advanced36 min studyLesson 606 in Physics

Choose collider coordinates without losing the mass

Derive rapidity addition and transverse invariance, distinguish pseudorapidity from true rapidity for massive particles, and connect coordinate densities to detector geometry.

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What you’ll understand

  • Write a massive four-momentum using transverse mass and rapidity.
  • Apply a longitudinal boost as an additive rapidity shift.
  • Identify when pseudorapidity equals rapidity and when the approximation fails.

At a hadron collider the partonic collision can be boosted by an unknown amount along the beam. Coordinates that respond simply to that boost help separate the event's internal geometry from its longitudinal motion. Transverse momentum and true rapidity have precisely such transformation rules. Pseudorapidity is convenient because a detector can infer it from direction alone, but it is not interchangeable with rapidity for a generic massive particle.

Parameterize the longitudinal mass shell

Choose the beam as the z axis and define pT=px2+py2p_T=\sqrt{p_x^2+p_y^2}. For an on-shell particle of mass m, define its single-particle transverse mass mT=m2+pT2m_T=\sqrt{m^2+p_T^2}. The mass shell becomes E2−pz2=mT2E^2-p_z^2=m_T^2, a hyperbola. Hyperbolic coordinates write E and longitudinal momentum using a rapidity y. This single-particle transverse mass is a coordinate quantity; later the module defines a different two-system transverse mass used to bound a parent mass. State which definition is being used.

A passive boost to a frame moving along positive z with rapidity ζ\zeta, where β=tanh⁡ζ\beta=\tanh\zeta, leaves transverse momentum unchanged and sends yy to y−ζy-\zeta. This follows immediately by applying the boost to E+pzE+p_z and E−pzE-p_z, which acquire opposite exponential factors. Differences of true rapidities therefore survive a common longitudinal boost. The simple shift applies to a fixed axis and a pure longitudinal boost, not to arbitrary rotations or transverse boosts.

mT=m2+pT2,E=mTcosh⁡y,pz=mTsinh⁡y,y=12log⁡E+pzE−pz,y′=y−ζm_T=\sqrt{m^2+p_T^2},\qquad E=m_T\cosh y,\quad p_z=m_T\sinh y,\quad y=\frac12\log\frac{E+p_z}{E-p_z},\qquad y'=y-\zeta
Use c = 1 and +−−−. The formulas assume a positive-energy on-shell particle with mT > 0. Rapidity and boost rapidity are dimensionless; pT and mT carry energy units. A boost of speed β along +z has ζ = artanh β.↔ Scroll wide equations horizontally.

Distinguish a direction coordinate from a velocity coordinate

Pseudorapidity is defined from the polar angle alone: η=−log⁡tan⁡(θ/2)\eta=-\log\tan(\theta/2). Equivalently it replaces E by the three-momentum magnitude in the rapidity logarithm. For a massless particle with nonzero transverse momentum, E equals that magnitude and the two coordinates agree exactly. For a massive particle, E is larger, so at positive longitudinal momentum true rapidity is smaller than pseudorapidity. A direction measurement alone cannot recover the missing mass-dependent energy.

The approximation becomes good when mass is negligible compared with the relevant momentum scale, and it can fail for soft heavy particles. A direction very close to the beam gives a large pseudorapidity because a small polar angle is being mapped to an unbounded coordinate. This does not mean the particle's speed exceeds light or that its energy has been measured. A particle exactly on the beam makes the angle coordinate singular, so finite-acceptance statements should specify the transverse and angular region rather than treating infinite pseudorapidity as an ordinary detector bin.

η=−log⁡tan⁡θ2=12log⁡∣p∣+pz∣p∣−pz=arsinh⁡pzpT,y=arsinh⁡pzm2+pT2\eta=-\log\tan\frac\theta2=\frac12\log\frac{|\mathbf p|+p_z}{|\mathbf p|-p_z}=\operatorname{arsinh}\frac{p_z}{p_T},\qquad y=\operatorname{arsinh}\frac{p_z}{\sqrt{m^2+p_T^2}}
The direction formulas require pT > 0 and a defined momentum direction. They make the massive difference explicit. Equality η = y holds exactly for massless particles in that domain; true rapidity, rather than generic massive pseudorapidity, has the simple longitudinal boost shift.↔ Scroll wide equations horizontally.

Use coordinate densities consistently

A histogram density depends on its coordinate measure. Uniform events per unit solid angle do not give a uniform pseudorapidity distribution; changing variables introduces a Jacobian. Detector acceptance is also commonly specified by pseudorapidity because it tracks geometry, while a theoretical boost-invariant quantity may use true rapidity. When comparing them, propagate the mass and momentum dependence rather than simply relabeling an axis. For reconstructed composite objects, their invariant masses can make the distinction relevant even when individual constituents are nearly massless.

