Transverse Velocity Transformation

Boost along $x$ and the Lorentz transformation does not touch $y$ . Not approximately, not to leading order — exactly: $y' = y$ , $z' = z$ , for every velocity, always. There is no transverse length contraction, and every textbook says so on the page where the boost is introduced. And yet the transverse velocity changes. A bullet fired straight across a moving train at $0.6c$ in the train's frame crosses the ground observer's field of view at $0.36c$ when the train is doing $0.8c$ — the same crosswise distance, a smaller crosswise speed. Nothing contracted. Something else did the work. That something is the denominator. Velocity is a ratio, $\mathrm{d}y/\mathrm{d}t$ , and relativity left the numerator alone and rebuilt the denominator from scratch. The elapsed time is dilated by $\gamma$ and mixed with position by the relativity of simultaneity, and the transverse velocity inherits both effects though the transverse coordinate inherits neither. So a $\gamma$ appears in a formula about a direction in which $\gamma$ has no business being — and that misplaced $\gamma$ is why a light ray tips forward when you run toward it, why a relativistic jet beams its radiation into a cone of half-angle $1/\gamma$ , and why the sky ahead of a fast starship collapses to a point.

Key takeaways

The convention, stated once and used everywhere. $S$ = lab; $S'$ moves at $+v\,\hat{x}$ ; $\beta = v/c$ , $\gamma = (1-\beta^2)^{-1/2}$ . Signature $(+,-,-,-)$ , timelike-positive, as fixed by The Minkowski Metric.
FORWARD ( $S \to S'$ ), the primary form: $u'_{\parallel} = \dfrac{u_{\parallel} - v}{1 - v u_{\parallel}/c^2}$ and $\boxed{u'_{\perp} = \dfrac{u_{\perp}}{\gamma\left(1 - v u_{\parallel}/c^2\right)}}$ . Both components share the same denominator, and only the transverse one carries the $1/\gamma$ .
INVERSE ( $S' \to S$ ) is the reciprocity corollary — send $v \to -v$ and swap primes: $u_{\perp} = u'_{\perp}/[\gamma(1 + v u'_{\parallel}/c^2)]$ . It is the same rule, not a second one.
The $\gamma$ does not come from a transverse contraction — there is none. It comes entirely from $\mathrm{d}t$ : time dilation supplies the $\gamma$ , the relativity of simultaneity supplies the $(1 - v u_{\parallel}/c^2)$ .
Set $u'_{\parallel} = 0$ and the simultaneity factor dies, leaving the pure time-dilation law $u_{\perp} = u'_{\perp}/\gamma$ — same distance, dilated clock, slower crossing.
Key idea: the transverse rule is exactly the rule that keeps light at $c$ . Feed it $u_x^2 + u_y^2 = c^2$ and the numerator factorises into $c^2$ times the square of the shared denominator, so the $c$ falls out untouched. The $1/\gamma$ is not decoration — it is the unique transverse coefficient that compensates the parallel enhancement, and any other value would break the second postulate.

The core insight

The transverse coordinate is invariant; the transverse velocity is not. That is not a contradiction — it is a statement about what a velocity is. $u_y = \mathrm{d}y/\mathrm{d}t$ has a Lorentz-invariant numerator and a Lorentz-variant denominator, so all the relativity lives downstairs. Read every factor in the transverse rule as a fact about clocks, never about rulers.

Where this lives today

Every neutrino beam on Earth is an application of this formula. At Fermilab, J-PARC and CERN, protons strike a target, the pions and kaons produced are focused by magnetic horns, and are then allowed to decay in flight down a long evacuated pipe. The decay is isotropic in the parent's rest frame — the parent has no idea which way the beam points — and yet the daughters emerge as a beam, because the transverse rule divides every transverse velocity by $\gamma$ while the parallel rule pushes every parallel velocity toward $c$ . A $90^\circ$ decay at $\gamma = 22$ comes out at $0.7^\circ$ : the collimation is free, and the $1/\gamma$ pays for it. The same $1/\gamma$ is why relativistic astrophysical jets look the way they do: radiation emitted isotropically in the plasma's rest frame is swept into a cone of half-angle $\sim 1/\gamma$ in ours — the headlight effect, treated in Relativistic Beaming. It is why a blazar whose jet points at us outshines an otherwise identical source by factors of thousands, and why synchrotron radiation from a storage ring emerges as a pencil of X-rays rather than a glow. In the purely transverse case the same $1/\gamma$ is the kinematic sibling of the transverse Doppler effect, measured by Ives and Stilwell in 1938.

