Part D · Minkowski · spine

Minkowski space: relativity as the geometry of a quadratic form

Homogeneity, isotropy and the group axioms leave one free constant, a speed; fix it, and every paradox of special relativity becomes a statement about one indefinite bilinear form.

needs 03, 14~2½ h slides + 5 h problemsproblems: ./mr problems 26

Part C ended with a crisis: Maxwell's equations are not invariant under the Galilean change of frame that unit 03 used without naming it. This unit is the resolution, and it is presented the way a mathematician would want it: not as two postulates followed by a page of algebra, but as a classification. We ask which groups of coordinate changes between inertial frames are possible at all, given only that space and time have no preferred point, space has no preferred direction, and changes of frame compose. The answer is a one-parameter family of groups, indexed by one constant $K$. $K = 0$ is Galileo. $K > 0$ is Lorentz, with an invariant speed $c = K^{-1/2}$. $K < 0$ is excluded because it lets a clock run backwards. Nature chose $K > 0$, and experiment fixes the value.

Once the group is in hand, the rest of the unit is geometry of the quadratic form it preserves. Relativity of simultaneity is the real content; time dilation and length contraction are corollaries of it. The causal structure — which parts of "before" survive a change of frame — is the geometry of the light cone. And the twin paradox, the most argued-about result in the subject, is a triangle inequality with the inequality reversed.

How to read this deck

Five sections: the group derived from structure; the form it preserves and the group of that form; simultaneity and its corollaries, with the diagrams drawn honestly; causal structure; and proper time, where the twin paradox is proved to be a theorem about the form. The same pair of twins is in every section, and at the end three independent computations of their ages agree.

Proofs are on the slides when they are one line and in these notes under Full proof when they are not. Anything not proved here is marked cited, with where its proof is. Terms with a dotted underline are glossary entries — click one, or press G. N opens these notes; R stacks the deck into one readable page.

26 · overview

One trip, three ways to see the same six years

the running example

One twin stays on Earth. The other flies at $0.8c$ to a star $4$ light-years away and straight back. Earth's clock reads $10$ years at the reunion; the traveller's reads $6$. Each thinks the other's clock runs slow, so why is the traveller the younger?

three answers, which must agree

1 Integrate proper time along each worldline: $10$ and $6$. 2 Count the one-a-year radio signals each twin receives: $10$ and $6$, and see when the Doppler shift flips. 3 Draw the traveller's lines of simultaneity on a correctly scaled diagram and watch them jump $6.4$ years at the turn.

the claim of the unit

Answer 1 is a theorem about the quadratic form $-c^2t^2 + x^2 + y^2 + z^2$, where time enters with the opposite sign to space: for it, the straight path between two events is the longest, not the shortest. The twin paradox is the reversed triangle inequality.

getting around

A dotted underline is a glossary term: click it, or press G. N opens the notes, where the longer proofs live. A solid underline, like the twins, shows the running example on hover.

Units. Throughout the running example, time is in years and distance in light-years, so $c = 1$ ly/yr and a light ray is a line at $45°$. Elsewhere $c$ is kept explicit and the time coordinate is $x^0 = ct$, a length, so all four coordinates have the same dimension.

The numbers. At $\beta = v/c = 0.8$: $\gamma = (1 - 0.64)^{-1/2} = 1/0.6 = 5/3$. Earth time for each leg $4/0.8 = 5$ yr; traveller time for each leg $5/\gamma = 3$ yr. The rapidity is $\operatorname{artanh} 0.8 = \tfrac12\ln\frac{1.8}{0.2} = \ln 3 = 1.0986$, and the Doppler factor $\sqrt{1.8/0.2} = 3$. Every one of these is derived in the deck before it is used; the list is here so you can check them as they arrive.

Why this example. It is the one result of special relativity that looks like a contradiction, which makes it the best test of whether the geometry has been understood. The usual textbook escape — "the traveller accelerates, so the situation is not symmetric" — is true but explains nothing, and section 5 shows why: the acceleration can be confined to an arbitrarily short stretch whose contribution to the traveller's age tends to zero, while the age difference stays at four years. What breaks the symmetry is that the traveller's worldline is not straight, and the geometry of the form says what that costs.

What this unit assumes of you. Linear algebra (bilinear forms, change of basis, determinants), the chain rule of unit 03, the line-integral habits of unit 04, and unit 11's second variation, which is used once in section 5. Nothing about four-vectors, momenta or energy: that is unit 27.

section 1

The group, from structure alone

Which changes of inertial frame are possible at all — before anyone mentions light.

26 · the group

Events, frames, and the question

definitions

An event is a point of spacetime: a place at an instant. An inertial frame is a coordinate system $(t, x, y, z)$ on the set of events in which every free particle moves on a straight line at constant speed. A worldline is the set of events a particle occupies.

the principle of relativity

No experiment distinguishes one inertial frame from another. So the coordinate changes between inertial frames form a group $G$, and the change to a frame moving at velocity $\vec v$ depends on $\vec v$ alone.

The question. What can $G$ be? Unit 03 answered, without saying so, with the Galilean boost $t' = t$, $x' = x - vt$. Unit 25 showed that Maxwell's equations are not invariant under it. So assume less, and see what is forced.

three structural assumptions

homogeneity: no event is special. isotropy: no spatial direction is special. group: frame changes compose and invert, depend continuously on $\vec v$, and $\vec v = 0$ is the identity.

Einstein's second postulate — the speed of light is independent of its source — is not among them. It will be needed for exactly one thing.

Why a group. If $f_{12}$ changes coordinates from frame 1 to frame 2 and $f_{23}$ from 2 to 3, then $f_{23}\circ f_{12}$ changes from 1 to 3, and frame 3 is inertial, so the composite is again an allowed change; inverses exist because frame 1 is inertial as seen from frame 2 as much as the other way round. This is the principle of relativity doing its first job.

Why "depends on $\vec v$ alone". Two frames with the same velocity relative to $S$ differ only by a translation and a rotation of their axes. Having set those aside by choosing coinciding origins and aligned axes, the only datum left distinguishing $S'$ from $S$ is its velocity.

History. The derivation in this section goes back to Ignatowski (1910) and Frank and Rothe (1911), who noticed that the relativity principle alone almost determines the transformation; it was rediscovered many times, most cleanly by Berzi and Gorini (J. Math. Phys. 10, 1969, 1518) and by Lévy-Leblond (Am. J. Phys. 44, 1976, 271, "One more derivation of the Lorentz transformation"). Einstein's 1905 paper used both postulates from the start.

A warning on the word "inertial". The definition is circular in a familiar way: free particles are those that move straight in inertial frames. Unit 03 met the same circle and broke it operationally — a frame is inertial if no Coriolis or centrifugal terms are needed. Unit 30 returns to the question in earnest, because gravity is exactly what makes it hard.

26 · the group

Homogeneity gives linearity

homogeneity, made precise

Let $f: \mathbb R^4 \to \mathbb R^4$ send an event's $S$-coordinates to its $S'$-coordinates. Homogeneity: the $S'$-separation of two events depends only on their $S$-separation, $$f(X + A) - f(X) = g(A)\quad\text{for all } X, A.$$

lemma

If $f$ is continuous and homogeneous, then $f(X) = f(0) + \Lambda X$ with $\Lambda$ linear.

proof

$g(A + B) = f(X + A + B) - f(X + A) + f(X + A) - f(X) = g(B) + g(A)$: $g$ is additive. Additive gives $g(qA) = q\,g(A)$ for rational $q$, and continuity extends it to real $q$. So $g$ is linear, and $f(X) = f(0) + g(X)$.

Choose the origins so the two frames' spatial origins coincide at $t = t' = 0$; then $f(0) = 0$ and $f = \Lambda$ is a $4\times 4$ matrix. Physically: straight worldlines go to straight worldlines, as they must if both frames are inertial.

Full proof

Additivity: set $X$ arbitrary and compute $g(A+B) = f(X + A + B) - f(X)$, then insert $\pm f(X + A)$: the first difference is $g(B)$ evaluated at base point $X + A$, the second is $g(A)$. So $g(A + B) = g(A) + g(B)$, and $g(0) = 0$.

Rational scaling: by induction $g(nA) = n\,g(A)$ for $n \in \mathbb N$; $g(-A) = -g(A)$ from $g(A) + g(-A) = g(0) = 0$; and $n\,g(A/n) = g(A)$ gives $g(A/n) = g(A)/n$. So $g(qA) = q\,g(A)$ for every $q \in \mathbb Q$.

Real scaling: $g$ is continuous because $f$ is. For real $\lambda$ take rationals $q_k \to \lambda$; then $g(q_kA) \to g(\lambda A)$ and $q_k g(A) \to \lambda g(A)$. So $g$ is $\mathbb R$-linear.

Is continuity needed? Yes, as a matter of mathematics: there are discontinuous additive maps $\mathbb R \to \mathbb R$ (built from a Hamel basis, so with the axiom of choice), and they are not linear. It is not needed as a matter of physics, where nobody proposes a coordinate change with a graph dense in the plane, and measurability would already do. Section 2 cites a theorem that removes even the assumption of homogeneity, at the price of assuming the causal order is preserved.

Why "straight lines go to straight lines" is the same thing. An affine map sends lines to lines. Conversely, a bijection of $\mathbb R^n$ ($n \ge 2$) sending lines to lines is affine — the fundamental theorem of affine geometry, cited (e.g. Berger, Geometry I, §2.6). Either route gives linearity; the route on the slide is shorter and uses the physical assumption directly.

26 · the group

Isotropy reduces it to two dimensions

Rotate both frames' axes so $\vec v = v\,\hat x$. Isotropy then forces the transverse coordinates to be left alone:

$$y' = y,\qquad z' = z,$$

and the $(t, x)$ block to decouple from them. What remains is a $2\times 2$ matrix $L(v)$ acting on $(t, x)$.

The origin of $S'$ moves along $x = vt$, and $x' = 0$ there, so the second row is fixed up to a factor:

$$x' = \gamma(v)\,(x - vt),\qquad t' = a(v)\,t + b(v)\,x.$$

two facts from isotropy

Mirror. Reflecting $x \mapsto -x$ in both frames turns a boost by $v$ into a boost by $-v$: $L(-v) = PL(v)P$, $P = \operatorname{diag}(1, -1)$. So $a$ and $\gamma$ are even in $v$, and $b$ is odd.

Reciprocity. $L(v)^{-1} = L(-v)$: if $S'$ moves at $v$ in $S$, then $S$ moves at $-v$ in $S'$. This one needs a proof, in the notes; it forces $a = \gamma$.

Full proof that $y' = y$, $z' = z$

A rotation by $\pi$ about the $x$-axis, $(y,z)\mapsto(-y,-z)$, applied to both frames, leaves the boost along $x$ unchanged (isotropy: nothing distinguishes the rotated picture). So $\Lambda$ commutes with $\operatorname{diag}(1,1,-1,-1)$, and a matrix commuting with that is block diagonal: $(t,x)$ and $(y,z)$ do not mix. The same argument with any rotation about $x$ shows the $(y,z)$ block commutes with $SO(2)$, and the reflection $y\mapsto -y$ (also a symmetry of the boost) makes it diagonal, so it is $k(v)\,I_2$. Reciprocity (below) gives $k(v)k(-v) = 1$, the mirror gives $k(-v) = k(v)$, so $k^2 = 1$, and $k(0) = 1$ with continuity gives $k \equiv 1$.

