Rapidity and the Additivity of Velocities

Inside every proton at the LHC are quarks and gluons carrying unknown, unequal fractions of the beam's momentum. When two of them collide, the collision's centre-of-momentum frame is therefore sliding along the beam axis at a speed nobody knows — different for every event, unmeasurable in principle, and different again for the next one. This ought to be fatal. Every angle you measure, every longitudinal momentum, every velocity is contaminated by an unknown boost that changes from event to event. And yet the LHC's data are clean, its detectors are segmented in a variable called $\eta$ , and its jet algorithms measure distances with a formula that looks like Pythagoras. Why does any of it work? Because there is a variable in which an unknown boost is a translation — and a translation cancels in a difference. Rapidity and Hyperbolic Boosts and Rapidity establish what rapidity is: the hyperbolic angle $\varphi = \operatorname{artanh}\beta$ that adds when boosts compose. This article establishes what it is for, and the claim to be proved is stronger than the additivity those articles own: > Boost along the beam by rapidity $\varphi_0$ , and every particle's rapidity shifts by exactly $\varphi_0$ — whatever its mass, whatever its transverse momentum. So rapidity differences are exactly invariant under the one transformation nobody can measure. Pseudorapidity, the coordinate stamped on the side of every collider detector, is what you get when you apply that theorem to something massless.

Key takeaways

Signature $(+,-,-,-)$ ; $S$ is the lab, $S'$ moves at $+v\hat{z}$ along the beam; $\beta = v/c$ , $\gamma = (1-\beta^2)^{-1/2}$ . The velocity rapidity is $\varphi = \operatorname{artanh}\beta$ , and its additivity under collinear boosts is derived and owned by Rapidity and Hyperbolic Boosts — recalled here, not re-derived.
The collider rapidity of a particle is a different object: $y = \dfrac12 \ln\dfrac{E + p_z c}{E - p_z c} = \operatorname{artanh}\!\left(\dfrac{p_z c}{E}\right)$ . It is defined for a massive particle with transverse momentum, and it equals $\varphi$ only for motion purely along the beam.
The shift theorem (owned here). Under a boost of rapidity $\varphi_0$ along $\hat{z}$ , $\;\boxed{y \to y - \varphi_0}\;$ exactly, for every $m$ and every $p_T$ , while $p_T$ and the azimuth $\phi$ are untouched. So $\Delta y$ , $\Delta\phi$ and $\Delta R = \sqrt{\Delta y^2 + \Delta\phi^2}$ are invariant under the unknown boost. Velocity differences are not.
Pseudorapidity is the massless rapidity. $\eta \equiv -\ln\tan(\theta/2) = \operatorname{artanh}(\cos\theta)$ , while $y = \operatorname{artanh}(\beta\cos\theta)$ . The only difference is the factor $\beta$ — so $\tanh y = \beta \tanh\eta$ , and $\eta = y$ exactly when $\beta = 1$ .
Key idea: rapidity is the variable in which the collider's one unmeasurable degree of freedom becomes an additive constant. Under the unknown boost every $y$ moves by the same amount, so every difference of $y$ stands still. Pseudorapidity is the version of this you can compute from a track's angle alone, without knowing its mass — which is why it is what detectors are built in, and why $|\eta| ＜ 2.5$ is written on the ATLAS inner tracker.

The core insight

Additivity is not the payoff; shift-invariance is. A quantity that transforms as $y \mapsto y + \text{const}$ under a symmetry you cannot measure is worth more than a quantity that transforms correctly under a symmetry you can. The unknown longitudinal boost is the collider's irreducible ignorance, and rapidity is the coordinate in which that ignorance is a choice of origin.

Where this lives today

Open any LHC paper and the kinematics is written in $(p_T, \eta, \phi)$ , never in $(v_x, v_y, v_z)$ . All three coordinates were chosen because of the shift theorem: $p_T$ and $\phi$ because a beam boost does not touch them, $\eta$ because it is the measurable stand-in for the one coordinate the boost merely translates. Jets are clustered with the anti- $k_t$ algorithm using $\Delta R = \sqrt{\Delta y^2 + \Delta\phi^2}$ at $R = 0.4$ ; isolation cones, $b$ -tagging and lepton–jet overlap removal use the same metric; and the reason a fixed $R$ means anything — the reason a jet does not change size when the parton frame slides — is the invariance proved above. The variable also fixed a piece of physics history. In the 1970s the question of whether hadron production was "boost-invariant" — whether the central region looked the same to every longitudinally boosted observer — was answered by the observation of a flat rapidity plateau, $\mathrm{d}N/\mathrm{d}y$ roughly constant near $y = 0$ . Because a boost translates $y$ rigidly, a flat plateau is a fixed point of the boost, and Bjorken built his 1983 hydrodynamic model of heavy-ion collisions on that symmetry. The variable did not merely make the plot readable; it made the symmetry visible, and the symmetry turned out to be the physics.

