Irodov 1.365: The Clock Transforms Too – Amit & Deepali Sinha Foundation

Problem Irodov 1.365

A frame $K'$ slides at a constant velocity $\vect V$ along the $x$ axis of a frame $K$ . In $K$ , a particle moves along a straight line with instantaneous speed $v$ and acceleration $w$ . Find the particle’s acceleration $w'$ as measured in $K'$ , in each of the following two cases:

the particle’s straight-line path runs along the direction of $\vect{V}$ ;
the particle’s straight-line path runs perpendicular to $\vect{V}$ .

Given

$\vect V$ along the shared $x$ axis
$\beta = V/c$
particle velocity $v$ and acceleration $w$ in $K$ , along one straight line

Find

Find

w'

, the particle’s acceleration in

K'

, in each of the two orientations

Solution

Step 1Collect the two transformation rules an acceleration needs

An acceleration is a velocity difference divided by a time difference, so both have to be transformed, and it is the second one that people forget. Take the axes in the standard configuration, with $\vect V$ along $x$ . The velocity components transform as

u_x' = \frac{u_x-V}{1-\dfrac{u_xV}{c^2}}, \qquad u_y' = \frac{u_y\sqrt{1-\beta^2}}{1-\dfrac{u_xV}{c^2}} ,

and the time interval between two neighboring events on the particle’s path transforms as

dt' = \frac{dt-\dfrac{V}{c^2}\,dx}{\sqrt{1-\beta^2}}

(1)

Step 2Case (a): the motion runs along the boost

Here $u_x = v$ and $u_y = 0$ , so the particle keeps to the $x$ axis in both frames and only the first transformation is needed.

Case (a). Velocity and acceleration both point along V, so the particle's own displacement enters the clock transformation. — Case (a). Velocity and acceleration both point along $\vect V$ , so the particle’s own displacement enters the clock transformation.

Put $dx = v\,dt$ into (1), so that the particle’s own displacement enters the clock relation.

dt' = \frac{dt-\dfrac{V}{c^2}\,v\,dt}{\sqrt{1-\beta^2}}

Taking $dt$ out of the numerator and writing $Vv/c^2$ as $\beta v/c$ puts the primed interval in the two ratios the answer will be built from.

dt' = \frac{dt\left(1-\dfrac{\beta v}{c}\right)} {\sqrt{1-\beta^2}}

(2)

Now differentiate the velocity transformation with respect to $v$ , since $v$ is what the acceleration changes. By the quotient rule,

\begin{aligned} \frac{du_x'}{dv} &= \frac{\left(1-\dfrac{vV}{c^2}\right)\cdot 1 -\left(v-V\right)\left(-\dfrac{V}{c^2}\right)} {\left(1-\dfrac{vV}{c^2}\right)^{2}} \\[4pt] &= \frac{1-\dfrac{vV}{c^2}+\dfrac{vV}{c^2}-\dfrac{V^2}{c^2}} {\left(1-\dfrac{vV}{c^2}\right)^{2}} \\[4pt] &= \frac{1-\beta^2}{\left(1-\dfrac{\beta v}{c}\right)^{2}} . \end{aligned}

The two cross terms in $vV/c^2$ cancel, which is what leaves the numerator free of $v$ altogether. The acceleration in $K$ is $w = dv/dt$ , so over the interval $dt$ the velocity in $K$ changes by $dv = w\,dt$ , and the derivative just found turns that into the change of the primed velocity.

\begin{aligned} du_x' &= \frac{du_x'}{dv}\,dv \\[4pt] &= \frac{\left(1-\beta^2\right)w\,dt} {\left(1-\dfrac{\beta v}{c}\right)^{2}} \end{aligned}

The acceleration in $K'$ is that change divided by the primed interval (2), and the common $dt$ cancels between them.

