Irodov 1.349: Two Simultaneities, Two Lengths, One Rod

Problem Irodov 1.349

A straight rod slides along a stationary ruler at a constant velocity $v$ directed along its own length, and every mark is read off that one ruler. In the first procedure the positions of the rod’s two ends are marked at one and the same instant of the ruler’s frame, and the two marks differ by $\Delta x_1 = 4.0$ m on the ruler. In the second procedure the positions of the same two ends are marked at one and the same instant of the rod’s frame, and those two marks differ by $\Delta x_2 = 9.0$ m on the same ruler. Writing $\beta = v/c$ , find the proper length $l_0$ of the rod and its velocity $v$ relative to the ruler.

Both markings in the ruler frame, the royal lines being the rod's ends. O and P are marked at one ruler instant, Δ x_1 apart; O and Q at one rod instant, on the tilted x' axis, Q dropped to x to give Δ x_2. — Both markings in the ruler frame, the royal lines being the rod’s ends. $O$ and $P$ are marked at one ruler instant, $\Delta x_1$ apart; $O$ and $Q$ at one rod instant, on the tilted $x'$ axis, $Q$ dropped to $x$ to give $\Delta x_2$ .

Given

$\Delta x_1 = 4.0$ m
$\Delta x_2 = 9.0$ m
$\beta = v/c$
$c = 3.00\times 10^{8}$ m/s

Find

the proper length

l_0

of the rod and its velocity

v

relative to the ruler

Solution

Step 1The ruler-simultaneous marking returns the contracted length

The first pair of marking events is simultaneous in the ruler frame $K$ , so $\Delta t = 0$ and their separation there is the reading difference $\Delta x_1$ . In the rod frame $K'$ those same two events sit at the rod’s two ends, a proper length $l_0$ apart. The Lorentz transformation from $K$ to $K'$ ties the two together.

\begin{align*}\Delta x' &= \frac{\Delta x - v\,\Delta t}{\sqrt{1-\beta^{2}}}, \\ l_0 &= \frac{\Delta x_1 - v\cdot 0}{\sqrt{1-\beta^{2}}}\end{align*}

\Delta x_1 = l_0\sqrt{1-\beta^{2}}

(1)

Step 2The rod-simultaneous marking returns a longer stretch of ruler

The second procedure from the rod: the rod at rest at its proper length l_0, the ruler streaming left, the marked stretch drawn contracted. It reads Δ x_2 on the ruler's graduations yet covers exactly the rod: the two brackets come out equal. — The second procedure from the rod: the rod at rest at its proper length $l_0$ , the ruler streaming left, the marked stretch drawn contracted. It reads $\Delta x_2$ on the ruler’s graduations yet covers exactly the rod: the two brackets come out equal.

The second pair of marking events is simultaneous in the rod frame $K'$ and separated there by the rod’s own proper length, so $\Delta t' = 0$ and $\Delta x' = l_0$ . Their ruler readings are their $x$ coordinates in the ruler frame $K$ , so the Lorentz transformation from $K'$ to $K$ delivers the answer directly.

\begin{align*}\Delta x_2 &= \frac{\Delta x' + v\,\Delta t'}{\sqrt{1-\beta^{2}}} \\ &= \frac{l_0 + v\cdot 0}{\sqrt{1-\beta^{2}}}\end{align*}

\Delta x_2 = \dfrac{l_0}{\sqrt{1-\beta^{2}}}

(2)

Step 3The product and the quotient separate the two unknowns

Equations (1) and (2) carry the two unknowns $l_0$ and $\beta$ in a form that a product and a quotient untangle at once, the product killing the radical and the quotient killing $l_0$ .

\begin{aligned} \Delta x_1\,\Delta x_2 &= l_0\sqrt{1-\beta^{2}}\cdot \frac{l_0}{\sqrt{1-\beta^{2}}} = l_0^{2}, \\ \frac{\Delta x_1}{\Delta x_2} &= \frac{l_0\sqrt{1-\beta^{2}}} {l_0/\sqrt{1-\beta^{2}}} = 1-\beta^{2} \end{aligned} \tag{3}

Taking the root of the first and putting in the data,

l_0 = \sqrt{(4.0\ \mathrm{m})(9.0\ \mathrm{m})}

l_0 = \sqrt{\Delta x_1\,\Delta x_2} = 6.0\ \mathrm{m}

(4)

Step 4The velocity from the ratio of the two readings

The second line of (3) already isolates $\beta^{2}$ , so only a square root and a multiplication by $c$ are left. Evaluating the ratio first,

\begin{align*}\beta^{2} &= 1-\frac{\Delta x_1}{\Delta x_2} = 1-\frac{4.0\ \mathrm{m}}{9.0\ \mathrm{m}} = 0.5556, \\ \beta &= 0.7454\end{align*}

Restoring the factor of $c$ that turns the dimensionless $\beta$ into a speed,

v = (0.7454)(3.00\times 10^{8}\ \mathrm{m/s})

v = c\sqrt{1-\Delta x_1/\Delta x_2} = 2.2\times 10^{8}\ \mathrm{m/s}

(5)

Check

Put the two answers back into the two measurements. With $l_0 = 6.0$ m and $\sqrt{1-\beta^{2}} = 0.6667$ , (1) gives $\Delta x_1 = (6.0)(0.6667) = 4.0$ m and (2) gives $\Delta x_2 = 6.0/0.6667 = 9.0$ m, both as stated. The units also check: each answer is built from lengths alone, so $l_0$ is a length and $\beta$ is a pure number.

Remark

The proper length is the geometric mean of the two ruler readings, and it must lie between them: the ruler-simultaneous reading is short of $l_0$ by the factor $\sqrt{1-\beta^{2}}$ and the rod-simultaneous reading is long by the same factor. Only the first of the two numbers is a length of the rod in any frame at all. The second is a stretch of the ruler, and it exceeds $l_0$ for the reason the second figure shows: in the rod’s frame it is the ruler’s graduations that are contracted, so it takes $\Delta x_2 = 9.0$ m of ruler markings to span the $6.0$ m the rod occupies there.