Part D · Minkowski · spine

Relativistic dynamics: four-vectors, and why colliders exist

Energy and momentum are the components of one vector whose Minkowski length is the mass; square the conservation law and a reaction threshold falls out in three lines, in any frame you like.

needs 26, 14, 16~2½ h slides + 5 h problemsproblems: ./mr problems 27

Unit 26 built the stage: an affine spacetime over a four-dimensional vector space $V$, a symmetric bilinear form $\eta$ of signature $(-,+,+,+)$, the Lorentz group preserving it, and proper time as the length of a timelike curve. It had no dynamics. This unit puts particles on the stage and asks what they do.

The answer has two halves. The mathematical half: a particle's worldline is a curve in Minkowski space, its velocity is the tangent vector normalised by proper time, and its momentum is that vector times the mass. Because these are vectors in $V$ — not triples of numbers that happen to transform in a certain way — every statement about them can be made without choosing a frame, and every scalar built from them can be evaluated in whichever frame is easiest. The physical half: the free action is $-mc^2$ times proper time, and everything — $p = \gamma m\vec v$, $E = \gamma mc^2$, $E^2 = p^2c^2 + m^2c^4$ — follows from it by the machinery of units 11 to 14, unchanged.

How to read this deck

One reaction runs through all four sections: $p + p \to p + p + p + \bar p$, the reaction by which the antiproton was discovered at Berkeley in 1955. Section 1 builds the objects it needs (four-vectors, proper time, four-velocity). Section 2 derives the dynamics from an action. Section 3 is the running example itself: its threshold, computed in two frames in three lines each, and why the answers differ by a factor of three. Section 4 takes acceleration seriously, which buys the hyperbolic worldline that unit 30 lives on, and shows why unit 16's rigid body has no relativistic successor.

Proofs are on the slides where they are one line and in these notes under Full proof where they are not; results imported from elsewhere are marked cited. Conventions throughout: $x^0 = ct$, $\eta = \mathrm{diag}(-1,1,1,1)$ in an inertial basis, Greek indices $0$ to $3$, Latin $1$ to $3$, Einstein summation as in unit 16.

27 · overview

One reaction, four sections

the question

A proton hits a proton and makes an antiproton: $$p + p \to p + p + p + \bar p .$$ Charge and baryon number force the three protons. How much beam energy does it take?

Newtonian intuition says $2m_pc^2$ of new mass needs $2m_pc^2$ of kinetic energy. It needs three times that on a fixed target — and exactly that when two beams collide head on.

how the answer is reached

1. Energy and momentum are one vector $p = (E/c,\ \vec p)$; add the two incoming ones to get the total $P$.

2. Its squared Minkowski length, $s = -c^2\eta(P,P)$, is the same in every frame and the same before and after.

3. Four protons come out, so $s \ge (4m_pc^2)^2$: a sum of four-momenta is at least as long as its parts put together.

4. Evaluate $s$ where it is easy: target at rest, $s = 2m_p^2c^4 + 2Em_pc^2$; head-on beams, $s = (2E)^2$. Solve each for $E$.

The plan. 1 four-vectors · 2 the action $-mc^2\!\int\mathrm{d}\tau$, giving $\vec p = \gamma m\vec v$, $E = \gamma mc^2$ · 3 invariant mass and thresholds: the reaction solved · 4 acceleration, horizons, and why nothing is rigid.

Getting around. A dotted underline is a glossary term: click it; G lists them all, N shows the notes. A solid underline, like the fixed-target frame, is a piece of the running example, shown on hover.

The sketch, step by step. Each step is a slide later on. Step 1 is section 1's four-vectors and section 2's "Four-momentum, and the mass shell", where $E/c$ and $\vec p$ turn out to be the components of $m$ times the four-velocity. Step 2 is "Why four-momentum is conserved" (the same before and after) and "Squaring, and the Mandelstam invariants" (the same in every frame, because $\eta$ is). Step 3 is "Invariant mass is not additive" and "Thresholds": the reversed triangle inequality of unit 26, applied to the four outgoing protons. Step 4 is "The antiproton threshold, in two frames", three lines per frame: $2m_p^2c^4 + 2Em_pc^2 \ge 16m_p^2c^4$ gives $E \ge 7m_pc^2$, kinetic energy $6m_pc^2$; $(2E)^2 \ge 16m_p^2c^4$ gives $E \ge 2m_pc^2$, kinetic energy $m_pc^2$ per beam. Hover over the running example to see the numbers now.

Why this example. It is the cleanest case in physics of an answer that Newtonian reasoning gets wrong by a large, round factor, and of a technique — evaluate an invariant in the frame where it is easiest — that makes a problem which looks like it needs a Lorentz transformation into one that does not. It also has a price tag: the factor of three, growing without bound at higher energies, is why particle physics abandoned fixed targets for colliders.

Where the numbers come from. $m_pc^2 = 938.27208816$ MeV (CODATA 2018). Every number in this deck was computed from that and the other constants quoted where they appear.

What the problem sheet adds. Compton scattering in four lines, the invariant mass of two photons, further thresholds (including the one that cuts off the cosmic-ray spectrum), a two-body decay, the rocket, a year at one $g$, a Born-rigid rod, and — the most instructive wrong answer in the course — the perihelion advance that special relativity alone predicts for Mercury.

section 1

Four-vectors

A velocity is a vector in $V$, and its components are a theorem.

27 · four-vectors

What unit 26 left on the table

Minkowski space, recalled

Spacetime $M$ is an affine space over a real 4-dimensional vector space $V$ with a symmetric, nondegenerate bilinear form $\eta$ of signature $(-,+,+,+)$. A basis $e_0,\dots,e_3$ with $\eta(e_\mu,e_\nu) = \eta_{\mu\nu} = \mathrm{diag}(-1,1,1,1)$ is an inertial frame.

$w\in V$ is timelike, null or spacelike as $\eta(w,w) \lt 0$, $= 0$, $\gt 0$.

two facts, used here

(a) The Lorentz group $O(1,3) = \{\Lambda\in GL(V) : \eta(\Lambda v,\Lambda w) = \eta(v,w)\}$ acts transitively on each sheet of $\{\eta(w,w) = -k^2\}$ via its orthochronous part (from 26's boosts; notes).
(b) Reversed triangle inequality: for future timelike $v,w$, $\ \sqrt{-\eta(v{+}w,v{+}w)} \ge \sqrt{-\eta(v,v)} + \sqrt{-\eta(w,w)}$ (unit 26).

Unit 26 was kinematics without dynamics: events, intervals, clocks. It never asked what makes a worldline bend, or what is conserved when two worldlines meet.

That needs vectors attached to particles: a velocity, a momentum. The question of this section is what kind of object those are, and the answer is: elements of $V$ itself.

sign convention

With $(-,+,+,+)$ a timelike vector has negative square. Books using $(+,-,-,-)$ flip every sign in this unit's $\eta$-products; the physics is identical.

Fact (a), proved. Unit 26 does not state it in this form, so here is the proof from its boosts. Let $w$ be future timelike with $\eta(w,w) = -k^2$. In an inertial basis $w = (w^0,\vec w)$ with $w^0 \gt |\vec w|$. Put $\vec\beta = \vec w/w^0$, so $|\vec\beta| \lt 1$; the boost with velocity $c\vec\beta$ (unit 26) takes $w$ to $(k,\vec 0)$. So every future timelike vector of square $-k^2$ is the image of $k e_0$ under a proper orthochronous Lorentz transformation, which is transitivity.

Fact (b) is unit 26's proof of the twin paradox, stated for vectors: if both are future timelike, pick the frame where $v = (v^0,\vec 0)$; then $-\eta(v,w) = v^0 w^0 \ge \sqrt{-\eta(v,v)}\sqrt{-\eta(w,w)}$ because $w^0 \ge \sqrt{(w^0)^2 - |\vec w|^2}$, and expanding $-\eta(v+w,v+w)$ gives the square of the right-hand side plus a nonnegative remainder. Equality holds iff $\vec w = \vec 0$ in that frame, that is iff $v$ and $w$ are parallel. Section 3 also needs it when one or both are null (future-pointing, nonzero), which the rest-frame argument does not cover as stated. If only $v$ is null, run it in the rest frame of $w$ instead; then $-\eta(v,w) = v^0w^0 \gt 0$ and the inequality is strict. If both are null, the right-hand side is $0$ and $-\eta(v+w,v+w) = -2\eta(v,w) = 2(v^0w^0 - \vec v\cdot\vec w) \ge 0$ by Cauchy–Schwarz, with equality iff $\vec v$ and $\vec w$ point the same way, that is iff $v$ and $w$ are parallel.

Why affine. Events are points; differences of events are vectors. A worldline is a curve in $M$, and its tangent at each point lives in $V$. That is why everything in this unit — velocity, acceleration, momentum — is an element of the same vector space.

27 · four-vectors

A four-vector is an element of a representation

definitions

A representation of a group $G$ on a vector space $W$ is a homomorphism $\rho: G\to GL(W)$.

A four-vector is an element of $V$, where $V$ carries the defining representation of $O(1,3)$: $\rho(\Lambda) = \Lambda$.

A Lorentz scalar is an element of the trivial representation $\rho(\Lambda) = 1$ on $\mathbb R$: a number every frame agrees on. $\eta(v,w)$ is one.

theorem: the transformation law

If $e'_\mu = e_\nu (\Lambda^{-1})^\nu{}_\mu$ is a second inertial basis, a vector with components $w^\nu$ in the first has components $w'^\mu = \Lambda^\mu{}_\nu\, w^\nu$ in the second.

Proof. $w = w'^\mu e'_\mu = w'^\mu e_\nu(\Lambda^{-1})^\nu{}_\mu$; compare with $w = w^\nu e_\nu$ and invert. It is unit 16's change-of-basis theorem with $\Lambda$ in place of $R$.

The physics: "four-vector" means a quantity whose values measured by two observers are related by the same $\Lambda$ that relates their coordinates. The definition says this without mentioning observers.

Why a representation. Unit 16 made the point for rotations: a tensor is a multilinear map, and "nine numbers that transform like this" is a theorem about that map's components. The same holds here. The Lorentz group acts on many vector spaces — on $V$ (four-vectors), on $V^*$ (covectors, unit 28's gradients), on $V\otimes V$ (unit 29's stress-energy), on $\Lambda^2 V^*$ (unit 28's field tensor), on $\mathbb R$ trivially (scalars). Each is a representation, and each "kind of relativistic quantity" is an element of one of them. This unit uses only two: $V$ and $\mathbb R$.

Full proof of the transformation law. The two bases are related by the matrix of $\Lambda^{-1}$ because a Lorentz transformation carries the old inertial basis to the new one: $e'_\mu = \Lambda^{-1}$-image, written in components as above. Then $w^\nu e_\nu = w = w'^\mu e'_\mu = w'^\mu (\Lambda^{-1})^\nu{}_\mu e_\nu$, and since the $e_\nu$ are a basis, $w^\nu = (\Lambda^{-1})^\nu{}_\mu w'^\mu$. Multiply by $\Lambda$: $w'^\mu = \Lambda^\mu{}_\nu w^\nu$. Nothing about $\eta$ was used: this is linear algebra. What $\eta$ adds is that $\Lambda$ ranges over $O(1,3)$ rather than all of $GL(V)$, so that $\eta_{\mu\nu}w^\mu w^\nu$ is the same number in both bases.

Checking a scalar is a scalar. $\eta(v,w)$ is defined without a basis, so it is invariant by construction. In components it reads $\eta_{\mu\nu}v^\mu w^\nu = -v^0w^0 + \vec v\cdot\vec w$, and the theorem plus $\Lambda^{\mathsf T}\eta\Lambda = \eta$ confirms it independently.

Warning. "Four numbers" is not enough: $(E/c, p_x, p_y, p_z)$ is a four-vector, but $(E, p_x, p_y, p_z)$ with $E$ in joules is not even dimensionally coherent, and $(p_x, p_y, p_z, m)$ is four numbers that transform like nothing in particular.

