Irodov 1.340: Small Contraction, Not Small Speed – Amit & Deepali Sinha Foundation

Problem Irodov 1.340

A straight rod travels at a constant velocity $v$ directed along its own length, relative to the inertial reference frame $K$ . Its proper length, the length measured in the frame in which the rod is at rest, is $l_0$ ; the length $K$ measures is $l$ . Find the value of $v$ for which $l$ falls short of $l_0$ by the fraction $\eta = 0.5\% = 0.005$ , that is, for which $l = (1-\eta)\,l_0$ . Take the speed of light to be $c$ .

Frame K at one instant, drawn to scale. Both rods have proper length l_0: the lower is at rest, the upper slides at the unknown v, so K measures it as the shorter l. The two dimension brackets differ by η. — Frame $K$ at one instant, drawn to scale. Both rods have proper length $l_0$ : the lower is at rest, the upper slides at the unknown $\vect v$ , so $K$ measures it as the shorter $l$ . The two dimension brackets differ by $\eta$ .

Given

$\eta = 0.5\% = 0.005$
$l = (1-\eta)\,l_0$
$c = 3.00\times10^{8}$ m/s

Find

the speed

v

of the rod relative to

K

Solution

Step 1The length

K

measures, in terms of the proper length

The rod moves along its own length, so the whole of it lies along the boost direction and the contraction rule applies to it as one length with no resolving into components. Writing $\beta = v/c$ , the length measured in $K$ is the proper length reduced by the contraction factor.

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

Step 2Turn the stated shortfall into a condition on

\beta

The problem fixes the ratio of the two lengths rather than either of them, so the next move is to write the stated shortfall as that ratio and let $l_0$ cancel. The condition is

l = (1-\eta)\,l_0 .

Setting this equal to (1) and dividing through by $l_0$ leaves the contraction factor alone on one side.

\sqrt{1-\beta^2} = 1-\eta

(2)

Step 3Solve the condition for the speed

Squaring (2) and expanding the right-hand side turns the condition into a plain quadratic statement about $\beta^2$ .

\begin{aligned} 1-\beta^2 &= (1-\eta)^2 \\[2pt] \beta^2 &= 1-\left(1-2\eta+\eta^2\right) \\[2pt] \beta^2 &= \eta\,(2-\eta) \end{aligned}

Putting the given $\eta$ into that product and taking its square root gives the numerical value of $\beta$ .

\eta\,(2-\eta) = 0.005\times 1.995 = 0.009975, \qquad \sqrt{0.009975} = 0.09987

The speed itself is $c$ times that number. Numerically,

v = 0.0999\,c

v = c\,\sqrt{\eta\,(2-\eta)} = 3.00\times10^{7}\ \text{m/s}

(3)

The speed demanded, v/c, against the fractional shortening η, over the range η = 0 to 0.02. The curve leaves the origin vertically. The marked point P is this problem's η = 0.005, already at v = 0.10c. — The speed demanded, $v/c$ , against the fractional shortening $\eta$ , over the range $\eta = 0$ to $0.02$ . The curve leaves the origin vertically. The marked point $P$ is this problem’s $\eta = 0.005$ , already at $v = 0.10c$ .

Step 4How steeply the required speed grows with the shortening

The plotted curve rises almost vertically out of the origin, and that shape is worth having in closed form, so expand the boxed result for a shortening much smaller than unity. With $\eta \ll 1$ the second factor differs from $2$ by only $\eta$ , so

\eta\,(2-\eta) \approx 2\eta ,

and the speed follows the square root of the shortening alone.

v \approx c\,\sqrt{2\eta}

(4)

At $\eta = 0.005$ this returns $0.100c$ against the exact $0.0999c$ , an error of one part in a thousand, so the square-root law is already accurate at the shortening the problem sets.

Check

Substitute the answer back into the contraction rule and see whether the stated shortfall comes out. With $\beta = 0.09987$ , $\beta^2 = 0.009975$ and $\sqrt{1-0.009975} = \sqrt{0.990025} = 0.99500$ , so $l = 0.995\,l_0$ , exactly $0.5\%$ short of the proper length as required. The limits also behave: $\eta \to 0$ gives $v \to 0$ , and $\eta \to 1$ , a rod contracted to nothing, gives $v \to c$ , the only speed at which that could happen.

Remark

The same equation answers a question about clocks. A clock moving at $\beta$ relative to $K$ advances by the factor $\sqrt{1-\beta^2}$ per unit of $K$ ’s time, which is the identical factor that appears in (1), so a clock that has fallen behind $K$ ’s clocks by the fraction $\eta$ of the elapsed time satisfies (2) word for word and demands exactly this same speed. One number, $\sqrt{1-\beta^2}$ , sets both the shortening of a rod along the motion and the slowing of a clock riding with it. It is also why the effect stayed hidden for so long: for $\beta \ll 1$ the small-shortening law (4) inverts to $\eta \approx \beta^2/2$ , so a body in low Earth orbit, at about $8$ km/s, has $\beta = 2.7\times10^{-5}$ and $\eta \approx 4\times10^{-10}$ , four tenths of a nanometer on a rod one meter long as measured from the ground.