The physical picture

Put a laser on the floor of a train, pointing straight up. In the train's frame $S'$ the beam is purely transverse: $u'_x = 0$ , $u'_y = c$ , crossing a height $h$ in a time $h/c$ . Now stand on the platform. In your frame $S$ the beam has picked up a forward component and travels a diagonal. Its speed is still $c$ — it must be — so the transverse part of that $c$ has to be less than $c$ , by exactly the amount needed to leave room for the forward part. Pythagoras in the lab: $u_x^2 + u_y^2 = c^2$ with $u_x = v$ forces $u_y = c\sqrt{1 - \beta^2} = c/\gamma$ . That is the light-clock, and it is the cleanest demonstration of where the $\gamma$ lives. The height $h$ did not change — $y' = y$ . What changed is the time: the beam takes longer to get there, because it is also going sideways. Same rise, longer run, dilated clock, reduced transverse speed. The $\gamma$ in the transverse velocity rule and the $\gamma$ in $\Delta t = \gamma \Delta \tau$ are the same $\gamma$ , arriving by the same route. The second factor appears only when the object is also moving along the boost. If the bullet has $u'_x \neq 0$ , the two ends of its transverse flight are events separated in $x'$ , and the two frames disagree about how much time elapsed between them — the relativity of simultaneity, owned by Relativity of Simultaneity (S1). That disagreement is the factor $(1 \pm v u'_{\parallel}/c^2)$ , and it is the same factor that sits in the denominator of the collinear rule derived in Relativistic Velocity Addition. Both components share it because both are divided by the same $\mathrm{d}t$ .

Figure

PUNCHLINE — the same γ, twice. LEFT: the light-clock. In the train frame S′ the beam goes straight up at speed c. In the lab S the train's motion tips it forward; the speed is still c, so the transverse part must shrink to c⁄γ. At β = 0.6 the beam climbs at 0.8c in the lab, while the SAME ceiling sits the SAME distance away — y′ = y exactly. RIGHT: the pion beam. A muon emitted at 90° in the pion rest frame at 0.2714c comes out in the lab at only 7.49° from the beam axis, because its transverse speed is divided by γ = 2.2942 while its parallel speed is boosted to 0.9c. Nothing contracted. The clock did all of it

The core equations

Take the boost from Lorentz Transformation Equations, which owns it. In differentials, $S \to S'$ :

\mathrm{d}x' = \gamma(\mathrm{d}x - v\,\mathrm{d}t), \qquad \mathrm{d}y' = \mathrm{d}y, \qquad \mathrm{d}z' = \mathrm{d}z, \qquad \mathrm{d}t' = \gamma\!\left(\mathrm{d}t - \frac{v\,\mathrm{d}x}{c^2}\right) \qquad (1)

where: $\gamma = 1/\sqrt{1-\beta^2}$ , $\beta = v/c$ . Physical reading: note which lines are trivial. Two of the four say nothing at all. All the content is in $\mathrm{d}t'$ — and a velocity divides by $\mathrm{d}t'$ . Divide the transverse line by the time line and you have the rule this article is named for:

\boxed{\ u'_{\perp} = \frac{u_{\perp}}{\gamma\left(1 - \dfrac{v\,u_{\parallel}}{c^2}\right)}\ } \qquad\text{alongside}\qquad u'_{\parallel} = \frac{u_{\parallel} - v}{1 - \dfrac{v\,u_{\parallel}}{c^2}} \qquad (2)

where: $u_{\parallel} = u_x$ is the component along the boost, $u_{\perp} = (u_y, u_z)$ the components across it. Physical reading: this is the FORWARD form, $S \to S'$ — this chapter's primary statement. The two rules are not independent laws; they are the same division by the same $\mathrm{d}t'$ , and the shared denominator $D \equiv 1 - v u_{\parallel}/c^2$ is the fingerprint of that shared origin. The $\gamma$ appears in exactly one of them. The inverse is not a new physical input. It is reciprocity: $S$ moves at $-v$ relative to $S'$ , so send $v \to -v$ and swap primes.

u_{\perp} = \frac{u'_{\perp}}{\gamma\left(1 + \dfrac{v\,u'_{\parallel}}{c^2}\right)}, \qquad u_{\parallel} = \frac{u'_{\parallel} + v}{1 + \dfrac{v\,u'_{\parallel}}{c^2}} \qquad (3)

where: primes now denote the input. Physical reading: (3) is a corollary of (2), and saying so is the point. The principle of relativity licenses the substitution — see Inverse Lorentz Transformation and Reciprocity — and (2) composed with (3) is the identity map, the check that they are one rule seen from two ends. Half the sign errors in relativistic kinematics are somebody quoting (3) while thinking (2). For a velocity in the $xy$ plane, divide the two components of (2). The shared denominator cancels:

\tan\theta' = \frac{u'_y}{u'_x} = \frac{u_y}{\gamma\,(u_x - v)}, \qquad\text{and for light}\quad (u_x, u_y) = c(\cos\theta, \sin\theta): \quad \tan\theta' = \frac{\sin\theta}{\gamma(\cos\theta - \beta)} \qquad (4)

where: $\theta$ is the angle of the velocity vector from the $+x$ axis in $S$ . Physical reading: $D$ cancels, so the direction depends on $\gamma$ and $\beta$ only — and the surviving $\gamma$ downstairs is the whole of relativistic aberration. That application belongs to Relativistic Aberration of Light; equation (4) is the machine it runs on. Finally, the identity that justifies the whole construction, proved as the owned result below:

u_x^2 + u_y^2 = c^2 \qquad\Longrightarrow\qquad u_x'^2 + u_y'^2 = c^2 \qquad (5)

where: both sides are speeds. Physical reading: a light ray stays a light ray — identically, for every $v$ and direction.