Full proof of reciprocity

$L(v)^{-1}$ is an allowed change of frame, so it is $L(\varphi(v))$ for some velocity $\varphi(v)$ — the velocity of $S$ as seen from $S'$. Directly from the matrix, the worldline $x = 0$ has $x' = -\gamma vt$ and $t' = at$, so $\varphi(v) = -\gamma v/a$: continuous, and $\varphi(v) < 0$ for small $v > 0$ because $\gamma, a \to 1$ as $v \to 0$. Three properties:

$\varphi(\varphi(v)) = v$, because inverting twice is the identity;
$\varphi(-v) = -\varphi(v)$: invert the mirror relation, $L(-v)^{-1} = PL(v)^{-1}P = PL(\varphi(v))P = L(-\varphi(v))$;
$\varphi$ is injective (it has an inverse, itself), so being continuous it is monotone.

If $\varphi$ were increasing, an increasing involution is the identity (if $\varphi(v) > v$ then $v = \varphi(\varphi(v)) > \varphi(v)$, a contradiction; similarly for $<$), contradicting $\varphi(v) < 0$ for small $v > 0$. So $\varphi$ is decreasing, and $\psi = -\varphi$ is increasing with $\psi(\psi(v)) = \varphi(\varphi(v)) = v$ by oddness. So $\psi$ is the identity and $\varphi(v) = -v$. Then $-\gamma v/a = -v$ gives $a = \gamma$.

This argument is Berzi and Gorini's, tidied. Many textbooks simply assert reciprocity; it is a theorem, and it uses isotropy essentially.

26 · the group

Only one constant is left free

theorem (Ignatowski, 1910)

Under homogeneity, isotropy and the group axioms, there is a constant $K$, the same for every $v$, with

$$t' = \gamma\,(t - Kvx),\qquad x' = \gamma\,(x - vt),\qquad \gamma = \frac{1}{\sqrt{1 - Kv^2}}.$$

proof

With $a = \gamma$ and $K(v) := -b/(\gamma v)$, $L(v) = \gamma\left(\begin{smallmatrix}1 & -Kv\\ -v & 1\end{smallmatrix}\right)$. Reciprocity, $L(v)^{-1} = L(-v)$, forces $\det L(v) = \gamma^2(1 - Kv^2) = 1$. The product $L(v)L(u)$ has diagonal entries $\gamma_v\gamma_u(1 + K(v)vu)$ and $\gamma_v\gamma_u(1 + K(u)uv)$; it is a boost, so they are equal, and $K(v) = K(u)$.

Check. $[Kvx] = [t]$ needs $[K] = \mathrm{s^2\,m^{-2}}$: $K^{-1/2}$ is a speed ✓. $v \to 0$: $\gamma \to 1$, identity ✓.

The structure leaves exactly one number undetermined. $K > 0$ gives the Lorentz transformation, with $c := K^{-1/2}$; the sign of $K$ decides which world we are in — drawn on the next slide.

Full proof

From the last slide, $L(v) = \begin{pmatrix}\gamma & b\\ -\gamma v & \gamma\end{pmatrix}$ with $\gamma$ even and $b$ odd. For $v \ne 0$ define $K(v) = -b/(\gamma v)$, which is even, so $L(v) = \gamma\begin{pmatrix}1 & -Kv\\ -v & 1\end{pmatrix}$. Its inverse is $\frac{1}{\det}\,\gamma\begin{pmatrix}1 & Kv\\ v & 1\end{pmatrix}$ with $\det = \gamma^2(1 - Kv^2)$, and $L(-v) = \gamma\begin{pmatrix}1 & Kv\\ v & 1\end{pmatrix}$. Reciprocity therefore says $\det = 1$, i.e. $\gamma^2 = 1/(1 - K(v)v^2)$.

Composition. $L(v)L(u) = \gamma_v\gamma_u\begin{pmatrix}1 + K_vvu & -K_uu - K_vv\\ -v - u & 1 + K_uuv\end{pmatrix}$. Every boost has equal diagonal entries, and the product of two boosts along $x$ is a boost along $x$ (it is an allowed change of frame; it keeps the transverse coordinates fixed; so it is some $L(w)$). Equal diagonal entries give $K_vvu = K_uuv$, so $K(v) = K(u)$ whenever $uv \ne 0$. $K$ is one number for the whole group.

Where the dimension check sits. $K$ has dimension $\mathrm{time}^2/\mathrm{length}^2$ because $Kvx$ is added to $t$; so $1/\sqrt K$ is a speed. Nothing in the three assumptions supplies a speed, which is why its value is a fact about the world, to be measured.

26 · the group

Three worlds, one per sign of $K$

Where a moving clock's first tick, $t' = 1$, lands in $S$ for several $v$ (dots, each on that frame's time axis $x = vt$). The boosts slide it along the curve $t^2 - Kx^2 = 1$: a line, a hyperbola inside the light lines, or a circle — and only on the circle can composing boosts carry it to $t < 0$ (dashed).

$K = 0$: Galileo, $t' = t$, $x' = x - vt$. $K > 0$: Lorentz, $c = K^{-1/2}$. $K < 0$: rotations of the $(t, x)$ plane — two boosts can turn a clock's time backwards. Excluded.

Why the curve. In $S$ the tick is at $(x, t) = (\gamma v, \gamma)$, by the inverse boost at $x' = 0$, $t' = 1$. So $t^2 - Kx^2 = \gamma^2(1 - Kv^2) = 1$ for every $v$: one curve per world.

What to look for. Each thin line is the time axis of a frame moving at $v$, and the dot on it is that frame's clock reading $1$. For Galileo the dots lie on the horizontal line $t = 1$ — absolute time — and $v$ is unbounded. For $K > 0$ ($K = 1$ drawn, so $c = 1$) they lie on the hyperbola $t^2 - x^2 = 1$, which approaches the light lines $x = \pm t$ but never reaches them: no frame moves at $c$. For $K < 0$ ($K = -1$ drawn) they lie on the circle $t^2 + x^2 = 1$; a single boost covers only the upper half, but boosts compose like rotations, and the lower half is reached.

Why $K < 0$ is excluded. Write $K = -1/k^2$. Then $L(v)$ is $\gamma\begin{pmatrix}1 & v/k^2\\ -v & 1\end{pmatrix}$; in the coordinates $(kt, x)$ it is the rotation by $\theta$ with $\tan\theta = v/k$. Rotations compose by adding angles, so four boosts at $v = k$ (each $\theta = \pi/4$) give a rotation by $\pi$: $t' = -t$ on the worldline $x = 0$. A clock at rest in $S$ would be seen, from an inertial frame, to run backwards; its successive ticks — timelike-related events on one worldline, each a cause of the next — would be reordered. Two boosts at $v = k$ already send the time axis to the space axis. The problems page works this world out.

Why the group closes for $K > 0$. The composite velocity is read off the bottom-left entry: $w = (u + v)/(1 + Kuv)$, and $\gamma_w = \gamma_u\gamma_v(1 + Kuv)$ consistently. If $|u|, |v| < c$ then $|w| < c$ (section 2 proves this cleanly with rapidity). So the set $|v| < c$ is closed, and it is the whole group: $\gamma$ is real only there.

26 · the group

The invariant speed, and the work each postulate does

the invariant speed

For $K = 1/c^2$ the composite velocity is $w = \dfrac{u + v}{1 + uv/c^2}$. Put $u = c$: $w = c$ for every $v$. Something moving at $c$ in one frame moves at $c$ in all of them. Proof: one line of algebra.

postulate	what it does here
relativity	makes the frame changes a group, gives reciprocity via isotropy, makes $K$ a constant
light speed independent of source	chooses $K > 0$ over $K = 0$, and names the invariant speed: light's

provenance of $c$

Unit 25 earned the value: the wave operator $\partial_t^2/c^2 - \nabla^2$ of Maxwell's theory is invariant under exactly this group with this $c$. If you have not done unit 25, you are taking $K = 1/c^2$ on credit here, with $c = 299\,792\,458\ \mathrm{m\,s^{-1}}$ (exact, by the definition of the metre).

the turn, in the outbound frame

$c = 1$, $v = 0.8$, $\gamma = 5/3$. $T = (t, x) = (5, 4)$: $t' = \tfrac53(5 - 0.8\cdot 4) = 3$, $x' = \tfrac53(4 - 0.8\cdot 5) = 0$. At the traveller's own position, after three years ✓.

Proof of invariance. $w = (c + v)/(1 + v/c) = c(c + v)/(c + v) = c$, valid for $v \ne -c$. The limit $uv \ll c^2$ returns Galileo's $w = u + v$, which is the special case every boxed result in this unit will be checked against.

What the second postulate really says. Einstein's wording (1905): light in empty space propagates with a definite speed $c$ independent of the state of motion of the emitting body. Combined with the first postulate this makes $c$ the same in every frame; the structural derivation shows that the whole content of that combination is "choose $K > 0$, and the invariant speed is light's". De Sitter's binary-star argument (unit 25) is the evidence that light's speed does not depend on its source.

On credit, honestly. If you skipped unit 25, the statement "$K > 0$ and $c$ is the speed of light" is here an experimental input, not a derivation. The course does not pretend otherwise. Unit 25's argument is that the one-dimensional wave operator changes form under a Galilean boost and keeps it under exactly the transformations of this slide with $K = 1/c^2$, $c$ being the constant in the wave equation. Nothing later in this unit depends on how you came by $K$; every result below uses only $K = 1/c^2 > 0$.

The example. The turnaround event $T$ is where the traveller is, so in the traveller's outbound frame it must have $x' = 0$; the computation confirms it, and gives the traveller's clock reading at the turn, $3$ years. That is the first appearance of the $6$-year total: the return leg is the mirror image.

section 2

The form, and the group of the form

What the Lorentz transformations preserve, and everything that follows from asking only that.

26 · the form

The interval is invariant

theorem

With $x^0 = ct$, every Lorentz transformation preserves the interval

$$s^2 = -(x^0)^2 + x^2 + y^2 + z^2 .$$

proof

$x^{0\prime} = \gamma(x^0 - \beta x)$, $x' = \gamma(x - \beta x^0)$ with $\beta = v/c$. Then $x'^2 - x^{0\prime 2} = \gamma^2(1 - \beta^2)(x^2 - x^{0\,2}) = x^2 - x^{0\,2}$, and $y, z$ are untouched.

So the physics — the group — has an invariant, and it is a quadratic form. Turn the logic round and define the setting by the form:

the programme

Spacetime is a 4-dimensional real vector space with a symmetric bilinear form of a certain signature. The Lorentz group is the group of that form. Every kinematic effect is a property of the form.

Check. $s^2$ is a length squared in every term ✓. On the turn: $-25 + 16 = -9$ in $S$ and $-9 + 0 = -9$ in $S'$ ✓.

