Irodov 1.345: Two Timings Fix One Proper Length – Amit & Deepali Sinha Foundation

Problem Irodov 1.345

A straight rod travels at a constant velocity directed along its own length, passing a mark that is a fixed point of the reference frame $K$ . Timed in $K$ , the passage of the rod over the mark, from the arrival of the rod’s leading end at the mark to the departure of its trailing end, lasts $\Delta t = 20$ ns. Timed instead in the reference frame $K'$ attached to the rod, in which the rod is at rest and the mark travels, the same passage of the mark from one end of the rod to the other lasts $\Delta t' = 25$ ns. Find the proper length $l_0$ of the rod, that is, its length measured in $K'$ .

Frame K at the instant the rod's leading end reaches the mark, which is fixed in K and carries the clock timing the passage, set to zero here. The rod slides right at v, and the length l beneath it is contracted. — Frame $K$ at the instant the rod’s leading end reaches the mark, which is fixed in $K$ and carries the clock timing the passage, set to zero here. The rod slides right at $v$ , and the length $l$ beneath it is contracted.

Given

$\Delta t = 20$ ns, timed in $K$
$\Delta t' = 25$ ns, timed in $K'$
velocity along the rod
$c = 3.00\times10^{8}$ m/s

Find

the proper length

l_0

of the rod

Solution

Step 1What the timing in

K

actually measures

In $K$ the mark is a single fixed point and the rod is the thing that moves, so the passage begins when the leading end arrives at the mark and ends when the trailing end leaves it. Between those two events the rod advances by exactly its own length as measured in $K$ , and that length is the contracted one. Writing $\beta = v/c$ , that length is

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

and the same length is covered at the constant speed $v$ in the stated time,

l = v\,\Delta t .

Equating the two readings of $l$ :

v\,\Delta t = l_0\sqrt{1-\beta^2}

(1)

The same event in the rod's own frame K'. The rod is at rest at its proper length l_0, dimensioned beneath it, while the mark, its clock still reading zero, sweeps left at v and has just reached the leading end. — The same event in the rod’s own frame $K'$ . The rod is at rest at its proper length $l_0$ , dimensioned beneath it, while the mark, its clock still reading zero, sweeps left at $v$ and has just reached the leading end.

Step 2What the timing in the rod’s frame measures

Move to $K'$ , where the rod is at rest and has its proper length $l_0$ , and the mark is the traveler. The mark enters at one end of the rod and leaves at the other, so it covers the whole proper length at speed $v$ in the time $\Delta t'$ :

v\,\Delta t' = l_0

(2)

Step 3Eliminate the speed, which was never asked for

The speed $v$ appears in both boxed relations and is not wanted, so divide (1) by (2) and let $v$ and $l_0$ cancel together:

\begin{aligned} \frac{v\,\Delta t}{v\,\Delta t'} &= \frac{l_0\sqrt{1-\beta^2}}{l_0}, \\[2pt] \frac{\Delta t}{\Delta t'} &= \sqrt{1-\beta^2}. \end{aligned}

Squaring and solving for $\beta$ turns the pair of timings straight into the velocity:

\beta^2 = 1-\left(\frac{\Delta t}{\Delta t'}\right)^{2},

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

(3)

Step 4The proper length from the rod-frame timing

Equation (2) already gives the proper length as $v\,\Delta t'$ , so substituting the speed (3) finishes the algebra. The numbers close exactly, the ratio of the two timings being the contraction factor itself:

\begin{aligned} \frac{\Delta t}{\Delta t'} &= \frac{20\ \text{ns}}{25\ \text{ns}} = 0.800, \\[2pt] \sqrt{1-(0.800)^{2}} &= \sqrt{0.360} = 0.600, \\[2pt] v &= 0.600\,c = 1.80\times10^{8}\ \text{m/s}, \\[2pt] c\,\Delta t' &= \left(3.00\times10^{8}\ \text{m/s}\right) \left(25.0\times10^{-9}\ \text{s}\right) = 7.50\ \text{m}. \end{aligned}

With the data,

l_0 = 7.50\ \text{m}\times 0.600

l_0 = c\,\Delta t' \sqrt{1-\left(\dfrac{\Delta t}{\Delta t'}\right)^{2}} = 4.5\ \text{m}

(4)

Check

Go back to $K$ and see whether the first figure’s dimension comes out right. The contracted length is $l = l_0\sqrt{1-\beta^2} = (4.5\ \text{m}) (0.800) = 3.6$ m, and the rod covers that at $1.80\times10^{8}$ m/s in $3.6/(1.80\times10^{8}) = 2.0\times10^{-8}$ s, which is the $20$ ns the problem gives. Dimensionally the answer is a speed times a time times a pure number, a length. The degenerate case also behaves: if the two timings were equal the contraction factor would be $1$ , forcing $v = 0$ , and (2) would then return $l_0 = 0$ , so equal timings describe no rod passing anything.

Remark

The same two numbers can be read as a time dilation instead, and they must give the same speed. In $K$ both events, the leading end reaching the mark and the trailing end leaving it, happen at one and the same place, the mark, so $\Delta t$ is the proper time between them, recorded by the single clock drawn at the mark in the first figure. The rod frame $K'$ , in which those two events happen at opposite ends of the rod, therefore assigns them the dilated separation $\Delta t' = \gamma\,\Delta t$ , giving $\gamma = 25/20 = 1.25$ and $\beta = \sqrt{1-1/\gamma^{2}} = 0.60$ , exactly the speed (3) returned. Length contraction and time dilation are not two independent facts to be applied in turn here: one set of data, read either way, has to yield one velocity.