Deriving the structure

Derivation AThe transverse rule, and the two distinct sources of its denominator

The formula is three lines of algebra, worth slowing down over, because a student who merely memorises the result invariably mis-remembers where the $\gamma$ sits. The goal is not the formula: it is to see that the numerator is inert while the denominator carries two physically separate effects — time dilation, contributing the $\gamma$ , and the relativity of simultaneity, contributing the $(1 - vu_{\parallel}/c^2)$ . The two can be switched on and off independently, and doing so is the cleanest test that the derivation has been understood rather than copied.

Step 1. Take the differentials of the boost, equation (1). The transverse line is $\mathrm{d}y' = \mathrm{d}y$ — exact, no $\gamma$ , no $\beta$ , no correction of any order. Why: this is the fact that generates all the confusion. There is genuinely no transverse contraction: a metre stick held perpendicular to the boost measures a metre in both frames. If the transverse velocity nevertheless transforms, the reason cannot be a ruler.

Step 2. Form the ratio, and watch the numerator do nothing:

u'_y = \frac{\mathrm{d}y'}{\mathrm{d}t'} = \frac{\mathrm{d}y}{\gamma\left(\mathrm{d}t - \dfrac{v\,\mathrm{d}x}{c^2}\right)}.

Why: the numerator is the invariant $\mathrm{d}y$ . Every relativistic effect in the transverse velocity is therefore, by construction, an effect on the elapsed time. That single observation is the article.

Step 3. Divide numerator and denominator by $\mathrm{d}t$ , and introduce $u_y = \mathrm{d}y/\mathrm{d}t$ , $u_x = \mathrm{d}x/\mathrm{d}t$ :

u'_y = \frac{\mathrm{d}y/\mathrm{d}t}{\gamma\left(1 - \dfrac{v}{c^2}\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)} = \boxed{\ \frac{u_y}{\gamma\left(1 - \dfrac{v u_x}{c^2}\right)}\ }

and the identical manipulation on $\mathrm{d}z' = \mathrm{d}z$ gives $u'_z$ with the same denominator. Why: the boost's own $\gamma$ — the one multiplying $\mathrm{d}t'$ — survives intact, because there was no second $\gamma$ upstairs to cancel it. In the parallel rule there is one: $\mathrm{d}x' = \gamma(\mathrm{d}x - v\,\mathrm{d}t)$ carries a $\gamma$ , the two cancel, and $u'_x$ comes out $\gamma$ -free. That asymmetry is the entire structural difference between the two rules.

Step 4. Separate the two effects by switching one off. Put $u_x = 0$ : the simultaneity term dies and

u'_y = \frac{u_y}{\gamma} \qquad (u_x = 0),

pure time dilation — the same $\mathrm{d}y$ over a $\mathrm{d}t'$ that is $\gamma$ times longer. Why: this is the light-clock of the figure, and it isolates the $\gamma$ with nothing else in the way. It is exactly here that no length has changed and the velocity has changed anyway.

Step 5. Now switch the other on. With $u_x \neq 0$ , the start and end of the transverse flight are events separated in $x$ , so the two frames disagree about the elapsed time by the simultaneity term $-v\,\mathrm{d}x/c^2$ inside $\mathrm{d}t'$ . Relative to Step 4, the transverse velocity is further multiplied by $1/(1 - v u_x/c^2)$ . Why: this factor is not new — it is bit-for-bit the denominator of the collinear rule that Relativistic Velocity Addition derives, and for the same reason: both velocities are ratios over the same $\mathrm{d}t'$ . One denominator, two rules. *Verified in verify_aq-sr-transverse-velocity-transformation.py: the forward rule is re-derived symbolically from the boost differentials in sympy; the $u_x = 0$ case collapses to $u_y/\gamma$ ; the composition of the forward rule (2) with the inverse rule (3) simplifies to the identity map on $(u_x, u_y)$ , proving they are one rule and not two; and the parallel rule is confirmed $\gamma$ -free while the transverse rule is not.*

The owned result: the transverse rule is the rule that keeps light at c

Here is the claim. The $1/\gamma$ in the transverse rule is not a curiosity to be explained away — it is forced. There is exactly one coefficient that lets the velocity transformation preserve $|\mathbf{u}| = c$ , and it is $1/\gamma$ . Change it and the second postulate breaks.

