Irodov 1.363: The Angle a Boost Bends – Amit & Deepali Sinha Foundation

Problem Irodov 1.363

In a frame $K$ , a particle moves with speed $v$ along a direction that makes an angle $\theta$ with the $x$ axis. A second frame $K'$ slides at a constant velocity $V$ along that same axis, in the positive $x$ direction, with the axes of $K$ and $K'$ kept parallel and the $x$ and $x'$ axes coinciding. Find the angle $\theta'$ that the particle’s velocity makes with the $x'$ axis, as measured in $K'$ .

The standard configuration: axes parallel, K' sliding along the shared x axis at V. The particle at P moves at velocity v in K, at angle θ to that axis. Drawn for V = 0.60c, v = 0.65c, θ = 55°. — The standard configuration: axes parallel, $K'$ sliding along the shared $x$ axis at $\vect V$ . The particle at $P$ moves at velocity $\vect v$ in $K$ , at angle $\theta$ to that axis. Drawn for $V = 0.60c$ , $v = 0.65c$ , $\theta = 55^\circ$ .

Given

$v$
$\theta$
$V$

Find

the angle

\theta'

between the particle’s velocity and the

x'

axis, measured in

K'

Solution

Step 1The angle in each frame is a ratio of components

In K the direction is fixed by the ratio of the two components of v, drawn here at the illustration's θ = 55°. The same definition, written against the axes of K', is what the rest of the solution has to evaluate. — In $K$ the direction is fixed by the ratio of the two components of $\vect v$ , drawn here at the illustration’s $\theta = 55^\circ$ . The same definition, written against the axes of $K'$ , is what the rest of the solution has to evaluate.

A direction is not something the Lorentz transformation acts on; what it acts on is the velocity, and the direction follows from the two components of that velocity. So resolve $\vect v$ on the axes of $K$ , and write the same definition of an angle in $K'$ .

\begin{aligned} v_x &= v\cos\theta \\ v_y &= v\sin\theta \end{aligned} \tag{1}

Measured against its own axes, the direction in $K'$ obeys the identical definition.

\tan\theta' = \frac{v'_y}{v'_x} \tag{2}

The problem is therefore entirely a matter of transforming the two components.

Step 2One step along the path, seen from the other frame

The two frames are in the standard configuration, so a boost along the shared axis relates them. Apply it to an infinitesimal displacement of the particle along its path, that is, to the interval between two neighboring events on the path, writing $\beta = V/c$ throughout.

\begin{aligned} dx' &= \frac{dx - V\,dt}{\sqrt{1-\beta^2}} \\ dy' &= dy \\ dt' &= \frac{dt - V\,dx/c^2}{\sqrt{1-\beta^2}} \end{aligned}

Step 3The longitudinal component in the primed frame

Divide the transformed displacement along the axis by the transformed time. The radicals are common to both and cancel, and dividing numerator and denominator by $dt$ turns the surviving derivatives into the components of (1).

\begin{aligned} v'_x &= \frac{dx'}{dt'} \\ &= \frac{dx - V\,dt}{dt - V\,dx/c^2} \\ &= \frac{dx/dt - V}{1 - (V/c^2)(dx/dt)} \end{aligned}

v'_x = \frac{v_x - V}{1 - v_x V/c^2}

(3)

Step 4The transverse component in the primed frame

The transverse displacement is left alone by the boost, yet the time it is divided by is not, so the transverse velocity does change. Divide exactly as before.

\begin{aligned} v'_y &= \frac{dy'}{dt'} \\ &= \frac{dy\sqrt{1-\beta^2}}{dt - V\,dx/c^2} \\ &= \frac{(dy/dt)\sqrt{1-\beta^2}}{1 - (V/c^2)(dx/dt)} \end{aligned}

v'_y = \frac{v_y\sqrt{1-\beta^2}}{1 - v_x V/c^2}

(4)

Step 5The angle in the primed frame

Form the ratio (2) out of (3) and (4). The factor $1 - v_x V/c^2$ is common to both components and cancels, which is what leaves the radical as the only trace of relativity in the result, and then (1) supplies the components in terms of $v$ and $\theta$ .

\begin{aligned} \tan\theta' &= \frac{v'_y}{v'_x} \\ &= \frac{v_y\sqrt{1-\beta^2}}{1 - v_x V/c^2}\cdot\frac{1 - v_x V/c^2}{v_x - V} \\ &= \frac{v_y\sqrt{1-\beta^2}}{v_x - V} \\ &= \frac{v\sqrt{1-\beta^2}\,\sin\theta}{v\cos\theta - V} \end{aligned}

Dividing numerator and denominator by the particle’s speed puts it in terms of the two ratios the data actually fix.

\tan\theta' = \frac{\sqrt{1-\beta^2}\,\sin\theta}{\cos\theta - V/v}\,,\qquad \beta = \frac{V}{c}

(5)

The velocity in K', resolved on the primed axes. The boxed formula at the illustration's numbers gives θ' = 118°, obtuse because vθ is smaller than V: in K' the particle drifts backwards along x'. — The velocity in $K'$ , resolved on the primed axes. The boxed formula at the illustration’s numbers gives $\theta' = 118^\circ$ , obtuse because $v\cos\theta$ is smaller than $V$ : in $K'$ the particle drifts backwards along $x'$ .

Check

Two cases, evaluated. (i) A light ray, $v = c$ : the denominator becomes $\cos\theta - \beta$ , and the boxed components then give ${v'_x}^2 + {v'_y}^2 = c^2$ , because their common numerator $(c\cos\theta - V)^2 + c^2\sin^2\theta\,(1-\beta^2)$ collapses to $c^2(1-\beta\cos\theta)^2$ , the exact square of their common denominator. The speed is still $c$ , as it must be, and the angle relation is the relativistic aberration formula for starlight. (ii) $\theta = 0$ : the numerator vanishes, so the motion in $K'$ is along the $x'$ axis, and which way it points is decided by the sign of $\cos\theta - V/v$ , here $1 - V/v$ . For $v > V$ it is positive and the particle still runs forward; for $v < V$ it is negative and the particle runs backwards in $K'$ , which is what a frame overtaking the particle has to see.

Remark

Let $c$ grow without bound and the radical goes to $1$ , leaving the ordinary velocity triangle. The whole relativistic content of the answer is thus the single factor $\sqrt{1-\beta^2}$ , and it multiplies the transverse part alone while the longitudinal composition keeps its Galilean form. Two things follow. The correction to the direction is second order in $V/c$ whereas the change of direction itself is first order in $V/v$ : for a transverse ray with $V = 0.1c$ and $v = c$ , the Galilean triangle gives an aberration of $5.71^\circ$ against the exact $5.74^\circ$ , which is why stellar aberration was measured and accounted for long before relativity existed. And since the factor shrinks the transverse component and never the longitudinal one, $|\tan\theta'|$ always comes out smaller than the velocity triangle predicts, so a boost tilts every direction a little further toward the line of the $x'$ axis: the beaming that concentrates a moving source’s radiation along its own line of motion.