---
title: "Relativistic Dynamics"
book: "University Physics — Year 3"
subject: physics
language: en
chapter: 5
exercises: 12
source: https://one-course.com/books/physics/5/en/chapter/5-relativistic-dynamics
---

# Chapter 5 — Relativistic Dynamics

In 1964 William Bertozzi filmed electrons racing down a linear accelerator: each burst was given a measured energy, and its speed was clocked over $8.4\,\mathrm{m}$ of flight. Newton predicted that $15\,\mathrm{MeV}$ electrons should fly at eight times the speed of light; the stopwatch said $0.9999c$ and refused to go further, however hard the electrons were pushed. Kinematics told us in the last chapter that $c$ is a limit; dynamics must now explain what happens to the pushing — where the work goes when speed no longer grows. The answer reorganises mechanics around a new momentum and a new energy, joined in the most famous equation in physics: energy stored is mass, mass is energy in residence, $E = mc^2$. This chapter builds the dynamics, then puts it to work where it lives daily — radioactive decays, particle collisions, and the accelerators that turn motion into new matter.

## 5.1 Momentum and energy, rebuilt

**Remark 5.1 (Why mv→m\vect vmv had to go).**

Classical momentum $m\vect v$ is conserved in every frame *if* velocities transform à la Galileo. They do not ([Proposition 4.10](https://one-course.com/books/physics/5/en/chapter/4-relativistic-kinematics#prop-b3-relativistic-kinematics-addition)): a collision that conserves $\sum m\vect v$ in one frame fails to in another, so the old momentum cannot be a law of nature — at best a slow-speed shadow of one. The repair is to measure the velocity with the particle’s *own* clock: $\dd\vect r/\dd\tau =
\gamma\,\vect v$ transforms cleanly, and $m\,\dd\vect r/\dd\tau$ turns out to be conserved in all frames at once.

**Definition 5.2 (Relativistic momentum and energy).**

A particle of mass $m$ and velocity $\vect v$ ($\gamma = 1/\sqrt{1 -
v^2/c^2}$) carries the momentum and the energy

$$
\vect p = \gamma m\vect v , \qquad
E = \gamma mc^2 .
$$

At rest, $E_0 = mc^2$: the *rest energy*. The kinetic energy is what motion adds, $E_k = E - mc^2 = (\gamma - 1)mc^2$, which for $v
\ll c$ reduces to the familiar $\tfrac12 mv^2$ (expand $\gamma$). In every isolated process, total $\vect p$ and total $E$ — rest energies included — are conserved.

**Theorem 5.3 (The energy–momentum relation).**

For any particle,

$$
E^2 = (pc)^2 + (mc^2)^2 ,
\qquad
\vect v = \frac{\vect p\,c^2}{E} :
$$

the combination $E^2 - p^2c^2 = m^2c^4$ has the same value in every frame — it is to energy and momentum what the interval is to time and space. Two regimes: at low speed, $E \approx mc^2 + p^2/2m$; at high energy ($E \gg mc^2$), $E \approx pc$ and $v \to c$: pushing harder adds energy and momentum, almost no speed.

**Proof.** $E^2 - p^2c^2 = \gamma^2m^2c^4(1 - v^2/c^2) = m^2c^4$; and $pc^2/E =
\gamma mvc^2/\gamma mc^2 = v$. Frame invariance follows because $m$ and $c$ are frame-independent. ∎

**Proposition 5.4 (Massless particles).**

A particle of zero mass has $E = pc$ and moves at exactly $c$ in every frame; it cannot be slowed, only redshifted. The photon is the standard case: $E = h\nu$ and $p = h\nu/c = h/\lambda$ — the Planck–Einstein relations of the Year 1 volume, now seen as the $m = 0$ corner of mechanics.

**Proof.** Set $m = 0$ in [Theorem 5.3](#thm-b3-relativistic-dynamics-relation): $E = pc$ and $v = pc^2/E = c$. ∎

![Left: Bertozzi’s “ultimate speed” experiment — measured electron speeds (dots) follow the relativistic curve and saturate at c while Newton’s line sails past it. Right: the energy–momentum hyperbola; its offset at p = 0 is the rest energy, its asymptote the photon’s E = pc.](https://one-course.com/images/onecourse/chapters/physics-5/b3-relativistic-dynamics/fig-7dc98d0f54c8.svg)

*Left: Bertozzi’s “ultimate speed” experiment — measured electron speeds (dots) follow the relativistic curve and saturate at $c$ while Newton’s line sails past it. Right: the energy–momentum hyperbola; its offset at $p = 0$ is the [rest energy](#def-b3-relativistic-dynamics-momentum-energy), its asymptote the photon’s $E = pc$.*

**Example 5.5 (Orders of magnitude).**

Particle physics counts energy in electronvolts and masses in $\mathrm{MeV}/c^2$: electron $0.511$, proton $938.3$, muon $105.7$, pion $139.6$. An electron of kinetic energy $1\,\mathrm{MeV}$ has $\gamma =
2.96$ and $v = 0.94c$ — already fully relativistic, which is why electronics stays classical ($\mathrm{eV}$) and nuclear physics does not ($\mathrm{MeV}$). A $6.8\,\mathrm{TeV}$ LHC proton has $\gamma = 7250$ and trails a photon by only $2.9\,\mathrm{m}/\mathrm{s}$.