The physical picture

Rapidity's additivity in one dimension is settled: compose two collinear boosts and their rapidities add, exactly. That theorem — $\Lambda(\varphi_1)\Lambda(\varphi_2) = \Lambda(\varphi_1+\varphi_2)$ — is owned by Rapidity and Hyperbolic Boosts, which also proves that no finite chain of boosts reaches $c$ and gives the closed form for $N$ equal boosts. None of it is re-derived here. This article takes that fact out of the one-dimensional world and into the one experiments live in, where particles have transverse momentum and mass. Two things could go wrong, and neither does. The first is mass. A slow heavy particle and a fast light one have wildly different velocities — but the boost acts on the pair $(E, p_z)$ , and rapidity is a function of that pair alone. The mass enters $E$ , and then cancels. The second is transverse motion. A particle flying mostly sideways barely moves along the beam, and one might expect the beam boost to treat it differently. But a boost along $\hat{z}$ leaves $p_x$ and $p_y$ untouched — that is just $y'=y$ , $z'=z$ applied to momentum instead of position — so the transverse momentum is a spectator, and it too cancels. What survives is a bare translation. Draw the event on the $(y, \phi)$ cylinder — rapidity along the axis, azimuth around it — and the unknown boost slides the whole picture along the axis without deforming it. Distances along the cylinder do not change. That is why the LHC's jet-clustering distance is $\Delta R = \sqrt{\Delta y^2 + \Delta\phi^2}$ , and why $R = 0.4$ means the same thing in every event.

Figure

PUNCHLINE — the unknown boost is a TRANSLATION. Every particle in a collision is a point on the (y, φ) cylinder: rapidity along the beam axis, azimuth around it. LEFT: the event as produced, in the parton centre-of-momentum frame. RIGHT: the SAME event in the lab, after the unknown longitudinal boost of rapidity φ₀ = 0.693. Every rapidity has slid by exactly the same 0.693 — the two jets go from y = +1.0 and −1.0 to y = 1.693 and −0.307 — but Δy = 2.0 is untouched, the azimuths are untouched, p_T is untouched, and so ΔR is untouched. Velocity differences would have been wrecked. Rapidity differences do not notice

The core equations

The velocity rapidity, recalled — not derived here. Its derivation, its additivity theorem and the $N$ -boost ladder belong to Rapidity and Hyperbolic Boosts and Rapidity:

\beta = \tanh\varphi, \qquad \gamma = \cosh\varphi, \qquad \gamma\beta = \sinh\varphi, \qquad \varphi_1 \oplus \varphi_2 = \varphi_1 + \varphi_2 \quad\text{(collinear)} \qquad (1)

where: $\varphi$ is the hyperbolic angle of a boost. Physical reading: this is the input, taken as given. Everything below is what happens when the same question is asked about a particle in a detector rather than about a frame. The collider rapidity of a particle, which is what an experiment actually measures:

\boxed{\; y \;\equiv\; \frac12 \ln\!\left(\frac{E + p_z c}{E - p_z c}\right) \;=\; \operatorname{artanh}\!\left(\frac{p_z c}{E}\right) \;} \qquad (2)

where: $E$ is the particle's energy, $p_z$ its momentum along the beam. Physical reading: this is not $\varphi$ . The quantity $p_zc/E$ is $\beta_z$ , the component of the velocity along the beam — not the speed — so $y = \operatorname{artanh}\beta_z$ , coinciding with the frame rapidity $\varphi$ only for motion purely along $\hat{z}$ . A particle at rest has $y = 0$ ; a particle flying transversely also has $y = 0$ , however fast. Rapidity measures longitudinality, not speed. The two transverse quantities that a boost along $\hat{z}$ cannot touch:

p_T = \sqrt{p_x^2 + p_y^2}, \qquad \phi = \arctan\!\left(\frac{p_y}{p_x}\right), \qquad\text{and}\qquad m_T \equiv \sqrt{m^2c^2 + p_T^2} \qquad (3)

where: $\phi$ is the azimuth about the beam, $m_T$ the transverse mass. Physical reading: $p_x$ and $p_y$ are invariant under a $z$ -boost, exactly as $y$ and $z$ are under an $x$ -boost. With (2) this gives the collider's coordinate system $(p_T, \phi, y)$ , in which the unknown boost moves exactly one coordinate, additively. And the pseudorapidity — the portrait of this article:

\boxed{\; \eta \;\equiv\; -\ln\tan\frac{\theta}{2} \;=\; \operatorname{artanh}(\cos\theta) \;=\; \frac12\ln\!\left(\frac{|\mathbf{p}| + p_z}{|\mathbf{p}| - p_z}\right) \;} \qquad (4)

where: $\theta$ is the polar angle from the beam. Physical reading: compare (4) with (2) — one thing has changed, $E/c$ has become $|\mathbf{p}|$ . So $\eta$ is exactly the rapidity a particle would have if it were massless, and it is computable from the track's direction alone, with no mass hypothesis and no energy measurement. That is the entire reason it exists.