The proof, expanded. $x'^2 - (x^{0\prime})^2 = \gamma^2\big[(x - \beta x^0)^2 - (x^0 - \beta x)^2\big] = \gamma^2\big[x^2 - 2\beta xx^0 + \beta^2x^{0\,2} - x^{0\,2} + 2\beta xx^0 - \beta^2x^2\big] = \gamma^2(1 - \beta^2)(x^2 - x^{0\,2})$, and $\gamma^2(1 - \beta^2) = 1$. The cross terms cancel exactly; that is the reason the time coordinate had to be mixed with $x$ in the particular proportion $K v = v/c^2$.

Not just boosts. Rotations of the spatial axes preserve $x^2 + y^2 + z^2$ and leave $x^0$ alone, so they preserve $s^2$ too; and a general change between inertial frames with coinciding origins is a rotation, then a boost along $x$, then a rotation, so it preserves $s^2$.

History. Poincaré noticed the invariance of $x^2 + y^2 + z^2 - c^2t^2$ in 1905–06 and even wrote it with an imaginary fourth coordinate $ict$ to make it look Euclidean. Minkowski's 1908 Cologne lecture, "Raum und Zeit", made the form the primary object: "Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows." The $ict$ trick survived in textbooks into the 1970s; it hides exactly the sign this unit is about, and this course does not use it.

The sign convention. This course uses $(-,+,+,+)$, as Misner–Thorne–Wheeler, Wald, Carroll and Schutz do; particle physicists usually use $(+,-,-,-)$. Nothing physical depends on the choice, but every formula with a sign in it does — see the warning on the classification slide.

26 · the form

Minkowski space, defined

Minkowski space

A 4-dimensional real vector space $V$ with a symmetric bilinear form $\eta: V\times V\to\mathbb R$ of signature $(-,+,+,+)$: some basis $(e_0, e_1, e_2, e_3)$ has $\eta(e_\mu, e_\nu) = \eta_{\mu\nu}$, $\eta_{\mu\nu} = \operatorname{diag}(-1, 1, 1, 1)$. Such a basis is a Lorentz orthonormal basis: an inertial frame.

The Minkowski form $\eta$ is the object; its matrix is what it looks like in one basis. Events form an affine space over $V$; differences of events are vectors.

summation convention

For $X = X^\mu e_\mu$, $Y = Y^\nu e_\nu$: $\ \eta(X, Y) = \eta_{\mu\nu}X^\mu Y^\nu$, a repeated upper-lower index summed over $0..3$. Notation, nothing more.

theorem: components change like this

If $e'_\alpha = \Lambda^\mu{}_\alpha\, e_\mu$ is another basis, then $\eta'_{\alpha\beta} = \Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta\,\eta_{\mu\nu}$, i.e. $\eta' = \Lambda^{\mathsf T}\eta\Lambda$.

Proof: $\eta'_{\alpha\beta} = \eta(e'_\alpha, e'_\beta) = \eta(\Lambda^\mu{}_\alpha e_\mu, \Lambda^\nu{}_\beta e_\nu)$; expand by bilinearity.

Definition first, transformation law second. The transformation law is a consequence of bilinearity, not a definition. This is the habit the course keeps for every tensor to come: unit 27's four-vectors, unit 28's field tensor, unit 32's metric. Here the only object is a bilinear form on one vector space, and that is linear algebra you already have. What unit 32 adds is a bilinear form on each tangent space of a manifold, varying from point to point; that is the metric, and this $\eta$ is its constant, flat case.

Signature is well defined. That the numbers of $-1$, $+1$ and $0$ entries in a diagonal matrix of $\eta$ do not depend on the diagonalising basis is Sylvester's law of inertia (cited: any linear algebra text, e.g. Lang, Linear Algebra, ch. V, or Axler's chapter on bilinear forms). So "signature $(-,+,+,+)$" is a property of $\eta$, not of a basis. In particular $\eta$ is non-degenerate: $\eta(X, Y) = 0$ for all $Y$ forces $X = 0$.

Affine versus linear. Spacetime itself has no origin; "the event where the traveller left" and "the turnaround" are points, and their difference, $(5, 4)$ in years and light-years, is a vector of $V$. An inertial frame is a choice of origin together with a Lorentz orthonormal basis. Section 1's homogeneity is precisely the statement that nothing depends on the origin.

Index positions. Components of vectors carry upper indices; the form's carry lower ones. This bookkeeping is the vector/covector distinction of unit 04's gradient slide, and unit 28 is where it stops being optional; here it is just the convention that makes the summation rule self-checking: an index summed over must appear once up and once down.

The proof, written out. $\eta(\Lambda^\mu{}_\alpha e_\mu, \Lambda^\nu{}_\beta e_\nu) = \sum_{\mu,\nu}\Lambda^\mu{}_\alpha\Lambda^\nu{}_\beta\,\eta(e_\mu, e_\nu)$, which in matrices is $(\Lambda^{\mathsf T}\eta\Lambda)_{\alpha\beta}$.

26 · the form

Timelike, null, spacelike: the light cone

classification of a nonzero $X \in V$

timelike if $\eta(X,X) < 0$; null if $\eta(X,X) = 0$; spacelike if $\eta(X,X) > 0$. The null vectors form the light cone. A timelike or null $X$ is future-directed if $X^0 > 0$.

Because $\eta$ is invariant, the classification is the same in every inertial frame: it is a property of the pair of events, not of the observer.

the twins

Out-leg $X = (5, 4)$: $-25 + 16 = -9$, timelike. A light ray $(4, 4)$: null. The separation from Earth-at-$t{=}5$ to $T$, $(0, 4)$: $+16$, spacelike.

The plane $(x, x^0)$, years and light-years. The cone $x^0 = \pm x$ splits it; each of the twins' three vectors, components $(x^0, x)$, ends where its $\eta(X, X)$ puts it.

Physical reading. Timelike separation: some inertial observer passes through both events (move at $\vec v = c\,\vec x/x^0$, which is slower than $c$). Null: a light signal can pass through both. Spacelike: no signal slower than or equal to light connects them; section 4 shows their time order is then not an invariant.

Reading the figure. Only the $x$ direction is drawn; with $y$ and $z$ restored, the two dashed lines are the cross-section of a double cone, and "spacelike" is the connected region outside it. The three arrows all end at $x = 4$: what decides the class is whether the time component beats the space component, $5 > 4$, $4 = 4$, $0 < 4$.

Is "future-directed" invariant? Only under the Lorentz transformations that do not reverse time — the orthochronous ones of "The Lorentz group and its four pieces". For the boosts derived in section 1 it is, because $x^{0\prime} = \gamma(x^0 - \beta x)$ and $|\beta x| < |x^0|$ for a timelike vector, so $x^{0\prime}$ has the sign of $x^0$. The cone of future timelike vectors is convex: if $X, Y$ are in it so is $sX + tY$ for $s, t \ge 0$ not both zero, because $\eta(sX + tY, sX + tY) = s^2\eta(X,X) + 2st\,\eta(X,Y) + t^2\eta(Y,Y) < 0$ by the lemma of the next slide.

26 · the form

Two future timelike vectors pair negatively

lemma

If $X, Y$ are future-directed timelike, $\eta(X, Y) < 0$. Proof: $\eta(X,Y) = -X^0Y^0 + \vec x\cdot\vec y \le -X^0Y^0 + |\vec x||\vec y| < 0$, by Cauchy–Schwarz in $\mathbb R^3$ and $|\vec x| < X^0$, $|\vec y| < Y^0$.

In the picture: the vectors $\eta$-orthogonal to $X$ form the line $X^\perp$, the mirror image of $X$ in the light line. It lies outside the cone, so it is spacelike, and the whole future cone sits on its negative side.

signs

In $(+,-,-,-)$ every inequality above flips: timelike means $\eta(X,X) > 0$. Check which convention a book uses before copying a formula.

$X$ and $Y$ are the twins' two legs, components $(x^0, x)$; $\eta(X, Y) = -25 - 16 = -41$. $Z = (4, 5)$ has $\eta(X, Z) = -20 + 20 = 0$: the line $X^\perp$ (dashed) is spacelike, and every future timelike vector is on the side where $\eta(X, \cdot) < 0$.

Reading the figure. $\eta(X, Y) = -X^0Y^0 + X^1Y^1$ vanishes for $Y$ on the line $X^0Y^0 = X^1Y^1$, which is $X$ with its two components swapped: the reflection of $X$ in the line $x^0 = x$. A timelike $X$ is steeper than the light line, so its mirror image is shallower — spacelike. The lemma says the future cone lies entirely on one side of that line, which is what the shading shows.

The lemma, with equality. The first inequality is equality iff $\vec x, \vec y$ are parallel and point the same way; the second is always strict. The lemma is what makes the reversed triangle inequality of section 5 have the right sign.

A lemma used in section 5: the orthogonal complement of a timelike vector is spacelike. If $X$ is timelike and $Z \ne 0$ with $\eta(X, Z) = 0$, then $\eta(Z, Z) > 0$. Proof: otherwise $\eta(aX + bZ, aX + bZ) = a^2\eta(X,X) + b^2\eta(Z,Z) \le 0$ on the 2-dimensional span $W$ of $X, Z$. But $W$ meets the 3-dimensional span of $e_1, e_2, e_3$ in a subspace of dimension at least $2 + 3 - 4 = 1$, and on that span $\eta$ is positive definite: contradiction. In the figure, $X^\perp$ is one line; in four dimensions it is a 3-dimensional hyperplane, and every direction in it is spacelike.

26 · the form

The Lorentz group and its four pieces

the Lorentz group $O(1,3)$

$\{\Lambda \in GL(V) : \eta(\Lambda X, \Lambda Y) = \eta(X, Y)\ \forall X, Y\}$. In an orthonormal basis: $\Lambda^{\mathsf T}\eta\Lambda = \eta$.

Two continuous invariants: $\det\Lambda = \pm1$ (take determinants: $\det(\Lambda)^2 = 1$), and whether $\Lambda$ maps the future cone to itself, which happens iff $\Lambda^0{}_0 \ge 1$; otherwise $\Lambda^0{}_0 \le -1$.

	$\Lambda^0{}_0 \ge 1$	$\Lambda^0{}_0 \le -1$
$\det = +1$	$SO^+(1,3)$: boosts, rotations	$PT$
$\det = -1$	$P = \operatorname{diag}(1,-1,-1,-1)$	$T = \operatorname{diag}(-1,1,1,1)$

theorem

$O(1,3)$ has exactly four connected components, the cells of the table. The identity's component, the proper orthochronous group $SO^+(1,3)$, is a normal subgroup; the others are its cosets $P\,SO^+$, $T\,SO^+$, $PT\,SO^+$.

$|\Lambda^0{}_0| \ge 1$. The $00$ entry of $\Lambda^{\mathsf T}\eta\Lambda = \eta$ reads $-(\Lambda^0{}_0)^2 + \sum_i(\Lambda^i{}_0)^2 = -1$, so $(\Lambda^0{}_0)^2 = 1 + \sum_i(\Lambda^i{}_0)^2 \ge 1$.

Orthochronous means $\Lambda^0{}_0 \ge 1$. $\Lambda e_0$ is timelike (the form is preserved), and its time component is $\Lambda^0{}_0$. The future timelike cone is convex, hence connected, and $\Lambda$ maps it continuously into the timelike vectors, which are the disjoint union of the future and past cones; so its image lies in one of them, the one containing $\Lambda e_0$. Write $C$ for the future cone. Maps sending $C$ into itself are closed under composition; they are closed under inverses too, because if $\Lambda^{-1}$ sent $C$ into $-C$ then $\Lambda$ would send $-C$ into $C$ and hence, being linear, $C$ into $-C$, which it does not. So the orthochronous transformations form a subgroup — the problems page asks for the argument in full.

