Irodov 1.347: The Muon Sees a Shorter Distance – Amit & Deepali Sinha Foundation

Problem Irodov 1.347

A muon is created at one point of the inertial reference frame $K$ and decays at another. In $K$ it moves at the constant velocity $v = 0.990c$ and the distance between the two points is $l = 3.0$ km; write $\beta = v/c$ . Find:

the muon’s proper lifetime, the interval between its birth and its decay as recorded by a clock traveling with it
the distance the muon covers in the frame $K$ , reckoned from the muon’s own standpoint

Frame K at the muon's birth. The lower row is K itself, birthplace to decay point a distance l, with two synchronized clocks; the muon is the clock above, traveling at v, ringed at the birthplace. All read zero, in microseconds. — Frame $K$ at the muon’s birth. The lower row is $K$ itself, birthplace to decay point a distance $l$ , with two synchronized clocks; the muon is the clock above, traveling at $\vect v$ , ringed at the birthplace. All read zero, in microseconds.

Given

$v = 0.990c$
$\beta = v/c = 0.990$
$l = 3.0$ km
$c = 3.00\times10^{8}$ m/s

Find

(a) the proper lifetime

\Delta t_0

of the muon, (b) the length

l'

of the same stretch of

K

in the muon’s rest frame

Solution

Step 1How long the journey lasts on

K

’s clocks

Everything in this step is measured inside $K$ , where the distance is $l$ and the speed is $v$ , so the elapsed time follows from the ordinary definition of speed with no relativity in it at all. Call that interval $\Delta t$ .

\Delta t = \frac{l}{v} = \frac{3.0\times10^{3}\ \text{m}} {0.990\left(3.00\times10^{8}\ \text{m/s}\right)} = 1.010\times10^{-5}\ \text{s} \tag{1}

Step 2What the muon’s own clock records

The muon is present at both events, so a clock carried with it is at one and the same place at both, and its reading is the proper time between them. It runs slow relative to $K$ ’s clocks by the factor $\sqrt{1-\beta^2}$ , which the given speed fixes.

\sqrt{1-\beta^2} = \sqrt{1-(0.990)^{2}} = \sqrt{0.0199} = 0.1411

Applying that factor to (1) gives the proper lifetime.

\Delta t_0 = \Delta t\,\sqrt{1-\beta^2} = \left(1.010\times10^{-5}\ \text{s}\right)(0.1411)

\Delta t_0 = \frac{l}{v}\sqrt{1-\beta^2} = 1.4\ \mu\text{s}

(2)

The same row at the other end of the journey, the instant of the decay, the gold rings marking the muon at the decay point. The two synchronized K clocks have run through 10.1 μs, the muon's own clock only 1.4 μs. — The same row at the other end of the journey, the instant of the decay, the gold rings marking the muon at the decay point. The two synchronized $K$ clocks have run through $10.1$ $\mu$ s, the muon’s own clock only $1.4$ $\mu$ s.

Step 3The same stretch of

K

measured in the muon’s frame

The journey in the muon's rest frame K'. The muon does not travel; the stretch of K sweeps past at v, measured shorter, the bracket l' against the previous figure's l, both drawn at one scale. The instant is the decay. — The journey in the muon’s rest frame $K'$ . The muon does not travel; the stretch of $K$ sweeps past at $\vect v$ , measured shorter, the bracket $l'$ against the previous figure’s $l$ , both drawn at one scale. The instant is the decay.

In the muon’s rest frame the muon goes nowhere and the pair of points of $K$ , the birthplace and the decay point, is what moves. That pair is a length at rest in $K$ , so $l$ is its proper length and the muon’s frame measures the contracted value.

l' = l\,\sqrt{1-\beta^2} = \left(3.0\ \text{km}\right)(0.1411)

l' = l\,\sqrt{1-\beta^2} = 0.42\ \text{km}

(3)

Check

The muon’s frame must be able to tell the story on its own terms, and it does. In $K'$ the ground sweeps past at $v$ for the whole proper lifetime, so the stretch that goes by is $v\,\Delta t_0$ . With $v = 2.970\times10^{8}$ m/s and $\Delta t_0 = 1.425\times10^{-6}$ s, that product is $423$ m, and the boxed contracted length is $0.4232$ km, the same number. The two answers are also consistent with each other by construction, both being the same factor $0.1411$ applied to a $K$ quantity. As $\beta \to 0$ that factor tends to $1$ and both boxes return the classical readings, $\Delta t_0 \to l/v$ and $l' \to l$ .

Remark

Both frames account for the muon reaching its decay point, and neither needs the other’s language. In $K$ the muon is short lived but its clock runs slow, so it survives $10.1$ $\mu$ s of $K$ ’s time and covers the full $3.0$ km. In $K'$ the muon lives only its $1.4$ $\mu$ s, but there is only $0.42$ km of ground to be crossed. The same number, $\sqrt{1-\beta^2} = 0.1411$ , does duty as a time dilation in the first account and as a length contraction in the second, and each frame multiplies its own pair of numbers to get a consistent journey. What the two accounts genuinely disagree about is simultaneity. The two $K$ clocks read alike in $K$ , as the first figure shows, but in $K'$ they are out of step by $lv/c^2 = 9.9$ $\mu$ s, the clock at the decay point leading, that being the rear clock of $K$ ’s row as the row sweeps past in $K'$ . So in $K'$ , at the moment the muon decays beside the second mark, that clock reads $10.1$ $\mu$ s while the clock left behind at the birthplace reads only $0.2$ $\mu$ s, which is the muon’s own $1.4$ $\mu$ s slowed by the same factor $0.1411$ . Every disagreement in this problem traces back to that offset.