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Relativity as measurement · A course with experiments

L3 · Chapter 7

Four-momentum and the mass every frame agrees on

Observers in different frames disagree about a particle’s energy and momentum, just as they disagree about times and distances. The two quantities change together, like time and position under a boost, and one combination of them comes out the same in every frame. For a group of particles that combination is the invariant mass. It decides whether a collision can make a new particle, and it is how a detector identifies what decayed.

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Energy and momentum change together

A particle of mass mm moving with velocity vv has energy E=γmE=\gamma m and momentum p=γmvp=\gamma m v. A frame moving at β\beta along the same line mixes them with the same boost that Chapter 3 applied to (t,x)(t,x). This is the one-dimensional case of Eq. (49.1) in the PDG kinematics review.

Time and position kept the interval fixed; energy and momentum keep E2−p2E^2-p^2 fixed. For one particle that number is its mass squared, which is why any frame can identify the particle.

E′=γ (E−βp),p′=γ (p−βE),E′2−p′2=E2−p2=m2E'=\gamma\,(E-\beta p),\qquad p'=\gamma\,(p-\beta E),\qquad E'^2-p'^2=E^2-p^2=m^2

The mass of a system is not the sum of its masses

Add the energies and the momenta of all particles in a system, then form the same combination. The result MM is the invariant mass of the system. Energy and momentum are each conserved in a collision, so MM is the same before and after, and it is the same in every frame.

In the frame where the total momentum is zero, the centre-of-momentum frame, MM is simply the total energy. It counts every particle’s rest energy and also the kinetic energy of their motion relative to each other. That extra part is what a collision can spend on making new particles.

M2=(∑iEi)2−(∑ipi)2,βcm=∑ipi∑iEiM^2=\Bigl(\sum_i E_i\Bigr)^{2}-\Bigl(\sum_i p_i\Bigr)^{2},\qquad \beta_{\mathrm{cm}}=\frac{\sum_i p_i}{\sum_i E_i}

Worked example

Motion turns into mass

Two lumps of mass 1 meet and stick. One is at rest; the other arrives with momentum 2. The moving lump has E1=1+4=2.236E_1=\sqrt{1+4}=2.236, so the total energy is 3.236 and the total momentum 2.

The merged lump has mass 2.544, not 2. Its extra 0.544 was kinetic energy of the two lumps relative to each other; after the merger it is internal energy and weighs like any other mass. The merged lump moves off at 0.618, the velocity of the centre-of-momentum frame.

M=(2.236+1)2−22=2.544,βcm=23.236=0.618M=\sqrt{(2.236+1)^2-2^2}=2.544,\qquad \beta_{\mathrm{cm}}=\frac{2}{3.236}=0.618

Worked example

The PDG example: a kaon on a resting proton

The PDG review works this case. A kaon of momentum 0.80 GeV/c strikes a proton at rest. With the listed masses mK=0.493677m_K=0.493677 and mp=0.938272m_p=0.938272 GeV, the kaon’s energy is 0.9401 GeV. Eq. (49.3) then gives the centre-of-momentum energy and Eq. (49.6) the momentum of each particle in that frame.

The review quotes 1.699 GeV and 0.442 GeV/c, and the numbers below reproduce them. In the laboratory the total energy is 1.878 GeV; the centre-of-momentum frame moves at 0.426 with γcm=1.105\gamma_{\mathrm{cm}}=1.105 (Eqs. 49.4 and 49.5).

Ecm=(mK2+mp2+2EKmp)1/2=1.699 GeVpcm=plab mpEcm=0.80×0.9381.699=0.442 GeV/c\begin{gathered}E_{\mathrm{cm}}=\bigl(m_K^2+m_p^2+2E_K m_p\bigr)^{1/2}=1.699\,\mathrm{GeV}\\ p_{\mathrm{cm}}=\frac{p_{\mathrm{lab}}\,m_p}{E_{\mathrm{cm}}}=\frac{0.80\times0.938}{1.699}=0.442\,\mathrm{GeV}/c\end{gathered}

How much beam energy makes an antiproton

Charge and baryon number are conserved, so a proton beam on a hydrogen target cannot make an antiproton alone. Something must carry the antiproton’s missing charge and baryon number, and the lightest choice is one more proton. The lightest final state is three protons and one antiproton, all at rest in the centre-of-momentum frame. That needs M=4mp=3.753 GeVM=4m_p=3.753\,\mathrm{GeV}.

For a beam proton of energy EE on a proton at rest, M2=2mp2+2EmpM^2=2m_p^2+2Em_p. Setting this to 16mp216m_p^2 gives E=7mpE=7m_p: a kinetic energy of 6mp=5.63 GeV6m_p=5.63\,\mathrm{GeV} and a momentum of 6.50 GeV/c. Only 3.75 GeV of the 6.57 GeV total becomes mass. The rest stays as motion of the whole system, because momentum must be conserved.

This threshold is our own derivation from the formulas above. Colliding two beams head-on avoids most of that waste, since the centre-of-momentum frame is then the laboratory.