## 5.2 Mass is energy in residence

**Theorem 5.6 (Mass–energy equivalence).**

The mass of a body is the total energy of its contents, divided by $c^2$, measured in its rest frame: heat a body, wind its spring, excite its atoms, and it weighs more by $\Delta E/c^2$; bind its parts together and it weighs *less* by the binding energy over $c^2$ — the *[mass defect](#thm-b3-relativistic-dynamics-emc2)*. Conversely, [rest energy](#def-b3-relativistic-dynamics-momentum-energy) can be released: whenever the final masses total less than the initial ones, the difference $\Delta m\,c^2$ emerges as kinetic energy or radiation.

**Proof.** *Admitted at this level.* ∎

**Example 5.7 (The ledger of c2c^2c2).**

$c^2 = 9 \times 10^{16}\,\mathrm{J}/\mathrm{kg}$: one gram of mass difference is $9 \times 10^{13}\,\mathrm{J}$, the energy of a small city for a day. Chemical bonds (a few eV per molecule) shift masses by parts in $10^{10}$ — forever unweighable, which is why chemistry never noticed. Nuclear binding shifts them by nearly $1\%$: helium weighs $0.7\%$ less than its four hydrogen ingredients, and that missing fraction, streaming from the Sun as light, costs it $\Delta m = L_\odot/c^2 =
4.3 \times 10^{9}\,\mathrm{kg}$ every second — four million tonnes of sunshine. Annihilation settles the whole account: an electron meeting a positron leaves nothing but two photons of $511\,\mathrm{keV}$ each, the signal by which PET scanners watch a living brain.

**Remark 5.8 (What “conversion” means).**

Nothing material “turns into” energy: total energy was conserved all along. What changes is its *residence* — from [rest energy](#def-b3-relativistic-dynamics-momentum-energy), which weighs, to kinetic energy and radiation, which fly. The deep statement of $E = mc^2$ is that inertia itself is an energy content: a box of hot gas resists acceleration more than the same box cold.

## 5.3 Collisions and decays

**Definition 5.9 (Four-momentum and invariant mass).**

Bundle a particle’s energy and momentum into its *four-momentum* $P = (E/c, \vect p)$. For a system of particles, sum componentwise: $E_{\text{tot}} = \sum E_i$, $\vect p_{\text{tot}} = \sum\vect p_i$. The system’s *invariant mass* $M$ is defined by

$$
M^2c^4 = E_{\text{tot}}^2 - \|\vect p_{\text{tot}}\|^2c^2 :
$$

the same number in every frame, equal to the total energy (over $c^2$) in the *centre-of-momentum frame* where $\vect p_{\text{tot}} = \vect 0$. Note that $M$ exceeds the sum of the parts’ masses whenever they move relative to each other — two photons flying apart have $M > 0$ though each is massless.

**Method 5.10 (Solving a relativistic collision).**

(1) Write [four-momentum](#def-b3-relativistic-dynamics-four-momentum) conservation: $E$ and $\vect p$ together, never one alone. (2) Compute the invariant $E^2 - p^2c^2$ of whatever bundle is convenient — it can be evaluated in the easiest frame and used in any other. (3) Thresholds: a reaction is possible when the [invariant mass](#def-b3-relativistic-dynamics-four-momentum) of the initial state reaches the summed rest masses of the final one. (4) Decays at rest: momenta of the products are opposite and equal; energies follow from the masses alone. (5) Only at the very end, if asked, convert to speeds via $v = pc^2/E$.

**Example 5.11 (The pion’s fingerprint).**

A charged pion at rest decays, $\pi^+ \to \mu^+ + \nu$ (the neutrino effectively massless). Momentum conservation makes the products back-to-back with equal $p$; energy conservation reads $m_\pi c^2 =
\sqrt{p^2c^2 + m_\mu^2c^4} + pc$. Solving:

$$
pc = \frac{(m_\pi^2 - m_\mu^2)c^2}{2m_\pi} = 29.8\,\mathrm{MeV} ,
$$

so every muon from a pion decaying at rest is born with exactly $4.1\,\mathrm{MeV}$ of kinetic energy — a monoenergetic line, and its observation (Powell, 1947) is how the pion’s mass was first weighed. Two-body decays always produce such lines; three-body decays smear them into spectra, which is precisely how the neutrino was first suspected in nuclear $\beta$ decay.

**Proposition 5.12 (Thresholds: the tyranny of the fixed target).**

To create new particles, only the *centre-of-momentum* energy $Mc^2$ is available. A beam of energy $E$ striking a target of mass $m_{\text{t}}$ at rest yields

$$
M^2c^4 = m_{\text{b}}^2c^4 + m_{\text{t}}^2c^4 +
2\,E\,m_{\text{t}}c^2 :
$$

the useful energy grows only as $\sqrt E$ — the rest is wasted on pushing the debris forward. Two identical beams colliding head-on yield $Mc^2 = 2E$: all of it useful. This single formula is why the frontier machines are colliders ([Problem 5.1](#pb-b3-relativistic-dynamics-1)).

