PHYS3036 · Condensed Matter and Particle Physics
Relativistic Kinematics & Angular Momentum
This particle-physics chapter of University of Sydney PHYS3036 supplies two calculational toolkits. Relativistic kinematics uses energy–momentum four-vectors and the invariant mass E² = (pc)² + (mc²)² to reconstruct short-lived particles from their decay products and to explain why colliders use colliding beams. Angular momentum covers spin-½, combining spins via Clebsch–Gordan coefficients, and the spin of composite particles. Both are provided-data-sheet topics on the exam, so the skill is applying the formulae cleanly.
What this chapter covers
- 01Energy–momentum relation E² = (pc)² + (mc²)²; E = mc² at rest; Lorentz factor γ = (1 − β²)^(−1/2) with β = v/c
- 02Four-momentum P = (E/c, p) and its Lorentz transformation between frames
- 03Invariant mass of a system: (Mc²)² = (ΣE)² − (Σp c)² — the same in every frame; reconstructing a decaying particle from its products
- 04For a photon |p|c = E (massless); two-photon invariant mass (Mc²)² = 2E₁E₂(1 − cos θ)
- 05Centre-of-momentum energy √s and why colliding beams beat fixed targets (more usable energy)
- 06Spin and spin-½; combining angular momenta with Clebsch–Gordan coefficients (½ ⊗ ½ = 1 ⊕ 0)
- 07Spin of a meson (two spin-½ quarks + orbital L); the note that 'spin is not conserved'
Reconstructing a neutral particle from two photons (invariant mass)
- +1Total energy: ΣE = E₁ + E₂ = 5 + 5 = 10 GeV. For the momenta, each photon has |p|c = E, so |p₁|c = |p₂|c = 5 GeV, at 60° to each other. [+1]
- +1Squared total momentum: (Σpc)² = (p₁c)² + (p₂c)² + 2(p₁c)(p₂c)cos θ = 5² + 5² + 2·5·5·cos 60°. [+1]
- +1Evaluate with cos 60° = 0.5: (Σpc)² = 25 + 25 + 50·0.5 = 25 + 25 + 25 = 75 GeV². And (ΣE)² = 10² = 100 GeV². [+1]
- +1Invariant mass: (Mc²)² = (ΣE)² − (Σpc)² = 100 − 75 = 25 GeV² ⇒ Mc² = √25 = 5 GeV. [+1]
Key terms
- Energy–momentum relation
- E² = (pc)² + (mc²)²: the relativistic link between a particle's energy, momentum and rest mass; reduces to E = mc² at rest.
- Four-momentum
- The four-vector P = (E/c, p) whose components mix under Lorentz transformations but whose 'length' (the invariant mass) is frame-independent.
- Invariant mass
- The frame-independent mass of a system, (Mc²)² = (ΣE)² − (Σpc)²; used to reconstruct a decaying particle from its decay products.
- Lorentz factor (γ)
- γ = (1 − β²)^(−1/2) with β = v/c; equals E divided by the rest energy, so a larger γ means a speed closer to c.
- Centre-of-momentum energy (√s)
- The total energy in the frame where the net momentum is zero; the energy actually available to create new particles, maximised by colliding beams.
- Clebsch–Gordan coefficients
- The coefficients that combine two angular-momentum states into states of definite total angular momentum (e.g. ½ ⊗ ½ = 1 ⊕ 0); provided on the exam data sheet.
Relativistic Kinematics & Angular Momentum FAQ
What is invariant mass and why is it useful?
It is the mass of a system computed from (Mc²)² = (ΣE)² − (Σpc)², and it has the same value in every reference frame. That frame-independence is exactly why detectors use it: by measuring the energies and directions of a short-lived particle's decay products in the lab, you reconstruct the parent's mass and identify it — the standard way new particles are found.
Why do accelerators use colliding beams instead of fixed targets?
Because colliding beams put far more energy into the centre-of-momentum frame, where particle creation actually happens. In a fixed-target collision much of the beam energy goes into the momentum of the recoiling products and is wasted; with two beams meeting head-on the net momentum is near zero, so almost all the energy is available as √s to make new, heavier particles.
How do I combine two spin-½ particles?
Use Clebsch–Gordan coefficients: ½ ⊗ ½ = 1 ⊕ 0, i.e. two spin-½ particles combine into a spin-1 triplet or a spin-0 singlet. For a meson made of two spin-½ quarks in an orbital state L, the total spin is 0 or 1 and the total angular momentum follows from combining that with L. The exam provides the Clebsch–Gordan tables, so you apply them rather than memorise them.
Which formulae are on the exam data sheet for this chapter?
The particle-physics data sheet includes the particle-property table, constants (c, ℏ), and Clebsch–Gordan coefficients, and the module supplies the relativistic-kinematics relations. So the marks come from setting problems up correctly — choosing four-vectors, applying the invariant-mass formula, combining spins — not from recall. Confirm exactly what the sheet contains on Canvas.
Exam move
Split your practice between the two toolkits. For kinematics, always write the four-momenta first, then reach for the invariant mass (Mc²)² = (ΣE)² − (Σpc)² — it is the workhorse for reconstructing decays — and keep the photon shortcut (Mc²)² = 2E₁E₂(1 − cos θ) ready. Rehearse why colliding beams maximise √s. For angular momentum, drill ½ ⊗ ½ = 1 ⊕ 0 and using the Clebsch–Gordan tables from the data sheet to get meson spins. Because the exam supplies the sheet, focus on clean setup and unit-consistent arithmetic (work in GeV throughout) rather than memorisation. These are reliable, quick-scoring questions, so keep them warm across the semester. When a kinematics setup stalls, ask Sia to lay out the four-vectors and check each squared-momentum term.
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