Irodov 1.341: Each Side Shrinks by Its Angle – Amit & Deepali Sinha Foundation

Problem Irodov 1.341

Each of the three sides of a triangle has proper length $a$ , so the triangle is equilateral in the frame in which it is at rest. A second reference frame moves relative to the triangle with constant velocity $V$ , the velocity lying in the plane of the triangle. Write $\beta = V/c$ , where $c$ is the speed of light. Find the perimeter of the triangle as measured in that moving frame when the velocity is directed:

along one of its angle bisectors
along one of its sides

In each case examine what the result becomes for $V \ll c$ and for $V \to c$ .

Given

$a$
$V$
$\beta = V/c$

Find

(a) the perimeter

P_a

with

\vect V

along a bisector, (b) the perimeter

P_b

with

\vect V

along a side, and both limits

V \ll c

and

V \to c

Solution

Step 1The length of one side at an arbitrary angle

A boost shortens lengths along its own direction and leaves lengths across it alone, so a side must not be treated as a single length: it has to be split into a piece along $\vect V$ and a piece across $\vect V$ . Take one side, of proper length $a$ , making an angle $\theta$ with the velocity in the triangle’s own rest frame.

One side of the triangle as its own rest frame sees it: a segment of proper length a inclined at θ to the velocity V of the other frame, resolved into a leg aθ along V and a leg aθ across it. — One side of the triangle as its own rest frame sees it: a segment of proper length $a$ inclined at $\theta$ to the velocity $\vect V$ of the other frame, resolved into a leg $a\cos\theta$ along $\vect V$ and a leg $a\sin\theta$ across it.

The two legs of that right triangle are the side’s components,

a_\parallel = a\cos\theta, \qquad a_\perp = a\sin\theta .

Each leg is now a length along a single fixed direction, so the contraction rule applies to each separately: the one along $\vect V$ is multiplied by $\sqrt{1-\beta^2}$ and the one across $\vect V$ is untouched.

a'_\parallel = a\cos\theta\,\sqrt{1-\beta^2}, \qquad a'_\perp = a\sin\theta .

The moving frame is an ordinary Euclidean frame, so the side’s length there, call it $l_\theta$ , follows from Pythagoras applied to the two contracted legs.

\begin{aligned} l_\theta^2 &= a'^2_\parallel + a'^2_\perp \\[2pt] &= a^2\cos^2\theta\left(1-\beta^2\right) + a^2\sin^2\theta \\[2pt] &= a^2\left(\cos^2\theta + \sin^2\theta - \beta^2\cos^2\theta\right) \end{aligned}

Collecting the first two terms with $\cos^2\theta+\sin^2\theta = 1$ and taking the square root gives the rule the whole problem runs on:

l_\theta = a\sqrt{1-\beta^2\cos^2\theta}

(1)

Step 2Perimeter with the velocity along a bisector

Everything is now geometry: all that is needed is the angle each of the three sides makes with $\vect V$ . A bisector of an equilateral triangle is also the altitude dropped on the opposite side, so that opposite side lies across $\vect V$ , and the other two sides, which make $60^\circ$ with it, lie at $30^\circ$ to $\vect V$ .

The velocity along a bisector (gray dashed), at β = 0.8. Dashed is the proper triangle, every side a; solid is what the moving frame measures: the transverse side keeps its length a, the two slanted sides become l_30. — The velocity along a bisector (gray dashed), at $\beta = 0.8$ . Dashed is the proper triangle, every side $a$ ; solid is what the moving frame measures: the transverse side keeps its length $a$ , the two slanted sides become $l_{30}$ .

The three angles to the velocity are therefore

\theta_1 = 90^\circ, \qquad \theta_2 = \theta_3 = 30^\circ .

Substituting each into (1) gives the three sides as the moving frame measures them.

\begin{aligned} l_{90} &= a\sqrt{1-\beta^2\cos^2 90^\circ} = a \\[2pt] l_{30} &= a\sqrt{1-\beta^2\cos^2 30^\circ} \\[2pt] &= a\sqrt{1-\frac{3}{4}\beta^2} \\[2pt] &= \frac{a}{2}\sqrt{4-3\beta^2} \end{aligned}

The perimeter is their sum, one side of the first kind and two of the second:

P_a = l_{90} + 2\,l_{30} .