**Proof.** Evaluate $E_{\text{tot}}^2 - p_{\text{tot}}^2c^2$ in the lab: $(E + m_{\text{t}}c^2)^2 - p^2c^2$ with $p$ the beam momentum; expand and use $E^2 - p^2c^2 = m_{\text{b}}^2c^4$. For the collider, $\vect p_{\text{tot}} = \vect 0$ and $Mc^2 = E_{\text{tot}}$. ∎

![Making matter from motion: on a fixed target, momentum conservation forces the products to keep flying, and the useful centre-of-momentum energy grows only as √ E; in a head-on collision it is all of 2E.](https://one-course.com/images/onecourse/chapters/physics-5/b3-relativistic-dynamics/fig-1e9d1998115f.svg)

*Making matter from motion: on a fixed target, momentum conservation forces the products to keep flying, and the useful centre-of-momentum energy grows only as $\sqrt E$; in a head-on collision it is all of $2E$.*

## 5.4 Force and the machines

**Proposition 5.13 (Relativistic equation of motion).**

Newton’s law survives in the form

$$
\vect F = \frac{\dd\vect p}{\dd t} = \frac{\dd(\gamma m\vect v)}
{\dd t} , \qquad
\frac{\dd E}{\dd t} = \vect F\cdot\vect v .
$$

A magnetic field, forever perpendicular to $\vect v$, changes no energy and bends the trajectory into a circle of radius

$$
r = \frac{p}{qB}
$$

— the same formula as in the Year 1 volume, with the relativistic $p$. But the circulation frequency $qB/\gamma m$ now *falls* as the particle gains energy: the cyclotron’s fixed-frequency push slips out of step near the MeV scale, and the high-energy machines — the *synchrotrons* — instead ramp their field and their frequency in synchrony with $\gamma$, holding the beam on one ring.

**Partial proof.** The force law is the definition of dynamics consistent with the new momentum ($\dd E/\dd t$ follows from $E^2 = p^2c^2 + m^2c^4$: $E\,\dot E = c^2\vect p\cdot\dot{\vect p}$). For the circle: $|\dd\vect p/\dd t| = qvB$ with $|\vect p|$ constant means the momentum vector turns at rate $qvB/p$, so the radius is $v/(qvB/p) = p/qB$. ∎

**Example 5.14 (Reading the LHC like an exercise).**

Protons of $E = 6.8\,\mathrm{TeV}$ in dipoles of $B = 8.3\,\mathrm{T}$: $p \approx E/c$ (ultrarelativistic), so $r = p/qB =
2.7\,\mathrm{km}$ — and indeed the ring’s magnetic bending radius is $2.8\,\mathrm{km}$, the $27\,\mathrm{km}$ circumference being mostly magnets. Head-on collisions provide $Mc^2 = 13.6\,\mathrm{TeV}$; to reach that on a fixed target would take a beam of $2E^2/m_{\text{p}}c^2 \approx
10^{17}\,\mathrm{eV}$ — a hundred-fold the reach of any machine ever built. The collider formula, not stronger magnets, is what bought the modern energy frontier.

![Charged particles crossing a cloud chamber. Tracks like these — curling in magnetic fields, appearing in pairs — turned E = mc2 from a formula into daily bookkeeping: the positron was discovered on such a photograph.](https://one-course.com/images/onecourse/chapters/physics-5/b3-relativistic-dynamics/img-3dff948dc296.jpg)

*Charged particles crossing a cloud chamber. Tracks like these — curling in magnetic fields, appearing in pairs — turned $E = mc^2$ from a formula into daily bookkeeping: the positron was discovered on such a photograph.*

## 5.5 Exercises

**Exercise 5.1 ★.**

An electron ($mc^2 = 0.511\,\mathrm{MeV}$) moves at $0.99c$. Compute $\gamma$, its momentum in $\mathrm{MeV}/c$, its total and kinetic energies. Same questions for a proton ($mc^2 = 938\,\mathrm{MeV}$) of kinetic energy $1\,\mathrm{GeV}$.

**Solution of Exercise 5.1.**

Electron: $\gamma = 7.09$; $pc = \gamma\beta\,mc^2 = 3.59\,\mathrm{MeV}$, so $p = 3.59\,\mathrm{MeV}/c$; $E = 3.62\,\mathrm{MeV}$, $E_k =
3.11\,\mathrm{MeV}$. Proton: $E = 938 + 1000 = 1938\,\mathrm{MeV}$, $\gamma
= 2.07$, $pc = \sqrt{E^2 - (mc^2)^2} = 1696\,\mathrm{MeV}$, $\beta =
pc/E = 0.875$.

**Exercise 5.2 ★.**

(a) How much energy sleeps in one gram of matter? (b) The Hiroshima explosion released about $6 \times 10^{13}\,\mathrm{J}$: what mass difference is that? (c) The Sun radiates $L = 3.8 \times 10^{26}\,\mathrm{W}$: how much mass does it shed per second, and what fraction of its $2 \times 10^{30}\,\mathrm{kg}$ in ten billion years? (d) Your phone battery stores $5 \times 10^{4}\,\mathrm{J}$: by how much is a charged phone heavier?

**Solution of Exercise 5.2.**

(a) $10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13}\,\mathrm{J}$. (b) $6 \times 10^{13}/
9 \times 10^{16} \approx 0.7\,\mathrm{g}$ — the bomb converted less than a gram. (c) $L/c^2 = 4.2 \times 10^{9}\,\mathrm{kg}/\mathrm{s}$; over $10^{10}\,\mathrm{yr}$ ($3.2 \times 10^{17}{}\,\mathrm{s}$): $1.3 \times 10^{27}{}\,\mathrm{kg}$, about $0.07\%$ of the Sun. (d) $5 \times 10^{4}/9 \times 10^{16} \approx
0.6\,\mathrm{ng}$ — real, and forever unweighable.