Full proof of the four components

At least four. $\det$ and $\operatorname{sign}\Lambda^0{}_0$ are continuous functions on $O(1,3)$ with values in $\{\pm1\}$, so they are constant on connected components; the four cells are therefore unions of components, and each is nonempty ($I, P, T, PT$).

At most four: $SO^+(1,3)$ is path-connected. Let $\Lambda \in SO^+$ and $u = \Lambda e_0$, a future unit timelike vector, $u = (\gamma, \gamma\vec\beta)$ with $|\vec\beta| < 1$. Let $B$ be the boost with velocity $\vec\beta$ (rotate $\vec\beta$ onto the $x$-axis, boost, rotate back): $Be_0 = u$. Then $R = B^{-1}\Lambda$ fixes $e_0$, so it preserves $e_0^\perp = \operatorname{span}(e_1,e_2,e_3)$ with its positive-definite restriction of $\eta$; it is an orthogonal $3\times3$ block with $\det = 1$, a rotation. So $\Lambda = BR$. Rotations are connected to $I$: a rotation $R$ of $\mathbb R^3$ has an axis, because $\det(R - I) = \det(R^{\mathsf T})\det(R - I) = \det(I - R^{\mathsf T}) = \det(I - R) = -\det(R - I)$ in odd dimension, so $\det(R - I) = 0$; on the plane orthogonal to the axis $R$ is a planar rotation by some $\theta$, and shrinking $\theta$ to $0$ is a path to $I$. Boosts are connected to $I$ by shrinking $\vec\beta$. So a path runs from $I$ to $\Lambda$. The other cells are $P\,SO^+$ etc., images of a connected set under a homeomorphism.

Normal. The identity component of any topological group is a normal subgroup (conjugation is continuous and fixes $I$). Physically only $SO^+(1,3)$ is a symmetry of everything: parity is violated by the weak interaction (Wu, 1957), and time reversal too — Christenson, Cronin, Fitch and Turlay found $CP$ violation in neutral-kaon decay in 1964, which is $T$ violation given $CPT$, and CPLEAR saw $T$ violation directly in 1998.

26 · the form

Boosts are hyperbolic rotations

rapidity

For $|\beta| < 1$ put $\varphi = \operatorname{artanh}\beta$. Then $\gamma = \cosh\varphi$, $\gamma\beta = \sinh\varphi$, and

$$B(\varphi) = \begin{pmatrix}\cosh\varphi & -\sinh\varphi\\ -\sinh\varphi & \cosh\varphi\end{pmatrix}.$$

theorem

$B(\varphi)B(\psi) = B(\varphi + \psi)$. The boosts along one axis form a one-parameter group isomorphic to $(\mathbb R, +)$.

Proof: multiply, and use $\cosh(\varphi+\psi) = \cosh\varphi\cosh\psi + \sinh\varphi\sinh\psi$ and its partner for $\sinh$.

A boost slides points along the hyperbolae $x^2 - (x^0)^2 = $ const, as a rotation slides them along circles (arrows: $B(\varphi)$, $\varphi > 0$). Dots: $e_0$ moved by $\varphi = 0, \pm0.5, \pm1$. The light lines are only rescaled.

Why $\gamma = \cosh\varphi$. $\cosh^2 - \sinh^2 = 1$ and $\tanh = \sinh/\cosh$ give $\cosh\varphi = 1/\sqrt{1 - \tanh^2\varphi} = 1/\sqrt{1 - \beta^2} = \gamma$, and $\sinh\varphi = \tanh\varphi\cosh\varphi = \beta\gamma$. So the boost of section 1, $x^{0\prime} = \gamma(x^0 - \beta x)$, $x' = \gamma(x - \beta x^0)$, is the matrix on the slide.

The product. $B(\varphi)B(\psi)$ has diagonal entries $\cosh\varphi\cosh\psi + \sinh\varphi\sinh\psi = \cosh(\varphi+\psi)$ and off-diagonal entries $-(\cosh\varphi\sinh\psi + \sinh\varphi\cosh\psi) = -\sinh(\varphi+\psi)$. Both identities follow from $\cosh a = (e^a + e^{-a})/2$, $\sinh a = (e^a - e^{-a})/2$ by expanding. The velocity formula is then $\tanh(\varphi+\psi) = (\tanh\varphi + \tanh\psi)/(1 + \tanh\varphi\tanh\psi)$, which is the $\sinh$ formula over the $\cosh$ formula divided through by $\cosh\varphi\cosh\psi$.

Reading the figure. The hyperbolae $x^2 - (x^0)^2 = \text{const}$ are the boosts' orbits, as circles are the rotations': a boost slides each point along its own hyperbola, the timelike ones (top and bottom) and the spacelike ones (left and right) alike. The light lines are the eigenvectors: $B(\varphi)(1, 1) = (\cosh\varphi - \sinh\varphi)(1, 1) = e^{-\varphi}(1, 1)$ and $B(\varphi)(1, -1) = e^{\varphi}(1, -1)$, components $(x^0, x)$. So a boost is a squeeze: it shrinks one light direction by $e^{-\varphi}$ and stretches the other by $e^{\varphi}$, keeping their product, which is $-\eta$ in null coordinates. Section 3's Doppler factor is this $e^{\varphi}$.

Only collinear boosts add. Two boosts in different directions do not compose to a boost. Proof: a pure boost is a symmetric matrix — its $0i$ and $i0$ entries are both $-\gamma\beta_i$, as the $2\times2$ case above shows — whereas the product $B_xB_y$ of a boost along $y$ and then one along $x$ has $(B_xB_y)^0{}_1 = -\gamma_x\beta_x$ and $(B_xB_y)^1{}_0 = -\gamma_x\gamma_y\beta_x$, which differ because $\gamma_y > 1$. By the decomposition $\Lambda = BR$ of "The Lorentz group and its four pieces" the product is therefore a boost followed by a non-trivial rotation — the Wigner rotation, whose accumulation along a curved path is Thomas precession. So rapidity is additive along one line, not in general; the set of all boosts is not a subgroup. Unit 27 meets the consequence.

A generator. Differentiating, $\frac{\mathrm d}{\mathrm d\varphi}B(\varphi) = K_xB(\varphi)$ with $K_x = \begin{pmatrix}0 & -1\\ -1 & 0\end{pmatrix}$ and $B(0) = I$: the boosts are the flow of a linear vector field, a one-parameter group in unit 02's sense, and the group property is uniqueness of solutions (unit 01). $K_x$ is the generator of boosts (the conventional letter; it has nothing to do with section 1's constant $K$); it is symmetric where a rotation generator is antisymmetric, which is the sign of $\eta$ showing up in the Lie algebra: $K^{\mathsf T}\eta + \eta K = 0$.

26 · the form

Velocity addition is tanh addition

velocity addition is tanh addition $$\beta_w = \tanh(\varphi + \psi) = \frac{\beta_u + \beta_v}{1 + \beta_u\beta_v}.$$

Since $|\tanh| < 1$ on $\mathbb R$, composing boosts can never reach $c$.

Rapidities add; speeds are their $\tanh$. Equal boosts are equal steps along the $\varphi$ axis, and $\tanh$ squashes the steps ever closer below $1$.

Check. $\beta_u = 1$: $\beta_w = (1 + \beta_v)/(1 + \beta_v) = 1$, the invariant speed ✓.

Repeated boosts of $0.5c$, each adding $\tfrac12\ln 3 = 0.549$ to $\varphi$: $\beta = 0.5, 0.8, 0.929, 0.976, 0.992$ (dots, $\beta = \tanh\varphi$). Galileo's sum (dashed) reaches $c$ after two.

26 · the form · example

The traveller's $0.8c$ is two half-light-speed boosts

adding by rapidity

1. $\varphi_{1/2} = \operatorname{artanh} 0.5 = \tfrac12\ln\dfrac{1 + 0.5}{1 - 0.5} = \tfrac12\ln 3 = 0.5493$.

2. Two such boosts: $\varphi = \ln 3 = 1.0986$.

3. $\beta = \tanh(\ln 3) = \dfrac{3 - 1/3}{3 + 1/3} = \dfrac{8}{10} = 0.8$.

4. Direct check: $\dfrac{0.5 + 0.5}{1 + 0.25} = 0.8$ ✓. And $\gamma = \cosh\ln 3 = \tfrac12(3 + \tfrac13) = \tfrac53$ ✓.

what the numbers are for

The traveller's rapidity is $\ln 3$, so $e^{\varphi} = 3$. Section 3 shows that $e^{\varphi}$ is the Doppler factor: that is why the signal counts come in threes.

Galileo would say two $0.5c$ boosts make $1.0c$. Rapidity says each successive $0.5c$ boost adds the same $0.549$ to $\varphi$, and $\beta = \tanh\varphi$ creeps towards $1$ without reaching it. The problems page does ten of them.

Step 1. $\operatorname{artanh}\beta = \tfrac12\ln\frac{1+\beta}{1-\beta}$, from solving $\beta = (e^{2\varphi} - 1)/(e^{2\varphi} + 1)$ for $e^{2\varphi} = (1+\beta)/(1-\beta)$. At $\beta = 0.5$ that is $\tfrac12\ln 3 = 0.549306$.

Step 3. $\tanh a = (e^a - e^{-a})/(e^a + e^{-a})$ at $a = \ln 3$: $(3 - \tfrac13)/(3 + \tfrac13) = (8/3)/(10/3) = 0.8$.

The physical meaning of "two boosts". Let frame $S_1$ move at $0.5c$ relative to Earth, and the traveller at $0.5c$ relative to $S_1$, in the same direction. Then the traveller moves at $0.8c$ relative to Earth. Each twin in $S_1$ — a third observer riding halfway — sees Earth recede at $0.5c$ and the traveller advance at $0.5c$: the symmetric description of the outbound leg.

Checks. Dimensionless on both sides ✓. The limit of two small boosts $\varepsilon$: $2\varepsilon/(1 + \varepsilon^2) = 2\varepsilon + O(\varepsilon^3)$, Galileo to second order ✓.

26 · the form

Causality alone forces the group

the Alexandrov–Zeeman theorem (cited)

Write $p \prec q$ if $q - p$ is a nonzero future-directed timelike or null vector. Let $f:\mathbb R^4\to\mathbb R^4$ be any bijection with $p \prec q \iff f(p) \prec f(q)$. Then $f$ is a composition of an orthochronous Lorentz transformation, a translation and a dilation $X \mapsto \lambda X$, $\lambda > 0$.

Zeeman, "Causality implies the Lorentz group", J. Math. Phys. 5 (1964) 490–493; Alexandrov proved the version for light-cone-preserving maps in the 1950s.

What it removes. No linearity, no continuity, no homogeneity: section 1's lemma is not needed. Preserving "could have caused" is already enough to force the group, up to the scale of the units.

needs three or more dimensions

In one space and one time dimension it is false: with $u = x^0 + x$, $w = x^0 - x$, any $(u, w)\mapsto(F(u), H(w))$ with $F, H$ increasing bijections preserves $\prec$. A problem on the sheet.

