Compare classical and relativistic momentum, derive p = gamma m v, and analyse the role of relativistic momentum in particle accelerators

What this dot point is asking

NESA wants you to know that classical $p = m v$ fails near $c$, that the correct relativistic momentum is $p = \gamma m v$, that energy and momentum are linked by $E^2 = (pc)^2 + (mc^2)^2$, and to explain how this shapes the design of particle accelerators.

The answer

Why classical momentum fails

Newtonian mechanics gives momentum as $p = m v$. Two clues that this must break at high speed:

1. Light has no rest mass, yet it carries momentum (radiation pressure, comet tails, solar sails). The classical expression $m v$ gives zero for $m = 0$.
2. Charged particles in cyclotrons fall out of phase with the accelerating voltage at high energies, contrary to the simple $r = m v / (q B)$ relation.

The resolution is that momentum must be modified at high speed so that conservation of momentum holds in all inertial frames consistent with Einstein's postulates.

Relativistic momentum

The correct expression for momentum of a particle of rest mass $m$ moving at velocity $\vec{v}$ is:

At low speeds $\gamma \to 1$ and we recover $\vec{p} = m \vec{v}$. As $v \to c$, $\gamma \to \infty$ and momentum grows without bound even though $v$ is capped at $c$.

This relation can be derived from a number of arguments: requiring momentum conservation in elastic collisions analysed from two different inertial frames, deriving the four-momentum from the four-velocity in spacetime, or demanding that Newton's second law $\vec{F} = d\vec{p}/dt$ produce a well-defined response with finite forces.

Total energy and the energy-momentum relation

The total relativistic energy is $E = \gamma m c^2$. Combining with $p = \gamma m v$, eliminating $\gamma$ and $v$:

This is the fundamental energy-momentum invariant of special relativity. Two important limits:

• Rest: $p = 0$, $E = m c^2$ (the rest energy).
• Massless particle (photon): $m = 0$, $E = p c$. Combined with $E = h f$ and $\lambda f = c$, this gives the photon momentum $p = h f / c = h / \lambda$.

The speed limit

The kinetic energy is $KE = (\gamma - 1) m c^2$. As $v \to c$, $KE \to \infty$, which means no finite amount of work can accelerate a massive particle to the speed of light. Massless particles travel at $c$ and cannot be accelerated or decelerated (they exist only at $c$ in vacuum).

Particle accelerators

The whole job of an accelerator is to push charged particles to extremely high energies for collision experiments. Relativistic momentum dominates the design.

Circular machines (cyclotron, synchrotron). A particle of momentum $p$ in a perpendicular magnetic field $B$ has radius:

In a cyclotron, $B$ is fixed and the radius grows with $p$. The angular frequency $\omega = q B / (\gamma m)$ decreases as $\gamma$ grows, so the AC accelerating voltage falls out of phase with the particle. This limits classical cyclotrons to non-relativistic energies (about $10$ MeV per nucleon for protons).

Synchrotrons fix the radius and ramp both $B$ and the AC frequency in step with the rising $\gamma$. The Large Hadron Collider keeps $r$ near $4.3$ km and ramps $B$ from about $0.5$ T to $8.3$ T while protons are accelerated from $450$ GeV to $7$ TeV. At $7$ TeV, $\gamma \approx 7460$, $v / c \approx 1 - 9 \times 10^{-9}$ - just a hair below light speed, but with enormous momentum.

Linear accelerators (linacs). A linac uses successive RF cavities to add small kicks to the particle's energy along a straight line. Relativistic momentum determines the spacing of the drift tubes: as $\gamma$ grows, $v$ saturates near $c$ but $p$ keeps increasing, so cavity spacings only need to grow modestly along the line.

Why this matters in collisions

The reachable physics is set not by the lab-frame energy but by the centre-of-mass energy $\sqrt{s}$ available to make new particles. For a fixed-target collision of a particle with rest energy $m c^2$ on a target of the same kind:

$\sqrt{s} \approx \sqrt{2 m c^2 \cdot E_{\text{lab}}}$ (for $E_{\text{lab}} \gg m c^2$),

which scales as $\sqrt{E_{\text{lab}}}$. For collider experiments (two beams meeting head-on), $\sqrt{s} = 2 E_{\text{beam}}$, scaling linearly with beam energy. This is why almost all modern high-energy machines are colliders rather than fixed-target.