**Exercise 5.3 ★.**

(a) What momentum does a $5\,\mathrm{mW}$ laser pointer’s beam carry per second, and what force does the pointer feel? (b) What force does sunlight ($1.4\,\mathrm{kW}/\mathrm{m}^{2}$) exert on a perfectly reflecting $100\,\mathrm{m}^{2}$ solar sail? (c) From rest, how fast is a $10\,\mathrm{kg}$ sail-craft moving after a month? (d) Why does a comet’s dust tail point away from the Sun?

**Solution of Exercise 5.3.**

(a) $p = P/c$ per second: force $F = P/c = 1.7 \times 10^{-11}\,\mathrm{N}$ of recoil. (b) Reflection doubles the transfer: $F = 2\Phi A/c =
2 \times 1400 \times 100/3 \times 10^{8} \approx 0.9\,\mathrm{mN}$. (c) $a
\approx 9 \times 10^{-5}\,\mathrm{m}/\mathrm{s}^{2}$; after $2.6 \times 10^{6}\,\mathrm{s}$: $v \approx
240\,\mathrm{m}/\mathrm{s}$ — slow to start, but the engine never runs dry. (d) Sunlight’s momentum (with the solar wind) pushes the dust continuously outward: the tail streams away from the Sun, not behind the comet.

**Exercise 5.4 ★.**

Practice with units. (a) Show that $\mathrm{MeV}/c^2$ is a mass unit and convert the electron mass to kilograms. (b) What is $1\,\mathrm{GeV}/\mathrm{c}$ in $\mathrm{kg}\,\mathrm{m}/\mathrm{s}$? (c) An [invariant mass](#def-b3-relativistic-dynamics-four-momentum) squared comes out as $s = 16\,\mathrm{GeV}^{2}$ (in $c = 1$ units): what centre-of-momentum energy is that? (d) Why do particle physicists set $c = 1$, and what must a reader re-insert to get SI numbers?

**Solution of Exercise 5.4.**

(a) $m = E_0/c^2$: $0.511 \times 10^{6} \times 1.6 \times 10^{-19}/9 \times 10^{16} =
9.1 \times 10^{-31}\,\mathrm{kg}$. (b) $10^{9} \times 1.6 \times 10^{-19}/3 \times 10^{8} =
5.3 \times 10^{-19}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$. (c) $\sqrt s = 4\,\mathrm{GeV}$ of centre-of-momentum energy. (d) With $c = 1$, energy, momentum and mass share one unit and every formula sheds its $c$’s; to return to SI, reinsert the unique power of $c$ that fixes the dimensions.

**Exercise 5.5 ★★.**

Bertozzi’s electrons had kinetic energies $0.5$, $1$, $4.5$ and $15\,\mathrm{MeV}$. (a) Compute Newton’s prediction $v^2 = 2E_k/m$ for each, in units of $c^2$. (b) Compute the relativistic $v^2/c^2$. (c) His time-of-flight over $8.4\,\mathrm{m}$ at $15\,\mathrm{MeV}$: what did the clock read, and what would Newton have predicted? (d) The electrons’ energy was also measured calorimetrically — by their heat on impact: why was that step the experiment’s real point?

**Solution of Exercise 5.5.**

(a) $2E_k/mc^2$: $1.96$, $3.9$, $17.6$, $59$ — absurd beyond the first. (b) $1 - (1 + E_k/mc^2)^{-2}$: $0.74$, $0.89$, $0.990$, $0.9989$. (c) $8.4/0.99946c = 28.0\,\mathrm{ns}$; Newton’s $7.7c$ would have read $3.7\,\mathrm{ns}$. (d) The calorimeter proved the electrons truly carried the full $15\,\mathrm{MeV}$ into the target: the energy was all there, and still the speed had ceased to grow — work now buys momentum and energy, not velocity.

**Exercise 5.6 ★★.**

The charged pion decays at rest: $\pi^+ \to \mu^+\nu$, $m_\pi c^2 =
139.6\,\mathrm{MeV}$, $m_\mu c^2 = 105.7\,\mathrm{MeV}$, $m_\nu \approx 0$. (a) Show $pc = (m_\pi^2 - m_\mu^2)c^4/2m_\pi c^2$ and evaluate it. (b) The muon’s kinetic energy and speed. (c) The neutrino’s energy. (d) In flight at $\gamma_\pi = 50$, what are the maximum and minimum lab energies of the decay muon (decay forward and backward)?

**Solution of Exercise 5.6.**

(a) $m_\pi c^2 = \sqrt{p^2c^2 + m_\mu^2c^4} + pc$; isolate the root and square: $pc = (m_\pi^2 - m_\mu^2)c^4/2m_\pi c^2 =
(139.6^2 - 105.7^2)/(2 \times 139.6) = 29.8\,\mathrm{MeV}$. (b) $E_\mu =
\sqrt{29.8^2 + 105.7^2} = 109.8\,\mathrm{MeV}$: $E_k = 4.1\,\mathrm{MeV}$, $\beta = 29.8/109.8 = 0.27$. (c) $E_\nu = pc = 29.8\,\mathrm{MeV}$. (d) Boosting: $E_{\text{lab}} = \gamma(E^* \pm \beta p^*c)$ with $\beta
\approx 1$: between $50 \times (109.8 - 29.8) = 4.0\,\mathrm{GeV}$ and $50 \times (109.8 + 29.8) = 7.0\,\mathrm{GeV}$.

