Irodov 1.344: How Fast to Lose Two Percent – Amit & Deepali Sinha Foundation

Problem Irodov 1.344

A clock travels at a constant velocity $v$ relative to the inertial reference frame $K$ . Over an interval that the clocks of $K$ record as $t = 5.0$ s, the traveling clock advances by less than that, falling behind by $\Delta t = 0.10$ s, so that its own reading increases by $t - \Delta t$ while $K$ ’s reading increases by $t$ . Find the speed $v$ . Take the speed of light to be $c$ .

The start of the interval in the frame K: K's synchronized clocks below, the traveling clock above, moving at v. Every reading is zero here, and the gold rings mark the pair side by side. Readings in seconds. — The start of the interval in the frame $K$ : $K$ ’s synchronized clocks below, the traveling clock above, moving at $\vect v$ . Every reading is zero here, and the gold rings mark the pair side by side. Readings in seconds.

The end of the interval in K. The traveling clock has reached K's third clock; every K clock now reads t = 5.0 s, the traveling clock only t-Δ t = 4.9 s. The gold rings mark the side-by-side pair. — The end of the interval in $K$ . The traveling clock has reached $K$ ’s third clock; every $K$ clock now reads $t = 5.0$ s, the traveling clock only $t-\Delta t = 4.9$ s. The gold rings mark the side-by-side pair.

Given

$t = 5.0$ s, recorded in $K$
$\Delta t = 0.10$ s
$c = 3.00\times10^{8}$ m/s

Find

the speed

v

of the clock relative to

K

Solution

Step 1What the traveling clock reads after an interval

t

K

’s time

A clock in motion relative to $K$ advances more slowly than $K$ ’s own clocks by the factor $\sqrt{1-\beta^2}$ , with $\beta = v/c$ . Over an interval that $K$ records as $t$ , the reading of the traveling clock therefore increases by

\tau = t\,\sqrt{1-\beta^2}. \tag{1}

Step 2Turn “slow by

\Delta t

” into a condition on

\beta

Being slow by $\Delta t$ means the traveling clock’s reading falls short of $K$ ’s by that amount, so the two readings at the end of the interval are related by

\tau = t - \Delta t .

Equating this with (1) and dividing through by $t$ leaves the contraction factor against the fractional lag.

\sqrt{1-\beta^2} = 1-\frac{\Delta t}{t}

(2)

Step 3Solve for the speed

Square (2) and expand, so that the fractional lag appears as a plain product.

\begin{aligned} 1-\beta^2 &= \left(1-\frac{\Delta t}{t}\right)^{2} \\[2pt] \beta^2 &= 1-\left(1-2\,\frac{\Delta t}{t}+\frac{\Delta t^2}{t^2}\right) \\[2pt] \beta^2 &= \left(2-\frac{\Delta t}{t}\right)\frac{\Delta t}{t} \end{aligned}

The data give the fractional lag and hence that product, whose square root is $\beta$ .

\frac{\Delta t}{t} = \frac{0.10\ \text{s}}{5.0\ \text{s}} = 0.0200, \qquad (2-0.0200)(0.0200) = 0.03960, \qquad \sqrt{0.03960} = 0.19900

Multiplying that number by $c$ gives the speed. Numerically,

v = 0.199\,c

v = c\,\sqrt{\left(2-\frac{\Delta t}{t}\right)\frac{\Delta t}{t}} = 5.97\times10^{7}\ \text{m/s}

(3)

Check

Run the answer forward and see whether the right-hand figure’s readings come out. With $\beta = 0.199$ the factor is $\sqrt{1-0.03960} = \sqrt{0.96040} = 0.98000$ , so over $t = 5.0$ s the traveling clock advances by $4.90$ s and is behind by $0.10$ s, which is the given $\Delta t$ . The limits are also right: $\Delta t \to 0$ forces $v \to 0$ , a clock at rest in $K$ keeping pace with $K$ ’s clocks, and $\Delta t \to t$ , a clock that does not advance at all, forces $v \to c$ .

Remark

Nothing here says $K$ ’s clocks are the fast ones. The relation is symmetric: in the traveling clock’s own frame it is each of $K$ ’s clocks that advances by only $\sqrt{1-\beta^2}$ per unit of that frame’s time, by exactly the same argument with the roles exchanged. The two statements are not in conflict because they compare different pairs of events. What the two figures actually show is one traveling clock read against two different clocks of $K$ , one at the start and another at the end, and that comparison relies on those two $K$ clocks having been synchronized in $K$ . In the traveling clock’s frame that same pair of $K$ clocks is not synchronized, and putting the offset back in is what turns the asymmetric-looking picture into the symmetric relation.