Why cited. Zeeman's proof is three pages of elementary but intricate geometry: he shows that $f$ maps light rays to light rays (a light ray is characterised order-theoretically: a chain along which no two points are chronologically related), then that it maps null planes to null planes, then that it is affine, and finally that an affine map preserving the cone is a Lorentz transformation times a dilation. The argument is worth reading once and is self-contained; it does not fit in speaker notes.

Why the dilation survives. $X\mapsto\lambda X$ scales every interval by $\lambda^2$ and keeps every sign, so it preserves $\prec$. Causal structure fixes the form only up to a positive multiple: it knows which way the cones point, not how long a second is. Clocks fix the scale.

Why 1+1 dimensions fail. In the null coordinates $u = x^0 + x$, $w = x^0 - x$, $q - p$ is future causal iff $\Delta u \ge 0$ and $\Delta w \ge 0$ (and not both zero): the cone is a quadrant. Any map applying increasing bijections to $u$ and $w$ separately preserves that, and $F(u) = u^3$ is not affine. In higher dimensions there are infinitely many null directions, and they are tied together by the rotations; that rigidity is what Zeeman's proof exploits.

The moral for this course. Special relativity can be founded on causal order alone. General relativity keeps the causal structure locally and drops everything global; unit 40's discussion of horizons is causal structure with the flatness removed.

section 3

Simultaneity, and what follows from it

The actual content of special relativity; dilation and contraction are corollaries.

26 · simultaneity

The relativity of simultaneity

the relativity of simultaneity

Two events with $\Delta t = 0$ and $\Delta x \ne 0$ in $S$ have, in $S'$,

$$\Delta t' = -\gamma\,\frac{v\,\Delta x}{c^2} \ne 0 .$$

In $S'$, the event further along $+x$ happens first.

The lines of constant $t'$ are $t - vx/c^2 = \text{const}$: in an $(x, ct)$ diagram they tilt up with slope $\beta$. They are exactly the vectors $\eta$-orthogonal to the observer's worldline.

$\beta = 0.5$, $c = 1$. $A$ and $B$ share $S$'s horizontal "now". $B$ is on the $x'$ axis, $t' = 0$; the line of constant $t'$ through $A$ (dashed) meets the $ct'$ axis at $t' = \gamma = 1.15$. So $\Delta t'_{B-A} = -\gamma v\Delta x/c^2 = -1.15$.

Proof. $t' = \gamma(t - vx/c^2)$, so $\Delta t' = \gamma(\Delta t - v\Delta x/c^2)$, and $\Delta t = 0$ leaves the formula. Checks: dimensions $[v\Delta x/c^2] = \mathrm{m\,s^{-1}\cdot m}/\mathrm{m^2\,s^{-2}} = \mathrm{s}$ ✓; $c\to\infty$ gives $\Delta t' = 0$, Galilean absolute simultaneity ✓; $\Delta x = 0$ gives $\Delta t' = 0$ — coincident events are simultaneous for everyone, which is why a collision is a frame-independent fact.

Reading the figure. $S'$ moves at $\beta = 0.5$, so its $ct'$ axis is the line $x = 0.5\,ct$ and its $x'$ axis, the set $t' = 0$, is $ct = 0.5\,x$: both tilt towards the light line by the same angle. Every line of constant $t'$ is parallel to the $x'$ axis. $A = (0, 1)$ and $B = (2, 1)$, as $(x, ct)$: $t'_A = \gamma(1 - 0.5\cdot0) = 1.155$, $t'_B = \gamma(1 - 0.5\cdot2) = 0$. The formula gives $\Delta t' = -\gamma\cdot0.5\cdot2 = -1.155$ ✓.

Einstein's train (1917, Relativity): lightning strikes the two ends of a train. The embankment observer, midway, sees the flashes together; the train's midpoint passenger, moving towards the front flash, sees it first and, knowing that light travels at $c$ in her frame and that she is midway, concludes that the front strike happened first. Both are right, each about their own simultaneity.

26 · simultaneity

Einstein synchronisation: simultaneity by radar

Einstein synchronisation

A clock at rest sends light at $t_1$ to a distant event $E$, and gets the reflection back at $t_2$. It defines $E$ to have happened at $t_E = \tfrac12(t_1 + t_2)$, at distance $\tfrac12 c(t_2 - t_1)$.

Apply that definition in each frame and you get each frame's own simultaneity, and they differ. No experiment picks one: the choice is a convention, and the group says all choices are equally good.

For a moving observer, the radar "now" is the tilted line of the previous slide: light out and back at $45°$ makes the triangle symmetric about it.

An observer at $\beta = 0.5$ times the echo from $E$: out at $\tau_1$, back at $\tau_2$, so $E$ is assigned the clock reading $\tfrac12(\tau_1 + \tau_2)$. That event and $E$ lie on a line of slope $\beta$ — the observer's now — not on $S$'s horizontal one.

Reading the figure. The observer's proper times are $\tau_1 = 0.8$, $\tau_2 = 3.2$, so $E$ is assigned $\tau = 2.0$ and distance $1.2$ (with $c = 1$). $E$ was built as $2u + 1.2\,n$, with $u = \gamma(1, \beta)$ the observer's unit time direction and $n = \gamma(\beta, 1)$ the unit spacelike vector $\eta$-orthogonal to it, components $(ct, x)$: radar recovers exactly those two numbers. $S$'s own clocks call $E$ simultaneous with $ct = 3.0$ at $x = 0$, a different "now".

Why radar simultaneity is $\eta$-orthogonality. Let the observer have unit future timelike direction $u$ (so $\eta(u,u) = -1$) through the origin, and let $E$ be an event. The outgoing and returning light rays meet the worldline at $\tau_1 u$ and $\tau_2 u$ (times in units with $c = 1$), with $E - \tau_1u$ and $E - \tau_2u$ null. Expand, using $\eta(u,u) = -1$: $0 = \eta(E - \tau u, E - \tau u) = \eta(E,E) - 2\tau\,\eta(E,u) - \tau^2$, a quadratic in $\tau$ whose two roots are $\tau_1, \tau_2$. Their sum is $-2\eta(E,u)$. So $\tfrac12(\tau_1 + \tau_2) = -\eta(E, u)$, and $E$ is radar-simultaneous with the origin iff $\eta(E, u) = 0$. The problems page asks for this proof in full, and for the radar distance $\sqrt{\eta(E,E)}$ of such an $E$.

Is it only a convention? Reichenbach and Grünbaum argued that $\tfrac12$ could be any $\varepsilon\in(0,1)$ without observable consequence, since only round-trip light speeds are measurable. Malament (1977) proved that Einstein's $\varepsilon = \tfrac12$ is the only simultaneity relation definable from the causal structure and the observer's worldline alone. Either way, what is not a convention is that two frames' Einstein simultaneities differ.

26 · simultaneity

Time dilation and length contraction, as corollaries

time dilation

A clock at rest in $S'$ ticks at $\Delta x' = 0$. By the inverse boost, $\Delta t = \gamma(\Delta t' + v\Delta x'/c^2) = \gamma\,\Delta t'$: the moving clock's ticks are $\gamma$ times further apart in $S$.

evidence

Muons in the CERN storage ring at $\gamma = 29.3$: rest lifetime $2.197\ \mu$s, so $\gamma\tau = 64.4\ \mu$s predicted and seen to about $0.1\%$ (Bailey et al., Nature 268, 1977).

length contraction

A rod at rest in $S'$ has ends at $x' = 0$ and $x' = L_0$. Its length in $S$ is the separation of its ends at one $S$-instant. $x' = \gamma(x - vt)$ at fixed $t$ gives $\Delta x = L_0/\gamma$.

Both follow in one line from the boost. What they mean is clearer drawn, once diagrams are drawn honestly: "Length contraction is a slice of a strip", two slides on, cuts a rod's history two ways.

Checks. Dimensions: $\gamma$ is a pure number ✓. $v\to0$: $\gamma\to1$, no effect ✓. $v\to c$: $\gamma\to\infty$, the moving clock stops and the rod flattens — neither is reached, since $|v| < c$ for every frame.

The symmetry that looks like a contradiction. $S$ says $S'$'s clocks run slow; $S'$ says $S$'s clocks run slow. There is no contradiction because each statement compares one clock of one frame with a succession of clocks of the other, synchronised by the other's convention. The two comparisons are different experiments. The twins break the symmetry by bringing the same two clocks together twice: then a frame-independent comparison exists, and section 5 computes it.

The numbers. The muon lifetime is $2.1969811\ \mu$s. $29.3\times2.19698 = 64.37\ \mu$s. The ring had radius about $7$ m, so the muons' acceleration in the lab was $\gamma^2v^2/r \approx 1.1\times10^{19}\ \mathrm{m\,s^{-2}}$, about $10^{18}\,g$. That the lifetime still came out as $\gamma\tau$ is the evidence for the clock hypothesis of section 5: acceleration itself does not affect an ideal clock.

Contraction does not mean looks contracted. What an eye or camera records is light that left the object's parts at different times, arriving together. Penrose and Terrell (both 1959) showed a moving sphere still photographs as a circular outline: contraction plus light-travel delays combine into an apparent rotation. Length contraction is a statement about simultaneous positions, not about images.

26 · simultaneity

Minkowski diagrams that do not lie

$\beta = 0.6$. Primed axes (thick) and the calibration hyperbolae $-(x^0)^2 + x^2 = \mp1$ (curves), which cross each axis at its unit tick: the primed ticks $1'$ sit further out on paper than the unprimed $1$.

a Minkowski diagram, drawn right

The $ct'$ axis is the worldline $x = \beta\,ct$; the $x'$ axis is the simultaneity line $ct = \beta x$. Both lean towards the light line by the same angle.

the ticks are not where a ruler says

The paper is Euclidean; the form is not. The unit tick on the $ct'$ axis is at $(x, ct) = (\gamma\beta, \gamma)$, a Euclidean distance $\sqrt{(1+\beta^2)/(1-\beta^2)}$ from the origin: $1.46$ at $\beta = 0.6$, $2.13$ at $\beta = 0.8$. Find it where the hyperbola crosses, not with a ruler.

Why the hyperbola. The unit event on the $ct'$ axis has $S'$-coordinates $(ct', x') = (1, 0)$ and hence interval $-1$. The interval is invariant, so in $S$ it lies on $-(ct)^2 + x^2 = -1$, which also passes through $S$'s own unit tick $(0, 1)$. The hyperbola is the set of all events at unit proper time from the origin, in the future: the Lorentzian "unit circle". Likewise $-(ct)^2 + x^2 = +1$ calibrates both space axes.

Where the tick is. The boost with rapidity $\varphi$ carries $(0,1)$ — as $(x, ct)$ — to $(\sinh\varphi, \cosh\varphi) = (\gamma\beta, \gamma)$; at $\beta = 0.6$, $\gamma = 1.25$, that is $(0.75, 1.25)$. Its Euclidean distance from the origin is $\gamma\sqrt{1+\beta^2} = \sqrt{(1+\beta^2)/(1-\beta^2)}$: $\sqrt{1.36/0.64} = 1.458$ at $\beta = 0.6$ and $\sqrt{1.64/0.36} = 2.134$ at $\beta = 0.8$. A diagram with both axes ticked at the same ruler spacing is simply wrong, and most "paradoxes" drawn with such diagrams are artefacts of it.