**Exercise 5.7 ★★.**

[Invariant mass](#def-b3-relativistic-dynamics-four-momentum) in [action](https://one-course.com/books/physics/5/en/chapter/1-lagrangian-mechanics#def-b3-lagrangian-mechanics-action). (a) Two photons of $200\,\mathrm{MeV}$ each fly at $60^\circ$ to one another: compute the [invariant mass](#def-b3-relativistic-dynamics-four-momentum) of the pair. (b) A neutral pion ($m_{\pi^0}c^2 = 135\,\mathrm{MeV}$) decays into two photons: what is the [invariant mass](#def-b3-relativistic-dynamics-four-momentum) of the photon pair, in every frame? (c) Explain how an experiment “discovers” a particle as a bump in the invariant-mass distribution of its decay products. (d) Two photons of equal energy fly exactly parallel: [invariant mass](#def-b3-relativistic-dynamics-four-momentum)? What does this say about a light beam’s rest frame?

**Solution of Exercise 5.7.**

(a) For two massless quanta, $M^2c^4 = 2E_1E_2(1 - \cos\theta) =
2 \times 200^2 \times \tfrac12$: $M = 200\,\mathrm{MeV}/c^2$. (b) Exactly $m_{\pi^0}$ — [invariant mass](#def-b3-relativistic-dynamics-four-momentum) is the particle’s mass, in every frame. (c) Compute $M$ for every pair of photons in the event: random pairs spread smoothly, true daughters of a particle pile up at its mass — the bump. (d) $\theta = 0$: $M = 0$; a parallel beam of light has no rest frame — it moves at $c$ as a whole, like each of its photons.

**Exercise 5.8 ★★.**

LHC bookkeeping ($E = 6.8\,\mathrm{TeV}$ protons). (a) $\gamma$, and the speed deficit $c - v$ (use $c - v \approx c/2\gamma^2$). (b) The revolution frequency on the $27\,\mathrm{km}$ ring and the number of laps per second. (c) The stored energy of a beam of $3 \times 10^{14}$ protons, in kilograms of TNT ($4.2 \times 10^{6}\,\mathrm{J}/\mathrm{kg}$). (d) The mass-equivalent of the two beams’ kinetic energy, in micrograms — matter about to be made.

**Solution of Exercise 5.8.**

(a) $\gamma = 6.8 \times 10^{12}/9.38 \times 10^{8} = 7250$; $c - v \approx
c/2\gamma^2 = 2.9\,\mathrm{m}/\mathrm{s}$. (b) $f = v/L \approx 3 \times 10^{8}/27000 =
11.1\,\mathrm{kHz}$: eleven thousand laps per second. (c) $3 \times 10^{14}
\times 6.8\,\mathrm{TeV} = 3.3 \times 10^{8}\,\mathrm{J} \approx 78\,\mathrm{kg}$ of TNT — in a hair-thin beam. (d) $2 \times 3.3 \times 10^{8}/9 \times 10^{16} \approx
7\,\text{µ}\mathrm{g}$: the working stock of matter-to-be.

**Exercise 5.9 ★★.**

Colliding photons. (a) Show that a single photon in vacuum cannot decay into an electron–positron pair, however energetic (use the invariant). (b) Against a second photon of energy $\epsilon$ head-on, show pair creation needs $E\epsilon \ge (m_{\text{e}}c^2)^2$. (c) A $511\,\mathrm{keV}$ photon needs what partner? An X-ray of $50\,\mathrm{keV}$? (d) The universe is filled with starlight and the cosmic microwave background: what does (b) predict for the reach of TeV $\gamma$-rays across intergalactic space?

**Solution of Exercise 5.9.**

(a) A photon’s [invariant mass](#def-b3-relativistic-dynamics-four-momentum) is $0$; an $e^+e^-$ pair’s is at least $2m_{\text{e}}c^2$. The invariant cannot change in an isolated decay: forbidden. (Equivalently: in no frame can momentum balance.) (b) Head-on, $M^2c^4 = 4E\epsilon$; pair creation needs $Mc^2 \ge
2m_{\text{e}}c^2$, i.e. $E\epsilon \ge (m_{\text{e}}c^2)^2$. (c) Another $511\,\mathrm{keV}$ photon; for $50\,\mathrm{keV}$, a partner of $0.511^2/0.05 = 5.2\,\mathrm{MeV}$. (d) A TeV photon meets the $(m_{\text{e}}c^2)^2/E \sim 0.3\,\mathrm{eV}$ threshold on ordinary starlight: the sky itself absorbs TeV $\gamma$-rays over cosmological distances — the universe is not transparent at all energies.

**Exercise 5.10 ★★★.**

Compton, by four-momenta. A photon of wavelength $\lambda$ strikes an electron at rest and scatters at angle $\theta$. (a) Write [four-momentum](#def-b3-relativistic-dynamics-four-momentum) conservation and isolate the final electron’s invariant. (b) Derive

$$
\lambda' - \lambda = \frac{h}{m_{\text{e}}c}\,(1 - \cos\theta) .
$$

(c) Evaluate $h/m_{\text{e}}c$ and the maximal shift; why is the effect invisible with visible light but decisive for X-rays? (d) What did Compton’s 1923 measurement establish about light that the photoelectric effect had not?