Deriving the structure

Derivation ATHE SHIFT THEOREM — a beam boost moves every rapidity by the same amount, whatever the mass and whatever the transverse momentum

This is the article's owned result and the reason the variable exists. The additivity of collinear rapidities — recalled in (1) — is a statement about frames moving along a line. What an experiment needs is a statement about a particle, which has mass and is generally not moving along the beam at all. The claim is that the additivity survives intact: a longitudinal boost of rapidity $\varphi_0$ shifts the particle's rapidity by exactly $\varphi_0$ , and by the same $\varphi_0$ for every particle in the event, whatever it is and wherever it is going. Nothing in that is obvious — the mass and the transverse momentum both have to disappear for it to be true.

Step 1. Boost along $\hat{z}$ , FORWARD ( $S \to S'$ , with $S'$ moving at $+v\hat{z}$ ), and apply it to the four-momentum $p^\mu = (E/c,\ p_x,\ p_y,\ p_z)$ exactly as the chapter's convention applies it to $x^\mu$ :

\frac{E'}{c} = \gamma\!\left(\frac{E}{c} - \beta p_z\right), \qquad p'_z = \gamma\!\left(p_z - \beta \frac{E}{c}\right), \qquad p'_x = p_x, \qquad p'_y = p_y.

Why: the transverse components are untouched — $y'=y$ , $z'=z$ carried over to momentum. $p_T$ and the azimuth $\phi$ are already invariant, before any rapidity algebra happens at all.

Step 2. Form the new rapidity from definition (2) and divide numerator and denominator by $\gamma E/c$ :

\tanh y' = \frac{p'_z c}{E'} = \frac{\gamma\left(p_z c - \beta E\right)}{\gamma\left(E - \beta p_z c\right)} = \frac{\dfrac{p_z c}{E} - \beta}{1 - \beta\,\dfrac{p_z c}{E}} = \frac{\tanh y - \tanh\varphi_0}{1 - \tanh y \,\tanh\varphi_0},

using $\tanh y = p_z c/E$ from (2) and $\beta = \tanh\varphi_0$ from (1). Why: the $\gamma$ 's cancelled and the mass never appeared. Both $E$ and $p_z$ carry the mass and both sit inside the same ratio; dividing through, only the ratio survives — and the ratio is $\tanh y$ by definition. That is why $y$ , not the velocity, is the right variable: it is built from precisely the combination the boost acts on simply.

Step 3. Recognise the right-hand side. It is the $\tanh$ subtraction formula — the same hyperbolic identity that Rapidity and Hyperbolic Boosts uses to prove additivity, with one sign flipped. Hence

\tanh y' = \tanh(y - \varphi_0) \qquad\Longrightarrow\qquad \boxed{\;y' = y - \varphi_0\;}

since $\tanh$ is injective. ∎ Why: exact, not approximate — and $m$ and $p_T$ do not appear in it. A $500\ \mathrm{GeV}$ jet, a stray $140\ \mathrm{MeV}$ pion and a $Z$ boson in the same event all have their rapidities shifted by the identical $\varphi_0$ . That is false of their velocities, their angles and their longitudinal momenta, all of which move by amounts that depend on what they are.

Step 4. Harvest the invariants. If every $y$ shifts by the same constant then, for any two particles in the event, $\Delta y$ is invariant — and $\Delta\phi$ and $p_T$ were already invariant by Step 1. Hence so is the distance used by every jet algorithm at every hadron collider,

\Delta R \;\equiv\; \sqrt{(\Delta y)^2 + (\Delta\phi)^2}.

Why: this is the payoff, and it is strictly stronger than "rapidity is additive". The parton centre-of-momentum frame slides by an unknown amount in every single event, so any observable built from $y$ itself is contaminated; any observable built from $\Delta y$ is not. The anti- $k_t$ algorithm clusters with $\Delta R ＜ R$ (typically $R = 0.4$ ), and the reason $R = 0.4$ means the same thing in every event is this step.

Step 5. Note what happens to a distribution. The boost replaces $\mathrm{d}N/\mathrm{d}y$ by the same function evaluated at $y + \varphi_0$ : rigidly translated, never stretched or squeezed, because the Jacobian $\mathrm{d}y'/\mathrm{d}y = 1$ exactly. Why: plateaux stay plateaux and widths stay widths. The famous "rapidity plateau" of hadron production is a statement that survives the unknown boost, and it survives for this reason. *Verified in verify_aq-sr-rapidity-additive-velocity.py: sympy boosts a symbolic four-momentum with arbitrary $m$ and $p_T$ , forms $y'$ , and simplifies $y' - (y - \varphi_0)$ to exactly zero; $p'_x - p_x$ and $p'_y - p_y$ are zero identically; $\mathrm{d}y'/\mathrm{d}y = 1$ ; and numerically, 500 random four-momenta (masses from $0$ to $200\ \mathrm{GeV}$ , $p_T$ from $0$ to $300\ \mathrm{GeV}$ ) all shift by the identical $\varphi_0$ to $10^{-13}$ , while their velocities and angles do not.*

