Irodov 1.348: One Clock Times a Contracted Gap – Amit & Deepali Sinha Foundation

Problem Irodov 1.348

Two particles travel one behind the other along one straight line in a laboratory frame, both with the same speed $v = (3/4)c$ , and both strike a target that is at rest in that laboratory. The two impacts happen at the target a time $\Delta t = 50$ ns apart, as measured in the laboratory. Find the proper distance between the two particles before they hit, that is, their separation as measured in the inertial frame in which the two particles are at rest.

The laboratory frame at the instant the leading particle 2 reaches the target: the trailing particle 1 is still a laboratory distance l short of it, and the single laboratory clock standing at the target reads zero. — The laboratory frame at the instant the leading particle $2$ reaches the target: the trailing particle $1$ is still a laboratory distance $l$ short of it, and the single laboratory clock standing at the target reads zero.

Given

$v = (3/4)c$
$\Delta t = 50$ ns
$\beta = v/c$
$c = 3.00\times 10^{8}$ m/s

Find

the proper distance

l_0

between the two particles

Solution

Step 1The laboratory gap, read off one laboratory clock

World lines in the laboratory frame. The target's world line is the ct axis, and the two impacts are the upper dots on it, c Δ t apart. Both particles carry the slope β = 3/4, their constant gap l dimensioned below the axis. — World lines in the laboratory frame. The target’s world line is the $ct$ axis, and the two impacts are the upper dots on it, $c\,\Delta t$ apart. Both particles carry the slope $\beta = 3/4$ , their constant gap $l$ dimensioned below the axis.

The target does not move in the laboratory, so both impacts happen at one and the same laboratory point and one clock standing there times them. Put the origin of the laboratory time at the first impact and let $x_1$ and $x_2$ be the laboratory coordinates of the trailing and the leading particle, both advancing at the same speed $v$ .

\begin{align*}x_2(t) &= v t, \\ x_1(t) &= v t - l,\end{align*}

with $l$ the laboratory separation. The second impact is the instant $x_1 = 0$ , which happens at $t = \Delta t$ .

0 = v\,\Delta t - l

l = v\,\Delta t

(1)

Step 2The pair of particles is a rod, and the laboratory sees it short

The two particles hold a fixed distance apart in their own rest frame too, so they behave like the two ends of one rigid rod of proper length $l_0$ flying at speed $v$ . The laboratory fixes both particle positions at one laboratory instant, which is precisely the measurement that returns the contracted length.

l = l_0\sqrt{1-\beta^{2}}

l_0 = \dfrac{l}{\sqrt{1-\beta^{2}}}

(2)

Step 3Put the two gaps together and evaluate

The laboratory gap (1) is what (2) needs, so one substitution finishes the algebra. Putting in the data, with $\beta = 0.750$ and $\sqrt{1-\beta^{2}} = 0.6614$ ,

l_0 = \frac{(0.750)(3.00\times 10^{8}\ \mathrm{m/s}) (50\times 10^{-9}\ \mathrm{s})}{0.6614} = \frac{11.25\ \mathrm{m}}{0.6614}

l_0 = \dfrac{v\,\Delta t}{\sqrt{1-(v/c)^{2}}} = 17\ \mathrm{m}

(3)

Check

Count the same interval on the particles’ own clocks. The two impacts happen at one laboratory point, so $\Delta t$ is the proper time between them and the particles’ frame records the dilated interval $\Delta t/\sqrt{1-\beta^{2}} = 75.6$ ns. In that frame the target is what moves, at speed $v$ , and it runs from one particle to the other in that time, covering $v(75.6\ \mathrm{ns}) = 17$ m, which is the separation just found. The units also check: a speed times a time is a length.

Remark

The two impacts are events at the same place in the laboratory, so the laboratory interval $\Delta t$ between them is the shortest any inertial frame can report, and it is the only one a single clock can measure. Their separation is timelike, which is why every inertial frame agrees that the leading particle landed first, even though frames disagree about how far apart the particles were and about how long the wait was.