Go deeper · Derivation B · 3 minA light ray stays a light ray — and the numerator turns into the denominator squared

Derivation BA light ray stays a light ray — and the numerator turns into the denominator squared

Take a light ray in $S$ , with $u_x^2 + u_y^2 = c^2$ and no $z$ component. Transform both components with the forward rules of (2) — derived with no reference whatever to light — and compute the speed in $S'$ . If the rules are consistent with the invariance of $c$ , the answer must be exactly $c$ : not to first order, not to second, but identically, for every $\beta$ and every ray direction. The mechanism is a perfect factorisation — the numerator of $u'^2$ collapses into $c^2$ times the square of the shared denominator, which then cancels. A factorisation that exact does not survive perturbation, which is what makes it a statement about the coefficient $1/\gamma$ .

Step 1. Write both transformed components over the shared denominator $D \equiv 1 - v u_x/c^2$ :

u'_x = \frac{u_x - v}{D}, \qquad u'_y = \frac{u_y}{\gamma D} \qquad\Longrightarrow\qquad u'^2_x + u'^2_y = \frac{(u_x - v)^2 + u_y^2/\gamma^2}{D^2}.

Why: the $1/\gamma$ has been quarantined onto exactly one of the two squares in the numerator. Whatever happens next has to happen because of it.

Step 2. Impose the light condition and the definition of $\gamma$ : substitute $u_y^2 = c^2 - u_x^2$ and $1/\gamma^2 = 1 - v^2/c^2$ :

N = (u_x - v)^2 + \left(1 - \frac{v^2}{c^2}\right)\left(c^2 - u_x^2\right).

Why: these are the only two facts fed in — "the ray is light" and " $\gamma$ is $\gamma$ ". No third assumption is available, so if the result comes out right, it comes out right for structural reasons.

Step 3. Expand, and watch four terms annihilate:

N = \underbrace{u_x^2 - 2vu_x + v^2}_{(u_x-v)^2} \; + \; \underbrace{c^2 - u_x^2 - v^2 + \frac{v^2u_x^2}{c^2}}_{\text{the product}} \; = \; c^2 - 2vu_x + \frac{v^2u_x^2}{c^2}.

The $u_x^2$ terms cancel; the $v^2$ terms cancel. Why: two cancellations, both exact, neither arranged. A wrong transverse coefficient would leave a residue proportional to $u_x^2$ that no amount of goodwill could remove.

Step 4. Recognise the survivor as a perfect square:

N = c^2\left(1 - \frac{2 v u_x}{c^2} + \frac{v^2 u_x^2}{c^4}\right) = c^2\left(1 - \frac{v u_x}{c^2}\right)^{2} = c^2 D^2.

Why: <strong>there it is.</strong> The numerator is the denominator squared, times $c^2$ . Nothing in Step 1 suggested that the transverse suppression by $1/\gamma$ and the parallel shift by $-v$ would conspire to reassemble the shared denominator — but they do, exactly.

Step 5. Divide:

u'^2_x + u'^2_y = \frac{c^2 D^2}{D^2} = \boxed{\,c^2\,}

independent of $\beta$ , independent of the ray's direction, exactly. ∎ Why: be precise about the logical direction here. We did not build the transverse rule to preserve $c$ ; we derived it from the boost and then discovered that it does. Light-speed invariance is therefore not an extra postulate riding on top of the velocity transformation — it is a theorem about it, exactly as Velocity Composition and the Speed-of-Light Invariant shows for the collinear case. And since the ray's speed is pinned while its components are free to move, all a boost can do to light is <strong>turn it</strong> — which is aberration, and which is why the next article exists.

Step 6 (the coefficient is unique). Replace $1/\gamma$ by an unknown $\kappa(\beta)$ in the transverse rule and demand $|\mathbf{u}'| = c$ for every ray direction. Step 3 then leaves the residue $\left(\kappa^2 - 1 + \beta^2\right)\left(c^2 - u_x^2\right)$ , which vanishes for all $u_x$ only if $\kappa^2 = 1 - \beta^2$ , i.e. $\kappa = 1/\gamma$ . Why: so the $1/\gamma$ is not a result one could have got slightly wrong and still had a working theory. It is the unique transverse coefficient compatible with the invariance of $c$ — which is a much stronger statement than "the algebra happens to work", and it is this article's owned result. *Verified in verify_aq-sr-transverse-velocity-transformation.py: the numerator $N$ is expanded in sympy and shown to equal $c^2D^2$ identically; $u'^2_x + u'^2_y - c^2$ simplifies to exactly zero for symbolic $\beta$ and $\theta$ ; the uniqueness argument of Step 6 is solved symbolically for $\kappa$ and returns $\pm\sqrt{1-\beta^2}$ ; and the invariance is checked numerically on a grid of 40 boosts × 60 ray directions to $10^{-14}$ .*