Worked example: a proton at the LHC

At $E = 7$ TeV, $E_0 = m_p c^2 = 0.938$ GeV.

$\gamma = E / E_0 = 7000 / 0.938 = 7463$.

$v / c = \sqrt{1 - 1/\gamma^2} \approx 1 - 1/(2 \gamma^2) = 1 - 9.0 \times 10^{-9}$.

$p c = \sqrt{E^2 - (m c^2)^2} \approx E = 7$ TeV (the rest energy is negligible compared to total energy).

Each proton carries the kinetic energy of a mosquito in flight, but concentrated into a single subatomic particle.

Examples in context

Example 1. Australian Synchrotron 3 GeV electron beam. Electrons in the storage ring have total energy $E = 3.0 \text{ GeV}$ and rest energy $m c^2 = 0.511 \text{ MeV}$, so $\gamma = 3000 / 0.511 = 5870$ and $\beta = v/c = \sqrt{1 - 1/\gamma^2} \approx 1 - 1.45 \times 10^{-8}$. Relativistic momentum $p = \gamma m v = 5870 \times 9.11 \times 10^{-31} \times 3.0 \times 10^8 = 1.60 \times 10^{-18} \text{ kg m/s}$. From $E^2 = (p c)^2 + (m c^2)^2$, $(p c)^2 = 3000^2 - 0.511^2 = 9 \times 10^6 \text{ MeV}^2$, so $p c = 3000 \text{ MeV}$. The Newtonian estimate $p_{\text{cl}} = m v$ underestimates by a factor $\gamma = 5870$ - the synchrotron simply could not exist without relativity.

Example 2. Cosmic-ray proton hitting a Lucas Heights detector. An ultra-high-energy cosmic-ray proton arrives with total energy $E = 10^{20} \text{ eV}$. Proton rest energy is $938 \text{ MeV} = 9.38 \times 10^8 \text{ eV}$, so $\gamma = 10^{20} / 9.38 \times 10^8 = 1.07 \times 10^{11}$. The proton's speed is $\beta = 1 - 4.4 \times 10^{-23}$ (extraordinarily close to $c$). Its momentum is $p = \gamma m_p v \approx \gamma m_p c = 1.07 \times 10^{11} \times 938 \text{ MeV}/c = 1.0 \times 10^{20} \text{ eV}/c$. From the proton's frame, the Earth's diameter is contracted to $L = L_0/\gamma = 12{,}742 \text{ km} / 10^{11} = 1.3 \times 10^{-4} \text{ m} = 0.13 \text{ mm}$.

Try this

Q1. Write the equation for relativistic momentum and the energy-momentum relation, defining each symbol. [2 marks]

• Cue. $p = \gamma m v$ where $\gamma = 1/\sqrt{1 - v^2/c^2}$; $E^2 = (pc)^2 + (mc^2)^2$.

Q2. A proton of rest mass $m = 1.67 \times 10^{-27} \text{ kg}$ has $\gamma = 2.0$. Calculate (a) its speed and (b) its momentum. [4 marks]

• Cue. (a) $\beta = \sqrt{1 - 1/4} = 0.866$; $v = 2.60 \times 10^8 \text{ m/s}$. (b) $p = \gamma m v = 2.0 \times 1.67 \times 10^{-27} \times 2.60 \times 10^8 = 8.68 \times 10^{-19} \text{ kg m/s}$.

Q3. An electron is accelerated through a potential difference of $V = 1.0 \text{ MV}$. (a) Calculate its total energy and $\gamma$. (b) Find its speed as a fraction of $c$. (c) Explain why fixed-field cyclotrons fail at this energy but synchrotrons succeed. [2+2+2 marks]

• Cue. (a) $KE = 1.0 \text{ MeV}$, $E = 1.511 \text{ MeV}$, $\gamma = 2.96$. (b) $\beta = \sqrt{1 - 1/\gamma^2} = 0.941$. (c) Cyclotron frequency depends on $m$; as $\gamma m$ grows, the electron falls out of resonance. Synchrotrons ramp both $B$ and $f$.