Irodov 1.364: A Rod Tilted by Disagreeing Clocks – Amit & Deepali Sinha Foundation

Problem Irodov 1.364

Two inertial frames $K$ and $K'$ are in the standard configuration: their $x$ and $x'$ axes coincide, their $y$ and $y'$ axes are parallel, and $K'$ travels along the common $x$ axis in the positive direction at the constant speed $V$ . Write $\beta = V/c$ . A straight rod $AB$ lies parallel to the $x'$ axis, and in the frame $K'$ it slides at the constant speed $v'$ along the $y'$ direction, staying parallel to $x'$ as it goes. Find the angle $\theta$ that the rod makes with the $x$ axis as measured in the frame $K$ .

The rod as K' sees it: parallel to the x' axis, end A on the y' axis, spanning l' to end B, sliding along y' at v'. The frame K' is carried along the shared x axis at V. — The rod as $K'$ sees it: parallel to the $x'$ axis, end $A$ on the $y'$ axis, spanning $l'$ to end $B$ , sliding along $y'$ at $\vect v'$ . The frame $K'$ is carried along the shared $x$ axis at $\vect V$ .

Given

rod $AB$ parallel to $x'$
rod speed $v'$ along $y'$ in $K'$
$V$ along the shared $x$ axis
$\beta = V/c$

Find

the angle

\theta

between the rod and the

x

axis in

K

Solution

Step 1Describe the rod by its two ends in

K'

An orientation is a ratio of two separations, so the rod has to be handed to the algebra as a pair of end events rather than as an object. In $K'$ the two ends keep fixed abscissae $x_A'$ and $x_B'$ , separated by the rod’s own length $l' = x_B'-x_A'$ , and they share one ordinate that slides along $y'$ at the constant speed $v'$ , in the negative $y'$ sense as drawn.

x_B'-x_A' = l', \qquad y'(t') = y_0'-v't'

Step 2Transform one end event into

K

The rod’s orientation in $K$ must be read off two end positions taken at the same time in $K$ , so the transformation is needed in the direction that takes a $K'$ event to a $K$ event, and then it must be solved for the primed time that a given $K$ time corresponds to. The Lorentz transformation from $K'$ to $K$ for the standard configuration is

x = \frac{x'+Vt'}{\sqrt{1-\beta^2}}, \qquad y = y', \qquad t = \frac{t'+\dfrac{Vx'}{c^2}}{\sqrt{1-\beta^2}} .

Solving the time relation for $t'$ gives the primed instant at which an end sitting at $x'$ is caught by the $K$ clock reading $t$ .

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

(1)

Step 3The two separations at one instant of

K

Now fix one instant $t$ in $K$ and evaluate both ends there. For the $x$ separation, substitute (1) into the transformation of $x$ and collect the terms in $x'$ .

\begin{aligned} x &= \frac{x'+V\left(t\sqrt{1-\beta^2}-\dfrac{Vx'}{c^2}\right)} {\sqrt{1-\beta^2}} \\[3pt] &= \frac{x'\left(1-\beta^2\right)}{\sqrt{1-\beta^2}}+Vt \\[3pt] &= x'\sqrt{1-\beta^2}+Vt \end{aligned}

The common drift $Vt$ cancels in the difference between the ends, which leaves the familiar contraction of the rod’s span along the direction of motion.

\Delta x = x_B-x_A = l'\sqrt{1-\beta^2} \tag{2}

For the $y$ separation, the ordinate is untouched by the boost, but the two ends are caught at different primed times, and it is that difference which the rod’s own motion turns into a height difference.

\begin{aligned} t_B'-t_A' &= -\frac{V\left(x_B'-x_A'\right)}{c^2} = -\frac{Vl'}{c^2} \\[3pt] \Delta y &= y_B-y_A = -v'\left(t_B'-t_A'\right) \end{aligned}

\Delta y = \frac{v'Vl'}{c^2}

(3)

Step 4The tilt the two separations define

The rod in $K$ joins the two simultaneous end positions, so its inclination is the ratio of (3) to (2). Both carry the factor $l'$ , which therefore drops out and leaves an angle independent of how long the rod is.

The rod in K, the hypotenuse joining its two ends at one instant of K: Δ x is the contracted span, Δ y the height the clock disagreement builds up. The angle θ at A is the tilt asked for. — The rod in $K$ , the hypotenuse joining its two ends at one instant of $K$ : $\Delta x$ is the contracted span, $\Delta y$ the height the clock disagreement builds up. The angle $\theta$ at $A$ is the tilt asked for.

\begin{aligned} \tan\theta &= \frac{\Delta y}{\Delta x} \\[3pt] &= \frac{\dfrac{v'Vl'}{c^2}}{l'\sqrt{1-\beta^2}} \end{aligned}

Canceling $l'$ and writing the root out in terms of $V$ and $c$ gives the tilt.

\tan\theta = \frac{v'V}{c^2\sqrt{1-\left(V/c\right)^2}}

(4)

Check

The right-hand side is a speed times a speed divided by $c^2$ , so it is dimensionless, as a tangent must be. Two limits: with $V = 0$ the frames coincide and $\tan\theta = 0$ , the rod staying parallel to $x$ ; with $v' = 0$ the rod is at rest in $K'$ and again $\tan\theta = 0$ , since a rod at rest parallel to $x'$ is contracted but not turned. Both effects have to be present for a tilt to appear, which is what the product $v'V$ in the numerator says.

Remark

The tilt is not something the rod does; it is what simultaneity does. The height difference (3) is built entirely out of the clock disagreement $t_B'-t_A' = -Vl'/c^2$ between the two ends, which is negative: an observer in $K$ catches the leading end $B$ at an earlier moment of the rod’s own history than the trailing end $A$ . The rod is sliding in the negative $y'$ sense, so the end caught earlier has fallen less far and is still the higher of the two, and a rod moving sideways is therefore photographed at two different heights. Length contraction then sharpens the effect rather than causing it: without the factor $\sqrt{1-\beta^2}$ in $\Delta x$ the tangent would be only $v'V/c^2$ . There is a second casualty of the same disagreement. In $K'$ the rod is exactly perpendicular to its own velocity, while in $K$ its velocity has components $\left(V,\,-v'\sqrt{1-\beta^2}\right)$ and the rod points along $\left(\sqrt{1-\beta^2},\,v'V/c^2\right)$ ; the scalar product of those two is $V\left(c^2-v'^2\right)\sqrt{1-\beta^2}\big/c^2$ , which is not zero for any $v' < c$ so long as $V \neq 0$ ; at $V = 0$ the two frames coincide and the right angle survives. So in the frame $K$ the rod is not perpendicular to the direction it is traveling in, and a right angle between two directions is not something the boost preserves.