Limiting cases

Case	Result	Meaning
$v \ll c$ , $u \ll c$	$u'_{\perp} \to u_{\perp}$	$\gamma \to 1$ , $D \to 1$ : frame-independent, as Galileo assumed. The leading correction is first order ( $-vu_{\parallel}/c^2$ , from simultaneity), not second — a point usually missed
$u_{\parallel} = 0$	$u'_{\perp} = u_{\perp}/\gamma$	pure time dilation; the light-clock; the only place the $\gamma$ appears naked
$u_{\perp} = 0$	$u'_{\perp} = 0$	collinear motion stays collinear; the parallel rule does all the work
$\|\mathbf{u}\| = c$	$\|\mathbf{u}'\| = c$	the owned result: light stays light, exactly
$\gamma \to \infty$ , $u_{\parallel} ＞ 0$	$u'_{\perp} \to 0$	transverse motion is crushed, everything swept forward — relativistic beaming, half-angle $\sim 1/\gamma$ , owned by Relativistic Beaming
$c \to \infty$	$u'_{\perp} = u_{\perp}$	Galileo is the exact $c \to \infty$ limit, not an approximation to relativity

Common trapFORWARD or INVERSE — pick one, say which, and never mix them

This is the single largest source of wrong signs in relativistic kinematics, and it is entirely avoidable. This chapter's primary form is FORWARD, $S \to S'$ , with $S'$ moving at $+v\hat{x}$ :

	parallel	transverse
FORWARD $S \to S'$	$u'_{\parallel} = \dfrac{u_{\parallel} - v}{1 - v u_{\parallel}/c^2}$	$u'_{\perp} = \dfrac{u_{\perp}}{\gamma(1 - v u_{\parallel}/c^2)}$
INVERSE $S' \to S$	$u_{\parallel} = \dfrac{u'_{\parallel} + v}{1 + v u'_{\parallel}/c^2}$	$u_{\perp} = \dfrac{u'_{\perp}}{\gamma(1 + v u'_{\parallel}/c^2)}$

Both are correct. They differ by $v \to -v$ and a swap of primes, and that substitution is the principle of relativity. The rule for staying out of trouble: write down which frame the quantity you were GIVEN lives in, then pick the row that starts there. Given a particle's velocity in a source's rest frame and want the lab? That is $S' \to S$ : the bottom row, the $+$ signs. Given a star's direction in the Sun's frame and want the Earth's? That is $S \to S'$ : the top row.

A base-error note, kept as teaching material. The source article for this page stated the inverse rule as its headline formula while declaring " $S'$ moves at $+v$ " — internally consistent, but the opposite of the convention every neighbouring article in this chapter uses. It has been converted here. That is exactly the silent mismatch that produces an aberration formula with a sign error two articles downstream.

Common trapTwo traps, and the malformed card that was discarded

Trap 1 — putting the $\gamma$ in the wrong rule. The parallel rule has no $\gamma$ : the one from $\mathrm{d}x'$ and the one from $\mathrm{d}t'$ cancel. The transverse rule has one, because nothing upstairs cancels it. This is backwards from the naive guess (the boost is along $x$ — surely the $x$ rule is the relativistic one?) and it catches almost everybody once.

Trap 2 — the malformed direction formula. The base's sixth equation card printed $\tan\theta = \sin\theta'/[\gamma(\cos\theta' + \beta/(u'/c))]$ — incoherent, reducing to nothing sensible as $u' \to c$ . The correct general statement is $\tan\theta = u'_y/[\gamma(u'_x + v)]$ for any speed; only when $u' = c$ may you set $u'_x = c\cos\theta'$ , $u'_y = c\sin\theta'$ and get $\tan\theta = \sin\theta'/[\gamma(\cos\theta' + \beta)]$ . The angle-only form exists only for light — for a massive particle $|\mathbf{u}'|$ is a third variable and does not cancel. That card is discarded and rewritten as equation (4).

Worked examples

1 · A real beam: where the muons from a pion beam actually go

The most-used transverse-velocity calculation in experimental physics is not a thought experiment about trains. It is: a charged pion decays in flight, $\pi^+ \to \mu^+ \nu_\mu$ , the muon emitted isotropically in the pion's rest frame — where does it come out in the lab? This is what makes a neutrino beam. Take a pion at $\beta = 0.9$ and a muon emitted at exactly $90^\circ$ in the pion frame.

Step 1 — identify. In the pion rest frame $S'$ the two-body decay is fixed by kinematics: the muon's momentum is $p = 29.79\ \mathrm{MeV}/c$ (PDG), with $m_\mu c^2 = 105.658\ \mathrm{MeV}$ . So

E_\mu = \sqrt{(29.79)^2 + (105.658)^2}\ \mathrm{MeV} = \sqrt{887.44 + 11163.6} = 109.78\ \mathrm{MeV}, \qquad \beta'_\mu = \frac{pc}{E} = \frac{29.79}{109.78} = \mathbf{0.2714}.

