Irodov 1.351: Simultaneous for Them, Not for Us – Amit & Deepali Sinha Foundation

Problem Irodov 1.351

Two unstable particles travel one behind the other along one straight line in a frame $K$ , both with the same velocity $v = 0.990c$ , and in that frame the distance between them is $l = 120$ m. In the inertial frame in which the two particles are at rest, the two decays happen at one and the same instant. Writing $\beta = v/c$ , find the time interval $\Delta t$ between the two decays as observed in $K$ , and say which of the two particles, the leading one or the trailing one, decays later in $K$ .

The pair as K sees them at one instant of its own time: both run along the same line at v, particle 2 ahead of particle 1, separated by the distance l that K measures. — The pair as $K$ sees them at one instant of its own time: both run along the same line at $\vect v$ , particle $2$ ahead of particle $1$ , separated by the distance $l$ that $K$ measures.

Given

$v = 0.990c$
$\beta = v/c$
$l = 120$ m in $K$
$c = 3.00\times 10^{8}$ m/s

Find

the interval

\Delta t

between the decays in

K

, and which particle decays later in

K

Solution

Step 1The separation the particles themselves would measure

The distance $l$ is the separation of the two particles in $K$ , so it is a length measured by fixing both positions at one instant of $K$ , and a pair of bodies moving together at speed $v$ is contracted by exactly that measurement. Their proper separation $l_0$ , the distance in the frame $K_0$ in which both are at rest, is therefore the longer one.

\begin{align*}l &= l_0\sqrt{1-\beta^{2}}, \\ \frac{l}{\sqrt{1-\beta^{2}}} &= l_0\end{align*}

l_0 = \dfrac{l}{\sqrt{1-\beta^{2}}}

(1)

Step 2Transform the pair of decay events into

K

In $K_0$ the two decays are one event each, simultaneous and a proper distance apart, so $\Delta t_0 = 0$ and $\Delta x_0 = l_0$ , taking the leading particle first in both differences. The Lorentz transformation from $K_0$ to $K$ carries that pair of events across, and (1) then removes $l_0$ .

\begin{align*}\Delta t &= \frac{\Delta t_0 + v\,\Delta x_0/c^{2}}{\sqrt{1-\beta^{2}}} \\ &= \frac{0 + v l_0/c^{2}}{\sqrt{1-\beta^{2}}} \\ &= \frac{v}{c^{2}\sqrt{1-\beta^{2}}}\cdot\frac{l}{\sqrt{1-\beta^{2}}} \\ &= \frac{v\,l}{c^{2}\,(1-\beta^{2})}\end{align*}

One power of $c$ pairs with $v$ to make $\beta$ and the other is left carrying the units, which is the form the data go into.

\Delta t = \frac{(120\ \mathrm{m})(0.990)} {(3.00\times 10^{8}\ \mathrm{m/s})(1-0.9801)} = \frac{118.8\ \mathrm{m}}{(3.00\times 10^{8}\ \mathrm{m/s})(0.0199)}

\Delta t = \dfrac{l\beta}{c\,(1-\beta^{2})} = 20\ \mu\mathrm{s}

(2)

Step 3Which decay comes later in

K

World lines in K, drawn at β = 0.6 for legibility. Particle 1 decays at O, particle 2 at the upper dot, both on the dashed x' axis, the particles' line of constant time. — World lines in $K$ , drawn at $\beta = 0.6$ for legibility. Particle $1$ decays at $O$ , particle $2$ at the upper dot, both on the dashed $x'$ axis, the particles’ line of constant time.

Only the sign of the transformed interval is left to read. Both $v$ and $\Delta x_0$ are positive with the leading particle taken first, so $\Delta t$ came out positive, and $\Delta t$ is the leading particle’s decay time minus the trailing particle’s.

\Delta t = t_{\text{lead}} - t_{\text{trail}} > 0

The leading particle decays later in the frame

K

Check

Two decays that are simultaneous in some inertial frame must be separated by a spacelike interval, and the numbers say they are. The same transformation gives their separation in $K$ as $\Delta x = l_0/\sqrt{1-\beta^{2}} = l/(1-\beta^{2}) = 6.03$ km, while $c\,\Delta t = (3.00\times 10^{8})(1.99\times 10^{-5}) = 5.97$ km, so $\Delta x$ exceeds $c\,\Delta t$ . Their ratio is $c\,\Delta t/\Delta x = \beta = 0.990$ , which is below unity for every sub-luminal $v$ , so no signal could have run from one decay to the other.

Remark

The stagger is the relativity of simultaneity magnified twice over, each magnification a factor $1/\sqrt{1-\beta^{2}}$ : the transformation carries one in front of its $v\,\Delta x_0/c^{2}$ term, so the $K_0$ simultaneity is staggered in $K$ by $vl_0/(c^{2}\sqrt{1-\beta^{2}})$ , and $l_0 = l/\sqrt{1-\beta^{2}}$ supplies the other, which is how the product $1/(1-\beta^{2})$ , about $50$ here, gets in. Inside $K$ alone the same result is the bare $\Delta t = v\,\Delta x/c^{2}$ with $\Delta x$ the decays’ own separation in $K$ : the particles stay $l$ apart at one instant of $K$ , but the later decay waits while its particle runs a further $v\,\Delta t$ , so $\Delta x = l + v\,\Delta t = l/(1-\beta^{2})$ , the check’s $6.03$ km. The order is not absolute either: since the two decays are spacelike separated, a frame moving along the positive $x$ direction faster than the particles themselves reverses it and reports the trailing particle decaying later, while the particles’ own frame sits exactly on the boundary and calls the two simultaneous.