Irodov 1.359: The Gap Closes Faster Than Anything Moves

Problem Irodov 1.359

In a laboratory frame of reference, two particles travel toward each other along one straight line: particle 1 at $v_1 = 0.50c$ and particle 2, coming from the opposite direction, at $v_2 = 0.75c$ .

Find the rate at which the distance between the two particles decreases, as measured in the laboratory frame.
Find the velocity of one particle as measured in the rest frame of the other.

World lines in the laboratory frame K: particle 1 at β_1 = +0.50, particle 2 at β_2 = -0.75. Both are steeper than the dotted light cone, so both are slower than light. — World lines in the laboratory frame $K$ : particle 1 at $\beta_1 = +0.50$ , particle 2 at $\beta_2 = -0.75$ . Both are steeper than the dotted light cone, so both are slower than light.

Given

$v_1 = 0.50c$
$v_2 = 0.75c$
both measured in the laboratory frame $K$ , motion along one line, opposite senses

Find

(a) the rate at which the particles approach in

K

; (b) the velocity of one particle in the rest frame of the other

Solution

Step 1Ask what closes the gap in the laboratory frame

Part (a) is a question about one frame only. Put particle 1 at $x_1(t)$ moving in the $+x$ sense and particle 2 at $x_2(t)$ moving in the $-x$ sense, both coordinates and the time all belonging to $K$ . Their separation is $s = x_2 - x_1$ , and differentiating it with respect to the laboratory time gives

\frac{ds}{dt} = \frac{dx_2}{dt} - \frac{dx_1}{dt} = -v_2 - v_1 .

The separation shrinks at the rate $v_1+v_2$ , so the approach velocity is

v = v_1 + v_2 = 0.50c + 0.75c = 1.25c

(1)

Step 2Change frames to get a genuine velocity

The relative velocity asked for in part (b) is a different object: it is the velocity of particle 2 as measured in the rest frame of particle 1. That requires a Lorentz boost with $V = v_1$ , and in the boosted frame $K'$ the $x$ component of a velocity $u$ transforms as

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

Particle 2 has $u = -v_2$ in $K$ , and the boost speed is $V = +v_1$ , so

u' = \frac{-v_2-v_1}{1+\dfrac{v_1v_2}{c^2}}

(2)

The same pair in K', the rest frame of particle 1. Particle 1's world line is now vertical; particle 2's has slope β' = -0.91 and still lies inside the light cone. — The same pair in $K'$ , the rest frame of particle 1. Particle 1’s world line is now vertical; particle 2’s has slope $\beta' = -0.91$ and still lies inside the light cone.

Step 3Take the magnitude and put the numbers in

The minus sign in (2) only says that particle 2 still runs the other way in $K'$ , so the relative speed is the magnitude of the composition law above. Substituting the two speeds,

\frac{v_1+v_2}{1+\dfrac{v_1v_2}{c^2}} = \frac{1.25c}{1+0.50\times 0.75} = \frac{1.25c}{1.375} .

v = \frac{v_1+v_2}{1+v_1v_2/c^2} = 0.91c

(3)

Check

The composition law can never break the light barrier here. Writing $\beta_i = v_i/c$ ,

c - v = c\left(1 - \frac{\beta_1+\beta_2}{1+\beta_1\beta_2}\right) = c\,\frac{\left(1-\beta_1\right)\left(1-\beta_2\right)} {1+\beta_1\beta_2} ,

and this is strictly positive whenever both betas are below unity. For the given numbers the right-hand side is $0.09c$ , so $v = 0.91c$ , which reproduces the answer by a second route. In the non-relativistic limit the product $\beta_1\beta_2$ drops out of the denominator and part (b) collapses onto part (a): the two are one question in Galilean physics and two questions here.

Remark

The two answers are not competing values of one quantity. The $1.25c$ belongs to the laboratory observer watching two objects; the $0.91c$ belongs to one of the objects watching the other. They sit side by side in one problem precisely because the phrase “how fast are they approaching” hides the choice of who is asking.