecting the terms in $v^2$ solves that for the speed.

v = \frac{w't}{\sqrt{1+\left(\dfrac{w't}{c}\right)^{2}}}

(2)

Step 2The dilation factor along the flight

The integrand of the given formula is the square root that (1) was built on, so it does not have to be assembled from $v$ again: rearranging (2) shows what the root is worth at every instant. Square the boxed speed, divide by $c^2$ and subtract from $1$ .

\begin{aligned} 1-\frac{v^2}{c^2} &= 1-\frac{\left(w't/c\right)^{2}}{1+\left(w't/c\right)^{2}} \\[3pt] &= \frac{1}{1+\left(\dfrac{w't}{c}\right)^{2}} \end{aligned}

\sqrt{1-\frac{v^2}{c^2}} = \frac{1}{\sqrt{1+\left(\dfrac{w't}{c}\right)^{2}}}

(3)

The dilation factor against w't/c, over a boost ending at w'τ/c=10.3. The rocket's elapsed time is the shaded area under the curve, in units of c/w'. The factor starts at 1 and falls to a tenth. — The dilation factor (3) against $w't/c$ , over a boost ending at $w'\tau/c=10.3$ . The rocket’s elapsed time is the shaded area under the curve, in units of $c/w'$ . The factor starts at $1$ and falls to a tenth.

Step 3Integrate the dilation factor over the boost

Putting (3) into the definition of $\tau_0$ gives an integral in $t$ alone. The substitution $x = w't/c$ , for which $dx = w'\,dt/c$ , makes it dimensionless and carries the limits with it.

\begin{aligned} \tau_0 &= \int_0^{\tau}\frac{dt}{\sqrt{1+\left(\dfrac{w't}{c}\right)^{2}}} \\[3pt] &= \frac{c}{w'}\int_0^{a}\frac{dx}{\sqrt{1+x^2}}, \qquad a = \frac{w'\tau}{c} \end{aligned}

The antiderivative is a logarithm, which the following differentiation confirms before it is used.

\frac{d}{dx}\ln\left(x+\sqrt{1+x^2}\right) = \frac{1+\dfrac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}} = \frac{1}{\sqrt{1+x^2}}

Evaluating that antiderivative between the limits, and noting that it vanishes at the lower one, leaves the boost time in closed form.

\tau_0 = \frac{c}{w'} \ln\left[\frac{w'\tau}{c}+\sqrt{1+\left(\frac{w'\tau}{c}\right)^{2}}\right]

(4)

Step 4Put the numbers in

Two quantities are needed: the time scale $c/w'$ that stands outside the logarithm, and the dimensionless $a$ inside it. The acceleration is $w' = 10g$ , which is $98$ m/s $^2$ .

\frac{c}{w'} = \frac{3.00\times10^{8}}{98} = 3.06\times10^{6}\ \mathrm{s}, \qquad a = \frac{98\times3.156\times10^{7}}{3.00\times10^{8}} = 10.31

The logarithm’s argument follows, and with it the answer, converted into months by dividing by one twelfth of a year, $2.63\times10^{6}$ s.

a+\sqrt{1+a^2} = 10.31+10.36 = 20.67, \qquad \ln 20.67 = 3.029

Numerically,

\tau_0 = 3.06\times10^{6}\times3.029

\begin{aligned} \tau_0 &= \frac{c}{w'}\ln\left(a+\sqrt{1+a^2}\right) \\ &= 9.27\times10^{6}\ \mathrm{s} = 3.5\ \text{months} \end{aligned}

(5)

Check

The prefactor $c/w'$ is a speed over an acceleration and so is a time, and the logarithm’s argument is dimensionless, so the boxed expression is a time. The result must be smaller than $\tau$ , since the integrand never exceeds $1$ , and it is: the ratio $\tau_0/\tau$ comes to $0.29$ . The non-relativistic limit also comes out right. For $a \ll 1$ the logarithm $\ln\left(a+\sqrt{1+a^2}\right)$ tends to $a$ , so $\tau_0 \to \left(c/w'\right)a$ , which is $\tau$ itself: a slow rocket’s crew ages exactly as the Earth does.

Remark

Contrast how the two clocks respond to a longer burn. Once the rocket is fast the distance it covers grows in proportion to the geocentric time $\tau$ , while the boxed $\tau_0$ grows only like $\ln\left(2w'\tau/c\right)$ , because the area under the curve in the figure is being added in ever thinner slices. Distance and geocentric time are nearly equal in light units at these speeds, so the crew’s aging goes like the logarithm of the distance crossed. Put a number on it. Reaching the galactic center, some $30\,000$ light-years away, needs $a \approx w'l/c^{2}$ , which is about $3.1\times10^{5}$ ; its logarithm $\ln 2a$ is $13.3$ , so $\tau_0$ is roughly $0.097\times13.3$ years, about $1.3$ years of shipboard time, against the $30\,000$ years that pass on Earth. This is not a way of outrunning light: the ship still arrives after the light does. It is the same time dilation as in a decaying muon, read from the traveling clock’s side.