Irodov 1.353: Each Rod Sees the Other Lag – Amit & Deepali Sinha Foundation

Problem Irodov 1.353

Two rods $AB$ and $A'B'$ lie parallel to each other, one alongside the other, and both have the same proper length $l_0$ . The rod $A'B'$ slides along at a constant velocity $v$ relative to $AB$ , in the direction that runs from its own end $A'$ toward its own end $B'$ , which is also the direction from $A$ toward $B$ . A clock is mounted at each of the four ends, and the two clocks of a rod are synchronized with each other in that rod’s own rest frame: $A$ with $B$ , and separately $A'$ with $B'$ . The instant at which the clock $B'$ comes opposite the clock $A$ is taken as the zero of the time count in each of the two frames. Writing $\beta = v/c$ , determine:

the readings of the clocks $B$ and $B'$ at the moment the two of them are opposite each other
the readings of the clocks $A$ and $A'$ at the moment the two of them are opposite each other

The apparatus in the frame of the rod AB: A'B' slides to the right, from A toward B, with a clock at each of the four ends. The upper rod is drawn contracted. — The apparatus in the frame of the rod $AB$ : $A'B'$ slides to the right, from $A$ toward $B$ , with a clock at each of the four ends. The upper rod is drawn contracted.

Given

proper length $l_0$ each
relative velocity $v$
$\beta = v/c$
counts start when $B'$ passes $A$

Find

(a)

t_B

and

t_{B'}

when

B

meets

B'

, (b)

t_A

and

t_{A'}

when

A

meets

A'

Solution

Step 1The two frames, and the one event that starts both counts

Call $K$ the rest frame of $AB$ and $K'$ the rest frame of $A'B'$ , and call $E$ the event at which $B'$ is opposite $A$ . Each rod keeps its proper length $l_0$ in its own frame and is seen contracted from the other, by the one factor that will run through every step below. Write $l$ for the length either rod has when it is the moving one.

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

(1)

At $E$ the clock $A$ and the clock $B'$ are both set to zero, and with them so are their synchronized partners: $B$ reads zero at the same instant of $K$ , and $A'$ reads zero at the same instant of $K'$ .

Part (a): the instant the clock B' comes opposite the clock B, in the frame of AB. Since E the end B' has traveled the whole proper length l_0 of AB. The dropper joins the two clocks asked for. — Part (a): the instant the clock $B'$ comes opposite the clock $B$ , in the frame of $AB$ . Since $E$ the end $B'$ has traveled the whole proper length $l_0$ of $AB$ . The dropper joins the two clocks asked for.

Step 2Part (a): what the clock

B

reads

Work in $K$ , where $AB$ is at rest and stretches its proper length $l_0$ from $A$ to $B$ , and where $B'$ moves at the steady speed $v$ . Between $E$ and the meeting of $B'$ with $B$ , the end $B'$ therefore covers exactly $l_0$ .

l_0 = v\,t \tag{2}

Dividing by the speed gives the elapsed $K$ time. The clock $B$ is synchronized with $A$ in $K$ and was reading zero at $E$ , so its reading at this meeting is that elapsed $K$ time itself.

t_B = \dfrac{l_0}{v}

(3)

Step 3Part (a): what the clock

B'

reads

The clock $B'$ is present at both events, at $E$ and at its meeting with $B$ , so what it accumulates between them is the proper time along its own path, and that is the $K$ interval shortened by the factor of (1).

t_{B'} = t\sqrt{1-\beta^{2}}

Putting in the elapsed $K$ time from (2) finishes it.

t_{B'} = \dfrac{l_0}{v}\sqrt{1-\beta^{2}}

(4)

Part (b): the instant the clock A' comes opposite the clock A, drawn now in the frame of A'B'. Here A'B' is at rest at full length and AB is the contracted rod, sliding to the left. — Part (b): the instant the clock $A'$ comes opposite the clock $A$ , drawn now in the frame of $A'B'$ . Here $A'B'$ is at rest at full length and $AB$ is the contracted rod, sliding to the left.

Step 4Part (b): what the clock

A

reads

Stay in $K$ for the meeting of $A'$ with $A$ . At the event $E$ the end $B'$ was at $A$ , and $A'$ trails $B'$ by the contracted length (1), so that is the gap $A'$ has to close at speed $v$ .

\begin{align*}l &= v\,t_A, \\ l_0\sqrt{1-\beta^{2}} &= v\,t_A\end{align*}

The clock $A$ has been running in $K$ since it read zero at $E$ , so the elapsed $K$ time this equation delivers is its reading.

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

(5)

Step 5Part (b): what the clock

A'

reads

For the primed clock, move into $K'$ , where $A'B'$ is at rest at its proper length $l_0$ and the rod $AB$ streams past to the left at speed $v$ . At the event $E$ , which is the zero of $K'$ time as well, the end $A$ is level with $B'$ , so $A$ has the whole proper length of $A'B'$ to travel before it reaches $A'$ .

l_0 = v\,t'

The clock $A'$ is synchronized with $B'$ in $K'$ and read zero at $E$ , so the elapsed $K'$ time this equation gives on division by $v$ is its reading.

t_{A'} = \dfrac{l_0}{v}

(6)

Check

Get $t_{B'}$ a second way, from the Lorentz transformation rather than from a proper-time argument. Put the origin of $K$ at $A$ , so $E$ is the event $(x = 0,\ t = 0)$ and the meeting of $B'$ with $B$ is $(x = l_0,\ t = l_0/v)$ . Then $t' = (t - vx/c^{2})/\sqrt{1-\beta^{2}} = (l_0/v)(1-\beta^{2})/\sqrt{1-\beta^{2}} = (l_0/v)\sqrt{1-\beta^{2}}$ , which is (4). As $\beta\to 0$ all four readings collapse to $l_0/v$ , the single Newtonian answer in which the two rods never differ in length and all four clocks agree.

Remark

The two meetings are not two views of one event: they happen at different places and at different times, and their separation is spacelike, since in $K$ they are $l_0$ apart while the distance light could cover in the time between them, $c(t_B - t_A) = (l_0/\beta)(1-\sqrt{1-\beta^{2}})$ , is smaller than $l_0$ for every $\beta < 1$ . That is why the two frames may and do disagree about their order: in $K$ the meeting of $A$ with $A'$ comes first, at $(l_0/v)\sqrt{1-\beta^{2}}$ against $l_0/v$ , while in $K'$ it is the meeting of $B$ with $B'$ that comes first, by the same pair of numbers with the primes exchanged. Neither rod is entitled to call the other’s clocks slow in any absolute sense: each frame compares one clock of the other rod against its own synchronized pair, and those are two different comparisons.