Irodov 1.361: Pythagoras Loses a Relativistic Term – Amit & Deepali Sinha Foundation

Problem Irodov 1.361

In a laboratory frame of reference two particles travel along mutually perpendicular straight lines, one at the speed $v_1$ and the other at the speed $v_2$ , and both speeds are large enough that relativistic kinematics is required. Find their relative velocity.

The laboratory frame K: particle 1 along x at v_1, particle 2 along y at v_2. The frame K' rides with particle 1, at its origin, so v_1 is also the velocity of K'. — The laboratory frame $K$ : particle 1 along $x$ at $\vect v_1$ , particle 2 along $y$ at $\vect v_2$ . The frame $K'$ rides with particle 1, at its origin, so $\vect v_1$ is also the velocity of $K'$ .

Given

speeds $v_1$ and $v_2$
both measured in $K$
directions mutually perpendicular

Find

The relative velocity of the two particles

Solution

Step 1Say which frame the answer lives in, and set the axes

Relative velocity is not a laboratory quantity: it is the velocity of one particle measured in the rest frame of the other, so the answer belongs to a frame that is itself moving. Lay the laboratory axes along the two given directions, with particle 1 traveling along $x$ and particle 2 along $y$ , and call $K'$ the frame that rides with particle 1. Then $K'$ moves relative to the laboratory at

V = v_1 \quad\text{along the } x \text{ direction,}

and the velocity components the two particles carry in the laboratory frame $K$ are as simple as they can be.

\text{particle 1: } (v_1,\,0), \qquad \text{particle 2: } (0,\,v_2)

Step 2Transform particle 2’s velocity into

K'

For a boost of speed $V$ along $x$ , a velocity with laboratory components $(u_x, u_y)$ appears in $K'$ with the components below, the transverse one carrying the extra root because the clock rate changes while $y$ does not.

u_x' = \frac{u_x-V}{1-\dfrac{u_xV}{c^{2}}}, \qquad u_y' = \frac{u_y\sqrt{1-\dfrac{V^{2}}{c^{2}}}}{1-\dfrac{u_xV}{c^{2}}}

Particle 2 has $u_x = 0$ , which is the simplification the perpendicular arrangement buys: the common denominator becomes exactly $1$ and both expressions collapse.

\begin{aligned} v_x' &= \frac{0-v_1}{1-0}, \\[4pt] v_y' &= \frac{v_2\sqrt{1-\dfrac{v_1^{2}}{c^{2}}}}{1-0} . \end{aligned}

v_x' = -v_1, \qquad v_y' = v_2\sqrt{1-\frac{v_1^{2}}{c^{2}}}

(1)

Step 3Combine the components into the relative speed

Particle 2's velocity in K', the rest frame of particle 1. Its component along x' points backwards at v_1; its component along y' is the reduced transverse one. The relative speed is the length of the parent vector. — Particle 2’s velocity in $K'$ , the rest frame of particle 1. Its component along $x'$ points backwards at $v_1$ ; its component along $y'$ is the reduced transverse one. The relative speed is the length of the parent vector.

Squaring and adding the two components of (1) gives the square of the relative speed, and multiplying out the second square brings the correction term into view.

\begin{aligned} v^{2} &= v_x'^{2}+v_y'^{2} \\[4pt] &= v_1^{2}+v_2^{2}\left(1-\frac{v_1^{2}}{c^{2}}\right) \\[4pt] &= v_1^{2}+v_2^{2}-\frac{v_1^{2}v_2^{2}}{c^{2}} \end{aligned}

Every term is a squared speed, so the last one can be written as the square of the single quantity $v_1v_2/c$ , and the root is taken with the positive sign because a speed is being asked for.

v = \sqrt{v_1^{2}+v_2^{2}-\left(\frac{v_1v_2}{c}\right)^{2}}

(2)

Check

First, the answer must not depend on which particle is called the observer. Boosting instead into particle 2’s frame, with $V = v_2$ along $y$ , particle 1 has no motion along $y$ in the laboratory, so the same two rules give components $-v_2$ along $y'$ and $v_1\sqrt{1-v_2^{2}/c^{2}}$ along $x'$ , whose squares sum to $v_1^{2}+v_2^{2}-v_1^{2}v_2^{2}/c^{2}$ again: the boxed expression is symmetric in $v_1$ and $v_2$ , as it must be. Second, the result never reaches $c$ , since

c^{2}-v^{2} = c^{2}-v_1^{2}-v_2^{2}+\frac{v_1^{2}v_2^{2}}{c^{2}} = \frac{\left(c^{2}-v_1^{2}\right)\left(c^{2}-v_2^{2}\right)} {c^{2}} ,

a product of two positive numbers whenever both speeds are below $c$ . It equals zero exactly when one of the particles is a light signal, and then $v = c$ , which is the second postulate reappearing.

Remark

The Pythagorean $\sqrt{v_1^{2}+v_2^{2}}$ that the correction term spoils is not a wrong quantity, it is a different one. If the two particles set out from one point at one instant, their separation measured in the laboratory is $t\sqrt{v_1^{2}+v_2^{2}}$ , so that root is the rate at which the laboratory separation grows: a derivative of laboratory coordinates with respect to laboratory time, which nothing forbids from exceeding $c$ . At $v_1 = v_2 = 0.90c$ that separation rate is $1.27c$ , while the relative velocity boxed above is $0.98c$ . The two numbers answer different questions, and only the second is a speed some observer actually measures.