**Solution of Exercise 5.10.**

(a) $P_{e'} = P_\gamma + P_e - P_{\gamma'}$; square both sides (invariants): $m_{\text{e}}^2c^4 = m_{\text{e}}^2c^4 + 2P_\gamma\!
\cdot\!P_e - 2P_{\gamma'}\!\cdot\!P_e - 2P_\gamma\!\cdot\!P_{\gamma'}$. (b) With $P_\gamma\!\cdot\!P_e = E m_{\text{e}}$, $P_{\gamma'}\!\cdot\!P_e = E'm_{\text{e}}$ and $P_\gamma\!\cdot\!
P_{\gamma'} = EE'(1 - \cos\theta)/c^2$: $m_{\text{e}}c^2(E - E') = EE'(1 - \cos\theta)$, which in wavelengths ($E = hc/\lambda$) is the stated shift. (c) $h/m_{\text{e}}c =
2.43\,\mathrm{pm}$, at most $4.9\,\mathrm{pm}$: one part in $10^5$ of visible light (invisible), several percent of a $100\,\mathrm{pm}$ X-ray (measured). (d) That a photon collides like a billiard ball — carrying momentum $h/\lambda$ exchanged in *individual* events, not merely energy in lumps.

**Exercise 5.11 ★★★.**

The end of the cosmic-ray spectrum. The universe bathes in microwave photons of typical energy $\epsilon = 6 \times 10^{-4}\,\mathrm{eV}$. A proton of energy $E$ hitting one head-on can be excited to the $\Delta$ resonance ($m_\Delta c^2 = 1232\,\mathrm{MeV}$), losing energy each time. (a) Show the threshold condition is approximately $4E\epsilon =
(m_\Delta^2 - m_{\text{p}}^2)c^4$. (b) Compute the threshold $E$. (c) Above it, the universe is opaque to protons beyond some $100\,\mathrm{Mly}$: what does this predict for the observed cosmic-ray spectrum (the Greisen–Zatsepin–Kuzmin cut-off)? (d) Cosmic rays of $3 \times 10^{20}\,\mathrm{eV}$ have been recorded — “Oh-My-God” particles: what does their existence demand of their sources?

**Solution of Exercise 5.11.**

(a) $M^2c^4 = m_{\text{p}}^2c^4 + 4E\epsilon$ head-on (the proton ultrarelativistic); threshold at $M = m_\Delta$. (b) $E =
(1232^2 - 938^2)\,\mathrm{MeV}^{2}/(4 \times 6 \times 10^{-4}\,\mathrm{eV})
\approx 2.7 \times 10^{20}\,\mathrm{eV}$ for the mean photon — the thermal tail of the microwave background brings the effective cut-off to $\sim6 \times 10^{19}\,\mathrm{eV}$. (c) Protons above the cut-off lose energy to the $\Delta$ within $\sim10^{8}\,\mathrm{ly}$: the spectrum should end near $6 \times 10^{19}\,\mathrm{eV}$ for distant sources — the observed GZK suppression. (d) Their sources must be both extraordinary accelerators and cosmically *nearby* — within our supercluster — which is part of why their origin is still hunted.

**Exercise 5.12 ★★★.**

The photon rocket. A rocket of initial mass $m_{\text{i}}$ emits its exhaust as a collimated light beam and reaches speed $\beta c$. (a) Conserving energy and momentum between start and end, show

$$
\frac{m_{\text{i}}}{m_{\text{f}}} =
\sqrt{\frac{1 + \beta}{1 - \beta}} = \eu^{\varphi} ,
$$

the exponential of the rapidity — the relativistic Tsiolkovsky equation with the best possible exhaust. (b) Mass ratio to reach $0.9c$? And to reach $0.9c$ and *stop* at destination? (c) The beam must be made somehow: with matter–antimatter annihilation at perfect efficiency, how many kilograms of antimatter per kilogram of payload for the one-way $0.9c$ trip? (d) World antiproton production is nanograms per year: conclude, in one sentence, on photon rockets.

**Solution of Exercise 5.12.**

(a) Energy: $m_{\text{i}}c^2 = \gamma m_{\text{f}}c^2 + E_\ell$; momentum: $\gamma m_{\text{f}}\beta c = E_\ell/c$. Eliminate $E_\ell$: $m_{\text{i}} = \gamma(1 + \beta)m_{\text{f}} =
m_{\text{f}}\sqrt{(1+\beta)/(1-\beta)}$. (b) $\sqrt{19} = 4.4$; accelerating *and* braking squares it: $19$. (c) The consumed mass $m_{\text{i}} - m_{\text{f}} = 3.4\,m_{\text{f}}$ must be annihilated fuel, half of it antimatter: $1.7\,\mathrm{kg}$ of antimatter per kilogram delivered (one way, no braking). (d) At nanograms per year of world production, photon rockets remain arithmetic, not engineering.

![The discovery of antimatter (Carl D. Anderson, 1932, public domain): a single positron climbing through the cloud chamber’s lead plate, curving the wrong way for an electron and too tightly for a proton — E = mc2’s ledger caught running in reverse.](https://one-course.com/images/onecourse/chapters/physics-5/b3-relativistic-dynamics/img-b683dc787f01.jpg)

*The discovery of antimatter (Carl D. Anderson, 1932, public domain): a single positron climbing through the cloud chamber’s lead plate, curving the wrong way for an electron and too tightly for a proton — $E = mc^2$’s ledger caught running in reverse.*

## 5.6 Problem: Making matter — the antiproton and the Bevatron

**Problem 5.1.**

Weekend problem — how momentum conservation priced the antiworld

Dirac’s equations demanded, from 1931, that the proton have a mirror twin of opposite charge. Making one meant buying its [rest energy](#def-b3-relativistic-dynamics-momentum-energy) with beam energy — and momentum conservation set the price. This problem prices it, designs the machine that paid it, and follows the 1955 discovery. Data: $m_{\text{p}}c^2 = 938.3\,\mathrm{MeV}$; charge and *baryon number* (protons and neutrons $+1$, antiprotons $-1$) are conserved in every reaction.