Go deeper · Derivation B · 2 minPSEUDORAPIDITY IS THE RAPIDITY OF A MASSLESS PARTICLE — and the only difference is one factor of beta

Derivation BPSEUDORAPIDITY IS THE RAPIDITY OF A MASSLESS PARTICLE — and the only difference is one factor of beta

The shift theorem is beautiful and useless to a tracking detector, because $y$ requires $E$ , which requires the mass, which a silicon tracker measuring a curved track does not know. What a tracker does know, to exquisite precision, is the direction of the track — the polar angle $\theta$ . So the working variable of experimental collider physics is not $y$ ; it is a purely geometric stand-in built from $\theta$ alone. The goal is to prove exactly what that stand-in is, how it differs from $y$ , and when the difference matters — and the answer is a single factor of $\beta$ , which is about as clean as a physics approximation ever gets.

Step 1. Start from the definition and turn the logarithm into an $\operatorname{artanh}$ . Using $\tan(\theta/2) = \sqrt{(1-\cos\theta)/(1+\cos\theta)}$ ,

\eta \equiv -\ln\tan\frac{\theta}{2} = \frac12 \ln\frac{1+\cos\theta}{1-\cos\theta} = \operatorname{artanh}(\cos\theta).

Why: the mysterious $-\ln\tan(\theta/2)$ is nothing but $\operatorname{artanh}$ of the direction cosine. The logarithm was never the point — it is what $\operatorname{artanh}$ looks like written out.

Step 2. Now write $y$ the same way. With $p_z = |\mathbf{p}|\cos\theta$ and $\beta = |\mathbf{p}|c/E$ (the particle's actual speed):

y = \operatorname{artanh}\!\left(\frac{p_z c}{E}\right) = \operatorname{artanh}\!\left(\frac{|\mathbf{p}|c}{E}\cos\theta\right) = \boxed{\;\operatorname{artanh}(\beta\cos\theta)\;}

against $\eta = \operatorname{artanh}(\cos\theta)$ . The two differ by exactly one factor of $\beta$ , and by nothing else. ∎ Why: equivalently, $\tanh y = \beta\tanh\eta$ . Since $\beta ＜ 1$ for anything massive, $|y| ＜ |\eta|$ always: pseudorapidity systematically overstates rapidity, and overstates it more the slower the particle.

Step 3. Read off the limits. $\beta \to 1$ (massless, or just $E \gg mc^2$ ) gives $y \to \eta$ exactly; and at $\theta = 90^\circ$ , $\cos\theta = 0$ , so $y = \eta = 0$ for every $\beta$ . Why: the approximation is exact in two places — massless particles, and any particle at $90^\circ$ — and degrades in between, worst for slow particles at small angles. For a charged pion at $p_T = 1\ \mathrm{GeV}/c$ it is a $0.5\%$ effect; for a proton at the same $p_T$ , a $15\%$ effect. Worked example 2 computes both.

Step 4. Draw the consequence for the shift theorem. Because $\eta$ is $y$ with $\beta$ set to $1$ , and a boost changes $\beta$ , pseudorapidity is only approximately shift-invariant — exactly so for massless particles, approximately for fast ones, badly for slow ones. Detector geometry does not care: a calorimeter cell subtends a fixed range of $\theta$ , hence a fixed range of $\eta$ , forever. Why: $\eta$ is where the detector is; $y$ is where the physics is. Acceptance is quoted in $\eta$ because $\eta$ is a property of the apparatus; cross-sections are differential in $y$ because $y$ is what the unknown boost merely translates. Confusing them is the standard beginner's error, and the fix is the single factor of $\beta$ derived above. *Verified in verify_aq-sr-rapidity-additive-velocity.py: $-\ln\tan(\theta/2) - \operatorname{artanh}(\cos\theta)$ simplifies to zero symbolically; $y = \operatorname{artanh}(\beta\cos\theta)$ is proved symbolically from $p_z = |\mathbf{p}|\cos\theta$ and $\beta = |\mathbf{p}|c/E$ ; $\tanh y = \beta\tanh\eta$ is verified on $10^4$ random $(m, p_T, \eta)$ triples; $y \to \eta$ as $m \to 0$ ; $y = \eta = 0$ at $\theta = 90^\circ$ for every $\beta$ ; and $|y| \leq |\eta|$ on the whole grid.*

