Irodov 1.358: The Transverse Component Changes Too – Amit & Deepali Sinha Foundation

Problem Irodov 1.358

A particle travels in the $xy$ plane of an inertial frame $K$ , and in that frame the two components of its velocity are $v_x$ and $v_y$ . A second inertial frame $K'$ has its axes parallel to those of $K$ and slides relative to $K$ at the constant velocity $V$ along the positive direction of the $x$ axis. Find the magnitude $v'$ of the particle’s velocity as measured in $K'$ .

The standard configuration: the axes of K' parallel to those of K, the origins separating along the shared x direction at V. The particle at P carries the velocity v of frame K, with dashed components v_x and v_y. — The standard configuration: the axes of $K'$ parallel to those of $K$ , the origins separating along the shared $x$ direction at $\vect V$ . The particle at $P$ carries the velocity $\vect v$ of frame $K$ , with dashed components $v_x$ and $v_y$ .

Given

$v_x$ , $v_y$ in frame $K$
$\vect V$ along $+x$ of $K$
axes of $K'$ parallel to those of $K$

Find

v'

, the particle’s speed in

K'

Solution

Step 1Write the boost as a relation between neighboring events

Two neighboring points of the particle’s path are two events, and a velocity component is one coordinate difference over the time difference between them. So the object to transform is the set of differentials, not the velocity itself. For the standard configuration the Lorentz transformation from $K$ to $K'$ acts on them as follows, with the transverse coordinate untouched.

\begin{aligned} dx' &= \frac{dx-V\,dt}{\sqrt{1-\dfrac{V^{2}}{c^{2}}}}, \qquad dy' = dy, \\[6pt] dt' &= \frac{dt-\dfrac{V}{c^{2}}\,dx}{\sqrt{1-\dfrac{V^{2}}{c^{2}}}} . \end{aligned}

Step 2Divide out the two velocity components in

K'

Dividing $dx'$ by $dt'$ cancels the square root entirely, while dividing $dy'$ by $dt'$ leaves one factor of it in the numerator. In both quotients divide top and bottom by $dt$ to bring $v_x = dx/dt$ and $v_y = dy/dt$ into view.

\begin{aligned} v_x' = \frac{dx'}{dt'} &= \frac{dx-V\,dt}{dt-\dfrac{V}{c^{2}}dx} = \frac{\dfrac{dx}{dt}-V}{1-\dfrac{V}{c^{2}}\dfrac{dx}{dt}}, \\[6pt] v_y' = \frac{dy'}{dt'} &= \frac{dy\sqrt{1-\dfrac{V^{2}}{c^{2}}}}{dt-\dfrac{V}{c^{2}}dx} = \frac{\dfrac{dy}{dt}\sqrt{1-\dfrac{V^{2}}{c^{2}}}} {1-\dfrac{V}{c^{2}}\dfrac{dx}{dt}} . \end{aligned}

Naming the two derivatives gives the pair of transformation rules the rest of the problem uses.

v_x' = \frac{v_x-V}{1-\dfrac{v_xV}{c^{2}}}, \qquad v_y' = \frac{v_y\sqrt{1-\dfrac{V^{2}}{c^{2}}}}{1-\dfrac{v_xV}{c^{2}}}

(1)

Step 3Assemble the two components into a speed

The particle's velocity v' in K', with its components v_x' and v_y' along the primed axes. The speed asked for is the length of v'. — The particle’s velocity $\vect v'$ in $K'$ , with its components $v_x'$ and $v_y'$ along the primed axes. The speed asked for is the length of $\vect v'$ .

The speed is the magnitude of the velocity vector in $K'$ , and its two components share the same denominator, so the sum of their squares is a single fraction.

\begin{aligned} v'^{2} &= v_x'^{2}+v_y'^{2} \\[4pt] &= \frac{(v_x-V)^{2}}{\left(1-\dfrac{v_xV}{c^{2}}\right)^{2}} + \frac{v_y^{2}\left(1-\dfrac{V^{2}}{c^{2}}\right)} {\left(1-\dfrac{v_xV}{c^{2}}\right)^{2}} \\[4pt] &= \frac{(v_x-V)^{2}+v_y^{2}\left(1-\dfrac{V^{2}}{c^{2}}\right)} {\left(1-\dfrac{v_xV}{c^{2}}\right)^{2}} \end{aligned}

Taking the square root is unambiguous: the denominator $1-v_xV/c^{2}$ is positive for every pair of subluminal speeds, so it comes out of the root with its own sign and no modulus is needed.

v' = \frac{\sqrt{(v_x-V)^{2} +v_y^{2}\left(1-\dfrac{V^{2}}{c^{2}}\right)}} {1-\dfrac{v_xV}{c^{2}}}

(2)

Check

Send the particle at the speed of light in $K$ , so that $v_x^{2}+v_y^{2} = c^{2}$ , and expand the numerator using $v_y^{2} = c^{2}-v_x^{2}$ :

\begin{aligned} (v_x-V)^{2}+\left(c^{2}-v_x^{2}\right)\left(1-\frac{V^{2}}{c^{2}}\right) &= c^{2}-2v_xV+\frac{v_x^{2}V^{2}}{c^{2}} \\ &= c^{2}\left(1-\frac{v_xV}{c^{2}}\right)^{2} . \end{aligned}

The root is therefore $c\left(1-v_xV/c^{2}\right)$ , which cancels the denominator exactly and leaves $v' = c$ for every $V$ and for every split of the light speed between the two components. Two further limits: $V = 0$ returns $v' = \sqrt{v_x^{2}+v_y^{2}}$ , and letting $c$ grow without bound returns the Galilean $v' = \sqrt{(v_x-V)^{2}+v_y^{2}}$ .

Remark

The transverse component is the interesting one. Its coordinate obeys $y' = y$ , so a reader who transforms the coordinate and forgets the clock concludes that $v_y' = v_y$ , and (1) says otherwise. Nor is the correction always a reduction. The ratio $v_y'/v_y$ is $\sqrt{1-V^{2}/c^{2}}\big/\left(1-v_xV/c^{2}\right)$ , which for a particle with no motion along the boost ( $v_x = 0$ ) is the familiar shrinkage, and which exceeds unity exactly when $v_x/c > \left(1-\sqrt{1-V^{2}/c^{2}}\right)\big/(V/c)$ . At $V = 0.90c$ that threshold is $v_x = 0.63c$ , so at $v_x = V = 0.90c$ the transverse component measured in $K'$ is $2.3$ times the one measured in $K$ , while the total speed still stays under $c$ .