27 · four-vectors

Proper time is arc length

proper time

A worldline is a $C^2$ curve $\gamma:[a,b]\to M$ with timelike, future-pointing tangent $\gamma'$. Its proper time is $$\tau[\gamma] = \frac1c\int_a^b \sqrt{-\eta\big(\gamma'(\lambda),\gamma'(\lambda)\big)}\,\mathrm{d}\lambda .$$

reparametrisation invariance

If $\lambda = \phi(\sigma)$ with $\phi' \gt 0$, then $\tau[\gamma\circ\phi] = \tau[\gamma]$.

Proof. $(\gamma\circ\phi)' = \phi'\,\gamma'$, so the integrand gains a factor $\phi'$, which is the Jacobian of the substitution.

It is unit 11's arc length, $\int\sqrt{g(\gamma',\gamma')}$, with the sign flipped so that the root is real on timelike curves.

in an inertial frame

Parametrise by $t$: $\gamma(t) = (ct, \vec x(t))$, $\gamma' = (c, \vec v)$. Then $-\eta(\gamma',\gamma') = c^2 - v^2$ and $$\mathrm{d}\tau = \sqrt{1 - v^2/c^2}\;\mathrm{d}t = \mathrm{d}t/\gamma ,$$ with $\gamma = (1 - v^2/c^2)^{-1/2}$ unit 26's Lorentz factor.

Check. Dimensions: $\sqrt{\mathrm{m^2 s^{-2}}}\cdot\mathrm s/(\mathrm{m\,s^{-1}}) = \mathrm s$ ✓. At rest, $\mathrm{d}\tau = \mathrm{d}t$ ✓.

Two meanings of $\gamma$. The worldline $\gamma$ and the Lorentz factor $\gamma = (1-v^2/c^2)^{-1/2}$ share a letter by long tradition. Context always tells them apart: the worldline is a curve and carries an argument or a prime, the Lorentz factor is a number. In this unit the worldline appears only in section 1.

Existence of the proper-time parametrisation. $\tau(\lambda) = \frac1c\int_a^\lambda\sqrt{-\eta(\gamma',\gamma')}$ has derivative $\frac1c\sqrt{-\eta(\gamma',\gamma')} \gt 0$ everywhere, because $\gamma'$ is timelike. So $\tau$ is a $C^1$ bijection $[a,b]\to[0,\tau[\gamma]]$ with $C^1$ inverse (inverse function theorem in one variable), and $\gamma\circ\tau^{-1}$ is the same worldline parametrised by its own clock. Along it, $-\eta(\gamma',\gamma') = c^2$ identically: the analogue of unit-speed parametrisation.

Why "future-pointing". A timelike tangent could point to the past; $O(1,3)$ contains time reversal. Requiring $\gamma'^0 \gt 0$ in one inertial frame makes it so in all orthochronous frames (unit 26), and it is what "a particle" means physically.

27 · four-vectors

Two clocks between the same events

Two worldlines from event $O$ to event $Q$, ten units of $ct$ apart. A dot marks each unit of proper time (times $c$), as defined on the previous slide.

On the moving worldline $\mathrm{d}\tau = \mathrm{d}t/\gamma$, so its dots spread out where it is fast. It collects $c\tau = 7.46$; the one at rest collects $10$.

straight is longest

Among timelike curves with the same ends, the straight one has the largest proper time — unit 26's twin paradox. Euclidean arc length is the opposite: straight is shortest.

Every tangent stays inside the light cone: a worldline never runs at $45^\circ$ or flatter.

Same ends, different clocks. The curved worldline ($x = A\sin(\pi t/T)$, top speed $0.9c$) has fewer dots, spaced widest in the middle where $\gamma$ is largest: $\int\sqrt{1-v^2/c^2}\,\mathrm{d}t = 7.46$ against $10$.

What to look for in the figure. Time runs up, space across, with $c = 1$ in the drawing so light moves at $45^\circ$ (the dashed lines from $O$). The vertical worldline is a clock at rest; its dots are evenly spaced because $\mathrm{d}\tau = \mathrm{d}t$. The curved one leaves $O$ at $0.9c$, slows to rest at $ct = 5$, and returns. Near $O$ and $Q$, where it is fastest, $\gamma = 2.29$ and its dots are more than twice as far apart in $t$ as the resting clock's. In the middle, where it is momentarily at rest, they are nearly as close.

The numbers. $x(t) = A\sin(\pi t/T)$ with $T = 10$ and $A = 0.9T/\pi$, so $v(t) = 0.9c\cos(\pi t/T)$. The proper time is $\int_0^{10}\sqrt{1 - 0.81\cos^2(\pi t/10)}\,\mathrm{d}t = 7.459$, evaluated numerically; the dots are placed where the running integral reaches $1, 2, \dots, 7$.

The analogy, and where it breaks. In unit 11 the length of a curve in the plane was $\int\sqrt{\dot x^2 + \dot y^2}\,\mathrm{d}\lambda$, reparametrisation-invariant by exactly the argument of the previous slide, and straight lines minimised it. Here the straight line maximises it among timelike curves with the same ends — unit 26's twin paradox, which is fact (b) of "What unit 26 left on the table" integrated along the curve. That sign is why the action on "The only free action there is" carries a minus sign.

27 · four-vectors

Four-velocity, and the constraint it obeys

four-velocity

For a worldline parametrised by proper time, $u := \dfrac{\mathrm{d}\gamma}{\mathrm{d}\tau}\in V$.

the normalisation $$\eta(u,u) = -c^2 .$$

Proof. Differentiate the definition of $\tau$ along itself: $c = \sqrt{-\eta(u,u)}$.

components

$u = \dfrac{\mathrm{d}t}{\mathrm{d}\tau}\dfrac{\mathrm{d}\gamma}{\mathrm{d}t} = \gamma\,(c,\ \vec v)$. Check: $-\gamma^2c^2 + \gamma^2v^2 = -\gamma^2c^2(1 - v^2/c^2) = -c^2$ ✓.

The fourth component is not information. It is $\gamma c$, fixed by the other three — the four-vector is constrained, and the constraint is what the rest of this unit keeps differentiating.

limit

$v \ll c$: $u \approx (c,\ \vec v)$. The spatial part is the Newtonian velocity; the time part is the rate $c\,\mathrm{d}t/\mathrm{d}\tau \approx c$ at which a clock at rest ticks.

The next slide draws the constraint: one hyperbola holds every four-velocity there is.

27 · four-vectors

Every four-velocity on one hyperbola

Four components, three free: the set of possible four-velocities is the upper sheet of the hyperboloid $\eta(u,u) = -c^2$, a 3-dimensional surface in $V$, parametrised by $\vec v$ in the open ball $|\vec v| \lt c$.

The figure keeps one space direction. Each arrow is a four-velocity $\gamma(c, v)$; its tip lies on $(u^0)^2 - (u^1)^2 = c^2$.

As $v\to c$ the tip runs off along the light cone, but never reaches it: $\gamma$ grows without bound while $u$ stays timelike.

Every arrow has the same Minkowski length $c$, though on paper the fast ones look longer: the $\eta$-length of $\gamma(c, v)$ is $\sqrt{\gamma^2c^2 - \gamma^2v^2} = c$. At $v = 0.9c$ the time component is $\gamma c = 2.29c$.

What to look for in the figure. Paper lengths lie. The arrow for $v = 0.9c$ is drawn about three times as long as the one for $v = 0$, yet both have $\eta(u,u) = -c^2$: the hyperbola is the Minkowski "circle" of radius $c$. The numbers on it are $u = (1, 0)c$, $(1.25, 0.75)c$ at $v = 0.6c$, and $(2.294, 2.065)c$ at $v = 0.9c$, all from $\gamma(c,v)$.

The hyperboloid is the space of velocities. In Newtonian mechanics the space of velocities is $\mathbb R^3$, a vector space, and velocities add. Here it is the future sheet of a two-sheeted hyperboloid, and velocities do not add — they compose by Lorentz boosts, and unit 26 showed that along a line the composition is addition of rapidities. The hyperboloid with the metric that $\eta$ induces on it is hyperbolic space; that remark is not used, but it is why the velocity-addition formula looks like hyperbolic trigonometry.

27 · four-vectors

Acceleration is orthogonal to velocity

four-acceleration

$a := \mathrm{d}u/\mathrm{d}\tau \in V$.

theorem

$\eta(a,u) = 0$, and $a$ is spacelike or zero.

Proof. Differentiate $\eta(u,u) = -c^2$: $2\eta(a,u) = 0$. In the frame where $u = (c,\vec 0)$ — it exists by transitivity — orthogonality forces $a^0 = 0$, so $\eta(a,a) = |\vec a|^2 \ge 0$.

proper acceleration

$\alpha := \sqrt{\eta(a,a)}$: the acceleration an accelerometer carried by the particle reads.

In Newtonian mechanics a particle can accelerate along its velocity and speed up. Here the four-velocity cannot change its length, so it can only turn — on a hyperbola, not a circle.

$a$ is tangent to the hyperbola, so $u$ stays on it. "$\eta$-orthogonal" is not a right angle on paper: it means mirror images in the light cone, as $(\cosh\theta,\sinh\theta)$ and $(\sinh\theta,\cosh\theta)$ are.

What to look for in the figure. One space dimension, units with $c = 1$. Write $u = (\cosh\theta, \sinh\theta)$ with $\theta = 0.7$; then the tangent to the hyperbola at $u$ is $(\sinh\theta, \cosh\theta)$, and $\eta$ of the two is $-\cosh\theta\sinh\theta + \sinh\theta\cosh\theta = 0$. On paper, swapping the two components is reflection in the line $u^0 = u^1$, the light cone. So the Minkowski right angle is symmetric about the light cone rather than square: the faster the particle, the closer $u$ and $a$ both lean towards the cone, and at the cone itself (a null vector) the vector would be orthogonal to itself.

Why "what an accelerometer reads". In the instantaneous rest frame the particle is momentarily at rest, Newtonian mechanics is exact to first order in the velocity, and $a = (0, \mathrm{d}\vec v/\mathrm{d}t)$ there. An accelerometer is a Newtonian device used at rest, so it reads $|\mathrm{d}\vec v/\mathrm{d}t|$ in that frame, which is $\sqrt{\eta(a,a)}$ — a scalar, so every frame agrees on what the accelerometer says.

27 · four-vectors

The four-acceleration, in a frame

In a frame, from $u = \gamma(c,\vec v)$ and $\dot\gamma = \gamma^3\,\vec v\cdot\dot{\vec v}/c^2$, with dot $= \mathrm{d}/\mathrm{d}t$: $$a = \gamma\,\frac{\mathrm{d}u}{\mathrm{d}t} = \big(\gamma^4\,\vec v\cdot\dot{\vec v}/c,\ \ \gamma^2\dot{\vec v} + \gamma^4(\vec v\cdot\dot{\vec v})\,\vec v/c^2\big).$$

The time component is nonzero whenever the speed changes: a frame that sees the particle speed up sees $a$ tilt out of its own space, exactly as the figure on the previous slide showed.

checks

Orthogonal. $\eta(a,u) = -\gamma^5\vec v\cdot\dot{\vec v} + \gamma^3\vec v\cdot\dot{\vec v} + \gamma^5 v^2(\vec v\cdot\dot{\vec v})/c^2 = 0$ ✓.

At rest. $a = (0,\dot{\vec v})$ ✓: the Newtonian acceleration.

Dimensions. $\gamma^2\dot{\vec v}$ is $\mathrm{m\,s^{-2}}$, and so is $\vec v\cdot\dot{\vec v}/c$ ✓.

Two cases. Circular motion at constant speed: $\alpha = \gamma^2v^2/r$. Motion along a line: $\alpha = \gamma^3|\dot{\vec v}|$.

Full proof of the frame formula. $\gamma = (1-v^2/c^2)^{-1/2}$ gives $\dot\gamma = -\tfrac12(1-v^2/c^2)^{-3/2}\cdot(-2\vec v\cdot\dot{\vec v}/c^2) = \gamma^3\vec v\cdot\dot{\vec v}/c^2$. Then $\mathrm{d}u/\mathrm{d}t = (\dot\gamma c,\ \dot\gamma\vec v + \gamma\dot{\vec v})$, and multiplying by $\mathrm{d}t/\mathrm{d}\tau = \gamma$ gives the slide. The check expands $\eta(a,u) = -a^0u^0 + \vec a\cdot\vec u$ with $u = \gamma(c,\vec v)$: $-\gamma^4(\vec v\cdot\dot{\vec v}/c)\gamma c = -\gamma^5\vec v\cdot\dot{\vec v}$, and $\gamma\vec v\cdot(\gamma^2\dot{\vec v} + \gamma^4(\vec v\cdot\dot{\vec v})\vec v/c^2) = \gamma^3\vec v\cdot\dot{\vec v} + \gamma^5 v^2(\vec v\cdot\dot{\vec v})/c^2$. Using $\gamma^2 v^2/c^2 = \gamma^2 - 1$, the last term is $\gamma^5\vec v\cdot\dot{\vec v} - \gamma^3\vec v\cdot\dot{\vec v}$, and the three terms cancel.