2mp2+2Emp=(4mp)2 ⇒ E=7mp,T=E−mp=6mp=5.63 GeV2m_p^2+2Em_p=(4m_p)^2\ \Rightarrow\ E=7m_p,\qquad T=E-m_p=6m_p=5.63\,\mathrm{GeV}

Photons have no mass, but pairs of them do

A neutral pion at rest decays into two photons that fly apart with 67.5 MeV each. Give the pion momentum 0.3 GeV/c and, along the line of flight, the forward photon carries 0.314 GeV and the backward one only 0.0145 GeV. Their sum is still the pion’s energy, 0.329 GeV, and their invariant mass is still 0.135 GeV.

Each photon has zero mass, yet a pair can have mass. Two 0.1 GeV photons meeting head-on have total momentum zero and M=0.2 GeVM=0.2\,\mathrm{GeV}; the same two photons moving side by side have M=0M=0. A detector uses exactly this sum: it measures the energies and directions of decay products and computes their invariant mass. A peak in the histogram of that mass marks the particle that decayed. The particle collider Lab builds such a mass spectrum from muon pairs made by a teaching generator.

Eγ±=12 (Eπ±pπ),Eγ+=0.314 GeV,Eγ−=0.0145 GeV,M2=2E1E2 (1−cos⁡θ)E_\gamma^{\pm}=\tfrac12\,(E_\pi\pm p_\pi),\quad E_\gamma^{+}=0.314\,\mathrm{GeV},\quad E_\gamma^{-}=0.0145\,\mathrm{GeV},\qquad M^2=2E_1E_2\,(1-\cos\theta)

Where this leads

The rule that every frame agrees on the interval and on the invariant mass is the starting point of relativistic quantum field theory. In David Tong’s lecture notes, Chapter 1 asks every field theory to respect Lorentz invariance, and Chapter 2 normalises one-particle states so that the normalisation itself is the same in every frame. The next course on this path, on field theory, picks up from there.

An experiment to think through

Predict, change, explain

  1. Predict which of the four readouts stay fixed when only the viewing frame changes.
  2. In the collision Lab, start from “Kaon bounces off a proton” and drag “Velocity of the viewing frame” across its range. Then choose “Just enough energy for an antiproton”.
  3. Explain why “Total energy in this frame” and “Total momentum in this frame” change while “Invariant mass” stays fixed, and why “Energy left after making the products” is zero for the antiproton preset.
Watch a collision from moving frames See how a detector reconstructs a collision

The clock and twin Labs set light speed to one light-year per year, as these chapters do. The collision Lab uses PDG 2024 masses; the particle collider Lab draws its events from a teaching generator.

Try explaining it yourself

1. What beam kinetic energy is needed for a proton on a resting proton to make a neutral pion, in the reaction to two protons and one neutral pion?

Read an explained answer

The final state needs M=2mp+mπ=2.011 GeVM=2m_p+m_\pi=2.011\,\mathrm{GeV}. Then E=(M2−2mp2)/(2mp)=1.218 GeVE=(M^2-2m_p^2)/(2m_p)=1.218\,\mathrm{GeV}, so T=0.280 GeVT=0.280\,\mathrm{GeV} and p=0.777 GeV/cp=0.777\,\mathrm{GeV}/c.

2. A pion with p=0.3 GeV/cp=0.3\,\mathrm{GeV}/c decays. What is its Lorentz factor, and what is the energy of each photon in the pion’s rest frame?

Read an explained answer

Its energy is 0.329 GeV, so γ=0.329/0.135=2.44\gamma=0.329/0.135=2.44. In the rest frame each photon carries half the mass, 67.5 MeV.

3. Two lumps of mass 1 approach each other, each with momentum 2, and stick. What is the merged mass, and how does it compare with the fixed-target case in this chapter?

Read an explained answer

The total momentum is zero and each lump has energy 2.236, so M=4.472M=4.472. Hitting a resting lump with the same momentum gave only 2.544, because part of the energy had to stay as motion of the whole system.

Sources to read alongside

  1. Particle Data Group, Review of Particle Physics (2024), Section 49: Kinematics; Navas et al., Physical Review D 110, 030001 (2024)

    Equations (49.1)–(49.6) give the boost of energy and momentum, the centre-of-momentum energy and the fixed-target formulas; the worked kaon–proton example after Eq. (49.6) is recomputed in the course.

  2. Particle Data Group, 2024 particle listings: proton, neutral pion, charged kaon and muon

    Masses used in the course: proton 938.27208816 MeV, neutral pion 134.9768 MeV, charged kaon 493.677 MeV; muon mean life 2.1969811 microseconds. The pion, kaon and muon values come from the matching 2024 listing files.

  3. David Tong, Quantum Field Theory, University of Cambridge Part III lecture notes

    Course page and chapter list checked: Lorentz invariance in Chapter 1, causality and relativistic normalisation in Chapter 2. A next step after this course, not a source for its calculations.

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