Reading the diagram. The hyperbola is tangent to the $x'$ axis direction where it meets the $ct'$ axis: tangent lines to the calibration hyperbola are lines of simultaneity for the observer through the point of tangency. That is the picture of "$\eta$-orthogonal to the worldline": differentiate $\eta(X,X) = -1$ along the hyperbola to get $\eta(X, \dot X) = 0$.

Time dilation, read off. The $S$-simultaneity line through the $1'$ tick on the $ct'$ axis is horizontal at $ct = \gamma = 1.25$: the moving clock reads $1$ when $S$'s clocks read $1.25$. The $S'$-simultaneity line through $S$'s tick $(0, 1)$ — an event with $ct' = \gamma(1 - 0) = 1.25$ — hits the $ct'$ axis above $1'$, at $ct' = 1.25$: $S$'s clock reads $1$ when $S'$ says $1.25$ has passed, the same effect seen from $S'$. Both at once, with no contradiction, because the two lines are different lines.

26 · simultaneity

Length contraction is a slice of a strip

A rod of rest length $L_0 = 2$ moving at $\beta = 0.6$ ($\gamma = 1.25$), $c = 1$. Its history is the shaded strip. $S$'s "now" cuts it in $L_0/\gamma = 1.6$; the rod's own "now", of slope $\beta$, cuts it in a segment whose end lies on the calibration hyperbola $x^2 - (ct)^2 = 4$: proper length $2$.

both are about simultaneity

Dilation compares one clock with two of the other frame's, which must be synchronised; contraction measures two ends at one instant. Change the simultaneity and both change. Neither is a mechanism acting on clocks or rods.

There is one rod and one strip. "Its length" is the length of a slice, and which slice depends on whose "now" you cut with. Neither slice is the rod's true length; $L_0$ is the one cut by the rod's own simultaneity.

The proof of contraction, with the inverse direction. The ends' worldlines are $x = vt$ and $x = vt + L_0/\gamma$, since $x' = L_0$ means $\gamma(x - vt) = L_0$. At any fixed $t$ they are $L_0/\gamma$ apart.

Reading the figure. With $c = 1$: the rear is $x = 0.6\,t$, the front $x = 0.6\,t + 1.6$. The horizontal slice at $t = 0$ runs from $0$ to $1.6$. The rod's own simultaneity through the rear at the origin is the $x'$ axis, $t = 0.6\,x$; it meets the front where $x = 0.36\,x + 1.6$, i.e. at $(x, t) = (2.5, 1.5)$. That segment has interval $-(1.5)^2 + 2.5^2 = 4$, length $2 = L_0$, which is why its end sits on the hyperbola through $(2, 0)$: on paper it looks longer still ($\sqrt{8.5} = 2.92$), the same Euclidean distortion as the tick marks of the previous slide.

26 · simultaneity · example

The twins on the diagram: simultaneity jumps at the turn

Earth frame, years and light-years. Red dashed: the traveller's simultaneity through $T$, before and after the turn; grey: the same at each traveller year, meeting Earth at $0.6$-year steps.

where "now at home" goes

1. Outbound simultaneity through $T = (5, 4)$ has slope $\beta = 0.8$: $t - 5 = 0.8(x - 4)$. At $x = 0$: $t = 5 - 3.2 = 1.8$.

2. Inbound, $\beta = -0.8$: $t - 5 = -0.8(x - 4)$. At $x = 0$: $t = 5 + 3.2 = 8.2$.

3. Outbound, the traveller's "Earth now" advances from $0$ to $1.8$ in $3$ years of the traveller's clock: rate $0.6 = 1/\gamma$, dilation ✓. Inbound, $8.2 \to 10$ in $3$ years: again $0.6$.

4. At the turn it jumps from $1.8$ to $8.2$: $6.4$ years. Total $1.8 + 6.4 + 1.8 = 10$ ✓.

What the jump is. Nothing happens on Earth at the turn. The traveller changes frame, and with it the convention for which Earth events count as "now". Each leg, taken alone, is a perfectly good inertial frame in which Earth's clock runs slow; the $6.4$ years are the Earth events that neither leg's simultaneity ever labels as "now at home" during the trip. That is the third way of seeing the age difference, and it agrees with the others: $10 - 6 = 4$ years of difference, with the traveller's $6$ matched by $1.8 + 1.8 = 3.6$ Earth years seen "live" and $6.4$ swept over at the turn.

A frame-independent statement. The Earth events in the jump, $1.8 < t < 8.2$, are all spacelike to $T$: from $T = (5, 4)$ the Earth event $(t, 0)$ has interval $-(t-5)^2 + 16 > 0$ iff $|t - 5| < 4$, i.e. $1 < t < 9$. The jump lies inside that window, as it must: section 4 proves that only spacelike-separated events can have their time order changed, and these are exactly the ones that swap from "after $T$" to "before $T$".

Checks. Step 3's rate: the outbound frame's Earth clock reading at traveller time $\tau$ is $0.6\tau$, and $0.6\times3 = 1.8$ ✓. Limit: for a star at distance $L$ the two lines meet Earth at $L/v \mp \beta L/c$, so the jump is $2\beta L/c$ ($6.4$ yr here), which vanishes as $\beta\to0$, as does the age difference $2(L/v)(1 - 1/\gamma) \approx \beta L/c$ ✓.

Do not over-read the jump. An accelerated observer has no canonical simultaneity at all; the dashed lines are the simultaneities of two inertial frames that happen to share an event with the traveller. With a turnaround of finite duration the lines sweep over the interval continuously instead of jumping; nothing physical depends on how. Unit 30 makes this precise with Rindler coordinates.

26 · simultaneity

The pole and the barn

the paradox

A pole $20$ m long (rest length) runs at $\gamma = 2$ into a barn $10$ m long with a door at each end. In the barn's frame the pole is $10$ m: both doors can be shut at once with the pole inside. In the pole's frame the barn is $5$ m: the pole can never be inside. Which is it?

Both pictures are correct contraction arithmetic. The trouble is the word "inside", which means "inside at one instant" — and the two frames do not share instants.

The two frames' pictures, to one scale. Each is right about lengths; each "now" is a different set of events.

26 · simultaneity

The pole and the barn: when the doors shut

Barn frame, metres. Both door events (dots) lie on the barn's now, $ct = 0$, with the pole's strip between them. The pole's now through the back-door event (dashed, slope $\beta = 0.866$) meets the front door $8.66$ m of light-time before it shuts.

the resolution, in one line

"Both doors shut at once" is a statement about two events $10$ m apart with $\Delta t = 0$ — a spacelike pair. In the pole's frame they are not simultaneous: the back door shuts first, by $\gamma\beta\cdot10\ \mathrm{m}/c = 57.8$ ns.

No contradiction: the frames disagree about when, and "the pole is inside" means "inside at one instant". Nothing frame-independent — a collision, a broken pole — differs. The problems page writes the events out and transforms them.

The number. $\gamma = 2$ gives $\beta = \sqrt{1 - 1/4} = \sqrt3/2 = 0.866$. The two door events are $\Delta x = 10$ m apart with $\Delta t = 0$ in the barn frame, so in the pole frame $|\Delta t'| = \gamma\beta\,\Delta x/c = 2\cdot0.866\cdot10\ \mathrm m/c = 17.32\ \mathrm m/c = 57.8$ ns.

Reading the figure. The pole's rear is $x = \beta\,ct$ and its front $x = 10 + \beta\,ct$, in metres, so at $ct = 0$ it fills the barn exactly. The pole's line of simultaneity through the back-door event $(x, ct) = (10, 0)$ is $ct = \beta(x - 10)$; at the front door, $x = 0$, it gives $ct = -8.66$ m. In the pole frame, then, the front door shuts $\gamma\cdot 8.66 = 17.32$ m of light-time ($57.8$ ns) after the back door, the same number as the boost formula gives.

The consistency check from the pole's side. In the pole frame the barn ($5$ m) moves at $0.866c$ towards the pole ($20$ m). When the back door shuts, the pole's front is at it; the front door then still has $20 - 5 = 15$ m to travel to reach the pole's rear, taking $15/0.866 = 17.32$ m of light-travel time: the same $57.8$ ns. So in the pole frame the back door shuts, and $57.8$ ns later the front door shuts behind the pole's rear — the pole is never entirely inside at one instant of its own frame, and yet both doors do shut with no part of the pole in their way. Both descriptions contain the same events.

If the back door stays shut. Then the pole's front hits it, and the pole stops or breaks. The collision is an event, frame-independent. The news that the front has stopped travels back along the pole no faster than $c$, so the rear keeps moving and passes the front door, which shuts behind it, in every frame; the pole ends up compressed inside the barn. "Rigid pole" is the hidden false premise: unit 27 shows no body can be rigid in relativity.

26 · simultaneity

Doppler: the rapidity, exponentiated

collinear Doppler factor

A source recedes at $\beta$ and emits pulses a proper time $T_0$ apart. The receiver's pulses are $T$ apart, with

$$\frac{T}{T_0} = \sqrt{\frac{1+\beta}{1-\beta}} = e^{\varphi}.$$

Approaching: $e^{-\varphi}$. Frequencies go the other way.

Proof. Receiver at the origin; source on $x = vt$, emitting at $t = 0$ and $t = \gamma T_0$. The second pulse starts $v\gamma T_0$ further away, so arrives $\gamma T_0(1 + \beta)$ after the first. $\gamma(1+\beta) = \sqrt{(1+\beta)/(1-\beta)}$.

Check. $\beta\to0$: $e^\varphi \approx 1 + \beta$, the classical Doppler shift ✓. $\beta\to1$: infinite stretch, redshift to nothing ✓.

The proof, drawn for the traveller: $\beta = 0.8$, $T_0 = 1$ yr, $c = 1$. Ticks at $t = \gamma T_0 = 5/3$ and $10/3$; each extra $\beta\gamma T_0$ of distance adds that much light-time. Arrivals at $3$ and $6$: $T = \gamma(1 + \beta)T_0 = 3T_0$.

The proof, in full. Work in the receiver's frame, receiver at $x = 0$, source on $x = vt$, $v > 0$ (receding). The source's clock runs slow by $\gamma$, so pulses separated by proper time $T_0$ are emitted at $t = 0$ (at $x = 0$) and $t = \gamma T_0$ (at $x = v\gamma T_0$). The second pulse needs an extra $v\gamma T_0/c$ to reach the receiver, so the arrival times are $0$ and $\gamma T_0 + \gamma\beta T_0$. Hence $T = \gamma(1 + \beta)T_0 = \frac{1 + \beta}{\sqrt{(1-\beta)(1+\beta)}}T_0 = \sqrt{\frac{1+\beta}{1-\beta}}\,T_0$. And $e^{2\varphi} = (1+\beta)/(1-\beta)$ from the rapidity slide.