The two special cases. Circular motion at constant speed: $\vec v\cdot\dot{\vec v} = 0$, so $a = (0,\gamma^2\dot{\vec v})$ and $\alpha = \gamma^2v^2/r$ (problem P1). Linear motion: $\dot{\vec v}\parallel\vec v$, so $\vec a = \gamma^2\dot{\vec v}(1 + \gamma^2v^2/c^2) = \gamma^4\dot{\vec v}$ and $\eta(a,a) = \gamma^8\dot v^2(1 - v^2/c^2) = \gamma^6\dot v^2$: $\alpha = \gamma^3|\dot{\vec v}|$, which section 4 uses.

section 2

The action is proper time

One functional, forced by symmetry, and units 11 to 14 run on it unchanged.

27 · action

The only free action there is

Ask for a functional of worldlines $S[\gamma] = \int F(\gamma')\,\mathrm{d}\lambda$ that is

(i) Lorentz-invariant: $F(\Lambda w) = F(w)$, so every inertial observer derives the same law;
(ii) reparametrisation-invariant: $F(kw) = kF(w)$ for $k \gt 0$, because the parameter is not physical;
(iii) translation-invariant: no explicit dependence on the event $\gamma$.

theorem

On future timelike $w$, (i) and (ii) force $F(w) = K\sqrt{-\eta(w,w)}$ for a constant $K$.

Proof. Put $k = \sqrt{-\eta(w,w)}$. By transitivity $w = \Lambda(k e_0)$, so $F(w) = F(k e_0) = kF(e_0)$.

the free-particle action $$\begin{aligned}S[\gamma] &= -mc\int\sqrt{-\eta(\gamma',\gamma')}\,\mathrm{d}\lambda\\ &= -mc^2\int\mathrm{d}\tau .\end{aligned}$$

$K = -mc$: the magnitude is fixed by the Newtonian limit ("Not $T - V$"), the sign by the twin paradox — straight worldlines maximise $\tau$, so with the minus sign they minimise $S$.

Check. $[mc^2\tau] = \mathrm{J\,s}$, the unit of action ✓.

Full proof. Let $w$ be future timelike and $k = \sqrt{-\eta(w,w)} \gt 0$. Then $w/k$ and $e_0$ are both future timelike with square $-1$, so by fact (a) there is a proper orthochronous $\Lambda$ with $\Lambda e_0 = w/k$. Invariance gives $F(w/k) = F(\Lambda e_0) = F(e_0)$, and homogeneity gives $F(w) = kF(w/k) = kF(e_0)$. So $F(w) = F(e_0)\sqrt{-\eta(w,w)}$, with $K := F(e_0)$. Nothing about differentiability was assumed; the form of $F$ is forced pointwise.

What the theorem does not say. It classifies Lagrangians that depend only on the tangent. Allowing second derivatives (curvature of the worldline) or couplings to external fields opens other possibilities; unit 28 adds the electromagnetic coupling $q\int A$, which is Lorentz-invariant and reparametrisation-invariant but depends on the event through the potential. For a free particle, though, proper time is the only candidate, and that is a theorem, not a guess.

Why reparametrisation invariance. A worldline is a set of events; the parameter is a bookkeeping device. If $S$ depended on it, two descriptions of the same history would have different actions and the variational principle would pick out a preferred clock — which is exactly what relativity denies.

History. Planck (1906) wrote the free Lagrangian $-mc^2\sqrt{1-v^2/c^2}$; the form $-mc^2\int\mathrm{d}\tau$ as a statement about Minkowski geometry is Minkowski's (1908) and was made standard by Landau and Lifshitz, The Classical Theory of Fields, §8, which derives it by the same symmetry argument.

27 · action · example

Euler–Lagrange gives $\vec p = \gamma m\vec v$

the derivation, every step

1. Parametrise by $t$ in an inertial frame: $\mathrm{d}\tau = \sqrt{1-v^2/c^2}\,\mathrm{d}t$, so $$S = \int L\,\mathrm{d}t,\qquad L(\vec x,\vec v) = -mc^2\sqrt{1 - v^2/c^2}.$$

2. $L$ does not depend on $\vec x$: $\partial L/\partial x^i = 0$.

3. $\dfrac{\partial L}{\partial v^i} = -mc^2\cdot\dfrac{-2v^i/c^2}{2\sqrt{1-v^2/c^2}} = \dfrac{m v^i}{\sqrt{1-v^2/c^2}} = \gamma m v^i$.

4. Unit 11's Euler–Lagrange equations, one per $i$: $$\frac{\mathrm{d}}{\mathrm{d}t}\big(\gamma m\vec v\big) = \vec 0 .$$

5. So the canonical momentum is $\vec p = \gamma m\vec v$, conserved; since $\vec v\mapsto\gamma m\vec v$ is injective, $\vec v$ is constant: free worldlines are straight.

checks

Dimensions. $[\gamma m v] = \mathrm{kg\,m\,s^{-1}}$ ✓.

Limit. $v \ll c$: $\gamma \to 1$, $\vec p \to m\vec v$ ✓.

Limit. $v\to c$: $|\vec p|\to\infty$ at bounded speed — unit 14's "nearly degenerate" column.

Number. A 6.8 TeV LHC proton (total energy): $\gamma = 7247.4$, $1 - v/c = 9.52\times10^{-9}$, so it trails light by $2.85\ \mathrm{m\,s^{-1}}$.

This is half the checkpoint. Close the deck and do it on paper once.

Step 5, the injectivity. $f(\vec v) = \gamma(v)m\vec v$ is $\vec v$ times the positive, strictly increasing function $m\gamma(v)$ of $v = |\vec v|$, so it preserves direction and $|f| = m v\gamma(v)$ is strictly increasing in $v$. Hence $f$ is injective on the ball $|\vec v| \lt c$ — in fact a bijection onto $\mathbb R^3$ — and constant $\vec p$ means constant $\vec v$.

The LHC number. The quoted beam energy is the total energy, $E = 6.8$ TeV (as on the collider slide), so $\gamma = E/m_pc^2 = 7247.36$ (using $E = \gamma m c^2$ from the Noether slide), and $1 - \beta = 1 - \sqrt{1-\gamma^{-2}} = 9.5194\times10^{-9}$, which times $c$ is $2.854\ \mathrm{m\,s^{-1}}$. Reading 6.8 TeV as kinetic energy instead moves $\gamma$ by one, in the fifth figure. A proton with twice the energy would trail light by a quarter as much: $1-\beta\approx 1/2\gamma^2$.

With the proper-time parameter instead. Use $\lambda$ arbitrary and $L = -mc\sqrt{-\eta(\gamma',\gamma')}$. The Euler–Lagrange equation is $\frac{\mathrm{d}}{\mathrm{d}\lambda}\big(mc\,\eta_{\mu\nu}\gamma'^\nu/\sqrt{-\eta(\gamma',\gamma')}\big) = 0$, and in the proper-time parametrisation this is $\frac{\mathrm{d}}{\mathrm{d}\tau}(m u_\mu) = 0$: all four components, including the energy, conserved at once. The same equations, but the Hamiltonian of this form is degenerate; the slide "With any parameter, $H \equiv 0$" deals with that.

27 · action

Not $T - V$, and why that matters

Expand for $v \ll c$: $$L = -mc^2 + \tfrac12 mv^2 + \tfrac18 m\frac{v^4}{c^2} + O\!\left(\frac{v^6}{c^4}\right).$$ The constant drops out of the Euler–Lagrange equations (unit 12's gauge freedom with $f = -mc^2t$); the second term is Newton's $T$; the rest is new.

what is new

Every Lagrangian from unit 12 on had $T = \tfrac12 g(\dot q,\dot q)$ quadratic in velocity — a metric on configuration space. This $L$ is not quadratic, and it is not a metric on anything in $\mathbb R^3$. It is the length functional of a metric on spacetime.

with a force

$L = -mc^2/\gamma - V(\vec x)$ gives $$\frac{\mathrm{d}}{\mathrm{d}t}\big(\gamma m\vec v\big) = -\nabla V .$$

Steps 2 to 4 of the Euler–Lagrange example with $\partial L/\partial x^i = -\partial_i V$.

$\vec F = \mathrm{d}\vec p/\mathrm{d}t$ survives; $\vec F = m\vec a$ does not. A static $V$ singles out a frame, so this is a model, not a covariant law — good enough for problem P12, and replaced properly in unit 28.

The expansion. $\sqrt{1-x} = 1 - \tfrac12x - \tfrac18x^2 - \dots$ with $x = v^2/c^2$ (single-variable Taylor, with the remainder bounded on $|x|\le x_0 \lt 1$). So $-mc^2\sqrt{1-x} = -mc^2 + \tfrac12mv^2 + \tfrac18mv^4/c^2 + \dots$.

Why "the point of departure". In units 12 to 16, the kinetic energy was a quadratic form on velocities — a metric on configuration space, whose free motions are its geodesics. The relativistic particle turns this round: its configuration space $\mathbb R^3$ carries no metric making $L$ a kinetic energy, but the whole action is the length of the worldline in spacetime. Free motion is still geodesic motion — straight lines maximise $\tau$ — only now the geometry lives on spacetime and the length is not a square. Units 30 to 32 take this seriously: gravity will be a change in $\eta$ itself, and $-mc^2\int\mathrm{d}\tau$ will not change at all.

Why the potential model is not covariant. $V(\vec x)$ is a function on space in one frame; in another frame the source moves and the potential should depend on $t$ too, and there is no rule here saying how. A covariant interaction needs a field on spacetime, which is unit 28. For the sun and a planet, the model is the honest question "what does special-relativistic kinematics alone do to a Kepler orbit?", and P12 answers it.

The failure of $F = ma$. $\frac{\mathrm{d}}{\mathrm{d}t}(\gamma m\vec v) = \gamma m\dot{\vec v} + \gamma^3m(\vec v\cdot\dot{\vec v})\vec v/c^2$. Force and acceleration are parallel only when $\dot{\vec v}\perp\vec v$ (coefficient $\gamma m$) or $\dot{\vec v}\parallel\vec v$ (coefficient $\gamma^3 m$). The slide against relativistic mass returns to this.

27 · action

Energy from Noether: $E = \gamma mc^2$

$L$ does not depend on $t$. Unit 13: the energy function is conserved, $$h = \vec v\cdot\frac{\partial L}{\partial\vec v} - L = \gamma mv^2 + \frac{mc^2}{\gamma} = \gamma m\Big(v^2 + c^2 - v^2\Big)$$

the energy $$E = \gamma mc^2,\qquad E_0 = mc^2 \ \text{ at rest.}$$

rest energy $mc^2$ is energy a body has by having mass. The kinetic energy is $T = E - mc^2 = (\gamma - 1)mc^2$.

checks

Dimensions. $\mathrm{kg\,m^2s^{-2}} = \mathrm J$ ✓.

Limit. $T = mc^2(\tfrac12v^2/c^2 + \tfrac38v^4/c^4 + \dots) \to \tfrac12mv^2$ ✓.

Number. Kinetic energy $6m_pc^2$ means $\gamma = 7$: the threshold beam moves at $0.98974c$.

Why is the additive constant $mc^2$ physical, when unit 12 said constants in $L$ are not? Because $E$ is the time component of a four-vector (the mass-shell slide), and a constant added to one component of a vector in every frame is not Lorentz-invariant.

The algebra. $\vec v\cdot\partial L/\partial\vec v = \gamma m v^2$, and $-L = mc^2\sqrt{1-v^2/c^2} = mc^2/\gamma$. So $h = \gamma m v^2 + mc^2/\gamma = \gamma m\,(v^2 + c^2/\gamma^2) = \gamma m\,(v^2 + c^2 - v^2) = \gamma mc^2$.

