Irodov 1.354: Same Moment, Two Different Answers – Amit & Deepali Sinha Foundation

Problem Irodov 1.354

Two long chains of clocks, $K$ and $K'$ , lie along a common line and move relative to each other at a constant velocity $v$ ; each chain is mutually synchronized in its own rest frame, with a fixed proper spacing $l_0$ between neighboring clocks. At the instant one particular clock of the moving chain, $A'$ , passes directly opposite one particular clock of the stationary chain, $A$ , both $A$ and $A'$ are set to read zero, and that instant is taken as the start of the time count. Sketch the approximate hand positions of every clock in both chains at that moment, first as judged “in terms of the $K$ clocks”, and then as judged “in terms of the $K'$ clocks”.

Two chains of seven clocks. The upper chain, frame K', slides to the right at v; the instant drawn is the one at which A' is level with A. — Two chains of seven clocks. The upper chain, frame $K'$ , slides to the right at $v$ ; the instant drawn is the one at which $A'$ is level with $A$ .

Given

two chains of clocks, each synchronized in its own rest frame
relative velocity $v$
proper spacing $l_0$ between neighbors
$A$ and $A'$ both set to zero as they pass

Find

Draw what every clock reads at that moment, first as

K

judges the moment, then as

K'

judges it

Solution

Step 1Deciding what the question is asking

“At this moment” is not one question but two, and that is the whole content of the problem. A moment is a set of events, one at each clock, and the set is different in the two frames. “In terms of the $K$ clocks” means the set of events that all carry the same reading on the $K$ chain, that is, the slice $t = 0$ of frame $K$ . “In terms of the $K'$ clocks” means the slice $t' = 0$ of frame $K'$ . The two slices agree at $A$ and $A'$ , where they were defined to agree, and nowhere else.

Step 2The transformation that carries a time between the frames

Put the $x$ axis along the chains, with the origin of both frames at the event “ $A'$ passes $A$ ”, and let $K'$ move at $+v$ . Then the Lorentz transformation of the time coordinate is

t' = \gamma\left(t - \frac{vx}{c^{2}}\right), \qquad \gamma = \frac{1}{\sqrt{1 - v^{2}/c^{2}}}, \tag{1}

and its inverse, obtained by exchanging primed for unprimed and reversing the sign of $v$ ,

t = \gamma\left(t' + \frac{vx'}{c^{2}}\right). \tag{2}

Step 3Where the moving clocks are on the slice

t = 0

Take the slice $t = 0$ . Number the $K'$ clocks by $n$ , counting from $A'$ and taking $n$ positive in the direction of motion, so that the $n$ -th one sits permanently at $x'_{n} = n\,l_{0}$ in its own frame, where $l_{0}$ is the proper spacing of neighboring clocks. In $K$ that chain is moving, so its spacing is contracted and

x_{n} = \frac{x'_{n}}{\gamma} = \frac{n\,l_{0}}{\gamma}.

Now read the reading of that clock off (1) with $t = 0$ :

\begin{aligned} t'_{n} &= \gamma\left(0 - \frac{v x_{n}}{c^{2}}\right), \\ &= -\gamma\,\frac{v}{c^{2}}\cdot\frac{n\,l_{0}}{\gamma}. \end{aligned}

The two factors of $\gamma$ cancel, leaving

t'_{n} = -\,n\,\frac{v l_{0}}{c^{2}} .

Writing $\tau = v l_{0}/c^{2}$ for the stagger between neighbors:

t'_{n} = -\,n\tau

(3)

Step 4Reading the answer off, in terms of the

K

clocks

Every $K$ clock reads $0$ , because $t = 0$ is what the slice was defined to be and the $K$ chain is synchronized in $K$ . The $K'$ clocks are staggered by $\tau$ per clock, and the sign in (3) says that the ones ahead in the direction of motion ( $n > 0$ ) read negative, that is, they lag, while the ones at the rear ( $n < 0$ ) read positive and so run ahead. The $K'$ chain is also drawn shorter, by the factor $1/\gamma$ , because that is how $K$ measures a moving chain.

Step 5Now the other slice, in terms of the

K'

clocks

Repeat the argument with the frames exchanged. On the slice $t' = 0$ every $K'$ clock reads $0$ . Number the $K$ clocks by $m$ from $A$ , again positive in the direction in which $K'$ moves, so that the $m$ -th one sits permanently at $x_{m} = m\,l_{0}$ in $K$ . In $K'$ the $K$ chain is the moving one, so its spacing is contracted and

x'_{m} = \frac{x_{m}}{\gamma} = \frac{m\,l_{0}}{\gamma}.

Putting $t' = 0$ into (2),

t_{m} = \gamma\left(0 + \frac{v}{c^{2}}\cdot\frac{m\,l_{0}}{\gamma}\right) = +\,m\tau,

with the same cancellation of $\gamma$ . The sign is now positive, so on this slice the $K$ clocks ahead in the $+x$ direction are the ones that read ahead. That is the same physical rule as before and not a different one: in $K'$ the chain that is moving is $K$ , and it moves in the $-x$ direction, so its rear is on the right, and it is again the rear of the moving chain whose hands are advanced.

In terms of the K' clocks. Now the K' chain reads 0 throughout and is drawn at full length, while K is the contracted chain moving left, its hands stepping forward by τ toward the right, which is its rear. — In terms of the $K'$ clocks. Now the $K'$ chain reads $0$ throughout and is drawn at full length, while $K$ is the contracted chain moving left, its hands stepping forward by $\tau$ toward the right, which is its rear.

\begin{aligned} \text{slice } t = 0 : &\quad t'_{n} = -\,n\tau \\ \text{slice } t' = 0 : &\quad t_{m} = +\,m\tau \end{aligned} \qquad \tau = \frac{v l_{0}}{c^{2}}

(4)

In words, and this is what the two panels draw: whichever chain is the moving one shows its rear clocks ahead, by $v l_0/c^2$ from each clock to the next.

Check

Transform one of the drawn readings back and see whether it returns. Take the $K'$ clock next in front of $A'$ : on the first panel it sits at $x' = l_{0}$ and reads $t' = -\tau$ . Feeding those into (2) gives $t = \gamma(-v l_{0}/c^{2} + v l_{0}/c^{2}) = 0$ , so that event really is on the slice $t = 0$ , as the panel claims. The clock $A'$ itself gives $n = 0$ and $t' = 0$ in both panels, which is the time origin the problem fixes, so neither drawing contradicts the other where they must agree. Dimensionally $[v l_{0}/c^{2}] = (\text{m/s})(\text{m})/(\text{m}^2/\text{s}^2) = \text{s}$ , a time, as a clock reading must be.

Remark

The two panels are mirror images of each other, and the principle of relativity requires that they be: neither chain is the one that is “really” moving, so any statement true of $K'$ as seen from $K$ must hold for $K$ as seen from $K'$ with the direction of motion reversed. Reading the first panel as “the moving clocks are simply out of step” is safe; reading it as “the moving clocks are slow” is not, since the stagger drawn here carries no factor of $\gamma$ and is a fact about simultaneity rather than about rates. The rates only show up when a single clock is watched over an interval, which is a different picture from this one.