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Topic 9 of 10

Relativistic Collisions & Decays

In relativistic mechanics, both energy and momentum are conserved in collisions and decays — encapsulated in conservation of four-momentum $P^\mu_{\rm total}=$ const. The invariant mass $M^2c^4=P_\mu P^\mu$ is the key tool for finding threshold energies and understanding what products are kinematically allowed.

Key Concepts

Four-Momentum Conservation

P^\mu_{\rm initial}=P^\mu_{\rm final}

. This is four equations (energy + three momentum components), encapsulating both relativistic energy and momentum conservation simultaneously.

Invariant Mass

M^2c^4=P_\mu P^\mu

is invariant — the same in all frames. It equals the total rest energy available in the CM frame. For a single particle at rest:

Mc^2=

rest mass.

Threshold Energy

The minimum lab energy for a reaction to occur. At threshold, all products are at rest in the CM frame. Using

s=(p_1+p_2)^\mu(p_1+p_2)_\mu=(\sum m_f)^2c^4

Two-Body Decay

Parent of mass

M

at rest decays to

m_1+m_2

. Conservation gives

E_1=(M^2+m_1^2-m_2^2)c^4/(2Mc^2)

and the momentum

|\vec p_1|=|\vec p_2|

Key Equations

Invariant mass squared

S = P_\mu P^\mu = E_{\rm cm}^2/c^2

The Mandelstam variable s; equals square of CM energy.

Threshold condition

\sqrt{s_{\rm th}} = \sum_f m_f c^2

At threshold, all final-state particles are at rest in the CM frame.

Two-body decay energy

E_1 = \frac{(M^2 + m_1^2 - m_2^2)c^4}{2Mc^2}

Energy of decay product 1 in the rest frame of parent M.

Worked Example

Pion Decay

Problem

A neutral pion ( $m_\pi=135$ MeV/c²) at rest decays into two photons. Find each photon's energy.

Solution

By symmetry, both photons have equal energy (momentum conservation gives equal and opposite momenta).

E_\gamma = \frac{m_\pi c^2}{2} = \frac{135}{2} = 67.5\text{ MeV}

Check: total energy $=2\times67.5=135$ MeV = rest energy ✓. Total momentum = 0 ✓.

Answer Each photon has

E_\gamma=67.5

MeV.

Practice

Exercises

7 problems

1 of 7

Watch a π⁰ at rest decay into two back-to-back photons. By momentum conservation, both photons have equal energy. Find the energy of each photon (in MeV) when $m_\pi c^2 = 135$ MeV.

pion decay

Photon energy

E_γ = MeV

m_π c² = 135 MeV. Pion at rest, massless photons.

2 of 7

A K⁰ meson ($Mc^2 = 494$ MeV) at rest decays into π⁺ and π⁻ (each $m_1c^2 = 140$ MeV). Watch the animation and find the momentum of each pion (in MeV/c).

kaon decay

Pion momentum

p = MeV/c

Mc² = 494 MeV (kaon), m_π c² = 140 MeV (each pion)

3 of 7

Two protons ( $m_pc^2=938$ MeV) collide head-on, each with energy $E=5000$ MeV. Find the CM energy $\sqrt{s}=2E$ (in GeV).

Unlock Exercise 3