The kinetic energy expansion. $\gamma = (1-x)^{-1/2} = 1 + \tfrac12x + \tfrac38x^2 + \dots$ with $x = v^2/c^2$, so $(\gamma-1)mc^2 = \tfrac12mv^2 + \tfrac38mv^4/c^2 + \dots$.

The number. $\gamma = 7$ gives $\beta = \sqrt{1 - 1/49} = \sqrt{48}/7 = 0.989743$.

On the constant. In Newtonian mechanics the zero of energy is a convention, because energy is a lone scalar. In relativity energy is $c$ times the time component of $p$, and adding $\Delta$ to it in one frame while leaving $\vec p$ alone changes the four-vector to $p + (\Delta/c)e_0$; in a boosted frame this is a different shift of both energy and momentum. So the zero of energy is fixed by the requirement that $(E/c,\vec p)$ transform as a vector, and that fixes $E = mc^2$ at rest. Einstein's 1905 paper "Does the inertia of a body depend upon its energy content?" reached $\Delta m = \Delta E/c^2$ by a thought experiment with light emission; the argument here is the modern one.

27 · action

Four-momentum, and the mass shell

four-momentum $$p := m\,u = \big(\gamma mc,\ \gamma m\vec v\big) = \big(E/c,\ \vec p\big)\in V .$$

The Euler–Lagrange and Noether results are the components of one vector: energy is its time part, momentum its space part.

the mass shell $$\eta(p,p) = -m^2c^2 \iff E^2 = p^2c^2 + m^2c^4 .$$

Proof. $\eta(p,p) = m^2\eta(u,u) = -m^2c^2$, and $\eta(p,p) = -E^2/c^2 + p^2$.

So the mass is the Minkowski length of the momentum: $$m = \frac1c\sqrt{-\eta(p,p)} .$$ A Lorentz scalar. Every observer agrees on it, while each disagrees about $E$ and $\vec p$.

checks

Dimensions. $E^2$, $p^2c^2$, $m^2c^4$ all $\mathrm{J^2}$ ✓.

Limits. $p = 0$: $E = mc^2$ ✓. $pc \gg mc^2$: $E \approx pc$, the massless case.

Number. Threshold beam: $E = 6567.9$ MeV, $pc = \sqrt{E^2 - (m_pc^2)^2} = 6500.5$ MeV.

27 · action

The mass shell, drawn

Fix $m$. The four-momenta a particle of that mass can have fill one sheet of the hyperboloid $\eta(p,p) = -m^2c^2$: its mass shell.

A boost moves $p$ along its shell and never off it. Rest is the bottom point, $E = mc^2$; the threshold beam sits far up, where the shell has nearly merged with the light cone.

Lighter particles have shells closer to the cone, and $m = 0$ is the cone itself: "Massless particles move at $c$".

Mass is the shell's height on the energy axis, and every frame agrees on which shell a particle is on. At the threshold beam, $E = 7m_pc^2$ and $pc = 6.93\,m_pc^2$: $E \approx pc$ already.

Why "mass shell". For fixed $m \gt 0$, the set of allowed $p$ is the future sheet of the hyperboloid $\eta(p,p) = -m^2c^2$ in $V$ — a three-dimensional "shell" in a four-dimensional space. Particle physicists say a particle is "on shell" when its four-momentum lies there; the term matters in quantum field theory, where intermediate particles are not.

What to look for in the figure. Momentum along one axis, in units of $m_pc$; energy in units of $m_pc^2$. The shells are $E = \sqrt{m^2 + p^2}$ for $m = m_p$ and $m = 0.45\,m_p$, computed. At $p = \sqrt{48}$ the proton shell is at $E = 7$ while the cone is at $6.93$: a gap of one per cent, which is the statement $E \approx pc$ for $pc \gg mc^2$. A boost along $x$ slides every point along its own hyperbola — the transformation of the previous slide's notes is a hyperbolic rotation of this plane.

27 · action

Hamilton: $H = \sqrt{p^2c^2 + m^2c^4}$

Unit 14's Legendre transform of $L(\vec v) = -mc^2\sqrt{1-v^2/c^2}$ on the ball $|\vec v| \lt c$:

the relativistic Hamiltonian $$\begin{aligned}H(\vec p) &= \sup_{|\vec v| \lt c}\big(\vec p\cdot\vec v - L(\vec v)\big)\\ &= \sqrt{p^2c^2 + m^2c^4}.\end{aligned}$$

Hamilton's equations: $\dot{\vec x} = \partial H/\partial\vec p = \vec pc^2/H$, $\ \dot{\vec p} = 0$.

checks

Unit 14. $m = c = 1$, one dimension: its third column, $f^*(p) = \sqrt{1+p^2}$ ✓.

Limit. $H = mc^2 + p^2/2m - p^4/8m^3c^2 + \dots$: rest energy plus Newton's $H$ ✓.

Dimensions. $p^2c^2$ and $m^2c^4$ are both $\mathrm{J^2}$ ✓.

$H$ is the mass shell solved for the energy: the same hyperbola, read as a function of $\vec p$. The next slide reads off what it says about speed.

Full proof. By unit 14's smooth-duality theorem it suffices that $L$ be strictly convex with $\nabla L$ a bijection from the ball onto $\mathbb R^3$. Convexity: $L$ is $mc^2\,g(|\vec v|)$ with $g(s) = -\sqrt{1-s^2/c^2}$ convex and increasing on $[0,c)$, and a convex increasing function of a norm is convex; strictness follows from the Hessian below. The Hessian: $\partial^2L/\partial v^i\partial v^j = \gamma m\,\delta_{ij} + \gamma^3 m\,v_iv_j/c^2$ (differentiate $\gamma m v^i$ using $\partial\gamma/\partial v^j = \gamma^3v_j/c^2$). On $\vec v^\perp$ it is $\gamma m$; along $\vec v$ it is $\gamma m + \gamma^3mv^2/c^2 = \gamma^3m$. Both positive. The gradient map $\vec v\mapsto\gamma m\vec v$ was shown bijective onto $\mathbb R^3$ in the Euler–Lagrange example's notes. Then the supremum is attained where $\vec p = \gamma m\vec v$, and $H = \vec p\cdot\vec v - L = \gamma mv^2 + mc^2/\gamma = \gamma mc^2 = E$; by the mass shell, $E = \sqrt{p^2c^2 + m^2c^4}$.

Check of Hamilton's equations. $\partial H/\partial p_i = p_ic^2/H = \gamma mv_ic^2/(\gamma mc^2) = v_i$ ✓, consistent with the velocity formula in the mass-shell slide's notes.

The expansion. $\sqrt{m^2c^4 + p^2c^2} = mc^2(1 + y)^{1/2}$ with $y = p^2/m^2c^2$, $= mc^2(1 + \tfrac12y - \tfrac18y^2 + \dots)$.

27 · action

Momentum without bound, speed below $c$

nearly degenerate

$L$ is strictly convex on the ball, but its Hessian eigenvalues, $\gamma m$ across and $\gamma^3 m$ along $\vec v$, blow up at $|\vec v| = c$. The map $\vec v\mapsto\vec p$ sends a bounded ball onto all of $\mathbb R^3$.

Read backwards: $|\dot{\vec x}| = |\vec p|c^2/\sqrt{p^2c^2 + m^2c^4} \lt c$ for every $\vec p$. The speed limit is a property of $H$.

The two curves agree at small $p$ and part company near $p = mc$. Seven times the rest energy buys $0.9897c$; doubling the momentum again buys almost nothing more.

What to look for in the figure. $|\vec v|/c = q/\sqrt{1+q^2}$ with $q = |\vec p|/mc$, plotted from Hamilton's equation; the dotted line is Newton's $v = p/m$, which crosses $c$ at $q = 1$. The marked point is the threshold beam, $q = \sqrt{48} = 6.93$, $v = 0.98974c$, the same number as on the Noether slide. At large $q$ the gap below $c$ is $1 - v/c \approx 1/2q^2$: momentum keeps growing, speed barely moves.

"Nearly degenerate", from unit 14. The transform exists and is involutive, so nothing is lost. What is peculiar is that an unbounded momentum space corresponds to a bounded velocity space: momentum can be pumped in without limit, and the speed only creeps towards $c$. That is the whole content of "nothing is accelerated past $c$", and here it is a property of a convex function.

27 · action

With any parameter, $H \equiv 0$ and the mass shell is a constraint

Keep the parameter free: $L(\gamma') = -mc\sqrt{-\eta(\gamma',\gamma')}$ on the four components of $\gamma'$.

The momentum is a covector: $$p_\mu = \frac{\partial L}{\partial\gamma'^\mu} = \frac{mc\,\eta_{\mu\nu}\gamma'^\nu}{\sqrt{-\eta(\gamma',\gamma')}},$$ i.e. $p_\mu = \eta(m u, \cdot\,)$: unit 16's $\flat$ map, applied to the four-momentum.

theorem

(1) $\ p_\mu\gamma'^\mu - L = 0$ identically. (2) $\ \eta^{\mu\nu}p_\mu p_\nu = -m^2c^2$ for every $\gamma'$.

a degenerate Lagrangian, as promised

(1) is Euler's theorem: $L$ is homogeneous of degree 1, so $\gamma'^\mu\partial_\mu L = L$. The Hessian has $\gamma'$ in its kernel — scaling $\gamma'$ changes no momentum.

(2) says four momenta obey one relation: $H$ lives on the mass shell and is zero there. Unit 14's second column, $f(v) = |v|$, in four dimensions.

The dynamics is not lost: it is carried by the constraint, and choosing $\lambda = t$ solves it for $p_0$ — giving back the Hamilton slide's $H$.

Full proof. Write $N = \sqrt{-\eta(\gamma',\gamma')}$. Then $\partial N/\partial\gamma'^\mu = -\eta_{\mu\nu}\gamma'^\nu/N$ and $p_\mu = -mc\,\partial N/\partial\gamma'^\mu = mc\,\eta_{\mu\nu}\gamma'^\nu/N$. (1): $p_\mu\gamma'^\mu = mc\,\eta(\gamma',\gamma')/N = -mcN = L$, so $p_\mu\gamma'^\mu - L = 0$. (2): $\eta^{\mu\nu}p_\mu p_\nu = m^2c^2\,\eta_{\alpha\beta}\gamma'^\alpha\gamma'^\beta/N^2 = -m^2c^2$, using $\eta^{\mu\nu}\eta_{\mu\alpha}\eta_{\nu\beta} = \eta_{\alpha\beta}$. The Hessian kernel: differentiate $p_\mu(k\gamma') = p_\mu(\gamma')$ (degree $0$) in $k$ at $k = 1$, giving $\gamma'^\nu\,\partial p_\mu/\partial\gamma'^\nu = 0$.

Euler's theorem, proved. If $F(kw) = kF(w)$ for all $k \gt 0$, differentiate in $k$ at $k = 1$: $w^\mu\partial_\mu F(w) = F(w)$.

Lowering an index. Unit 16 showed an inner product gives an isomorphism $V\to V^*$, $w\mapsto w^\flat = g(w,\cdot)$. Its proof used only that $g$ is nondegenerate — injectivity is "$g(w,\cdot) = 0$ implies $w = 0$", and a linear injection between spaces of equal finite dimension is bijective. So $\eta$, though indefinite, gives an isomorphism too, and $p_\mu = \eta_{\mu\nu}p^\nu$ is the covector $p^\flat$. In components the only change is a sign: $p_0 = -E/c$, $p_i = p^i$. Unit 28 makes this unavoidable; here it is enough to notice that the canonical momentum of unit 12 was always a covector, and $\eta$ is what identifies it with the vector $mu$.

How the dynamics survives. In the reparametrisation-invariant form the Euler–Lagrange equations are $\mathrm{d}p_\mu/\mathrm{d}\lambda = 0$, four equations of which one is redundant (it follows from the constraint). Solving the constraint for $p_0$ in terms of $\vec p$ gives $-p_0c = \sqrt{p^2c^2+m^2c^4}$: the energy, now as the Hamiltonian conjugate to $t$. Dirac's theory of constrained systems (cited in unit 14) makes this systematic; general relativity's Hamiltonian is zero on its constraint surface for the same reason.