**Part I — The [four-momentum](#def-b3-relativistic-dynamics-four-momentum) toolkit.**

1. For one particle, show $E^2 - p^2c^2 = m^2c^4$ from the definitions of $E$ and $\vect p$ .
2. For a system, why is $E_{\text{tot}}^2 -  p_{\text{tot}}^2c^2$ the same in all frames, and what does it equal in the centre-of-momentum frame?
3. A proton beam of energy $E$ hits a proton at rest: show $M^2c^4 = 2m_{\text{p}}^2c^4 + 2E\,m_{\text{p}}c^2$ .
4. Check the two limits: at low energy $Mc^2 \to  2m_{\text{p}}c^2$ ; at high energy $Mc^2 \approx  \sqrt{2E\,m_{\text{p}}c^2}$ — the square-root law.
5. A reaction is allowed only if $Mc^2$ reaches the summed rest masses of the products, with all products at rest in the centre-of-momentum frame at threshold: justify this last clause.
6. Why can the kinetic energy of the products in the lab never be recovered for particle creation on a fixed target?

**Part II — Pricing the antiproton.**

7. In $p + p$ collisions, the cheapest reaction making an antiproton $\bar p$ must also make an *extra* proton: write it, and justify with charge and baryon number.
8. Show the threshold requires $Mc^2 = 4m_{\text{p}}c^2$ .
9. Deduce the required beam energy $E = 7m_{\text{p}}c^2$ and kinetic energy $T = 6m_{\text{p}}c^2$ ; evaluate $T$ .
10. At threshold, what fraction of the beam’s kinetic energy actually became new rest mass ( $2m_{\text{p}}c^2$ of it)? Where is the rest?
11. In a head-on $p$ – $p$ collider, what kinetic energy *per beam* would the same reaction need? Compare the two designs.
12. In 1954 no collider existed (beams were too thin to hit each other): what practical fact forced the fixed-target choice, and what did it cost in beam energy?

**Part III — Designing the Bevatron.** The machine built at Berkeley reached $T = 6.2\,\mathrm{GeV}$ — comfortably above threshold — with dipole fields of $B =
1.56\,\mathrm{T}$.

13. Compute the beam’s total energy and momentum at top energy.
14. Compute the bending radius $r = p/qB$ and compare with the machine’s $15.2\,\mathrm{m}$ .
15. Compute the proton’s speed and the revolution frequency on the $2\pi r$ ring.
16. At injection the protons arrive at $10\,\mathrm{MeV}$ kinetic: compute their speed, and explain why the accelerating frequency had to *sweep* during each cycle (which two quantities change as $E$ grows?).
17. With about $1.5\,\mathrm{kV}$ gained per turn, how many turns and how long does one acceleration cycle take?
18. The name “Bevatron” came from BeV, billions of electronvolts: state in one line what physics fixed the design figure of $6$ and a fraction BeV.

**Part IV — The discovery, 1955.**

19. A magnet bends each candidate on a circle: explain why this selects particles by *momentum* ( $r = p/qB$ ), not by energy or speed — and why a momentum-selected beam still mixes species.
20. The beam on a copper target produced torrents of negative pions ( $m_\pi c^2 = 139.6\,\mathrm{MeV}$ ) among which a few antiprotons hid. The spectrometer selected negatives of momentum $p = 1.19\,\mathrm{GeV}/c$ : compute the speed of an antiproton and of a pion at that momentum.
21. Over the $12\,\mathrm{m}$ flight path, compute the two times of flight: what time resolution did the discovery need?
22. A second, independent signature used Čerenkov counters, firing only above a speed threshold: which particle was arranged to fire it, and why does redundancy matter for a discovery?
23. Chamberlain and Segrè counted about one antiproton per $44\,000$ pions: why did the threshold argument guarantee the rate would be tiny even above threshold?
24. An antiproton eventually meets a proton and annihilates: how much energy is released per event, and into what, typically?
25. Summarise the named result: conservation of [four-momentum](#def-b3-relativistic-dynamics-four-momentum) priced the antiproton at $T = 6m_{\text{p}}c^2 =  5.6\,\mathrm{GeV}$ on a fixed target; a $6.2\,\mathrm{GeV}$ , $15\,\mathrm{m}$ synchrotron paid it, and the antiworld became laboratory fact — the Nobel Prize of 1959.

