Irodov 1.342: Only One Leg of the Rod Shrinks – Amit & Deepali Sinha Foundation

Problem Irodov 1.342

A straight rod moves through a laboratory frame at constant velocity $v = c/2$ . In that laboratory frame the rod’s length is measured to be $l = 1.00$ m, and the rod is tilted at $\theta = 45^\circ$ to its own direction of motion. Find the rod’s proper length $l_0$ , that is, its length in the frame in which it is at rest.

Given

$v = c/2$
$l = 1.00$ m
$\theta = 45^\circ$

Find

Find the proper length

l_0

of the rod

Solution

Step 1Split the rod into an along-motion piece and an across-motion piece

A boost treats the two directions differently, so the first move is to stop thinking of the rod as one length and resolve it. Write $\beta = v/c$ and take the $x$ axis along the velocity. In the laboratory frame the rod’s two components are read straight off the right triangle in the laboratory-frame sketch below:

l_x = l\cos\theta, \qquad l_y = l\sin\theta .

The rod as the laboratory measures it: length l at θ to the motion. — The rod as the laboratory measures it: length $l$ at $\theta$ to the motion.

The same rod in its own rest frame: the along-motion leg is longer by γ, the across-motion leg is unchanged. — The same rod in its own rest frame: the along-motion leg is longer by $\gamma$ , the across-motion leg is unchanged.

Step 2Undo the contraction on one leg only

Each component is now a length along a fixed axis, so the standard contraction rule applies to it separately. The component along the boost is contracted and the component across it is not, so the laboratory values follow from the proper ones as

l_x = l_{0x}\sqrt{1-\beta^2}, \qquad l_y = l_{0y} .

Solving each for the proper component gives the two legs of the rest-frame sketch above, with $\gamma = 1/\sqrt{1-\beta^2}$ :

l_{0x} = \frac{l\cos\theta}{\sqrt{1-\beta^2}}, \qquad l_{0y} = l\sin\theta

(1)

Step 3Reassemble the proper length from its two legs

The rest frame is an ordinary Euclidean frame, so Pythagoras applies to (1) without any further relativity:

\begin{aligned} l_0^2 &= l_{0x}^2 + l_{0y}^2 \\[2pt] &= \frac{l^2\cos^2\theta}{1-\beta^2} + l^2\sin^2\theta \\[2pt] &= \frac{l^2\left[\cos^2\theta + \left(1-\beta^2\right)\sin^2\theta \right]}{1-\beta^2} \\[2pt] &= l^2\,\frac{1-\beta^2\sin^2\theta}{1-\beta^2} . \end{aligned}

The last line used $\cos^2\theta + \sin^2\theta = 1$ to collect the bracket.

Step 4Put the numbers in

Evaluate $\beta$ and $\sin\theta$ for the given data before substituting them into the bracket found above.

\beta = \frac{v}{c} = \frac{1}{2}, \qquad \sin^2\theta = \sin^2 45^\circ = \frac{1}{2} .

Substituting these values and reducing the fraction,

\frac{1-\beta^2\sin^2\theta}{1-\beta^2} = \frac{1 - \dfrac{1}{4}\cdot\dfrac{1}{2}}{1-\dfrac{1}{4}} = \frac{7/8}{3/4} = \frac{7}{6} , \qquad \sqrt{\frac{7}{6}} = 1.080 .

With $l = 1.00$ m this gives the proper length:

l_0 = l\,\sqrt{\frac{1-\beta^2\sin^2\theta}{1-\beta^2}} = 1.08\ \mathrm{m}

(2)

Check

Two limits fix the formula. At $\theta = 0$ the rod lies along the motion and it returns $l_0 = l/\sqrt{1-\beta^2}$ , the plain contraction formula; at $\theta = 90^\circ$ it returns $l_0 = l$ , since a transverse length is not affected by the boost at all. In between, $\beta^2\sin^2\theta \le \beta^2$ makes the fraction at least $1$ for every angle, so $l_0 \ge l$ always, which a proper length must satisfy because it is the largest length any frame can measure.

Remark

The orientation changes as well as the length. From (1), $\tan\theta_0 = l_{0y}/l_{0x} = \sqrt{1-\beta^2}\, \tan\theta$ , which for these numbers is $\tan\theta_0 = 0.866$ , or $\theta_0 = 41^\circ$ . A rod that the laboratory sees at $45^\circ$ to its motion therefore lies at only $41^\circ$ to that direction in its own rest frame.