27 · action

The case against relativistic mass

the old bookkeeping

Many older texts write $m_{\text{rel}} = \gamma m$ — relativistic mass — so that $\vec p = m_{\text{rel}}\vec v$ looks Newtonian, and say "mass increases with speed".

But $\vec F = \mathrm{d}\vec p/\mathrm{d}t$ gives $$\vec F = \gamma m\,\dot{\vec v} + \gamma^3m\,\frac{(\vec v\cdot\dot{\vec v})\,\vec v}{c^2},$$ so $\vec F = m_{\text{rel}}\dot{\vec v}$ holds only across the motion. Along it the "mass" would be $\gamma^3m$. One inertia per direction is not a mass.

this course's usage

$m$ means the invariant $\frac1c\sqrt{-\eta(p,p)}$, always. It does not change with speed, because it is a length.

$\gamma mc^2$ already has a name: $E$. Calling $E/c^2$ a mass gives one quantity two names and hides that it is a component, frame-dependent, not a property of the particle.

$E = mc^2$ is therefore the rest energy. The general statement is the mass shell.

Where the formula comes from. $\frac{\mathrm{d}}{\mathrm{d}t}(\gamma m\vec v) = \dot\gamma m\vec v + \gamma m\dot{\vec v}$ with $\dot\gamma = \gamma^3\vec v\cdot\dot{\vec v}/c^2$. Split $\dot{\vec v}$ into parts along and across $\vec v$: the across part gets coefficient $\gamma m$; the along part gets $\gamma m + \gamma^3mv^2/c^2 = \gamma^3m$ (the same algebra as the Hessian on the Hamilton slide — not a coincidence, since the force is the rate of change of $\partial L/\partial\vec v$). Early relativists called these the transverse and longitudinal masses (Lorentz 1904, Abraham 1902); the concept did not survive because it is not a single number.

The argument, briefly. Einstein wrote to Lincoln Barnett in 1948 that it is "not good to introduce the concept of the mass $M = m/(1-v^2/c^2)^{1/2}$ of a moving body for which no clear definition can be given". L. B. Okun, "The concept of mass", Physics Today 42 (6), 31 (1989), is the standard modern statement.

What is true in the old slogan. A system's invariant mass does include the kinetic energy of its parts — the next section's box of hot gas. That is a statement about the length of a sum of four-vectors, not about any part getting heavier.

27 · action

Massless particles move at $c$

The mass shell makes sense at $m = 0$: a massless particle has a nonzero, future-pointing, null four-momentum, $\eta(p,p) = 0$.

theorem

$E = |\vec p|c$; its velocity $\partial H/\partial\vec p$ has $|\vec v| = c$; and it has no rest frame.

Proof. $H = |\vec p|c$ gives $\partial H/\partial\vec p = c\,\vec p/|\vec p|$. A rest frame would need $\Lambda p = (E'/c,\vec 0)$ with $\eta = -E'^2/c^2 \ne 0$; but $\Lambda$ preserves $\eta(p,p) = 0$.

the photon — cited input

Light comes in quanta of energy $E = hf = hc/\lambda$, with $h = 6.626\times10^{-34}$ J s. This is quantum physics, not derivable here: cited, Einstein 1905 on the photoelectric effect.

Its four-momentum is $\frac hc f(1,\hat n)$. Frequency transforms as a component: that is unit 26's Doppler formula, recovered in one line.

Limit. $v = pc^2/\sqrt{p^2c^2+m^2c^4} \to c$ as $m\to0$ ✓.

Why not "a particle at speed $c$ has $\gamma = \infty$, so $m$ must be $0$". That argument divides by zero. The honest route is the one on the slide: define a particle by its four-momentum on a mass shell, allow the shell with $m = 0$ (the light cone), and derive the speed from $H$. Nothing in the definition requires a proper-time parametrisation, which a null worldline does not have ($\mathrm{d}\tau = 0$ along it).

Doppler in one line. A photon moving along $+x$ has $p = \frac{hf}{c}(1,1,0,0)$. Boost with velocity $\beta c$ along $x$: $E' = \Gamma(E - \beta c\,p_x) = \Gamma E(1 - \beta)$, so $f'/f = \Gamma(1-\beta) = \sqrt{(1-\beta)/(1+\beta)}$ — unit 26's longitudinal Doppler factor for a receiver moving away from the source, obtained from the transformation law alone.

On the citation. $E = hf$ is used in this unit and on the problem sheet only to convert a wavelength into a four-momentum (Compton scattering, P4) and to name the energy of a CMB photon (P6). Everything relativistic about those problems follows from $\eta(p,p) = 0$; the constant $h$ only converts units.

section 3

Systems, collisions, thresholds

Add the four-vectors, square the sum, and pick the frame last.

27 · systems

Invariant mass is not additive

invariant mass

For particles with four-momenta $p_1,\dots,p_n$, the total is $P = \sum p_i$ and $$Mc^2 := c\sqrt{-\eta(P,P)} .$$ The frame where $\vec P = \vec 0$ is the centre-of-momentum frame; there $Mc^2 = \sum E_i$.

theorem

$M \ge \sum m_i$, with equality iff all the $p_i$ are parallel — for massive particles, all at rest in one frame.

Proof. Unit 26's reversed triangle inequality, fact (b), applied $n-1$ times.

Two particles of mass 1, moving apart. Their sum $P$ lands on the shell of mass $2.81$, well above the dashed shell of mass $1 + 1$: in Minkowski geometry the long way round is the short way.

Full proof. For two future timelike or null vectors fact (b) gives $\sqrt{-\eta(p_1+p_2,p_1+p_2)} \ge \sqrt{-\eta(p_1,p_1)} + \sqrt{-\eta(p_2,p_2)}$, with equality iff they are parallel. The sum of future-pointing vectors is future-pointing, so induction on $n$ gives $\sqrt{-\eta(P,P)} \ge \sum\sqrt{-\eta(p_i,p_i)} = c\sum m_i$. For the centre-of-momentum frame: if $P$ is timelike, fact (a) gives a frame where $P = (Mc,\vec 0)$, and there $-\eta(P,P) = (P^0)^2 = (\sum E_i/c)^2$.

What to look for in the figure. Units with $c = 1$ and both masses $1$: $p_1 = (1.562,\ 1.2)$ and $p_2 = (1.281,\ -0.8)$ as (energy, momentum), each on the mass-1 shell. Their sum is $P = (2.843,\ 0.4)$, and $M = \sqrt{2.843^2 - 0.4^2} = 2.81$. The dashed hyperbola is the shell of mass $2$, where $P$ would land only if the two moved together. The picture is the reversed triangle inequality: the broken path $p_1$ then $p_2$ is shorter, in $\eta$-length, than the straight $P$.

27 · systems

A box of hot gas has more mass

A box of hot gas has more mass than its molecules. In its rest frame $Mc^2 = \sum\gamma_im_ic^2 \gt \sum m_ic^2$. The excess is the thermal energy over $c^2$.

Each molecule is one arrow of the previous figure; the faster they move about, the higher the shell their total lands on.

Limit. Two photons: $M = 0$ only if they travel in the same direction (P2).

a kettle

Heat 1 kg of water through 80 K, specific heat $4186\ \mathrm{J\,kg^{-1}K^{-1}}$: $\Delta E = 3.349\times10^5$ J, so $\Delta M = \Delta E/c^2 = 3.73\times10^{-12}$ kg. Real, and four parts in $10^{12}$.

mass is not conserved; energy is

Two particles that stick together make one heavier particle. What is conserved is $P$, hence $M$ — not $\sum m_i$.

"Weighs more". What is proved here is that the box has more inertia: its total four-momentum is $MU$ with $U$ the box's four-velocity, so a force on the box changes its momentum as if its mass were $M$. That it also weighs more — that gravity couples to $M$ and not to $\sum m_i$ — is the equivalence principle, which is unit 30's subject. Eötvös-type experiments compare the free fall of materials whose nuclear binding energies — a much larger fraction of their mass than a kettle's heat — differ, and find no difference.

The kettle. $1\times4186\times80 = 334\,880$ J; divided by $c^2 = 8.98755\times10^{16}\ \mathrm{m^2s^{-2}}$ is $3.726\times10^{-12}$ kg. No balance can see it; nuclear binding energies (about $0.9\%$ of the mass of an iron-56 nucleus) are where the effect is large.

Mass is not conserved, energy is. Two particles colliding to make one heavier particle: $\sum m$ before is less than $M$ after. Nothing is violated; what is conserved is $P$, and hence $M$. The sum of rest masses is simply not a conserved quantity, and this is what "mass is not additive" means.

27 · systems

Why four-momentum is conserved

Input: in a collision, total 3-momentum after equals total 3-momentum before — in every inertial frame (momentum conservation plus the principle of relativity).

theorem: four-momentum conservation

Then total energy is conserved too: $\Delta P = 0$ as a vector in $V$.

Proof. Let $\Delta P = P_{\text{after}} - P_{\text{before}}$, with $\Delta\vec P = 0$ in frame $S$. Boost with velocity $\beta c$ along $x$: $\Delta P'^1 = \Gamma(\Delta P^1 - \beta\,\Delta P^0) = -\Gamma\beta\,\Delta P^0$. This vanishes by hypothesis, so $\Delta P^0 = 0$.

So energy and momentum conservation are one law. Newtonian mechanics conserves mass and momentum separately; relativity replaces mass by energy, rest energy included, and finds it is the time component of the same vector.

what is not claimed

Kinetic energy is not conserved in general, and neither is $\sum m_i$. Only the vector $P$.

Noether's route — invariance of the total action under spacetime translations — needs an interacting action; that is unit 29's, for fields.

What the proof uses. Only the transformation law for four-vectors and one boost with $\beta \ne 0$. It shows that 3-momentum conservation cannot hold in all frames unless energy — rest energy included — is conserved as well. Einstein's own 1905 route to $E = mc^2$ was different (light emitted, seen from two frames: the Noether slide's notes); this one — demand a conservation law in every frame and see what else it forces — has the shape of G. N. Lewis and R. C. Tolman's 1909 collision argument.

Why not Noether here. Unit 13's theorem needs one action whose symmetry gives the conserved charge. For free particles it gives each particle's $p$ separately, which says nothing about collisions. An action for particles interacting at a distance, in a Lorentz-invariant way, does not exist in general: the no-interaction theorem (Currie, Jordan and Sudarshan, Rev. Mod. Phys. 35, 350 (1963), cited) shows that a Hamiltonian system of point particles with Lorentz-covariant worldlines is free. Interactions in relativity are mediated by fields, whose action (unit 29) is translation invariant, and Noether's theorem then gives the conserved total four-momentum including the field's share. Until then, the theorem on the slide is the honest justification.

The input is experimental. Momentum conservation in collisions is Newton's third law, checked in every frame one cares to use; the principle of relativity says the same laws hold in all inertial frames. Particle physics has confirmed four-momentum conservation to the precision of every missing-mass measurement ever made — it is how the neutrino was inferred.

27 · systems

Squaring, and the Mandelstam invariants

The technique. Write conservation as a vector equation, isolate the four-momentum you know least about, and apply $\eta(\cdot,\cdot)$ to both sides. Each $\eta(p_i,p_i) = -m_i^2c^2$ is known; cross terms $\eta(p_i,p_j)$ are scalars, so evaluate each in the frame where it is easiest.

Mandelstam invariants, for $a + b \to c + d$ $$\begin{aligned}s &= -c^2\eta(p_a{+}p_b,\,p_a{+}p_b)\\ t &= -c^2\eta(p_a{-}p_c,\,p_a{-}p_c)\\ u &= -c^2\eta(p_a{-}p_d,\,p_a{-}p_d)\end{aligned}$$

theorem $$s + t + u = (m_a^2 + m_b^2 + m_c^2 + m_d^2)\,c^4 .$$

proof in the notes (N)

$\sqrt s$ is the energy available in the centre-of-momentum frame: $s = (E_a + E_b)^2$ there. $t$ measures momentum transfer.

Check. Units $\mathrm{J^2}$ throughout ✓. Two particles at rest together: $s = (m_a + m_b)^2c^4$, the least possible ✓.

