Irodov 1.350: Two Rods, One Clock, One Gap – Amit & Deepali Sinha Foundation

Problem Irodov 1.350

Two rods, identical when at rest and each of proper length $l_0$ , move toward each other parallel to a common straight axis, passing each other. Work in the rest frame of one of the rods. At one instant the right-hand ends of the two rods coincide; a time $\Delta t$ later, as measured in that frame, the left-hand ends of the two rods coincide. Find the velocity $v$ of one rod relative to the other.

Given

proper length $l_0$ for each rod
time interval $\Delta t$ between the two coincidences, measured in the rest frame of rod 1

Find

Find

v

, the velocity of one rod relative to the other

Solution

Step 1Sit on one rod, where only the other one is short

Everything is measured in the rest frame of rod 1, so in that frame rod 1 keeps its proper length while rod 2 sweeps past at the relative speed $v$ and is contracted. Writing $\beta = v/c$ , the two drawn lengths are

l_1 = l_0, \qquad l_2 = l = l_0\sqrt{1-\beta^2} .

First coincidence: the right ends line up, the clock on rod 1 reads t_1. Rod 2 is drawn contracted to l = l_0sqrt(1-β^2). — First coincidence: the right ends line up, the clock on rod 1 reads $t_1$ . Rod 2 is drawn contracted to $l = l_0\sqrt{1-\beta^2}$ .

Second coincidence, a time Δ t = t_2-t_1 later in that frame: rod 2 has slid a distance l_0 - l to the left and the left ends now line up. — Second coincidence, a time $\Delta t = t_2-t_1$ later in that frame: rod 2 has slid a distance $l_0 - l$ to the left and the left ends now line up.

Step 2Turn the two coincidences into one distance

Put the moving rod’s right end level with the stationary rod’s right end: that is the first event. Rod 2 then keeps sliding, and its left end reaches the stationary rod’s left end when rod 2 has advanced by exactly the difference of the two lengths. The gap it must close is therefore

\Delta x = l_1 - l_2 = l_0 - l_0\sqrt{1-\beta^2} .

Rod 2 covers this at the constant speed $v$ in the stated time, so

v\,\Delta t = l_0\left(1-\sqrt{1-\beta^2}\right)

(1)

Step 3Solve for the speed hidden on both sides

Equation (1) carries $v$ both openly and inside the radical, so isolate the radical first:

\sqrt{1-\frac{v^2}{c^2}} = 1 - \frac{v\,\Delta t}{l_0} .

Squaring removes the root, and the $1$ on each side cancels:

\begin{aligned} 1-\frac{v^2}{c^2} &= 1 - \frac{2v\,\Delta t}{l_0} + \frac{v^2\,\Delta t^2}{l_0^2} \\[2pt] \frac{v^2}{c^2} + \frac{v^2\,\Delta t^2}{l_0^2} &= \frac{2v\,\Delta t}{l_0} . \end{aligned}

Every term carries one factor of $v$ at least, and $v \ne 0$ , so divide through by $v$ and collect it:

v\left(\frac{1}{c^2} + \frac{\Delta t^2}{l_0^2}\right) = \frac{2\,\Delta t}{l_0} . \tag{2}

Step 4Read off the relative velocity

Divide (2) by the coefficient of $v$ :

v = \frac{2\,\Delta t/l_0}{\dfrac{1}{c^2}+\dfrac{\Delta t^2}{l_0^2}} .

Multiplying numerator and denominator by $l_0^2/\Delta t^2$ clears the small quantities out of the denominator and gives the relative velocity:

v = \frac{2l_0/\Delta t}{1+\left(\dfrac{l_0}{c\,\Delta t}\right)^2}

(3)

Check

Write $k = l_0/(c\,\Delta t)$ , so that the answer reads $v/c = 2k/(1+k^2)$ . Since $(k-1)^2 \ge 0$ gives $1+k^2 \ge 2k$ , this is at most $1$ for every positive $k$ , with equality only at $k = 1$ : the formula can never return a superluminal relative velocity, whatever $\Delta t$ is fed to it. A worked case confirms the algebra: at $v = 0.6c$ the contraction factor is $0.8$ , so (1) gives $\Delta t = 0.2\,l_0/(0.6c) = l_0/(3c)$ , hence $k = 3$ and $2k/(1+k^2) = 6/10 = 0.6$ , returning the speed it started from.

Remark

The non-relativistic limit is the interesting one, because there is nothing left of the effect: as $\beta \to 0$ the right-hand side of (1) vanishes like $\beta^2$ , so $\Delta t \to 0$ and the two coincidences become simultaneous. In Galilean physics both rods are $l_0$ long in every frame and the ends meet in pairs at one instant, so the whole of $\Delta t$ is contraction and nothing else.