Emitted at $90^\circ$ in $S'$ : $u'_x = 0$ , $u'_y = 0.2714\,c$ . We are given the source-frame velocity and want the lab: that is $S' \to S$ , so we use the INVERSE row.

Step 2 — apply. The pion has $\beta = 0.9$ , so $\gamma = 1/\sqrt{0.19} = \mathbf{2.2942}$ . With $u'_x = 0$ the simultaneity factor is exactly $1$ :

u_x = \frac{0 + 0.9c}{1 + 0} = \mathbf{0.9\,c}, \qquad u_y = \frac{0.2714\,c}{2.2942 \times 1} = \mathbf{0.1183\,c}.

The transverse speed was divided by $\gamma$ and by nothing else.

Step 3 — the boundary number. The lab emission angle:

\tan\theta = \frac{0.1183}{0.9} = 0.13144 \qquad\Longrightarrow\qquad \theta = \mathbf{7.49^\circ}.

A muon that went sideways in the pion's frame comes out within seven and a half degrees of the beam axis. That is the whole principle of a neutrino beam line: you do not have to collect the decay products, because relativity collects them for you. The muon's lab speed is $\sqrt{0.9^2 + 0.1183^2} = \mathbf{0.9077\,c}$ — faster than the pion that made it, since the parallel component was boosted while only the transverse one was suppressed.

Step 4 — push the boost. Take a $\beta = 0.999$ pion ( $\gamma = 22.366$ ). The same $90^\circ$ muon comes out at $u_y = 0.2714c/22.366 = 0.012134c$ , $u_x = 0.999c$ , so

\theta = \arctan\frac{0.012134}{0.999} = \mathbf{0.696^\circ},

well inside the beaming cone $1/\gamma = 0.0447\ \mathrm{rad} = 2.56^\circ$ . Every doubling of $\gamma$ roughly halves the angle: the $1/\gamma$ in the transverse rule is the collimation of the beam, and nothing else contributes.

2 · A light ray at $90^\circ$ , and the invariance you can check by hand

Now the case that motivates everything downstream. In the lab $S$ , starlight arrives travelling straight "up": $\theta = 90^\circ$ , so $u_x = 0$ , $u_y = c$ . What is its velocity in a frame $S'$ moving at $\beta = 0.6$ along $+x$ — and is it still light?

Step 1 — identify. Given the ray in $S$ , want $S'$ : FORWARD, top row. $\gamma = 1/\sqrt{0.64} = \mathbf{1.25}$ , and $D = 1 - vu_x/c^2 = \mathbf{1}$ — the simultaneity factor is off, because the ray has no parallel component in $S$ .

Step 2 — apply.

u'_x = \frac{0 - 0.6c}{1} = \mathbf{-0.6\,c}, \qquad u'_y = \frac{c}{1.25 \times 1} = \mathbf{0.8\,c}.

Step 3 — the boundary number. Is it still light?

|\mathbf{u}'| = c\sqrt{(-0.6)^2 + (0.8)^2} = c\sqrt{0.36 + 0.64} = \mathbf{c}. \;\checkmark

Exactly $c$ , with the components landing on the $(3,4,5)$ triple scaled by $0.2$ . The $1/\gamma = 0.8$ that suppressed the transverse component is precisely the $0.8$ Pythagoras needs. The owned result, arithmetically visible.

Step 4 — read off the aberration. The new direction is

\theta' = \operatorname{atan2}(0.8,\ -0.6) = 180^\circ - 53.13^\circ = \mathbf{126.87^\circ}.

The ray that went straight up in the lab goes up and backward in $S'$ — exactly right: an observer running in $+x$ sees the star's light arriving more from ahead, so the ray's propagation direction tips away from the motion. (Beware: the star appears shifted toward $+x$ ; the ray tips toward $-x$ . Same statement, opposite arrows — the commonest confusion in the subject, and Relativistic Aberration of Light draws it.) Cross-check with (4): $\tan\theta' = 1/[1.25 \times (-0.6)] = -1.3333$ , and the quadrant fixed by $u'_y ＞ 0 ＞ u'_x$ gives $126.87^\circ$ ✓.

