Irodov 1.360: Compose the Speeds, Then Contract – Amit & Deepali Sinha Foundation

Problem Irodov 1.360

Two rods have the same proper length $l_0$ . In a laboratory frame they both lie along one common axis and travel lengthwise straight toward each other, each with the speed $v$ measured in that laboratory frame. Find the length of one rod as measured in the reference frame attached to the other rod.

The laboratory frame K: two rods of proper length l_0 on one axis, closing on each other at the same speed v. — The laboratory frame $K$ : two rods of proper length $l_0$ on one axis, closing on each other at the same speed $v$ .

Given

proper length $l_0$ each
motion lengthwise
senses opposite
speed $v$ each in $K$
$\beta = v/c$

Find

Length of one rod in the rest frame of the other

Solution

Step 1Get the speed of one rod relative to the other

Length contraction is governed by the rod’s speed relative to the frame that measures it, and here that frame is the other rod, not the laboratory, so the first job is a velocity composition. Take $K$ with rod 1 at $+v$ and rod 2 at $-v$ along the shared axis, and boost to rod 1’s frame $K'$ , which travels at $V = +v$ relative to $K$ . The composition rule for a velocity along the boost is

u' = \frac{u-V}{1-\dfrac{uV}{c^{2}}}

and rod 2 carries $u = -v$ into it while the boost carries $V = +v$ , so both signs conspire to make the denominator larger than one.

u' = \frac{-v-v}{1-\dfrac{(-v)(v)}{c^{2}}} = \frac{-2v}{1+\dfrac{v^{2}}{c^{2}}}

Its magnitude is the speed rod 2 has in rod 1’s frame, and writing $\beta = v/c$ compresses it.

u = \frac{2v}{1+\beta^{2}}

(1)

Step 2Turn that speed into a contraction factor

Both rods in K', the rest frame of rod 1. Rod 1 keeps its proper length l_0; rod 2 sweeps past at the composed speed u and measures shorter, l. The two lines share one axis. Proportions: v = 0.50c. — Both rods in $K'$ , the rest frame of rod 1. Rod 1 keeps its proper length $l_0$ ; rod 2 sweeps past at the composed speed $u$ and measures shorter, $l$ . The two lines share one axis. Proportions: $v = 0.50c$ .

The contraction factor is $\sqrt{1-u^{2}/c^{2}}$ , so the quantity to build is $u^{2}/c^{2}$ from (1) and then its complement. Putting the complement over the common denominator $(1+\beta^{2})^{2}$ makes the numerator collapse.

\begin{aligned} 1-\frac{u^{2}}{c^{2}} &= 1-\frac{4\beta^{2}}{\left(1+\beta^{2}\right)^{2}} \\[4pt] &= \frac{\left(1+\beta^{2}\right)^{2}-4\beta^{2}} {\left(1+\beta^{2}\right)^{2}} \\[4pt] &= \frac{1+2\beta^{2}+\beta^{4}-4\beta^{2}} {\left(1+\beta^{2}\right)^{2}} = \frac{\left(1-\beta^{2}\right)^{2}}{\left(1+\beta^{2}\right)^{2}} \end{aligned}

Both numerator and denominator are perfect squares of positive quantities, since $\beta < 1$ , so the root can be taken factor by factor.

\sqrt{1-\frac{u^{2}}{c^{2}}} = \frac{1-\beta^{2}}{1+\beta^{2}}

(2)

Step 3Apply the contraction in the rod’s rest frame

Rod 2 has the proper length $l_0$ in its own rest frame and moves at the speed $u$ relative to rod 1, so the length rod 1 measures is $l_0$ times the factor just found in (2).

l = l_0\sqrt{1-\frac{u^{2}}{c^{2}}}

Substituting the factor gives the answer, and because the setup is unchanged by swapping the labels 1 and 2, the same number is what rod 2 measures for rod 1.

l = l_0\,\frac{1-\beta^{2}}{1+\beta^{2}}, \qquad \beta = \frac{v}{c}

(3)

Check

The composed speed never breaks the light barrier, which the answer needs if it is to stay real and positive. From (1), $c-u = c\left(1-2\beta/(1+\beta^{2})\right) = c\left(1-\beta\right)^{2}/\left(1+\beta^{2}\right)$ , which is strictly positive for every $\beta<1$ . Numerically, at $v = 0.50c$ the relative speed is $u = 2(0.50)c/1.25 = 0.80c$ , whose contraction factor $\sqrt{1-0.64} = 0.60$ agrees with the boxed form $(1-0.25)/(1+0.25) = 0.60$ . And $\beta \to 0$ returns $l = l_0$ , the Galilean case where all three frames measure the same rod.

Remark

Each rod is shorter in the other’s frame than it is in the laboratory. In the laboratory each measures $l_0\sqrt{1-\beta^{2}}$ , and the ratio of the boxed length to that is $\sqrt{1-\beta^{2}}/(1+\beta^{2})$ , below unity for every $\beta > 0$ , simply because $u$ exceeds $v$ . There is also a tidy way to carry the result: writing $\gamma_v$ for the laboratory Lorentz factor of one rod and $\gamma_u$ for the factor belonging to the relative speed, the boxed answer says $\gamma_u = \left(1+\beta^{2}\right)/\left(1-\beta^{2}\right) = 2\gamma_v^{2}-1$ , so the relative-motion factor is fixed by the laboratory one alone. At $v = 0.50c$ that reads $\gamma_u = 2(4/3)-1 = 5/3$ , and $l = 3l_0/5$ .