Full proof of $s+t+u$. Write $\langle\cdot,\cdot\rangle$ for $-c^2\eta$ so that $\langle p_i,p_i\rangle = m_i^2c^4$. Expand: $s = m_a^2c^4 + m_b^2c^4 + 2\langle p_a,p_b\rangle$, $t = m_a^2c^4 + m_c^2c^4 - 2\langle p_a,p_c\rangle$, $u = m_a^2c^4 + m_d^2c^4 - 2\langle p_a,p_d\rangle$. Sum: $3m_a^2c^4 + m_b^2c^4 + m_c^2c^4 + m_d^2c^4 + 2\langle p_a, p_b - p_c - p_d\rangle$. By conservation $p_b - p_c - p_d = -p_a$, so the last term is $-2m_a^2c^4$. Total: $(m_a^2+m_b^2+m_c^2+m_d^2)c^4$.

Why the factor $-c^2$. It makes $s$ an energy squared and positive for timelike sums, so that $\sqrt s$ reads in GeV. Particle physicists set $c = 1$ and use signature $(+,-,-,-)$, where the same definition is just $s = (p_a+p_b)^2$.

Why "squaring" works. Conservation is four equations. Squaring throws most of them away and keeps one scalar equation — but that scalar equation is exactly the one that does not depend on the frame, and very often it is the only one the question needs. The unknown directions of outgoing particles drop out because they only ever appear inside $\eta$-products with vectors whose frame you can choose.

History. Stanley Mandelstam introduced $s, t, u$ in 1958 for dispersion relations; the physics of $s$ as available energy is older.

27 · systems

Thresholds

theorem: the threshold

The reaction $a + b \to f_1 + \dots + f_n$ is kinematically possible iff $$s \ge \big(m_{f_1} + \dots + m_{f_n}\big)^2c^4 ,$$ and at equality all products are at rest in the centre-of-momentum frame.

Proof. $s$ is conserved (it is built from $P$) and invariant. After the reaction $s = M_f^2c^4$ with $M_f \ge \sum m_{f_i}$ by the invariant-mass theorem, with equality iff the products share one velocity. Conversely, that configuration exists.

The recipe. Compute $s$ from the initial state in the easiest frame; demand the inequality; read off the energy.

For the running example: $n = 4$, all masses $m_p$, so $$s \ge (4m_pc^2)^2 = (3.753\ \text{GeV})^2 .$$

No frame has been chosen yet. The next slide chooses two.

Full proof. Necessity: $s$ equals $-c^2\eta(P,P)$ with $P$ the conserved total, so computing it from the final state gives $M_f^2c^4$, and the invariant-mass theorem gives $M_f \ge \sum m_{f_i}$. Sufficiency: if $s \ge (\sum m_{f_i})^2c^4$ and $P$ is future timelike, go to the centre-of-momentum frame, where $P = (\sqrt s/c,\vec 0)$. If equality holds, put every product at rest; if not, give the products momenta summing to zero with energies summing to $\sqrt s$ — possible because the total energy $\sum\sqrt{p_i^2c^2 + m_i^2c^4}$ of, say, two groups moving apart with equal and opposite momenta increases continuously from $\sum m_ic^2$ to $\infty$. "Kinematically possible" means exactly this: four-momentum conservation permits it. Whether it happens is a question about interactions and conserved charges.

Why three protons and not one. Baryon number is conserved: two baryons in, so the antiproton (baryon number $-1$) must come with three protons. Charge: $+2$ in, $+3 - 1 = +2$ out ✓. The cheapest final state containing an antiproton is therefore four nucleon masses.

The number. $4\times938.272 = 3753.09$ MeV.

27 · systems · example

The antiproton threshold, in two frames

fixed target

Beam $p_a = (E/c,\vec p)$, target $p_b = (m_pc,\vec 0)$.

1. $\eta(p_a,p_a) = \eta(p_b,p_b) = -m_p^2c^2$; $\ \eta(p_a,p_b) = -(E/c)(m_pc) = -Em_p$.

2. $s = -c^2\big(-2m_p^2c^2 - 2Em_p\big) = 2m_p^2c^4 + 2Em_pc^2$.

3. $2m_p^2c^4 + 2Em_pc^2 \ge 16m_p^2c^4 \iff E \ge 7m_pc^2$.

Kinetic energy $T = E - m_pc^2 \ge 6m_pc^2 = \mathbf{5.630}$ GeV.

collider

Beams $(E/c,\ \vec p)$ and $(E/c,\ -\vec p)$.

1. $P = (2E/c,\ \vec 0)$.

2. $s = -c^2\eta(P,P) = (2E)^2$.

3. $4E^2 \ge 16m_p^2c^4 \iff E \ge 2m_pc^2$.

$T \ge m_pc^2 = \mathbf{0.938}$ GeV per beam, $1.877$ GeV in all: one third of the fixed-target figure.

27 · systems · example

Both frames on the mass shells

Each proton's four-momentum ends on the proton shell. The total $P$ must reach the shell of mass $4m_p$, the dashed hyperbola $s = (4m_pc^2)^2$.

Collider: two arrows of energy $2$, mirror images, sum straight up: momentum $0$, energy $4$.

Fixed target: the target adds only $(0, 1)$, so the beam must climb to energy $7$ to lift the sum onto the hyperbola, at momentum $\sqrt{48}$, energy $8$.

check

Fixed-target threshold: products share $\gamma_{\text{cm}} = (E + m_pc^2)/4m_pc^2 = 2$. Energy $4\times2m_pc^2 = 8m_pc^2 = 7m_pc^2 + m_pc^2$ ✓. Dimensions: every term in $s$ is $\mathrm{J^2}$ ✓. No Lorentz transformation was used.

Both totals end on the same hyperbola: they are one invariant, and a boost slides one point onto the other. The fixed-target sum spends its height climbing sideways — the momentum the products must keep.

Where the missing energy goes. On a fixed target the beam brings momentum $pc = \sqrt{48}\,m_pc^2 = 6500.5$ MeV, and momentum is conserved: the products cannot be at rest in the lab. At threshold they move together with $\gamma = 2$, $v = 0.866c$, carrying total kinetic energy $4(\gamma-1)m_pc^2 = 4m_pc^2$. Of the beam's $6m_pc^2$, only $2m_pc^2$ became new mass; two thirds went into moving the debris. In the collider the total momentum is zero, the products can be at rest, and all the kinetic energy becomes mass.

The check, done differently. The centre-of-momentum frame moves with velocity $\vec P c^2/(P^0c)$: here $\beta_{\text{cm}} = pc/(E + m_pc^2) = 6500.5/7506.2 = 0.86603 = \sqrt3/2$, so $\gamma_{\text{cm}} = 2$, agreeing with the slide.

What to look for in the figure. Units of $m_p$ with $c = 1$. The proton shell is $E = \sqrt{1 + p^2}$; the dashed hyperbola is $E = \sqrt{16 + p^2}$. Collider beams: $(E, p) = (2, \pm\sqrt3)$, on the proton shell since $4 - 3 = 1$. Fixed target: beam $(7, \sqrt{48})$, since $49 - 48 = 1$, plus target $(1, 0)$; total $(8, \sqrt{48})$ with $64 - 48 = 16$. The point $(4, 0)$ is the same total seen from the centre-of-momentum frame, and the boost with $\gamma = 2$ carries one to the other along the hyperbola.

27 · systems

Why colliders exist

The two formulas from the running example, for any beam energy $E \gg mc^2$:

	$s$	$\sqrt s$ grows like
fixed target	$2m^2c^4 + 2Emc^2$	$\sqrt{E}$
collider	$4E^2$	$E$

At threshold the ratio was 3. At high energy it is unbounded.

the LHC's figure

Two 6.8 TeV beams: $\sqrt s = 13.6$ TeV. On a fixed proton target the same $\sqrt s$ needs $E = (s - 2m_p^2c^4)/2m_pc^2 = 9.86\times10^{16}$ eV — about $14\,500$ times the energy of one LHC beam.

The first proton–proton collider, CERN's ISR, followed in 1971. Every energy-frontier machine since has collided beams.

The asymptotics, honestly. For $E \gg mc^2$ the fixed-target $s \approx 2Emc^2$, so $\sqrt s \approx \sqrt{2Emc^2}$, while the collider gives $\sqrt s = 2E$. The ratio of beam energies needed for equal $\sqrt s$ is $E_{\text{fixed}}/E_{\text{coll}} \approx 2E_{\text{coll}}/mc^2$, which grows without bound. No limit theorem beyond "the dominant term" is needed.

The LHC number. $s = (13.6\times10^6\ \text{MeV})^2$; $E = (s - 2\times938.272^2) /(2\times938.272)\ \text{MeV} = 9.856\times10^{10}$ MeV $= 9.86\times10^{16}$ eV, against $6.8\times10^{12}$ eV per LHC beam: a factor of $1.45\times10^4$ in beam energy, or about $7200$ against the two beams' combined $13.6$ TeV. Cosmic rays do reach such energies, which is why air-shower physics at $10^{17}$ eV probes collider-scale $\sqrt s$.

27 · systems

Available energy against beam energy

the Bevatron, 1955

Berkeley's Bevatron ran at about 6.2 GeV — chosen to clear the antiproton threshold. On a fixed target: $\sqrt s = \sqrt{2m_p^2c^4 + 2(7.138\text{ GeV})(0.938\text{ GeV})} = 3.893$ GeV, $140$ MeV above the $3.753$ GeV needed.

Chamberlain, Segrè, Wiegand and Ypsilantis found the antiproton there in 1955.

On log axes the table's growth laws are straight lines: slope $1$ for colliding beams, slope $\tfrac12$ for a fixed target.

Near threshold the two lines are a factor of $3.5$ apart in beam energy; at the LHC's $\sqrt s$ they are $14\,500$ apart. The gap is the beam momentum the products must carry away.

What to look for in the figure. Both curves are computed from the table, with $m_pc^2 = 0.938$ GeV and $E$ the total energy of one beam. The dashed horizontal line is the antiproton threshold, $\sqrt s = 3.753$ GeV. It crosses the collider line at $E = 2m_pc^2 = 1.877$ GeV and the fixed-target curve at $E = 7m_pc^2 = 6.568$ GeV: the factor $3.5$ in total energy is the factor $3$ in kinetic energy. The Bevatron, at $E = 7.138$ GeV, sits just to the right of the fixed-target crossing — too close to separate at this scale. Along the top, the LHC point ($E = 6.8$ TeV, $\sqrt s = 13.6$ TeV) and the fixed-target energy giving the same $\sqrt s$, $9.86\times10^{7}$ GeV.

The Bevatron. "BeV" was the American name for GeV. The machine was designed in the late 1940s with antiproton production as a stated goal, and ran at about 6.2 GeV. With $E = 6.2 + 0.938 = 7.138$ GeV, $\sqrt s = \sqrt{2(0.938)^2 + 2(7.138)(0.938)} = 3.893$ GeV. The target in the discovery experiment was copper, whose bound nucleons move (Fermi motion) and so lower the effective threshold a little further; the margin above threshold also increased the production rate, which rises steeply from zero at threshold. O. Chamberlain, E. Segrè, C. Wiegand and T. Ypsilantis, "Observation of antiprotons", Phys. Rev. 100, 947 (1955); Chamberlain and Segrè shared the 1959 Nobel prize.

27 · systems · example

A decay: $\pi^+\to\mu^+\nu_\mu$

the muon's energy, by squaring

Pion at rest: $p_\pi = (m_\pi c,\vec 0)$. Treat the neutrino as massless. Conservation: $p_\nu = p_\pi - p_\mu$.

1. Isolate the unknown-direction particle and square: $$0 = \eta(p_\nu,p_\nu) = -m_\pi^2c^2 - m_\mu^2c^2 - 2\eta(p_\pi,p_\mu).$$

2. In the pion's rest frame $\eta(p_\pi,p_\mu) = -m_\pi E_\mu$.

3. $E_\mu = \dfrac{(m_\pi^2 + m_\mu^2)c^2}{2m_\pi} = 109.778$ MeV.

4. $p_\mu c = \sqrt{E_\mu^2 - m_\mu^2c^4} = \dfrac{(m_\pi^2 - m_\mu^2)c^2}{2m_\pi} = 29.792$ MeV.

With $m_\pi c^2 = 139.570$ MeV, $m_\mu c^2 = 105.658$ MeV. The muon's kinetic energy is $4.120$ MeV and its speed $0.2714c$. The neutrino carries $E_\nu = p_\mu c = 29.792$ MeV.

checks

Energy. $109.778 + 29.792 = 139.570$ ✓.