Figure

The velocity circle. Any light ray in S sits on the circle |u| = c; the transverse rule maps it to a point on the SAME circle in S′ — that is the owned result, drawn. At β = 0.6 the ray at θ = 90° (0, c) maps to (−0.6c, 0.8c), i.e. θ′ = 126.87°: the transverse component is cut by exactly 1⁄γ = 0.8, the parallel component is shifted by −0.6, and the two land back on the circle because 0.6² + 0.8² = 1. Grey arrows: where every 30° of the sky goes. Everything crowds toward θ′ = 180°, i.e. toward the BACK of the moving frame — which is the same fact as stars crowding toward the FRONT

Physics of the portrait

Origins · 2 minWolfgang Pauli and how this was discovered

Wolfgang Pauli

1900–1958 · Vienna · Munich · Göttingen · Copenhagen · Hamburg · Zürich

The full velocity transformation — parallel and transverse together, in general vector form, with the shared denominator displayed — is not a stand-alone result in Einstein's 1905 paper; it is scattered through the literature of the next fifteen years. The article that gathered it, and everything else, into a systematic whole was written by a twenty-one-year-old. Wolfgang Pauli was born in Vienna on 25 April 1900. Arnold Sommerfeld, who had him as a student in Munich and had been commissioned to write the relativity article for the Encyklopädie der mathematischen Wissenschaften, handed the job to him instead. The result — Relativitätstheorie, 1921 — ran to 237 pages with some four hundred references, covering special relativity, general relativity and the unified-field programme with a critical judgement that has aged remarkably well. It is still in print. In December 1921 Einstein wrote to Max Born: "Whoever studies this mature and comprehensive work might not believe that the author is a man of twenty-one years old." Pauli's own contributions are larger: the exclusion principle (1925), for which he took the 1945 Nobel Prize, and the postulate of the neutrino (1930) — the particle later produced in beams by exactly the pion decay computed above. Ehrenfest nicknamed him die Geissel Gottes, the scourge of God. He died in Zürich on 15 December 1958. The kinematics on this page is a small corner of an article he wrote before he had a doctorate.

History

1905

— Einstein derives the velocity transformation in "Zur Elektrodynamik bewegter Körper", giving both the parallel and the perpendicular components.

1908

— Minkowski's "Raum und Zeit" recasts the transformation geometrically; the velocity rule becomes a statement about four-vectors, not coordinates.

1921

— Pauli, aged twenty-one, publishes Relativitätstheorie in the Encyklopädie der mathematischen Wissenschaften: 237 pages, and the first systematic treatment of the full velocity transformation in general vector form.

1930

— Pauli postulates the neutrino to rescue energy conservation in beta decay.

1962

— Lederman, Schwartz and Steinberger use a pion-decay-in-flight beam at the Brookhaven AGS to show that the muon neutrino is distinct from the electron neutrino; the beam's collimation is the 1/γ of the transverse velocity rule.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed., §14 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.2 · Taylor & Wheeler, Spacetime Physics, 2nd ed., ch. 3–4. Primary source: Pauli, Relativitätstheorie, Encyklopädie der mathematischen Wissenschaften V.19 (Teubner, 1921); English translation Theory of Relativity (Pergamon, 1958; Dover reprint).

Provenance. Signature $(+,-,-,-)$ ; boost convention FORWARD, $S'$ at $+v\hat{x}$ . Two base defects found and repaired. (i) The base gave the inverse rule as its headline result while declaring the forward convention — converted here, with the inverse kept explicitly as the reciprocity corollary (3). (ii) The base's sixth equation card, $\tan\theta = \sin\theta'/[\gamma(\cos\theta'+\beta/(u'/c))]$ , is malformed; discarded and replaced by (4). Every number recomputed in verify_aq-sr-transverse-velocity-transformation.py: $\gamma(0.8) = 5/3$ , $0.6c/\gamma = 0.36c$ ; the pion chain $p_\mu = 29.79\ \mathrm{MeV}/c$ , $E_\mu = 109.78\ \mathrm{MeV}$ , $\beta'_\mu = 0.2714$ , $\gamma(0.9) = 2.2942$ , $u_y = 0.1183c$ , $|\mathbf{u}| = 0.9077c$ , $\theta = 7.49^\circ$ , and $0.696^\circ$ at $\beta = 0.999$ ; the ray $(0,c) \to (-0.6c, 0.8c)$ with $|\mathbf{u}'| = c$ , $\theta' = 126.87^\circ$ ; and both derivations symbolically.

Builds on. Lorentz Transformation Equations · Relativistic Velocity Addition · Time Dilation · Relativity of Simultaneity (S1) · The Minkowski Metric. Leads to. Relativistic Aberration of Light · Relativistic Beaming · Thomas–Wigner Rotation.

If there is no transverse length contraction, where exactly does the γ come from?

From the clock, and only from the clock. Write $u'_y = \mathrm{d}y'/\mathrm{d}t'$ . The numerator is $\mathrm{d}y' = \mathrm{d}y$ — invariant, no $\gamma$ , nothing. The denominator is $\mathrm{d}t' = \gamma(\mathrm{d}t - v\,\mathrm{d}x/c^2)$ — and there is the $\gamma$ , in front of the whole bracket. It is the same $\gamma$ as in time dilation, arriving by the same mechanism, and it never touched a ruler. The cleanest demonstration is the light-clock: the beam crosses the same height $h$ in both frames (because $y' = y$ ), but the crossing takes longer in the frame where the clock moves, so the transverse speed is smaller by exactly the dilation factor. If you catch yourself explaining this with a contracted ruler, you have explained something that does not happen.