Limiting cases

Case	Result	Meaning
$p_T = 0$	$y = \varphi$	the collider rapidity collapses to the velocity rapidity; only then are they the same object
$\beta \to 1$	$y \to \eta$	pseudorapidity is the massless rapidity
$\theta = 90^\circ$	$y = \eta = 0$	exact for every mass and speed; "central" means $y = 0$
$\theta \to 0$	$\eta \to +\infty$	the beam pipe is at infinite pseudorapidity — which is why no detector covers it
$\beta \ll 1$	$y \approx \beta\cos\theta$	the two diverge completely: a slow particle has $y \approx 0$ while $\eta$ can be huge
boost $\varphi_0$ along $\hat{z}$	$y \to y - \varphi_0$ ; $p_T, \phi$ fixed	the shift theorem: $\Delta y$ , $\Delta\phi$ , $\Delta R$ all invariant

Common trapThree different quantities are all called "rapidity". Keep them apart.

symbol	name	definition	belongs to
$\varphi$	velocity rapidity	$\operatorname{artanh}\beta$ , $\beta$ = the speed	a frame. Adds under composition (Rapidity and Hyperbolic Boosts)
$y$	collider rapidity	$\tfrac12\ln\dfrac{E+p_zc}{E-p_zc} = \operatorname{artanh}\beta_z$	a particle. Shifts by $\varphi_0$ under a beam boost
$\eta$	pseudorapidity	$-\ln\tan(\theta/2) = \operatorname{artanh}\cos\theta$	a direction. Equals $y$ only when $\beta = 1$

$y \neq \varphi$ unless $p_T = 0$ . $y$ is $\operatorname{artanh}$ of the longitudinal component of the velocity, not of the speed. A particle moving at $0.99c$ straight across the beam has $y = 0$ .
$\eta \neq y$ unless $\beta = 1$ . The exact bridge is $\tanh y = \beta\tanh\eta$ — no other correction, and no approximation in that statement.
Coverage is quoted in $\eta$ ; physics is done in $y$ . A cut $|\eta| ＜ 2.5$ says where the silicon is; $\mathrm{d}\sigma/\mathrm{d}y$ says what the collision did.

Common trapRapidities add along a LINE. They are not vectors.

Every additivity statement on this page — including the shift theorem — is for boosts along a single axis, which at a collider is the beam. Rapidity is emphatically not a vector composing by the parallelogram rule. Boost along $x$ , then along $y$ , and the result is not a boost of rapidity $\sqrt{\varphi_x^2 + \varphi_y^2}$ ; it is a boost composed with a spatial rotation — the Thomas–Wigner rotation, owned by Thomas–Wigner Rotation. The geometry underneath is hyperbolic, and the rotation angle is the defect of a hyperbolic triangle: precisely the amount by which naive vector addition of rapidities fails. Collider physics need not worry, and it is worth saying why: the unknown boost is always along the beam, so the shift theorem applies exactly, with no rotation anywhere. The one-dimensionality is not an approximation — it is a fact about how the machine is built.

Worked examples

1 · The boost nobody can measure, and the number that survives it

Two $7\ \mathrm{TeV}$ protons collide. Inside them, a quark carrying a fraction $x_1 = 0.20$ of one beam's momentum strikes a quark carrying $x_2 = 0.05$ of the other's. Find the rapidity of the parton centre-of-momentum frame, the collision energy, and show that a measurement made in the lab can still recover a parton-frame quantity exactly.

Step 1 — identify the unknown boost. Treat the partons as massless (they are, to a part in $10^5$ , at these energies). Their total energy and longitudinal momentum are $E = (x_1+x_2)E_{\text{beam}}$ and $p_zc = (x_1-x_2)E_{\text{beam}}$ , so (2) gives the parton CM rapidity directly:

\varphi_0 = \frac12 \ln\frac{E + p_z c}{E - p_z c} = \frac12\ln\frac{2x_1}{2x_2} = \frac12 \ln\frac{x_1}{x_2} = \frac12\ln 4 = \mathbf{0.693147}.

This is the number nobody can measure. $x_1$ and $x_2$ are drawn from parton distribution functions; they differ in every event.

Step 2 — apply. The collision energy is $\sqrt{\hat{s}} = \sqrt{x_1 x_2 s}$ with $\sqrt{s} = 14\ \mathrm{TeV}$ :

\sqrt{\hat{s}} = \sqrt{0.20 \times 0.05} \times 14\,000\ \mathrm{GeV} = 0.1 \times 14\,000 = \mathbf{1400\ \mathrm{GeV}} = 1.4\ \mathrm{TeV}.

Suppose the collision produces two back-to-back jets at $y^* = +1.0$ and $y^* = -1.0$ in the parton CM frame. Each jet has $E^* = \sqrt{\hat{s}}/2 = 700\ \mathrm{GeV}$ and, being massless, transverse momentum

p_T = \frac{E^*}{\cosh y^*} = \frac{700}{\cosh 1} = \frac{700}{1.543081} = \mathbf{453.64\ \mathrm{GeV}/c}.