**Solution of Problem 5.1.**

**1.** $E^2 - p^2c^2 = \gamma^2m^2c^4(1 - \beta^2) = m^2c^4$. **2.** $E_{\text{tot}}$ and $\vect p_{\text{tot}}$ transform like one particle’s $E$ and $\vect p$ (they are sums), so the same algebra gives an invariant; where $\vect p_{\text{tot}} = \vect 0$ it is the total energy squared: $Mc^2 = E_{\text{CM}}$. **3.** $(E + m_{\text{p}}c^2)^2 - p^2c^2 = m_{\text{p}}^2c^4 +
2Em_{\text{p}}c^2 + m_{\text{p}}^2c^4$. **4.** $E \to m_{\text{p}}c^2$ gives $M = 2m_{\text{p}}$; $E \gg m_{\text{p}}c^2$ gives $Mc^2 \to \sqrt{2Em_{\text{p}}c^2}$. **5.** Any relative motion of the products adds kinetic energy in the CM frame on top of their rest masses; the minimum $Mc^2$ — the threshold — leaves them all at relative rest. **6.** The lab momentum $\vect p_{\text{tot}} \neq \vect 0$ must survive the collision: the products are condemned to move, and their kinetic energy is locked away from mass-making. **7.** $p + p \to p + p + p + \bar p$: charge $+2 \to +2$; baryon number $+2 \to 1 + 1 + 1 - 1 = +2$. Making $\bar p$ alone, or with anything less than a full extra baryon, breaks one of the two. **8.** Four proton masses at relative rest: $Mc^2 =
4m_{\text{p}}c^2$. **9.** $16m_{\text{p}}^2c^4 = 2m_{\text{p}}^2c^4 +
2Em_{\text{p}}c^2$: $E = 7m_{\text{p}}c^2$, $T = 6m_{\text{p}}c^2 =
5.63\,\mathrm{GeV}$. **10.** New rest mass: $2m_{\text{p}}c^2$ out of $T =
6m_{\text{p}}c^2$ — one third; the other two thirds fly on as the products’ kinetic energy, protected by momentum conservation. **11.** Head-on, $Mc^2 = 2(T' + m_{\text{p}}c^2) =
4m_{\text{p}}c^2$: $T' = m_{\text{p}}c^2 = 0.94\,\mathrm{GeV}$ per beam — six times less than the fixed-target beam, twelve times less total kinetic energy. **12.** Beams of the 1950s were far too tenuous to collide with each other usefully; a solid target offers $10^{22}$ protons per cubic centimetre. The price: $5.6\,\mathrm{GeV}$ instead of $2 \times 0.94\,\mathrm{GeV}$. **13.** $E = 6.2 + 0.94 = 7.14\,\mathrm{GeV}$; $pc =
\sqrt{7.14^2 - 0.938^2} = 7.08\,\mathrm{GeV}$. **14.** $p = 7.08 \times 10^{9} \times 1.6 \times 10^{-19}/3 \times 10^{8} =
3.8 \times 10^{-18}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$: $r = p/qB = 15.1\,\mathrm{m}$ — the machine as built. **15.** $\beta = pc/E = 0.991$; orbit $2\pi r = 95\,\mathrm{m}$: $f = 3.1\,\mathrm{MHz}$. **16.** At $10\,\mathrm{MeV}$, $\beta = 0.145$: $f =
0.46\,\mathrm{MHz}$. As the energy climbs, the speed (hence $f$) and the momentum (hence the field needed at fixed radius) both change: field and radio-frequency must ramp together — the defining trick of the synchrotron. **17.** $6.19 \times 10^{9}/1500 \approx 4 \times 10^{6}$ turns; at an average megahertz-scale frequency, an acceleration cycle of the order of one to two seconds. **18.** The antiproton threshold $T = 6m_{\text{p}}c^2 =
5.6\,\mathrm{GeV}$, plus margin: the accountancy of this problem is literally what the machine’s energy — and name — were chosen for. **19.** $r = p/qB$ contains no mass: the magnet bends equal momenta equally, whatever the particle — so the selected beam still mixes pions, kaons and (rarely) antiprotons, all at $1.19\,\mathrm{GeV}/c$. **20.** $\bar p$: $E = \sqrt{1.19^2 + 0.938^2} =
1.51\,\mathrm{GeV}$, $\beta = 0.79$. $\pi^-$: $E = \sqrt{1.19^2 +
0.140^2} = 1.20\,\mathrm{GeV}$, $\beta = 0.993$. **21.** $t_{\bar p} = 12/(0.785c) = 51\,\mathrm{ns}$; $t_\pi =
40\,\mathrm{ns}$: the $11\,\mathrm{ns}$ gap demanded nanosecond timing — available, just, in 1955. **22.** The counters were set to fire on the fast pions and stay dark for the slow antiprotons (a velocity veto); with a $1$-in-$44000$ needle, only two independent signatures (time of flight *and* Čerenkov threshold) could exclude coincidence fakes. **23.** Just above threshold the products are born almost at relative rest: the reaction has almost no [phase space](https://one-course.com/books/physics/5/en/chapter/2-hamiltonian-mechanics#def-b3-hamiltonian-mechanics-hamiltonian), while pion production, far above *its* threshold, is copious — rarity was guaranteed by the same kinematics that set the price. **24.** $2m_{\text{p}}c^2 = 1.9\,\mathrm{GeV}$ per annihilation, typically into a handful of pions that decay onward to photons, muons and neutrinos. **25.** [Four-momentum](#def-b3-relativistic-dynamics-four-momentum) conservation priced the antiproton at $6m_{\text{p}}c^2 = 5.6\,\mathrm{GeV}$ on a fixed target; the $6.2\,\mathrm{GeV}$, $15.1\,\mathrm{m}$ Bevatron paid it; time of flight and a Čerenkov veto found one antiworld particle per $44000$ impostors — and the 1959 Nobel Prize certified the ledger.