Irodov 1.362: Two Gammas Multiply Across the Boost – Amit & Deepali Sinha Foundation

Problem Irodov 1.362

Two inertial frames $K$ and $K'$ are in the standard configuration: their $x$ and $x'$ axes coincide, their $y$ and $y'$ axes are parallel, and $K'$ slides along the common $x$ axis in the positive direction at a constant speed $V$ . Write $\beta = V/c$ . An unstable particle travels in $K'$ along the $y'$ axis at the constant speed $v'$ , and it decays after a proper lifetime $\Delta t_0$ , that is, after a time $\Delta t_0$ measured by a clock traveling with the particle itself. Find the distance the particle covers in the frame $K$ between its creation and its decay.

The two frames in the standard configuration, with K' carried along the shared x axis at V. The particle P sits on the y' axis and moves straight up it at v', the velocity measured in K'. — The two frames in the standard configuration, with $K'$ carried along the shared $x$ axis at $\vect V$ . The particle $P$ sits on the $y'$ axis and moves straight up it at $\vect v'$ , the velocity measured in $K'$ .

Given

$v'$ along the $y'$ axis of $K'$
$V$ along the shared $x$ axis
$\beta = V/c$
proper lifetime $\Delta t_0$

Find

the distance

s

covered in

K

Solution

Step 1The particle’s velocity as

K

measures it

The distance traveled in $K$ is a speed in $K$ times a time in $K$ , so both have to be transformed and neither may be taken from $K'$ . Start with the velocity. For the standard configuration, with $K'$ moving at $+V$ along $x$ , the components of a velocity measured in $K'$ transform into $K$ as

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

The particle runs along $y'$ , so its components in $K'$ are $u_x' = 0$ and $u_y' = v'$ . Both denominators become $1$ , and the two components in $K$ follow.

u_x = V, \qquad u_y = v'\sqrt{1-\beta^2}

(1)

The particle's velocity as K measures it, resolved on the axes of K: the along-boost component is the frame speed V, the transverse component the K' value v' cut by sqrt(1-β^2). The angle θ belongs to K alone. — The particle’s velocity as $K$ measures it, resolved on the axes of $K$ : the along-boost component is the frame speed $V$ , the transverse component the $K'$ value $v'$ cut by $\sqrt{1-\beta^2}$ . The angle $\theta$ belongs to $K$ alone.

Step 2The speed in

K

, and the Lorentz factor it carries

The distance needs the magnitude of that velocity and, through the time dilation, the combination $1-u^2/c^2$ as well. Squaring and adding the two components of (1) gives the first.

u^2 = u_x^2+u_y^2 = V^2+v'^2\left(1-\beta^2\right) \tag{2}

Dividing by $c^2$ and using $V^2/c^2 = \beta^2$ turns the second into a product of two independent factors.

\begin{aligned} 1-\frac{u^2}{c^2} &= 1-\frac{V^2}{c^2}-\frac{v'^2}{c^2}\left(1-\beta^2\right) \\[3pt] &= \left(1-\beta^2\right)-\frac{v'^2}{c^2}\left(1-\beta^2\right) \end{aligned}

Taking the common factor out shows that the two speeds enter separately.

1-\frac{u^2}{c^2} = \left(1-\beta^2\right)\left(1-\frac{v'^2}{c^2}\right)

(3)

Step 3The lifetime measured in

K

The particle’s own clock reads the proper interval $\Delta t_0$ between its creation and its decay, two events that happen at the same place in the particle’s rest frame. In $K$ that same pair of events is separated by the dilated time, with the Lorentz factor built on the speed $u$ that $K$ measures. Substituting (3) under the root gives it in terms of the data of the problem.

\begin{aligned} \Delta t &= \frac{\Delta t_0}{\sqrt{1-\dfrac{u^2}{c^2}}} \\[3pt] &= \frac{\Delta t_0} {\sqrt{\left(1-\beta^2\right) \left(1-\dfrac{v'^2}{c^2}\right)}} \end{aligned} \tag{4}

Step 4The distance covered in

K

The particle’s velocity in $K$ is constant, so the distance is simply that speed times the lifetime just found. Multiplying the speed of (2) by the time of (4) and carrying the speed under the root gives the result.

\begin{aligned} s &= u\,\Delta t \\[3pt] &= \sqrt{V^2+v'^2\left(1-\beta^2\right)}\; \frac{\Delta t_0}{\sqrt{\left(1-\beta^2\right) \left(1-\dfrac{v'^2}{c^2}\right)}} \end{aligned}

s = \Delta t_0\sqrt{\dfrac{V^2+\left(1-\beta^2\right)v'^2} {\left(1-\beta^2\right)\left(1-\dfrac{v'^2}{c^2}\right)}}

(5)

Check

Every factor under the root is dimensionless except the $V^2$ and $v'^2$ in the numerator, so the bracket has the dimensions of a speed squared and $s$ comes out as a speed times $\Delta t_0$ , a length. Two limits check the structure. With $v' = 0$ the particle is at rest in $K'$ and the formula gives $s = V\Delta t_0/\sqrt{1-\beta^2}$ , the frame speed times the dilated lifetime, which is the ordinary moving-clock result. With $V = 0$ the two frames coincide and it gives $s = v'\Delta t_0/\sqrt{1-v'^2/c^2}$ , the same result written for the other speed. The formula is therefore symmetric in the roles of the two speeds, as it must be: with $u_x' = 0$ the particle and the frame carry perpendicular motions and nothing distinguishes which of the two is called the boost.

Remark

Equation (3) says something sharper than the algebra it came from: for a velocity that is purely transverse in $K'$ , the Lorentz factors multiply, $\gamma_u = \gamma_V\gamma_{v'}$ . That clean product is special to the perpendicular case. In general the composition carries an extra bracket, $\gamma_u = \gamma_V\gamma_{u'}\left(1+u_x'V/c^2\right)$ , which reduces to the product only when the $K'$ velocity has no component along the boost, exactly the situation here. The consequence is the one that makes atmospheric muons reach the ground: $s$ has no upper bound as either speed approaches $c$ , so the distance covered in $K$ can exceed $c\,\Delta t_0$ by any factor, while the particle’s own clock still runs only $\Delta t_0$ and nothing in the picture travels faster than $c$ .