Step 3 — the boundary number. The lab sees the event boosted by $\varphi_0$ . By the shift theorem, each jet's rapidity moves by exactly that:

y_1 = 1.0 + 0.693147 = \mathbf{1.693147}, \qquad y_2 = -1.0 + 0.693147 = \mathbf{-0.306853}.

Neither number is the parton-frame value; both are contaminated by the unmeasurable $\varphi_0$ . But

\Delta y = y_1 - y_2 = \mathbf{2.000000} = y_1^* - y_2^*.

Exactly the parton-frame separation, recovered from the lab, without ever knowing $x_1$ or $x_2$ . And $p_T = 453.64\ \mathrm{GeV}/c$ is likewise unchanged by the boost — which is why every LHC analysis triggers, cuts and plots on $p_T$ and $\Delta y$ , and never on $p_z$ or the lab angle.

Step 4 — see what would have been lost. The jets' beam-axis velocities are $\beta_z = \tanh y$ . In the CM, $\tanh(\pm1) = \pm0.761594$ : a difference of $1.523188$ . In the lab, $\tanh(1.693147) = 0.934547$ and $\tanh(-0.306853) = -0.297571$ : a difference of $1.232118$ . The velocity difference moved by $19\%$ under a boost you cannot measure. The rapidity difference did not move at all. That contrast is the whole case for the variable.

2 · A pion and a proton at the same pseudorapidity are not at the same rapidity

A tracking detector reports two tracks, both with $p_T = 1\ \mathrm{GeV}/c$ and both at $\eta = 2.0$ . One is a charged pion ( $mc^2 = 139.57\ \mathrm{MeV}$ ), the other a proton ( $938.27\ \mathrm{MeV}$ ). The detector cannot tell them apart from the track alone. How much does the difference matter?

Step 1 — identify. At $\eta = 2.0$ the geometry is fixed: $\cos\theta = \tanh 2 = 0.964028$ , $\sin\theta = \operatorname{sech}2 = 0.265802$ , so $\theta = \mathbf{15.41^\circ}$ . Both tracks point the same way, and $p_z = p_T\sinh\eta$ , $|\mathbf{p}| = p_T\cosh\eta$ :

p_z = 1 \times \sinh 2 = \mathbf{3.62686\ \mathrm{GeV}/c}, \qquad |\mathbf{p}| = 1 \times \cosh 2 = \mathbf{3.76220\ \mathrm{GeV}/c}.

Step 2 — apply. The energies differ, and so therefore do the speeds:

E_\pi = \sqrt{3.76220^2 + 0.13957^2} = \mathbf{3.76478\ \mathrm{GeV}}, \qquad \beta_\pi = \frac{3.76220}{3.76478} = \mathbf{0.999313},

E_p = \sqrt{3.76220^2 + 0.93827^2} = \mathbf{3.87743\ \mathrm{GeV}}, \qquad \beta_p = \frac{3.76220}{3.87743} = \mathbf{0.970281}.

Step 3 — the boundary number. Apply $y = \operatorname{artanh}(\beta\cos\theta)$ , the bridge proved above:

y_\pi = \operatorname{artanh}(0.999313 \times 0.964028) = \mathbf{1.9907}, \qquad y_p = \operatorname{artanh}(0.970281 \times 0.964028) = \mathbf{1.6997}.

Two tracks at the same $\eta = 2.0$ ; rapidities differing by $0.29$ . For the pion the pseudorapidity is good to $0.5\%$ — it is already, at $1\ \mathrm{GeV}$ , effectively massless. For the proton it is wrong by $15\%$ , and an analysis that used $\eta$ as if it were $y$ would have mislocated it.

Step 4 — read the physics off. Both satisfy $|y| ＜ |\eta|$ , as the theorem requires, and the gap closes as $\beta \to 1$ : give the proton $p_T = 10\ \mathrm{GeV}/c$ and $\beta = 0.999689$ , so $y = 1.9958$ — pseudorapidity is now good to $0.21\%$ , a seventyfold improvement for a tenfold increase in $p_T$ . $\eta$ is a safe coordinate for a detector and a dangerous one for a measurement: exact for the ultra-relativistic particles that dominate high- $p_T$ physics, quietly wrong for everything slow.

Figure

The β-bridge, drawn. y = artanh(β cos θ) against η = artanh(cos θ). The diagonal (grey) is the massless limit, β = 1, where y and η coincide exactly. A charged pion with p_T = 1 GeV⁄c (blue) already sits essentially ON the diagonal — at η = 2 its rapidity is 1.9907, a 0.5% shortfall. A proton with the SAME p_T (orange) sags visibly: at η = 2 its rapidity is only 1.6997, wrong by 15%. Both curves pass exactly through the origin, because at θ = 90° every particle has y = η = 0 whatever its mass. And |y| ＜ |η| everywhere, always, because β ＜ 1

Physics of the portrait

Origins · 2 minPseudorapidity and how this was discovered

Pseudorapidity

1908–present · Minkowski's hyperbolic angle · the cosmic-ray emulsion plates · the ISR and the rapidity plateau · η stamped on every LHC detector