Oblique direction. If the source moves at $\beta$ and the line from source to receiver makes angle $\theta$ with its velocity (in the receiver's frame), the same argument gives $T = \gamma(1 - \beta\cos\theta)T_0$: between emissions the source closes the distance by $v\gamma T_0\cos\theta$ to first order in the ratio of that step to the distance, which is negligible for a distant source. At $\theta = \pi/2$: $T = \gamma T_0$, the transverse Doppler effect, which is pure time dilation and has no classical counterpart. Ives and Stilwell (1938) saw it in moving hydrogen ions (unit 25).

26 · simultaneity

Doppler factors multiply

rapidities add, Doppler factors multiply

Relay a signal through a chain of observers each receding from the last: the factors multiply, since $e^{\varphi}e^{\psi} = e^{\varphi+\psi}$. A second proof of velocity addition, needing only light rays.

For the traveller, $e^{\ln 3} = 3$. Outbound, one pulse a year arrives every $3$ years; inbound, every $4$ months.

The factor $e^{\varphi}$ is the one "Boosts are hyperbolic rotations" found on the light lines: a boost stretches one null direction by exactly the Doppler factor.

The traveller's $0.8c$ as two $0.5c$ steps: $B$ recedes from $A$ at $0.5c$, $C$ from $B$ at $0.5c$. Light from $C$'s yearly ticks passes $B$ every $\sqrt3$ years of $B$'s clock and reaches $A$ every $3$: $\sqrt3\cdot\sqrt3 = 3$.

Reading the figure. One light ray serves the whole chain, since everything moves at $c$ in every frame. $C$'s tick at $\tau = 1$ is the $A$-frame event $(t, x) = (5/3, 4/3)$; its light runs back along $x = 3 - t$ and crosses $B$'s worldline $x = t/2$ at $t = 2$, where $B$'s clock reads $2\sqrt{1 - 0.25} = \sqrt3$; it reaches $A$ at $t = 3$. Each $0.5c$ step has Doppler factor $\sqrt{1.5/0.5} = \sqrt3$, rapidity $\tfrac12\ln3$, and the product of the two is the traveller's $3$, rapidity $\ln 3$.

Bondi's k-calculus. Hermann Bondi built all of special relativity from the factor $k = e^{\varphi}$ and light signals (Relativity and Common Sense, 1964). The twin paradox by counting pulses, on the two slides that follow, is his.

The traveller's numbers. $\sqrt{1.8/0.2} = \sqrt9 = 3$. The traveller sends one pulse per year of their own clock; on the way out Earth receives them every $3$ years, on the way back every $1/3$ year.

26 · simultaneity · example

The twins by radio: when does the shift flip?

what the traveller receives

Earth sends a pulse each Earth year. Outbound ($3$ yr of the traveller's clock) they arrive $3$ yr apart: $1$ pulse. Inbound ($3$ yr) they arrive every $\tfrac13$ yr: $9$ pulses.

Total $10$: Earth aged $10$ years. The flip comes at the traveller's turn, half-way through the traveller's trip.

The flip is on the traveller's own worldline, at the turn: everything sent after Earth's year $1$ is met on the way back.

Earth frame. Earth's yearly pulses (orange, at $45°$): only year $1$'s reaches the traveller before $T$.

26 · simultaneity · example

The twins by radio: what Earth receives

what Earth receives

The traveller sends a pulse each of the traveller's years. Light from the turn at $(5, 4)$ reaches Earth at $t = 5 + 4 = 9$. For $9$ years, pulses arrive $3$ yr apart: $3$ pulses. In the last year, every $\tfrac13$ yr: $3$ pulses.

Total $6$: the traveller aged $6$ years. The flip comes at year $9$ of $10$.

the asymmetry, visible

Both twins see the other's clock slow by $3$ while separating and fast by $3$ while approaching. The traveller spends half the trip in each; Earth spends $9$ years seeing slow and $1$ seeing fast, because it learns of the turn only when its light arrives. No acceleration appears in the count.

Earth frame. The traveller's yearly pulses: three outbound ones reach Earth at years $3$, $6$, $9$; the light of the turn (red) arrives at year $9$; three inbound ones crowd into the last year.

Reading the figure. The red ray is the boundary: everything the traveller sends before the turn arrives by year $9$, everything after it arrives later. Earth's worldline crosses $T$'s future light cone at year $9$, and that crossing, not the turn itself, is when Earth can see the change.

Earth's side. The traveller's pulse at traveller time $\tau$ on the outbound leg is emitted at Earth-frame $(t, x) = (\tfrac53\tau, \tfrac43\tau)$ and arrives at $t = \tfrac53\tau + \tfrac43\tau = 3\tau$: one pulse every $3$ Earth years, up to $\tau = 3$, arriving at $t = 9$. Inbound, the pulse at $\tau = 3 + s$ is emitted at $(5 + \tfrac53 s,\ 4 - \tfrac43 s)$ and arrives at $9 + \tfrac13 s$: from $t = 9$ to $t = 10$, three pulses.

Why this is the cleanest resolution. The counting uses only the Doppler factor, which is symmetric between the twins — each sees the other redshifted by the same $3$ while separating. The asymmetry is where the flip happens: for the traveller it is a local event on their own worldline; for Earth it is where Earth's worldline meets the future light cone of $T$, at year $9$. Acceleration is not mentioned, and cannot matter, since the counts do not depend on how long the turn takes provided it is short.

Check. $1 + 9 = 10$ and $3 + 3 = 6$ agree with the proper times of section 5, and each twin's received count equals the other's elapsed age, as it must: every pulse sent is received, the moment the two are together again.

26 · simultaneity

Aberration and the headlight effect

aberration

A source moving at $\beta$ along $x$ emits light at angle $\theta'$ to $x$ in its own frame. In the lab the angle is $\theta$ with

$$\cos\theta = \frac{\cos\theta' + \beta}{1 + \beta\cos\theta'}.$$

Proof. The ray's direction is a null vector, $(1, \cos\theta', \sin\theta', 0)$ in the source frame. Boost by $-\beta$: $(\gamma(1 + \beta\cos\theta'),\ \gamma(\cos\theta' + \beta),\ \sin\theta',\ 0)$. It is still null, and $\cos\theta$ is its $x$ part over its time part.

Stellar aberration. The Earth's orbital $\beta \approx 29.8/299\,792 = 9.9\times10^{-5}$ shifts starlight by up to $\beta$ radians: $20.5''$, which Bradley measured in 1728.

Check. $\beta = 0$: $\theta = \theta'$ ✓. $\theta' = 0$: $\theta = 0$ ✓.

What the formula does to a whole sky of rays is the headlight effect, drawn on the next slide.

Why the boost is by $-\beta$. The lab sees the source moving at $+\beta$, so the lab's coordinates are obtained from the source's by the inverse boost, $x^0 = \gamma(x^{0\prime} + \beta x')$, $x = \gamma(x' + \beta x^{0\prime})$. Applied to $(1, \cos\theta', \sin\theta', 0)$ this gives the vector on the slide. Null-ness is preserved because $\eta$ is: check, $\gamma^2(\cos\theta' + \beta)^2 + \sin^2\theta' - \gamma^2(1 + \beta\cos\theta')^2 = \gamma^2(1 - \beta^2)(\cos^2\theta' - 1) + \sin^2\theta' = 0$ ✓.

The Doppler factor is in the same vector. Scaling the null direction by the frequency $\omega'$, its time component in the lab is $\gamma(1 + \beta\cos\theta')\omega'$: the observed frequency. It agrees with the oblique Doppler formula in the notes of "Doppler: the rapidity, exponentiated", $\omega = \omega'/(\gamma(1 - \beta\cos\theta))$, once the aberration formula is used to convert $\theta'$ to $\theta$ — a two-line identity the problems page asks for.

Stellar aberration. For a star at angle $\theta'$ in the Sun's frame, $\theta - \theta' \approx -\beta\sin\theta'$ to first order (a sign that depends on which way the observer, rather than the source, moves); the amplitude is $\beta$ radians. With the mean orbital speed $29.78\ \mathrm{km\,s^{-1}}$: $\beta = 9.934\times10^{-5}$ rad $= 20.49''$. Bradley found it while looking for stellar parallax, announced in 1729. First order in $\beta$ is also what the ether theories could reproduce; unit 25 describes why they could not reproduce it together with Fizeau's experiment.

26 · simultaneity

The headlight cone, drawn

Thirteen rays every $15°$, emitted isotropically in the source frame (left), seen in the lab with the source moving right at $\beta = 0.8$ (right). The forward seven (blue), which filled $90°$, are squeezed into $\theta \le \arccos 0.8 = 36.9°$.

the headlight effect

Emission that is isotropic in the source frame puts half its light ($\theta' < 90°$) into the lab cone $\cos\theta > \beta$, i.e. $\sin\theta < 1/\gamma$. At $\gamma = 10$: a cone of half-angle $5.74°$. Synchrotron light is beamed for this reason.

Each ray keeps its order: aberration is a monotone map of $[0°, 180°]$ onto itself. It only redistributes the rays, crowding the forward half into a cone that shrinks like $1/\gamma$.

section 4

Causal structure

Which parts of "before" survive a change of frame, and why nothing outruns light.

26 · causality

What survives of "before"

theorem

(a) If $q - p$ is timelike or null, every orthochronous Lorentz transformation preserves the sign of $\Delta t$.

(b) If $q - p$ is spacelike, there are inertial frames in which $\Delta t' > 0$, $= 0$ and $< 0$.

proof

the causal future and the elsewhere

$J^+(p) = \{q : q - p \text{ future causal or } 0\}$, and $J^-(p)$ likewise. The elsewhere of $p$ is everything outside $J^+(p)\cup J^-(p)$: events whose order relative to $p$ is a matter of frame.

$q$ is spacelike to $p$. Through $p$, three frames' lines of simultaneity put $q$ after $p$, level with it, or before it. No such line can pass above an event in $J^+(p)$.

Proof of (a), with null separation. For timelike $q - p$ this is the definition of orthochronous (the future timelike cone maps to itself). A future null vector is a limit of future timelike vectors; its image is a limit of future timelike vectors, so it has $\Delta t' \ge 0$, and $\Delta t' = 0$ would make it (being null) the zero vector, which it is not.

Proof of (b), in full. Rotate so that $\Delta\vec x = (\Delta x, 0, 0)$ with $\Delta x > 0$; spacelike means $\Delta x > c|\Delta t|$. The boost with $v = c^2\Delta t/\Delta x$ has $|v| < c$ and gives $\Delta t' = 0$; any larger $v < c$ gives $\Delta t' < 0$, and any smaller $v > -c$ gives $\Delta t' > 0$. The frame in which the two events are simultaneous is unique up to transverse motion, and in it their distance is the proper distance $\sqrt{\eta(q - p, q - p)}$.

Reading the figure. With $c = 1$, $p$ at the origin and $q = (x, ct) = (2, 0.6)$, spacelike since $2 > 0.6$. The frame moving at $v$ has simultaneity lines $ct = (v/c)\,x$ through $p$: for $v = -0.5c$ the line is below $q$ ($\Delta t' > 0$), for $v = c^2\Delta t/\Delta x = 0.3c$ it passes through $q$, for $v = 0.7c$ it is above ($\Delta t' < 0$). For an event inside the shaded cones every such line, having slope between $-1$ and $1$, leaves it on the same side: part (a).