Azimuthal angle and transverse momentum retain their values under a longitudinal boost. Together with true rapidity differences, they provide useful descriptions of collider configurations. A distance defined with pseudorapidity differences is approximately boost stable for ultrarelativistic objects, but that approximation has a domain. Also distinguish a boost of the complete event from an unknown change in the constituents' collision energy: an invariant mass remains fixed in the first operation and can vary in the second. Coordinate convenience does not remove the need to reconstruct or model the relevant four-vectors.

A massive track has two different rapidities

A particle has mass 3 GeV3\,\mathrm{GeV}, transverse momentum 4 GeV4\,\mathrm{GeV}, and longitudinal momentum 12 GeV12\,\mathrm{GeV}. Find its energy, true rapidity, and pseudorapidity.

  1. The transverse mass is mT=32+42=5 GeVm_T=\sqrt{3^2+4^2}=5\,\mathrm{GeV}. Hence E=mT2+pz2=25+144=13 GeVE=\sqrt{m_T^2+p_z^2}=\sqrt{25+144}=13\,\mathrm{GeV}.

  2. True rapidity is y=12log⁡[(13+12)/(13−12)]=12log⁡25=log⁡5≈1.6094y=\tfrac12\log[(13+12)/(13-12)]=\tfrac12\log25=\log5\approx1.6094. Its calculation uses the energy and therefore the mass.

  3. Direction instead gives η=arsinh⁡(12/4)=log⁡(3+10)≈1.8184\eta=\operatorname{arsinh}(12/4)=\log(3+\sqrt{10})\approx1.8184. The larger positive value is consistent with replacing E by the smaller momentum magnitude in the logarithm.

The result

The energy is thirteen GeV, true rapidity is about 1.609, and pseudorapidity is about 1.818. A direction-only reconstruction would not justify identifying these two coordinates.

Explore the model

Interactive lab

Interactive lab · spacetime

Verify the hyperbolic transformation algebra

Reset and set β = 0.6 and ct = 1. The fixed event position is x = 0.6. Predict transformed ct′ = 0.8, x′ = 0, and interval 0.64. Compare after a fresh Reset with β = 0 and ct = 1, where the original components remain. This illustrates the same two-component Lorentz algebra used for E and pz, but the readouts are spacetime coordinates. The lab has no pT, mass, rapidity, pseudorapidity, or detector acceptance controls.

Verify the hyperbolic transformation algebraA fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Current readings: Lorentz factor γ: 1.12; Event x′: 0.168; Event ct′: 0.817. All parameters are adjustable with the labeled controls below.−4.69−2.3502.354.69−2.2−1.101.12.2x or x′ · c = 1ct or ct′EE′ct² − x² = 0.64
Original eventTransformed eventLight cone
Lorentz factor γ
1.12
Event x′
0.168
Event ct′
0.817
t = 0 s

A fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forLongitudinal boost algebra preserves a hyperbola while changing its component coordinates.

Check your understanding

For a generic massive particle with nonzero transverse momentum, which statement about a common longitudinal boost is exact?

Try it yourself

A massless particle has pT=3 GeVp_T=3\,\mathrm{GeV} and pz=4 GeVp_z=4\,\mathrm{GeV}. Find y and η, then apply a passive longitudinal boost of rapidity ζ=log⁡3\zeta=\log3. Compute the new energy and longitudinal momentum.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The initial energy is five GeV. Use cosh(log3) = 5/3 and sinh(log3) = 4/3.

Compare with the solution

Initially E=5 GeVE=5\,\mathrm{GeV} and y=η=12log⁡[(5+4)/(5−4)]=log⁡3y=\eta=\tfrac12\log[(5+4)/(5-4)]=\log3. The boost gives E′=(5/3)(5)−(4/3)(4)=3 GeVE'=(5/3)(5)-(4/3)(4)=3\,\mathrm{GeV} and pz′=(5/3)(4)−(4/3)(5)=0p_z'=(5/3)(4)-(4/3)(5)=0. Transverse momentum stays three GeV, so the mass shell remains massless and y′=0y'=0. The direction is now transverse, giving η′=0\eta'=0 as well. This exact agreement uses zero mass; it would not extend automatically to the massive worked example.

Take it with you

  • True rapidity is the hyperbolic coordinate of energy and longitudinal momentum.
  • Pseudorapidity depends only on direction and equals rapidity exactly for massless particles.
  • State the coordinate measure, mass approximation, and boost axis when comparing collider distributions.

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