This card is a coordinate, not a person, because pseudorapidity has no discoverer — it has a lineage. The hyperbolic angle of a boost is Minkowski's (1908); the name rapidity is Alfred Robb's (Optical Geometry of Motion, 1911) — that history is told in Rapidity and Hyperbolic Boosts. What happened next was practical. Cosmic-ray physicists measuring hundreds of tracks in nuclear emulsion needed a variable in which the tracks of one event could be compared, but could not measure the momentum of most of them — only the angle. And a very fast particle's rapidity, they realised, is a function of its angle alone. So they used the angle, in the one combination that transforms under a longitudinal boost the way rapidity does: $\eta = -\ln\tan(\theta/2)$ . It is the massless limit of the rapidity, and it is measurable with a protractor. The convention crystallised at the CERN Intersecting Storage Rings in the 1970s — the first hadron collider, and the first machine in which the unknown longitudinal boost of the parton collision was the dominant practical nuisance — where the flat "rapidity plateau" of produced hadrons only made sense plotted in this variable. Today $\eta$ is engineering. ATLAS's silicon inner tracker covers $|\eta| ＜ 2.5$ ; its liquid-argon calorimetry runs to $|\eta| = 4.9$ ; CMS's forward hadron calorimeter reaches $|\eta| \simeq 5$ . These are not physics choices — they are where the beam pipe is. In angle, $|\eta| = 2.5$ is $\theta = 9.39^\circ$ and $|\eta| = 5$ is $\theta = 0.77^\circ$ : the coordinate stretches the forward direction precisely where the particles pile up, and it does so with the one stretching a beam boost leaves alone.

History

1908

— Minkowski presents the boost as a rotation through a hyperbolic angle, the quantity later called rapidity.

1911

— Alfred Robb, in Optical Geometry of Motion, names the hyperbolic angle of a boost the rapidity.

1949

— Cosmic-ray emulsion experiments begin measuring hundreds of tracks per event by angle alone, creating the practical need for a purely angular stand-in for rapidity.

1971

— The CERN Intersecting Storage Rings, the first hadron collider, begins operation; the unknown longitudinal boost of the parton collision becomes the central practical problem of collider kinematics.

1972

— The flat central rapidity plateau of hadron production is established at the ISR, an observation that only takes its simple form in rapidity variables.

2008

— The LHC starts up with detectors segmented in pseudorapidity: the ATLAS inner tracker at |η| ＜ 2.5, its forward calorimetry to |η| = 4.9.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed., §2.7 · Taylor & Wheeler, Spacetime Physics, 2nd ed., ch. 3. For the collider variables: Halzen & Martin, Quarks and Leptons, ch. 1–2 · Ellis, Stirling & Webber, QCD and Collider Physics · Cacciari, Salam & Soyez, "The anti- $k_t$ jet clustering algorithm", JHEP 04 (2008) 063 · Bjorken, Phys. Rev. D 27, 140 (1983).

Provenance. Signature $(+,-,-,-)$ ; boost convention FORWARD ( $S'$ at $+v\hat{z}$ ). Two of the base's three derivation cards were out of band and have been discarded: the uniformly accelerated rocket ( $\varphi = \alpha\tau/c$ ) belongs to the S6 accelerated-frames chapter, and the Bondi factor $k = e^\varphi$ to the Doppler chapter. The base's surviving card — rapidity additivity from the $\tanh$ addition formula — is already owned by Rapidity and Hyperbolic Boosts and Rapidity, which also own the $N$ -boost ladder; both are recalled by name and neither is re-derived. This article owns the kinematic payoff instead: the shift theorem for a massive particle with transverse momentum, and the exact $\beta$ -bridge to pseudorapidity. Every number recomputed in verify_aq-sr-rapidity-additive-velocity.py: $\varphi_0 = \tfrac12\ln 4 = 0.693147$ ; $\sqrt{\hat{s}} = 1400\ \mathrm{GeV}$ ; $p_T = 700/\cosh 1 = 453.64\ \mathrm{GeV}/c$ ; lab rapidities $1.693147$ , $-0.306853$ , $\Delta y = 2.000000$ ; the velocity-difference degradation $1.523188 \to 1.232118$ ( $19\%$ ); $\tanh 2 = 0.964028$ , $\operatorname{sech}2 = 0.265802$ , $\sinh 2 = 3.62686$ , $\cosh 2 = 3.76220$ , $\theta = 15.41^\circ$ ; $E_\pi = 3.76478$ , $\beta_\pi = 0.999313$ , $y_\pi = 1.9907$ ; $E_p = 3.87743$ , $\beta_p = 0.970281$ , $y_p = 1.6997$ ; the proton at $p_T = 10\ \mathrm{GeV}/c$ giving $\beta = 0.999689$ , $y = 1.9958$ ; $\eta = 2.5 \Rightarrow \theta = 9.39^\circ$ , $\eta = 4.9 \Rightarrow 0.85^\circ$ , $\eta = 5 \Rightarrow 0.77^\circ$ .