Why does the parallel rule have no γ, when the boost is along that very axis?

Because the parallel numerator has one of its own. $\mathrm{d}x' = \gamma(\mathrm{d}x - v\,\mathrm{d}t)$ carries a $\gamma$ ; $\mathrm{d}y' = \mathrm{d}y$ carries none. Both are divided by $\mathrm{d}t' = \gamma(\dots)$ , which carries a $\gamma$ . In the parallel rule the two cancel, leaving the clean quotient $(u_x - v)/(1 - vu_x/c^2)$ . In the transverse rule there is nothing to cancel with, so the $\gamma$ survives. The asymmetry is an accident of which coordinate the boost mixes with time — and it is a useful memory hook, being the reverse of what everyone guesses.

Can the transverse velocity ever increase under a boost?

Yes — the simultaneity factor can beat the $1/\gamma$ . In $u'_{\perp} = u_{\perp}/[\gamma(1 - vu_{\parallel}/c^2)]$ , a particle moving against the boost has $1 - vu_{\parallel}/c^2 ＞ 1$ , which suppresses $u'_\perp$ further. But with $u_{\parallel} ＞ 0$ and close to $c$ the bracket becomes small, and $u'_\perp$ can exceed $u_\perp$ despite the $1/\gamma$ . At $v = 0.9c$ with $u_x = 0.9c$ , $u_y = 0.1c$ : $\gamma = 2.294$ , $D = 0.19$ , and $u'_y = 0.1c/(2.294 \times 0.19) = 0.229c$ — more than doubled. This is beaming seen from the other frame, and a good check that you have not memorised "transverse velocities always shrink".

The formula has a 1/γ. Does that mean transverse motion is time-dilated but longitudinal motion is not?

No — every motion is time-dilated, and that is precisely why the $\gamma$ 's cancel in the parallel rule. Time dilation is not directional; it is a property of the clock, and it enters both rules identically through $\mathrm{d}t'$ . What differs is the numerator. The boost rebuilds the denominator once, then hands the same denominator to every component; the components differ only in what happens upstairs. The vector form says it best: $\mathbf{u}'_{\perp} = \mathbf{u}_{\perp}/[\gamma(1 - \mathbf{v}\cdot\mathbf{u}/c^2)]$ and $u'_{\parallel} = (u_{\parallel} - v)/(1 - \mathbf{v}\cdot\mathbf{u}/c^2)$ — one denominator, written once.

Is the transverse rule enough to derive aberration, or do I need the Doppler shift too?

It is enough for the direction — that is the point of equation (4), and Relativistic Aberration of Light runs with it. Take a photon, resolve its velocity into components, transform both with the rules on this page, take the arctangent: you have the aberration formula, with the $\gamma$ inherited straight from the transverse rule. The frequency is a separate question needing a separate object — the null four-vector $k^\mu$ , whose time component gives the Doppler factor. The two are siblings (both fall out of boosting the same $k^\mu$ ), but a velocity has no frequency in it. Textbooks routinely derive both from one boost in a single breath and leave the impression that direction and frequency are one result. They are two results from one four-vector.

How large is the transverse effect in everyday terms — is it ever measurable at low speed?

The $\gamma$ part is second order ( $\gamma^{-1} \approx 1 - \beta^2/2$ ) and hopeless at ordinary speeds. But the simultaneity part is first order in $\beta$ — the factor $(1 - vu_{\parallel}/c^2)$ — so the leading relativistic correction to a transverse velocity is $O(vu_{\parallel}/c^2)$ , not $O(v^2/c^2)$ . It is still tiny for a train. Where it becomes an engineering fact is in accelerators, where $\gamma$ is routinely $10^3$ – $10^4$ : the transverse angular spread of a beam in the lab is smaller than its rest-frame spread by exactly $1/\gamma$ , and this adiabatic damping of the emittance is designed into every machine. At the LHC, $\gamma = 7460$ .

Transverse Velocity Transformation

Where this lives today

The physical picture

The core equations

Deriving the structure

The owned result: the transverse rule is the rule that keeps light at c

Limiting cases

Worked examples

1 · A real beam: where the muons from a pion beam actually go

2 · A light ray at $90^\circ$ , and the invariance you can check by hand

Physics of the portrait

Going further

Before this

Where this leads

Transverse Velocity Transformation

Where this lives today

The physical picture

The core equations

Deriving the structure

The owned result: the transverse rule is the rule that keeps light at c

Limiting cases

Worked examples

1 · A real beam: where the muons from a pion beam actually go

2 · A light ray at 90∘90^\circ90∘, and the invariance you can check by hand

Physics of the portrait

Going further

Before this

Where this leads

2 · A light ray at $90^\circ$ , and the invariance you can check by hand