Limit. $m_\mu\to m_\pi$: $p_\mu\to0$, no energy to spare ✓. $m_\mu\to0$: $E_\mu = m_\pi c^2/2$, two massless particles share equally ✓.

Nothing in the working depended on the direction the muon took. Every decay at rest gives the same numbers.

Step 4, simplified. $E_\mu^2 - m_\mu^2c^4 = \frac{c^4}{4m_\pi^2}\big[(m_\pi^2+m_\mu^2)^2 - 4m_\pi^2m_\mu^2\big] = \frac{c^4}{4m_\pi^2}(m_\pi^2 - m_\mu^2)^2$.

Numbers. $m_\pi c^2 = 139.57039$ MeV, $m_\mu c^2 = 105.6583755$ MeV (PDG). $E_\mu = (139.57039^2 + 105.65838^2)/(2\times139.57039) = 109.7782$ MeV; $p_\mu c = (139.57039^2 - 105.65838^2)/(2\times139.57039) = 29.7921$ MeV; $\beta_\mu = p_\mu c/E_\mu = 0.27138$.

The general two-body formula, for $M\to m_1 + m_2$ at rest, is $|\vec p|c = \frac{c^2}{2M}\sqrt{[M^2 - (m_1+m_2)^2][M^2 - (m_1-m_2)^2]}$; problem P7 derives it and applies it to $\Lambda\to p\pi^-$.

Why the neutrino is taken massless. Neutrino masses are below about $0.8$ eV (KATRIN, 2022), and the correction to $E_\mu$ is of order $m_\nu^2c^4/m_\pi c^2$, a few times $10^{-9}$ eV. The single muon momentum of $29.79$ MeV$/c$ is a standard calibration line in pion-decay experiments.

27 · systems

Two bodies, one momentum

A two-body decay at rest gives a single momentum — a line, not a spectrum. Momentum conservation makes the two products back to back with equal $|\vec p|$, and the squaring argument fixes that $|\vec p|$ from the masses alone.

With three products the momenta need only sum to zero, and the energy can be shared in a continuum of ways.

That is why the continuous spectrum of beta decay forced Pauli to posit a third particle.

Left: the decay in the pion's rest frame. Right: the two-body decay gives one spike at $29.79$ MeV$/c$; a three-body decay gives a spread (schematic shape).

What to look for in the figure. On the left the arrows are equal in length because the total momentum is zero, whatever the direction of the pair. On the right the spike is the computed $p_\mu c = 29.792$ MeV; the dashed hump is only a shape, standing for any decay with three products, where how the energy is split varies from decay to decay. Its real form depends on the interaction, which this unit does not touch.

Pauli, 1930. In beta decay the electron's energy is spread over a continuum. If the decay were two-body the squaring argument would force a single line, so either energy conservation failed (Bohr considered this) or a third, unseen particle shared the energy. Pauli proposed the particle; Fermi named it; Cowan and Reines detected it in 1956.

27 · systems

The relativistic rocket: fuel buys rapidity

A rocket of rest mass $M$ ejects exhaust at speed $w$ in its own instantaneous rest frame. In that frame, over one instant, it goes from rest to speed $\mathrm{d}v$ while its rest mass changes by $\mathrm{d}M \lt 0$.

Energy: $Mc^2 = (M + \mathrm{d}M)c^2 + \gamma_w\mu c^2$, with $\mu$ the exhaust's rest mass, so $\gamma_w\mu = -\mathrm{d}M$.
Momentum: $0 = M\,\mathrm{d}v - \gamma_w\mu w$, to first order.

So $M\,\mathrm{d}v = -w\,\mathrm{d}M$ in the rest frame, where $\mathrm{d}v = c\,\mathrm{d}\theta$ with $\theta$ the rapidity.

the relativistic rocket $$\theta = \frac wc\,\ln\frac{M_i}{M_f}$$

Check. Dimensionless both sides ✓. $w, v \ll c$: $v = c\theta = w\ln(M_i/M_f)$, Tsiolkovsky's equation ✓.

The lab speed is $v = c\tanh\theta$: rapidity is unbounded, speed is not.

Full derivation. Work in the rocket's rest frame at one instant. Before: four-momentum $(Mc,\vec 0)$. After a time $\mathrm{d}\tau$: rocket $(M+\mathrm{d}M)\big(\gamma(\mathrm{d}v)c,\ \gamma(\mathrm{d}v)\,\mathrm{d}v\big)$ and exhaust $\mu\gamma_w(c, -w)$ along the axis. To first order $\gamma(\mathrm{d}v) = 1$. The time component: $Mc = (M+\mathrm{d}M)c + \mu\gamma_wc$, so $\mu\gamma_w = -\mathrm{d}M$ (exact to first order; note $\mu \lt -\mathrm{d}M$: rest mass is spent on the exhaust's kinetic energy). The space component: $0 = M\mathrm{d}v - \mu\gamma_ww$ to first order, so $M\mathrm{d}v = \mu\gamma_ww = -w\mathrm{d}M$.

Why rapidity. A small speed increment $\mathrm{d}v$ measured in the instantaneous rest frame is a rapidity increment $\mathrm{d}\theta = \mathrm{d}v/c$ (because $\tanh\theta\approx\theta$ at small $\theta$), and unit 26 showed rapidities along one line add. So the increments add up in any frame: $c\,\mathrm{d}\theta = -w\,\mathrm{d}M/M$, integrated with $w$ constant, gives the slide. The lab speed is $v = c\tanh\theta$.

27 · systems

What the fuel buys

Put $\theta = (w/c)\ln(M_i/M_f)$ into $v = c\tanh\theta$. For a photon rocket, $w = c$: $$\frac vc = \frac{R^2 - 1}{R^2 + 1},\qquad R = \frac{M_i}{M_f}.$$

Number. Reaching $0.9c$ needs $R = e^{\operatorname{artanh}0.9} = \sqrt{19} = 4.36$; to stop at the far end, $19$.

Tsiolkovsky's straight line would pass $c$ at $R = e$. The real curve bends over and never gets there.

Each factor $e$ of mass ratio buys the same rapidity, $w/c$, but less and less speed. Halving the exhaust speed halves the rapidity: at $R = 10$, $0.82c$ instead of $0.98c$.

What to look for in the figure. Three curves, all from $v = c\tanh\theta$ with $\theta = (w/c)\ln R$: the photon rocket, $(R^2-1)/(R^2+1)$; exhaust at $w = c/2$, $\tanh(\tfrac12\ln R) = (R-1)/(R+1)$; and dotted, the Newtonian $v = w\ln R$ with $w = c$, which is the small-$\theta$ approximation and crosses $c$ at $R = e = 2.72$. At $R = 10$ the photon rocket reaches $99/101 = 0.980c$ and the slower exhaust $9/11 = 0.818c$.

Numbers. $\operatorname{artanh}0.9 = \tfrac12\ln19 = 1.4722$, so $e^\theta = \sqrt{19} = 4.359$; accelerating and then decelerating doubles $\theta$, so the ratio is $19$; there and back again, $361$. For comparison, chemical exhaust at $w = 4.5\ \mathrm{km\,s^{-1}}$ would need $M_i/M_f \approx 10^{42\,596}$ just to reach $0.9c$. Problem P9 has more.

section 4

Acceleration, and why nothing is rigid

Constant proper acceleration gives a hyperbola and a horizon; a rigid body gives a contradiction.

27 · acceleration

Hyperbolic motion

Motion along $x$ with constant proper acceleration $\alpha$. On the hyperboloid, write $u = c(\cosh\theta,\ \sinh\theta)$, $\theta$ the rapidity. Then $a = c\,\dot\theta\,(\sinh\theta,\ \cosh\theta)$ with dot $=\mathrm{d}/\mathrm{d}\tau$, and $\eta(a,a) = c^2\dot\theta^2$.

So $\alpha = c\dot\theta$: rapidity grows linearly in proper time, $\theta = \alpha\tau/c$. Integrate $u$ once more:

hyperbolic motion $$ct = \frac{c^2}{\alpha}\sinh\frac{\alpha\tau}{c},\qquad x = \frac{c^2}{\alpha}\cosh\frac{\alpha\tau}{c},$$ $$x^2 - c^2t^2 = \frac{c^4}{\alpha^2}.$$

Check. $c^2/\alpha$ is a length ✓. $\alpha\tau \ll c$: $x \approx \frac{c^2}{\alpha} + \tfrac12\alpha t^2$, Galileo ✓.

Lengths in units of $c^2/\alpha$. The pad's signals (at $45^\circ$) catch the traveller later and later; the one sent at $ct = c^2/\alpha$ runs along the horizon for ever.

Full derivation. In 1+1 dimensions every future timelike vector with $\eta(u,u) = -c^2$ is $c(\cosh\theta,\sinh\theta)$ for a unique $\theta$ (since $\cosh^2 - \sinh^2 = 1$ and $\sinh$ is a bijection of $\mathbb R$). Differentiating in $\tau$, $a = c\dot\theta(\sinh\theta,\cosh\theta)$; this is visibly $\eta$-orthogonal to $u$ ($-\cosh\sinh + \sinh\cosh = 0$), and $\eta(a,a) = c^2\dot\theta^2(-\sinh^2 + \cosh^2) = c^2\dot\theta^2$. So constant proper acceleration $\alpha \gt 0$ means $\dot\theta = \alpha/c$, and with $\theta(0) = 0$, $\theta = \alpha\tau/c$. Then $\mathrm{d}(ct)/\mathrm{d}\tau = c\cosh(\alpha\tau/c)$ and $\mathrm{d}x/\mathrm{d}\tau = c\sinh(\alpha\tau/c)$ integrate to the slide's formulas, with the constants chosen so that $t = 0$, $x = c^2/\alpha$ at $\tau = 0$. The origin is placed so that the hyperbola is centred on it; any other starting point is a translation.

The last-signal claim. A signal leaving the starting point $x = c^2/\alpha$ at time $t_e$ travels on $x = c^2/\alpha + c(t - t_e)$. Setting this equal to $\sqrt{c^4/\alpha^2 + c^2t^2}$ and squaring gives $ct\,(2c^2/\alpha - 2ct_e) = ct_e(2c^2/\alpha - ct_e)$. A reception must come after the emission, $t \gt t_e$, and a root with $t \gt t_e \gt 0$ exists iff $ct_e \lt c^2/\alpha$, that is $t_e \lt c/\alpha$ (for $t_e \ge 2c/\alpha$ there is a positive root, but it lies before $t_e$: an artefact of squaring). Signals sent later never arrive — although the sender is at rest and the traveller is finitely far away at every moment. The events that can never signal the traveller are exactly those with $ct \ge x$: the region beyond the null line through the hyperbola's centre. That line is the Rindler horizon.

The figure. Computed in units of $c^2/\alpha$: the traveller is $x = \sqrt{1 + (ct)^2}$, the dashed line is $x = ct$, and the dotted vertical line is the launch pad, at rest at $x = 1$. A signal sent at $ct_e$ meets the traveller at $ct = t_e(2 - t_e)/(2(1 - t_e))$ from the formula above: the one sent at $0.4$ arrives at $ct = 0.533$, the one sent at $0.8$ not until $ct = 2.4$. The last signal the pad can send leaves at $ct = 1$, and that signal is the horizon: it runs along the asymptote for ever.

27 · acceleration · example

A year at one $g$, and a horizon behind you

the traveller's first year

$\alpha = g = 9.80665\ \mathrm{m\,s^{-2}}$. Then $c^2/g = 9.165\times10^{15}$ m $= 0.9687$ ly and $c/g = 0.9687$ yr.

1. After $\tau = 1$ yr: $\theta = g\tau/c = 1.0323$.

2. Speed $v = c\tanh\theta = 0.7748c$; $\ \gamma = \cosh\theta = 1.5818$.

3. Distance $\frac{c^2}{g}(\cosh\theta - 1) = 0.5636$ ly.

4. Earth time $\frac cg\sinh\theta = 1.1873$ yr.

After 10 years of proper time: $\theta = 10.32$, distance $14\,735$ ly, $\gamma = 15\,211$.

the Rindler horizon

Mission control, at the launch point, has until $t = c/g = 0.9687$ yr to send its last message. Anything sent later never arrives, though the ship is always a finite distance away and never reaches $c$.