Builds on. Rapidity and Hyperbolic Boosts · Rapidity · Relativistic Velocity Addition · Velocity Composition and the Speed-of-Light Invariant. Leads to. Thomas–Wigner Rotation · Longitudinal Doppler Effect · Twin Paradox and the Doppler Ledger.

If rapidity is just artanh of the velocity, why does the collider definition look so different?

Because it is a different quantity, and the resemblance is a trap. The frame rapidity is $\varphi = \operatorname{artanh}\beta$ , with $\beta$ a speed. A particle's collider rapidity is $y = \operatorname{artanh}(p_zc/E) = \operatorname{artanh}\beta_z$ , with $\beta_z$ the component along the beam. They coincide only when $p_T = 0$ : a particle moving transversely at $0.99c$ has $\beta = 0.99$ but $y = 0$ . What makes $y$ useful is that it is built from exactly the pair $(E, p_z)$ that a longitudinal boost mixes — the mass and the transverse momentum are inert and drop out of the transformation law. That is the shift theorem, and it would be false for $\operatorname{artanh}\beta$ .

Why not just measure the longitudinal boost and correct for it?

You cannot. The boost is set by $x_1$ and $x_2$ , the momentum fractions of the two colliding partons, and these are not observable event by event — the rest of each proton escapes down the beam pipe with the information. All that is known is the probability distribution of $x$ , the parton distribution function. So the unknown boost is not a nuisance to be measured away; it is irreducible. The field's response was to reformulate every observable so that it does not depend on what cannot be known — and rapidity is the coordinate in which that reformulation is trivial, because the unknown enters as an additive constant and cancels in every difference.

Is pseudorapidity ever exactly equal to rapidity for a massive particle?

Yes, at exactly one place: $\theta = 90^\circ$ , where $\cos\theta = 0$ , so $y = \operatorname{artanh}(\beta \cdot 0) = 0$ and $\eta = 0$ for every mass and speed. Everywhere else $\tanh y = \beta\tanh\eta$ with $\beta ＜ 1$ , so $|y| ＜ |\eta|$ strictly. The gap grows with $|\eta|$ and shrinks with $\beta$ : negligible for a $1\ \mathrm{GeV}$ pion ( $\beta = 0.9993$ ), noticeable for a $1\ \mathrm{GeV}$ proton ( $\beta = 0.9703$ ), catastrophic for anything genuinely slow. The working rule is that $\eta$ may stand in for $y$ whenever $p \gg mc$ — nearly always true of the particles anyone is looking at, and nearly always false of the soft debris.

Does the shift theorem work for boosts that are not along the beam?

No, and it cannot. Every step used the fact that the boost is along $\hat{z}$ — that is what leaves $p_x$ , $p_y$ invariant and what makes the $(E, p_z)$ pair close under the transformation. A transverse boost mixes $p_x$ with $E$ , so $p_T$ is no longer inert and the rapidity does not shift by a constant. Worse, two boosts in different directions do not compose to a pure boost at all (Thomas–Wigner Rotation), so there is not even a well-defined "total rapidity" to shift by. The saving grace: the collider's unknown boost is along the beam by construction, since the beams are collinear and the partons inherit their longitudinal momentum from them.

What is the transverse mass, and why does it appear everywhere in collider formulae?

$m_T = \sqrt{m^2c^2 + p_T^2}$ is the piece of the four-momentum a longitudinal boost cannot touch, and it is rapidity's natural companion. Together they parametrise the four-momentum cleanly: $E = m_Tc\cosh y$ , $p_z = m_T\sinh y$ , with $p_T$ and $\phi$ as given. Under a beam boost, $m_T$ is unchanged and $y \to y - \varphi_0$ — the four-momentum transforms by shifting a single number in a single place, which is why every collider kinematics library stores particles as $(p_T, \eta, \phi, m)$ rather than in Cartesian components. Note too that $(E, p_z)$ traces a hyperbola of "radius" $m_Tc$ as $y$ varies: the invariant hyperbola of the boost group, seen in momentum space.

The rapidity plateau: what exactly is being claimed, and why is rapidity essential to seeing it?

The claim is that in the central region of a high-energy hadron collision $\mathrm{d}N/\mathrm{d}y$ is approximately flat. Because a longitudinal boost translates $y$ rigidly (Jacobian exactly $1$ ), a flat distribution is one the boost cannot change: the plateau is a fixed point of the boost group acting on the spectrum. That is a symmetry statement, and it is invisible in any other variable — plotted against $p_z$ or against the angle, the same data are a shapeless peak. Bjorken took the symmetry seriously in 1983 and built the standard hydrodynamic picture of heavy-ion collisions on it, assuming boost-invariant initial conditions in the central region. It remains the working assumption.