Irodov 1.355: The Point Where Both Clocks Agree – Amit & Deepali Sinha Foundation

Problem Irodov 1.355

A frame $K'$ slides along the $x$ axis of a frame $K$ in the positive $x$ direction with a constant relative velocity $V$ , the axes of the two frames kept parallel. Each frame carries its own array of clocks, synchronized throughout that frame, and at the instant the two origins $O$ and $O'$ coincide the clock at $O$ and the clock at $O'$ both read zero. There is one point of the $x$ axis at which the reading of the $K$ clock sitting there and the reading of the $K'$ clock passing there are the same at every moment. Writing $\beta = V/c$ , find the velocity $\dot x$ with which that point moves in the frame $K$ , and demonstrate that $\dot x < V$ .

The two frames in the standard configuration: the x and x' axes coincide, the y and y' axes parallel, and K' sliding along the shared axis at the constant V. The clocks at O and O' read zero as the origins pass. — The two frames in the standard configuration: the $x$ and $x'$ axes coincide, the $y$ and $y'$ axes parallel, and $K'$ sliding along the shared axis at the constant $\vect V$ . The clocks at $O$ and $O'$ read zero as the origins pass.

Given

$V$ , the velocity of $K'$ along $+x$
$\beta = V/c$
clocks at $O$ and $O'$ read zero when the origins meet

Find

the velocity

\dot x

of the point of equal readings, and a demonstration that

\dot x < V

Solution

Step 1Write down what “the same reading” means

A point of the $x$ axis carries a $K$ clock reading the $K$ time $t$ , and at each moment some $K'$ clock is passing it, reading the $K'$ time $t'$ of that same event. The Lorentz transformation gives that $K'$ reading from the event’s $K$ coordinates.

t' = \frac{t - Vx/c^{2}}{\sqrt{1-\beta^{2}}}

The point asked for is the one where the two readings coincide at every moment, so the condition to impose is $t' = t$ .

t = \frac{t - Vx/c^{2}}{\sqrt{1-\beta^{2}}}

Step 2Solve the condition for the position at each instant

Clearing the denominator and collecting the terms in $t$ leaves the position of the agreeing point as an explicit function of the time.

\begin{align*}t\sqrt{1-\beta^{2}} &= t - \frac{Vx}{c^{2}}, \\ \frac{Vx}{c^{2}} &= t\left(1-\sqrt{1-\beta^{2}}\right)\end{align*}

x = \dfrac{c^{2}t}{V}\left(1-\sqrt{1-\beta^{2}}\right)

(1)

Step 3Read off the velocity of that point

Everything on the right of (1) except $t$ is a constant, so the point moves uniformly along the $x$ axis and its velocity is that constant coefficient.

\dot x = \frac{dx}{dt} = \frac{c^{2}}{V}\left(1-\sqrt{1-\beta^{2}}\right)

One of the two powers of $c$ pairs with $V$ to build $\beta$ , which puts the result in terms of that one parameter.

\frac{c^{2}}{V} = \frac{c}{V/c} = \frac{c}{\beta}

\dot x = \dfrac{c}{\beta}\left(1-\sqrt{1-\beta^{2}}\right)

(2)

The ratio x/V that the boxed velocity makes, plotted against β. It starts at 1/2 for a slow boost and climbs toward the dotted asymptote at 1 as β approaches unity. — The ratio $\dot x/V$ that the boxed velocity makes, plotted against $\beta$ . It starts at $1/2$ for a slow boost and climbs toward the dotted asymptote at $1$ as $\beta$ approaches unity.

Step 4Demonstrate that the point lags the moving origin

Comparing $\dot x$ with $V$ directly means comparing $1-\sqrt{1-\beta^{2}}$ with $\beta^{2}$ , which a difference of two squares settles at once. Multiplying above and below by $1+\sqrt{1-\beta^{2}}$ ,

\begin{align*}1-\sqrt{1-\beta^{2}} &= \frac{\left(1-\sqrt{1-\beta^{2}}\right) \left(1+\sqrt{1-\beta^{2}}\right)}{1+\sqrt{1-\beta^{2}}} \\ &= \frac{1-(1-\beta^{2})}{1+\sqrt{1-\beta^{2}}} = \frac{\beta^{2}}{1+\sqrt{1-\beta^{2}}}\end{align*}

Putting that into (2),

\dot x = \frac{c}{\beta}\cdot\frac{\beta^{2}}{1+\sqrt{1-\beta^{2}}}

and $c\beta$ is $V$ itself. For any real relative motion $0 < V < c$ the radical is a positive number, so the denominator is strictly larger than one.

\dot x = \dfrac{V}{1+\sqrt{1-\beta^{2}}} < V

(3)

World lines from the coincidence event, at β = 0.6. The royal line is the origin O' of K', the gold line the point P of equal readings, the gray ct axis the origin O of K. The gold line is steeper. — World lines from the coincidence event, at $\beta = 0.6$ . The royal line is the origin $O'$ of $K'$ , the gold line the point $P$ of equal readings, the gray $ct$ axis the origin $O$ of $K$ . The gold line is steeper.

Check

The two boxed forms are the same number. At $\beta = 0.6$ the first gives $\dot x = (c/0.6)(1-0.8) = 0.333c$ and the second gives $\dot x = 0.6c/1.8 = 0.333c$ , which is the slope the chart above is drawn at. Dimensionally $c/\beta$ is a speed times a pure number, as $\dot x$ must be, and since the denominator of the second form lies between $1$ and $2$ , $\dot x$ is always between $V/2$ and $V$ , hence below $c$ .

Remark

In the frame $K$ the agreeing point sits between the two origins at every instant, and always at the same fraction of the way across: dividing (3) by $V$ gives $\dot x/V = 1/(1+\sqrt{1-\beta^{2}})$ , which is the fraction of the distance $O O'$ at which the point rides. For a slow boost that fraction is $1/2$ , so the clocks agree at the midpoint between the origins, which is the Newtonian picture recovering as it should: in that limit all the clocks of both frames agree everywhere, and the midpoint is the surviving trace of it. As $\beta$ climbs toward one the fraction grows toward the whole distance and the agreeing point crowds up against $O'$ itself. None of this makes the two frames agree about anything else: what has been found is a single moving place where two particular clocks show the same number, not a shared time.