P_a = a\left(1+\sqrt{4-3\beta^2}\right)

(2)

Step 3Perimeter with the velocity along a side

The same three-step count runs again with one different set of angles. One side now lies along $\vect V$ , and the other two, each making $60^\circ$ with it, are at $60^\circ$ to the velocity.

The velocity along a side, at β = 0.8. Dashed is the proper triangle of side a; solid is what the moving frame measures: the side along V contracts to l_0 and the slanted sides to l_60. The copies share a vertex. — The velocity along a side, at $\beta = 0.8$ . Dashed is the proper triangle of side $a$ ; solid is what the moving frame measures: the side along $\vect V$ contracts to $l_0$ and the slanted sides to $l_{60}$ . The copies share a vertex.

Here the three angles to the velocity are

\theta_1 = 0, \qquad \theta_2 = \theta_3 = 60^\circ ,

and (1) turns them into three lengths as before.

\begin{aligned} l_0 &= a\sqrt{1-\beta^2\cos^2 0} = a\sqrt{1-\beta^2} \\[2pt] l_{60} &= a\sqrt{1-\beta^2\cos^2 60^\circ} \\[2pt] &= a\sqrt{1-\frac{1}{4}\beta^2} \\[2pt] &= \frac{a}{2}\sqrt{4-\beta^2} \end{aligned}

Adding the one contracted side to the two slanted ones,

P_b = l_0 + 2\,l_{60} .

P_b = a\left(\sqrt{1-\beta^2}+\sqrt{4-\beta^2}\right)

(3)

Step 4The two limits

With both perimeters in closed form, the slow limit is a binomial expansion of each root to first order in $\beta^2$ .

\begin{aligned} \sqrt{4-3\beta^2} &= 2\sqrt{1-\frac{3\beta^2}{4}} \approx 2-\frac{3\beta^2}{4} \\[2pt] \sqrt{1-\beta^2} &\approx 1-\frac{\beta^2}{2} \\[2pt] \sqrt{4-\beta^2} &= 2\sqrt{1-\frac{\beta^2}{4}} \approx 2-\frac{\beta^2}{4} \end{aligned}

Putting the first line into the boxed $P_a$ and the last two into the boxed $P_b$ collapses both onto one and the same second-order form:

P_a \approx P_b \approx 3a\left(1-\frac{\beta^2}{4}\right)

(4)

At the other extreme the roots are evaluated directly at $\beta = 1$ .

P_a \to a\left(1+\sqrt{4-3}\right) = 2a, \qquad P_b \to a\left(0+\sqrt{4-1}\right) = \sqrt{3}\,a

In that limit every length along $\vect V$ has gone to zero, so the triangle collapses onto its own projection on the line across $\vect V$ and the perimeter becomes twice the length of that projection: the side $a$ in case (a), the altitude $a\sqrt{3}/2$ in case (b).

Check

Set $\beta = 0$ , where the frame is at rest with respect to the triangle: $P_a = a(1+\sqrt{4}\,) = 3a$ and $P_b = a(1+\sqrt{4}\,) = 3a$ , the proper perimeter, as both must give. Both perimeters also stay below $3a$ for every $0<\beta<1$ , as they must, since no side can exceed its proper length.

Remark

The two answers are far closer than the two very different pictures suggest: expanded in $\beta$ they agree not only at order $\beta^2$ but at order $\beta^4$ as well, parting company only at $\beta^6$ , where $P_a$ carries $-27a\beta^6/512$ against $P_b$ ’s $-33a\beta^6/512$ . The reason is a Fourier one. Each side contributes $a\sqrt{1-\beta^2\cos^2\theta}$ , whose expansion in $\beta^{2n}$ carries angular harmonics only up to $\cos 2n\theta$ , while the three side directions of an equilateral triangle sit $60^\circ$ apart, and summing over three such directions annihilates every harmonic except multiples of the sixth. The first term able to carry a sixth harmonic is the $\beta^6$ one, so the perimeter of an equilateral triangle is independent of the direction of the boost within its plane through fourth order, and only there does the choice of bisector or side begin to show: at $\beta = 0.9$ the two come out $2.253a$ and $2.222a$ .