\begin{aligned} w' &= \frac{du_x'}{dt'} \\[4pt] &= \frac{\left(1-\beta^2\right)w\,dt} {\left(1-\dfrac{\beta v}{c}\right)^{2}} \cdot\frac{\sqrt{1-\beta^2}} {dt\left(1-\dfrac{\beta v}{c}\right)} \end{aligned}

w' = \frac{du_x'}{dt'} = \frac{w\left(1-\beta^2\right)^{3/2}}{\left(1-\dfrac{\beta v}{c}\right)^{3}}

(3)

Step 3Case (b): the motion runs across the boost

Now $u_x = 0$ and $u_y = v$ throughout the motion, because the particle keeps to a straight line perpendicular to $\vect V$ .

Case (b). The particle's displacement along x is zero, so the clock transformation reduces to plain time dilation. In K' the particle also drifts backwards at V, which tilts v' but contributes nothing to w'. — Case (b). The particle’s displacement along $x$ is zero, so the clock transformation reduces to plain time dilation. In $K'$ the particle also drifts backwards at $V$ , which tilts $\vect v'$ but contributes nothing to $\vect w'$ .

The two transformations simplify at once. Since $u_x = 0$ the denominators become $1$ , so

u_x' = -V \quad\text{(a constant)}, \qquad u_y' = v\sqrt{1-\beta^2} ,

and since $dx = 0$ the clock relation (1) loses its second term and becomes plain time dilation.

dt' = \frac{dt}{\sqrt{1-\beta^2}}

The $x$ component of the primed velocity is a constant, so it contributes nothing to the acceleration, and only the transverse component is left to differentiate. It carries the constant factor $\sqrt{1-\beta^2}$ , so its change over the interval is that factor times the change $dv$ in $K$ , and dividing that by the primed interval just found gives the acceleration in $K'$ .

\begin{aligned} w' &= \frac{du_y'}{dt'} \\[4pt] &= \frac{\sqrt{1-\beta^2}\,dv}{dt/\sqrt{1-\beta^2}} \\[4pt] &= \left(1-\beta^2\right)\frac{dv}{dt} \end{aligned}

w' = w\left(1-\beta^2\right)

(4)

Check

Set $V = 0$ in both results and each returns $w' = w$ , as any transformation between coincident frames must. A sharper test for case (a): put $V = v$ , so that the particle is momentarily at rest in $K'$ and $w'$ is its proper acceleration. The formula then gives $w' = w\left(1-\beta^2\right)^{3/2}/\left(1-\beta^2\right)^{3} = w/\left(1-v^2/c^2\right)^{3/2}$ , which is exactly the standard relation between proper and coordinate acceleration in straight-line motion. For case (b), the speed in $K'$ comes out as $\sqrt{V^2+v^2\left(1-\beta^2\right)}$ , and this is below $c$ for every $v<c$ , since the condition reduces to $v^2\left(1-\beta^2\right) < c^2\left(1-\beta^2\right)$ .

Remark

The two answers behave quite differently as the motion proceeds. In case (b) nothing on the right-hand side depends on $v$ , so a constant $w$ in $K$ is a constant $w'$ in $K'$ ; combined with the fixed drift $u_x' = -V$ , the straight path of $K$ becomes a parabola in $K'$ . In case (a) the answer carries $v$ explicitly, so a constant $w$ in $K$ is emphatically not a constant $w'$ : the cube in the denominator makes $w'$ grow steadily as $v$ increases. It does not run away, though. At fixed $\beta$ the denominator tends to $\left(1-\beta\right)^{3}$ as $v \to c$ , so $w'$ climbs to the finite ceiling $w\left[\left(1+\beta\right)/\left(1-\beta\right)\right]^{3/2}$ , which is $5.20\,w$ at $\beta = 0.5$ and $82.8\,w$ at $\beta = 0.9$ . It could only diverge if $vV = c^2$ , which is out of reach while both speeds are below $c$ . The genuinely enormous values need $v$ and $V$ both near $c$ , as at $v = V$ , where the check above gives $w' = \gamma^{3}w$ .