Irodov 1.346: The Lifetime Ratio Is Gamma – Amit & Deepali Sinha Foundation

Problem Irodov 1.346

An unstable particle is created and later decays. A clock traveling with the particle records $\Delta t_0 = 10$ ns between the two events, which is the particle’s proper lifetime. The clocks of the laboratory frame, in which the particle moves at a constant velocity, record $\Delta t = 20$ ns between the same two events. Find the distance $s$ the particle covers in the laboratory frame between its creation and its decay. Take the speed of light to be $c$ .

Laboratory frame, ct upward and x across, with c Δ t and s dimensioned. B is the creation, D the decay, the gold segment between them the particle's worldline. — Laboratory frame, $ct$ upward and $x$ across, with $c\,\Delta t$ and $s$ dimensioned. $B$ is the creation, $D$ the decay, the gold segment between them the particle’s worldline.

Given

$\Delta t_0 = 10$ ns
$\Delta t = 20$ ns
$c = 3.00\times10^{8}$ m/s

Find

the distance

s

the particle traverses in the laboratory frame before decaying

Solution

Step 1The ratio of the two lifetimes fixes the speed

The particle is present at both events, so its own clock records the proper time between them, and the laboratory’s clocks record the dilated interval. Writing $\beta = v/c$ for the particle’s speed in the laboratory frame,

\Delta t = \frac{\Delta t_0}{\sqrt{1-\beta^2}} .

Solving this for the contraction factor turns the two given times into a pure number,

\sqrt{1-\beta^2} = \frac{\Delta t_0}{\Delta t},

and squaring and rearranging gives the speed itself.

\beta = \sqrt{1-\left(\frac{\Delta t_0}{\Delta t}\right)^{2}}

(1)

Step 2The distance covered in the laboratory frame

Distance is speed times elapsed time, and both must be read in the same frame, so the laboratory distance uses the laboratory time $\Delta t$ and not the proper time.

s = v\,\Delta t = c\,\beta\,\Delta t

Substituting the speed (1) gives the distance in terms of the two given lifetimes alone. The data supply the ratio and the length $c\,\Delta t$ ,

\frac{\Delta t_0}{\Delta t} = \frac{10\ \text{ns}}{20\ \text{ns}} = 0.500, \qquad \sqrt{1-(0.500)^{2}} = \sqrt{0.750} = 0.8660,

c\,\Delta t = \left(3.00\times10^{8}\ \text{m/s}\right) \left(20.0\times10^{-9}\ \text{s}\right) = 6.00\ \text{m},

and the product of the last two numbers is the answer.

s = c\,\Delta t\,\sqrt{1-\left(\frac{\Delta t_0}{\Delta t}\right)^{2}} = 5.2\ \text{m}

(2)

Step 3The same distance read as an invariant interval

The same chart with the dashed curve of events a proper time Δ t_0 after B, every one of them at the interval c Δ t_0 from B. It passes through D. — The same chart with the dashed curve of events a proper time $\Delta t_0$ after $B$ , every one of them at the interval $c\,\Delta t_0$ from $B$ . It passes through $D$ .

The answer can be rewritten so that the three lengths in the figure appear together. Squaring (2) and moving the term in $s$ across,

\begin{aligned} s^2 &= \left(c\,\Delta t\right)^{2} - \left(c\,\Delta t\right)^{2} \left(\frac{\Delta t_0}{\Delta t}\right)^{2} \\[2pt] &= \left(c\,\Delta t\right)^{2} - \left(c\,\Delta t_0\right)^{2} \end{aligned}

which is a Pythagorean relation with a minus sign, the spacetime interval between the creation and the decay.

\left(c\,\Delta t\right)^{2} - s^2 = \left(c\,\Delta t_0\right)^{2}

(3)

Every inertial frame assigns the same value to the left-hand side, and the frame that makes $s$ vanish is the particle’s own, where the whole of it is the proper lifetime. That is why the dashed curve of the figure passes through $D$ .

Check

Evaluate (3) on the numbers. With $c\,\Delta t = 6.00$ m and $s = 5.196$ m the left side is $36.0 - 27.0 = 9.00$ m $^2$ , and $c\,\Delta t_0 = 3.00$ m makes the right side $9.00$ m $^2$ as well. The limits behave too: if the two lifetimes were equal the particle would be at rest in the laboratory and the distance (2) would vanish, while $\Delta t_0 \to 0$ at fixed $\Delta t$ sends $s \to c\,\Delta t$ , the distance only a light signal could cover.

Remark

Without dilation one would guess that a particle living $10$ ns can cross at most $c\,\Delta t_0 = 3.0$ m, and the answer beats that bound. It is not a bound at all: writing (2) as $s = c\,\Delta t_0\,\beta/\sqrt{1-\beta^2}$ shows that at fixed proper lifetime the reach grows without limit as $\beta \to 1$ , since the divergent factor is the dilation itself. Here $\beta/\sqrt{1-\beta^2} = 1.73$ and the particle covers $1.73$ times the naive figure. This is exactly why muons made high in the atmosphere are detected at sea level: their proper lifetime would allow only a few hundred meters, and the laboratory frame, in which they are fast, grants them tens of kilometers.