The horizon is a null plane, $x = ct$, fixed in spacetime and invisible locally: nothing happens at it. Unit 30 will notice that this is exactly what a black hole's horizon looks like to someone hovering just outside it.

Check. $1/\sqrt{1 - 0.7748^2} = 1.582 = \gamma$ ✓. Dimensions: $c^2/g$ is m, $c/g$ is s ✓. Small $\tau$: distance $\approx\tfrac12 g\tau^2$ ✓.

The numbers. $c^2/g = 299792458^2/9.80665 = 9.16475\times10^{15}$ m; one light year is $9.46073\times10^{15}$ m (Julian year), so $c^2/g = 0.968715$ ly — and $c/g = 3.0570\times10^7$ s $= 0.968715$ yr, the same number because a light year is $c$ times a year. That coincidence is why "one $g$ for a year gets you to a good fraction of $c$" is a standard remark. $\theta = 3.15576\times10^7\times9.80665/299792458 = 1.03230$; $\tanh = 0.774827$, $\cosh = 1.581845$, $\sinh = 1.225657$. Distance $0.968715\times0.581845 = 0.563642$ ly; coordinate time $0.968715\times1.225657 = 1.187312$ yr. Ten years: $\cosh 10.3230 = 15\,211.5$, distance $0.968715\times15\,210.5 = 14\,735$ ly.

A second check: the event $(ct, x) = (1.187312,\ 0.968715 + 0.563642)$ ly must lie on the hyperbola $x^2 - c^2t^2 = (c^2/g)^2$. Indeed $1.532357^2 - 1.187312^2 = 2.348118 - 1.409710 = 0.938408$, and $0.968715^2 = 0.938409$ ✓.

Why this matters later. Unit 30 uses exactly this worldline as the model of an observer at rest in a gravitational field (equivalence principle), and the family of such worldlines with different $\alpha$ ("A Born-rigid rod: hyperbolas sharing a clock") as a coordinate system — Rindler coordinates — in which flat spacetime has a position-dependent clock rate. The horizon here costs nothing: it is a straight line in Minkowski space, and every property of it follows from the reversed triangle inequality and the null cone.

Honesty about fuel. A photon rocket accelerating at $g$ for ten proper years reaches $\theta = 10.32$ and so needs $M_i/M_f = e^{10.32} = 3.04\times10^4$, by the rocket slide. The kinematics is free; the engineering is not.

27 · acceleration

There are no rigid bodies

why unit 16 cannot be copied

A body rigid in unit 16's sense moves every point the instant one point is pushed. Push one end of a rigid rod at event $A$; the far end moves at event $B$ simultaneous with $A$ — spacelike-separated. Unit 26: then $B$ precedes $A$ in some frame, and cause follows effect. No relativistic body is rigid.

Born rigidity

The best available substitute (Born, 1909): the distance between neighbouring points, measured in their instantaneous common rest frame, stays constant.

$B$ lies outside $A$'s light cone. For an observer moving at $0.4c$, whose "now" is the dotted line, $B$ happens before $A$. A real rod answers no earlier than the light cone allows, and in practice at the speed of sound.

What to look for in the figure. The rod is at rest, so its ends are vertical worldlines and the shaded strip is the rod's history. The push is event $A$. A rigid rod would move its far end at $B$, on the horizontal line $t = 0$ through $A$. The tilted dotted line is the set $ct - \beta x = \text{const}$ through $B$, with $\beta = 0.4$: events simultaneous with $B$ for an observer moving at $0.4c$. It crosses the near end below $A$, so for that observer the far end moves before the push. The earliest causal response is where $A$'s light cone meets the far end; sound, drawn far too fast to fit, arrives later still.

How real bodies cope. A real rod is elastic: a push travels through it at the speed of sound, which is far below $c$, and the rod compresses and rings. Relativity does not forbid solids; it forbids solids with infinite sound speed, which is what unit 16's rigid body is. The continuum mechanics of unit 22 is what survives.

27 · acceleration

A Born-rigid rod: hyperbolas sharing a clock

The family $x^2 - c^2t^2 = X^2$, one hyperbola per material point $X \gt 0$, is Born-rigid: every line $ct = x\tanh\theta$ through the origin is simultaneous for all of them, and along it the separation is $X_2 - X_1$, constant.

the price

Point $X$ has proper acceleration $c^2/X$. The back must accelerate harder than the front. A rod whose front accelerates at $g$ cannot be longer than $c^2/g = 0.97$ ly: its back would need infinite acceleration.

Each dotted line through $O$ is one instant of the rod's own time; on each the hyperbolas cut the same lengths. The back's hyperbola bends hardest: the biggest proper acceleration.

Full proof of Born rigidity for the family. Point $X$ at rapidity $\theta$ is at $(ct,x) = X(\sinh\theta,\cosh\theta)$ with four-velocity $c(\cosh\theta,\sinh\theta)$. Its instantaneous simultaneity line is $\eta$-orthogonal to that four-velocity, so has direction $(\sinh\theta,\cosh\theta)$ — the direction of the position vector itself. Hence it is the line through the origin with $ct/x = \tanh\theta$, the same line for every $X$ at the same $\theta$. Along that line the points are $X(\sinh\theta,\cosh\theta)$, and the proper distance between $X_1$ and $X_2$ is $\sqrt{\eta}$ of the difference, $(X_2 - X_1)\sqrt{\cosh^2\theta - \sinh^2\theta} = X_2 - X_1$. Proper acceleration of point $X$ is $c^2/X$ by the hyperbolic-motion slide with $c^2/\alpha = X$.

What to look for in the figure. Hyperbolas for $X = 1, 1.5, 2, 2.5$ (any length unit), with the rod running from $X = 1$ (back) to $X = 2.5$ (front), thick on each of three simultaneity lines, $\theta = 0, 0.45, 0.9$. On paper the later copies of the rod look longer — the Euclidean length is $1.5\sqrt{\cosh^2\theta + \sinh^2\theta}$ — but the $\eta$-length is $1.5\sqrt{\cosh^2\theta - \sinh^2\theta} = 1.5$ every time. A smaller $X$ means a tighter hyperbola, and proper acceleration $c^2/X$ is the reciprocal of that hyperbola's "radius".

The rod. Front at $X_f = c^2/g$, back at $X_b = X_f - \ell$, with proper acceleration $c^2/(X_f - \ell)$, which diverges as $\ell\to X_f$. For a 100 m train at $1g$ at the front, the back's acceleration is larger by the factor $1/(1 - \ell g/c^2) \approx 1 + 1.09\times10^{-14}$: invisible, but not zero. Problem P11 works through it, and through Bell's spaceship paradox, where two ships with equal accelerations stretch the thread between them.

27 · acceleration

The Ehrenfest disc, and unit 16's missing successor

The Ehrenfest paradox (1909). Spin a disc up from rest to angular velocity $\omega$, Born-rigidly. Each rim element moves along the rim at $v = \omega R$, so its length along the rim contracts by $1/\gamma$; radial elements move perpendicular to their length and do not contract.

Born rigidity demands the proper length of every rim element be unchanged. Then the lab-measured circumference is $2\pi R/\gamma$ while the radius is still $R$ — but the rim is a circle of radius $R$ in the lab, whose circumference is $2\pi R$. Contradiction.

The disc in the lab, contraction exaggerated ($\gamma = 1.6$). Rulers that tiled the rim at rest now fall short of it: the rim must stretch or tear, so the motion cannot be Born-rigid.

Where the contradiction lives. Measured by rulers riding on the rim, the circumference of a rotating disc is $2\pi R\gamma$ (lab length $2\pi R$ divided by the contraction factor), while the radius measured by riding rulers is still $R$. So the rotating observers' spatial geometry has circumference-to-radius ratio $2\pi\gamma \gt 2\pi$: it is not Euclidean. Einstein cited exactly this in 1912–1916 as the moment he realised that accelerated frames — and therefore gravity, by the equivalence principle — need non-Euclidean geometry. Unit 30 takes it up.

What to look for in the figure. Schematic. Twelve rim rulers, each of the proper length that tiled the rim when the disc was at rest, are drawn with lab length shortened by $1/\gamma$; the radial rulers are untouched. To keep the rim in one piece the rim rulers would have to be stretched in their own rest frame — exactly the strain a real disc develops, and exactly what Born rigidity forbids.

What happens to a real disc. It deforms: the rim stretches (in its own rest frame) under stresses that an elastic material supplies. Spun fast enough, it breaks — long before any relativistic effect matters for a real material, since sound speeds are about $10^{-5}c$.

27 · acceleration

Rigid motion has three degrees of freedom, not six

Herglotz–Noether theorem — cited

A Born-rigid motion of an extended body in Minkowski space is determined by the motion of one of its points, except for uniform rotation already in progress: it has three degrees of freedom, not six. In particular a body at rest cannot be set rotating Born-rigidly.

G. Herglotz, Ann. Phys. 31, 393 (1910); F. Noether, Ann. Phys. 31, 919 (1910).

So unit 16's six-dimensional configuration space $SO(3)\times\mathbb R^3$ has no relativistic counterpart. Extended bodies become continua (unit 22), and their dynamics a field theory (unit 29).

The disc of the previous slide is the special case the theorem names: no Born-rigid path leads from rest to rotation.

27 · closing

The running example, closed

what the antiproton needed

Four-momentum is a vector in $(V,\eta)$ (section 1), whose length is $mc$ because the action is proper time (section 2). Conservation is a vector equation; its square is one scalar, $s$, conserved and invariant (section 3).

So $p + p\to 3p + \bar p$ needs $s \ge (4m_pc^2)^2$, and: $T \ge 5.630$ GeV on a target, $T \ge 0.938$ GeV per colliding beam. The Bevatron's 6.2 GeV cleared the first by 140 MeV of $\sqrt s$.

What comes next.

Unit 28 couples the particle to a field: $S = -mc^2\!\int\mathrm{d}\tau + q\!\int A$, and $\vec E, \vec B$ become one antisymmetric tensor.

Unit 30 takes section 4's hyperbola as an observer standing in a gravitational field, and asks which feature of gravity flat spacetime cannot hold.

Unit 37 finishes problem P12: special relativity gives one sixth of Mercury's perihelion advance.

The one idea to keep. Invariants first, frames last. Every result in section 3 was obtained by writing a vector equation, squaring it, and evaluating scalars in whatever frame made them trivial. The same habit — state it without coordinates, then compute in the best coordinates — is how general relativity is done, and it is why the course insisted in unit 16 that a tensor is a multilinear map rather than an array.

Reading. E. F. Taylor and J. A. Wheeler, Spacetime Physics (2nd ed., 1992), chapters 7 and 8, for momenta, collisions and thresholds with many worked numbers; it uses no calculus beyond this unit's. Landau and Lifshitz, The Classical Theory of Fields, §§8–13, for the action and the collision kinematics in a few pages; chapter 3 (§16 on) couples to fields and should wait for unit 28.

The instructive wrong answer. P12 solves the Kepler problem with $H = \sqrt{p^2c^2 + m^2c^4} - k/r$ and finds the orbit precesses by $\pi(k/Lc)^2$ per revolution: $7.16''$ per century for Mercury, against the $42.98''$ general relativity predicts and observation confirms. The missing five sixths are the curvature of space and the gravitational slowing of clocks, neither of which a potential in flat spacetime can represent.

27 · checkpoint

Checkpoint

you can now

Compute a reaction threshold by evaluating the invariant $s$ in whichever frame is easiest, and compare it with a published accelerator design energy; and derive $\vec p = \gamma m\vec v$ from the action $-mc^2\!\int\mathrm{d}\tau$ by the Euler–Lagrange equations of unit 11.

it is false if

you need a Lorentz transformation to find the antiproton threshold; or you cannot say why the fixed-target figure is three times the collider figure; or the derivation of $\vec p$ skips the step $\partial L/\partial v^i = \gamma m v^i$.

to check it

Close the deck. On paper: the $p + p\to p + p + p + \bar p$ threshold in both frames, compared with the Bevatron's 6.2 GeV; then $L = -mc^2\sqrt{1-v^2/c^2}$ to $\vec p = \gamma m\vec v$. Then ./mr problems 27.