Irodov 1.352: Two Marks Made at Two Different Times – Amit & Deepali Sinha Foundation

Problem Irodov 1.352

A rod $AB$ lies along the $x$ axis of a reference frame $K$ and slides along that axis at a constant velocity $v$ in the positive $x$ direction. Point $A$ is the rod’s leading (forward) end and point $B$ its trailing (rear) end.

An observer in $K$ records the coordinate $x_A$ of point $A$ at the moment $t_A$ , and separately records the coordinate $x_B$ of point $B$ at the moment $t_B$ . From $v$ and these four recorded values, find the rod’s proper length $l_0$ (its length as measured in its own rest frame).
Find what time interval $t_A - t_B$ should separate the two markings of the coordinates of the rod’s ends in the frame $K$ in order for the bare difference of the recorded coordinates, $x_A - x_B$ , to come out equal to the rod’s proper length $l_0$ .

The rod is at rest in K', which slides along the shared x axis at v. Its length there, l_0, is the proper length; A is the leading end. — The rod is at rest in $K'$ , which slides along the shared $x$ axis at $v$ . Its length there, $l_0$ , is the proper length; $A$ is the leading end.

Given

$v$
$\left(x_A,\,t_A\right)$ marked on the forward end
$\left(x_B,\,t_B\right)$ marked on the rear end
$\beta = v/c$

Find

(a) the proper length

l_0

; (b) the interval

t_A - t_B

that makes

x_A - x_B = l_0

Solution

Step 1Say what a length in

K

actually means

In $K$ the rod is moving, so its length there is the distance between its two ends taken at one and the same instant of $K$ . Call that contracted length $l$ ; it is related to the proper length by

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

The two markings we are given were not made at one instant, so their coordinate difference $x_A - x_B$ is not $l$ and certainly not $l_0$ . The worldline diagram below shows why: the two events $E_A$ and $E_B$ sit on the two ends’ world lines at different heights.

World lines of the two ends in K, both of slope dx/d(ct) = β. The markings E_A and E_B are made at different times, so the horizontal gap between them is not the rod's length. The dashed axes belong to the rod's own frame. — World lines of the two ends in $K$ , both of slope $dx/d(ct) = \beta$ . The markings $E_A$ and $E_B$ are made at different times, so the horizontal gap between them is not the rod’s length. The dashed axes belong to the rod’s own frame.

Step 2Slide the rear mark forward to the instant of the front mark

The rear end travels at the constant $v$ , so between $t_B$ and $t_A$ it moves by $v\left(t_A-t_B\right)$ . Its coordinate at the instant $t_A$ is therefore

x_B\bigl(t_A\bigr) = x_B + v\left(t_A-t_B\right) .

Both ends are now referred to the single instant $t_A$ , so subtracting gives a genuine length in $K$ :

l = x_A - x_B - v\left(t_A-t_B\right)

(1)

Step 3Convert that length to the rod’s own frame

The length in (1) is the contracted one, so dividing by the contraction factor returns the proper length:

l_0 = \frac{x_A-x_B-v\left(t_A-t_B\right)}{\sqrt{1-(v/c)^2}}

(2)

Step 4Choose the marking interval that makes the raw difference read

l_0

For part (b), turn (1) around and ask for the coordinate difference itself:

x_A - x_B = l + v\left(t_A-t_B\right) = l_0\sqrt{1-\beta^2} + v\left(t_A-t_B\right) .

The raw difference is the contracted length plus whatever ground the rod covered between the two markings, so we can inflate it to any value we like by choosing the delay. Setting $x_A - x_B = l_0$ and solving for the interval:

t_A-t_B = \left(1-\sqrt{1-(v/c)^2}\right)\frac{l_0}{v}

(3)

Marking the front end first and the rear end afterwards works too, and gives a second branch. Put $x_A - x_B = -l_0$ , which says that the rear mark is the larger coordinate, into the same relation:

v\left(t_A-t_B\right) = -l_0\left(1+\sqrt{1-\beta^2}\right), \qquad\text{hence}\qquad t_B-t_A = \left(1+\sqrt{1-\beta^2}\right)\frac{l_0}{v} .

Now it is the rear end that is marked later.

Check

Set $t_A = t_B$ in (1): the marks are then simultaneous in $K$ , $l = x_A - x_B$ , and part (a) reduces to the plain contraction formula, as it must. In part (b) the first branch’s delay goes to zero as $\beta \to 0$ , which is right, because in Galilean physics simultaneous marks already give the true length. The second branch goes to $2l_0/v$ in that same limit, which is also right: with no contraction the rear end must cover one rod length to reach the front end’s mark and a second one to get $l_0$ past it. In general the two travel distances are $l_0 - l$ and $l_0 + l$ , so the second branch is always the longer wait.

Remark

Part (b) has two answers, not one, because “difference of coordinates” fixes only a magnitude. Wait a little and the front end’s lead grows from $l$ up to $l_0$ ; wait a lot and the rear end overshoots the front end’s old mark until it is $l_0$ beyond it. Both are legitimate, and only the first is ever quoted in a lecture.