Chronological versus causal. Some books write $I^+(p)$ for the open cone (timelike only) and $J^+(p)$ for the closed one; in Minkowski space $J^+(p)$ is the closure of $I^+(p)$. The distinction matters in unit 40, where in curved spacetimes the two can behave differently.

26 · causality · example

The turn's elsewhere, on Earth's worldline

$T$ and Earth

Earth-at-$t{=}5$ is spacelike to $T$. Outbound frame: $\Delta t' = \tfrac53(0 - 0.8\cdot(-4)) = +5.33$ yr, after $T$. Inbound frame: $\tfrac53(0 + 0.8\cdot(-4)) = -5.33$, before. The elsewhere of $T$ on Earth is $1 < t < 9$ — the Doppler flip times.

The traveller's two frames disagree about the order of $T$ and Earth-at-year-$5$. That is the $6.4$-year jump of "The twins on the diagram", seen as an ordering — and the window's ends are exactly the radio flips.

Earth frame. $J^\pm(T)$ shaded; Earth between years $1$ and $9$ (orange) is $T$'s elsewhere. Earth-at-$5$ is above the outbound "now" through $T$ (to $1.8$) and below the inbound one (to $8.2$).

26 · causality

No superluminal signalling

theorem

Suppose signals could be sent at some speed $u > c$, in every inertial frame (relativity). Then some sequence of two such signals arrives at its sender's worldline before it was sent.

proof

Emission $p$ at the origin, reception $q = (\Delta t, u\Delta t)$: spacelike. In the frame with $c^2/u < v < c$, $\Delta t' = \gamma\Delta t(1 - uv/c^2) < 0$: the signal runs backwards in time. The receiver, at rest in that frame, replies with its own speed-$u$ signal backwards to the sender, arriving in the sender's past (details in the notes).

$c = 1$, $u = 2$, Bob at $v = 0.9$. Bob's "now" through $q$ passes below $p$, so for him the question arrives before it is sent. His reply (red), at $u$ in his frame, reaches Alice at $t = -0.45$.

Full proof: the tachyonic antitelephone

Units $c = 1$. Alice at rest at $x = 0$ sends a signal at speed $u > 1$ at $t = 0$, received by Bob at $q = (t_1, ut_1)$, $t_1 > 0$. Let Bob move at $v$ with $1/u < v < 1$. In Bob's frame $t'_q = \gamma t_1(1 - uv) < 0 = t'_p$: the signal arrived before it left. Bob replies at once, with a speed-$u$ signal aimed back along $-x$ in his frame. By the principle of relativity this is as allowed for Bob as Alice's was for her. In Bob's frame it travels forward in his time, from $q$ to Alice's worldline. Where does it land? Its Alice-frame velocity is, by velocity addition, $w = (-u + v)/(1 - uv)$; since $1 - uv < 0$ and $v - u < 0$, $w > 0$ — in Alice's frame the reply moves in the $+x$ direction while going backward in $t$, i.e. it travels from $q$ towards $x = 0$ with $\mathrm dt/\mathrm dx = 1/w$ and decreasing $x$ means decreasing $t$. It covers the distance $ut_1$ in Alice-time $-ut_1/w$, arriving at $t_1 - ut_1/w = t_1(1 - u/w)$. Now $w < u$ exactly when $v > 2u/(1 + u^2)$, a number between $1/u$ and $1$ (strictly less than $1$ because $2u < 1 + u^2$); for such $v$, $u/w > 1$ and the reply arrives at negative $t$: before Alice sent the question. That is the loop.

The figure's numbers. $u = 2$, $t_1 = 1$, so $q = (t, x) = (1, 2)$. Bob moves at $v = 0.9$, above the loop threshold $2u/(1 + u^2) = 0.8$. His line of simultaneity through $q$ is $t - 1 = 0.9(x - 2)$, which meets Alice's worldline at $t = -0.8$, below $p$: in his frame $p$ comes after $q$. The reply's Alice-frame velocity is $w = (-2 + 0.9)/(1 - 1.8) = 1.375$, and it arrives at $t = 1 - 2/1.375 = -0.455$.

26 · causality

What the antitelephone proves, and what it does not

what exactly is proved

One backwards-in-time signal is odd but not contradictory. The contradiction is the loop: a reply that reaches the sender before the question left, who then does not send it. Relativity plus "no causal loops" forbids signals faster than $c$. Neither alone does.

why it fails for timelike separation

If $u < c$, then $uv/c^2 < 1$ for every $|v| < c$, so $\Delta t'$ has the sign of $\Delta t$ in every frame — part (a) of the theorem on "What survives of 'before'".

Put together: if relativity holds, and nothing can arrive at its own past to stop itself being sent, then nothing travels faster than $c$. The light cone is the boundary of every possible influence.

section 5

Proper time, and the twin paradox as a theorem

The length of a timelike curve, and the inequality that makes straight lines the longest.

26 · proper time

Proper time: the length of a worldline

proper time

For a future-directed timelike curve $\lambda\mapsto X(\lambda)$, piecewise $C^1$, $\lambda\in[a, b]$:

$$\tau[X] = \frac1c\int_a^b\sqrt{-\eta_{\mu\nu}\,\dot X^\mu\dot X^\nu}\ \mathrm d\lambda .$$

Parametrised by the frame's $t$, with $\vec x(t)$: $\ \tau = \displaystyle\int\sqrt{1 - |\dot{\vec x}|^2/c^2}\ \mathrm dt$.

It is invariant twice over: under reparametrisation (as unit 03's arc length is), and under Lorentz transformations (the integrand is built from $\eta$).

the clock hypothesis (physics, not mathematics)

An ideal clock carried along a worldline reads its proper time, whatever its acceleration. The storage-ring muons, at about $10^{18}\,g$, obeyed it.

Check. $[\eta\,\dot X\dot X] = \mathrm m^2$ per unit of $\lambda$ squared, so the integral is a length; over $c$, a time ✓. At rest, $\tau = \Delta t$ ✓. Constant speed, $\tau = \Delta t/\gamma$: time dilation ✓.

Why timelike. $-\eta(\dot X, \dot X) > 0$ exactly when the tangent is timelike, so the square root is real. Future-directed timelike curves are the possible worldlines of massive particles: at every point the velocity is below $c$.

Reparametrisation invariance. If $\lambda = \phi(\sigma)$ with $\phi' > 0$, then $\mathrm dX/\mathrm d\sigma = \phi'(\sigma)\,\mathrm dX/\mathrm d\lambda$, the square root scales by $\phi'$, and the substitution rule absorbs it: $\int\sqrt{-\eta(X_\sigma, X_\sigma)}\,\mathrm d\sigma = \int\sqrt{-\eta(X_\lambda, X_\lambda)}\,\phi'\,\mathrm d\sigma = \int\sqrt{-\eta(X_\lambda, X_\lambda)}\,\mathrm d\lambda$. The same computation as unit 03's arc length, with $-\eta$ in place of the dot product.

The $t$-parametrisation. Along a future timelike curve $\mathrm dt/\mathrm d\lambda > 0$, so $t$ is a legitimate parameter. With $X = (ct, \vec x(t))$: $-\eta(\dot X, \dot X) = c^2 - |\dot{\vec x}|^2$, and dividing by $c$ gives the slide's formula.

The clock hypothesis is an assumption. Nothing in the Lorentz group says how a clock behaves when accelerated; that is a statement about clocks. It holds for clocks whose internal time scales are short compared with the time over which their velocity changes appreciably — atoms, muons — and fails for clocks that the acceleration damages (a pendulum clock on a centrifuge). Its evidence is the storage ring: proper acceleration $\gamma^2v^2/r$ at $\gamma = 29.3$, $r \approx 7$ m is $1.1\times10^{19}\ \mathrm{m\,s^{-2}} \approx 1.1\times10^{18}\,g$, and the lifetime was still $\gamma\tau$.

Unit 27 will take $-mc^2\tau$ as the action of a free particle, and the Euler–Lagrange equations of unit 11 applied to it give relativistic momentum. The next two slides say why that action has a maximum, not a minimum.

26 · proper time · example

The twins by proper time

integrating along each worldline

Earth. $\vec x(t) = 0$ for $0 \le t \le 10$: $\tau_E = \int_0^{10} 1\,\mathrm dt = 10$ yr.

Traveller, out. $\dot x = 0.8$ on $[0, 5]$: $\int_0^5\sqrt{1 - 0.64}\,\mathrm dt = 5\times0.6 = 3$ yr.

Traveller, back. $\dot x = -0.8$ on $[5, 10]$: again $3$ yr.

Total. $\tau_T = 6$ yr. The traveller is $4$ years younger.

As vectors. Out $X = (5, 4)$, back $Y = (5, -4)$: $\sqrt{25 - 16} = 3$ each. Straight $X + Y = (10, 0)$: $10$.

Earth frame. Dots: one-year ticks of each twin's own clock. Ten on the straight worldline, six on the bent one, though its ticks look further apart on paper ($5/3$ Earth years each).

The integrals. $\sqrt{1 - 0.8^2} = \sqrt{0.36} = 0.6$ exactly, so each leg is $0.6\times5 = 3$ years; the Earth leg is $10$. In the vector form, $\sqrt{-\eta(X, X)} = \sqrt{25 - 16} = 3$ with $c = 1$, the same computation in a different notation.

Reading the figure. The traveller's tick $n$ on the way out is at $(t, x) = (\tfrac53 n, \tfrac43 n)$, interval $-\tfrac{25}{9}n^2 + \tfrac{16}{9}n^2 = -n^2$ from the start: proper time $n$, as the clock hypothesis requires. The paper is Euclidean, so each traveller tick looks $\sqrt{(25 + 16)/9} = 2.13$ years long; in $\eta$ it is one year.

Three ways, one number. Proper time: $10$ and $6$. Radio counts: $10$ and $6$ — each twin's received count is the other's age. Simultaneity: $1.8 + 1.8$ years of Earth time seen "live" plus the $6.4$-year jump make $10$, against the traveller's $6$. The three computations use different structures — the length of a curve, light cones, and hyperplanes orthogonal to a worldline — and agree because all three are built from the one form $\eta$.

26 · proper time · example

It is not the acceleration

"it is because the traveller accelerates"

Replace the instant turn by one lasting Earth-time $\delta$. Its contribution to $\tau_T$ is at most $\delta$, since the integrand is at most $1$. As $\delta\to0$ it vanishes, and the $4$-year difference does not. Acceleration marks which twin's worldline is bent; it is not where the time goes.

The age difference is accumulated during the coasting legs, uniformly, because the bent path is shorter in the sense of $\eta$. The next slide proves that bent is always shorter.

The instant turn replaced by one lasting $\delta = 1$ or $2$ Earth years, speed going linearly from $+0.8c$ to $-0.8c$. The traveller's proper time grows only by $0.28\,\delta$; as $\delta\to0$ it returns to $6$, and the $4$-year gap stays.

The warning, quantitatively. A smooth turnaround in which the speed goes from $+0.8$ to $-0.8$ linearly in Earth-time over $[5 - \delta/2,\ 5 + \delta/2]$ contributes $\int\sqrt{1 - \beta(t)^2}\,\mathrm dt = \frac{\delta}{1.6}\int_{-0.8}^{0.8}\sqrt{1 - \beta^2}\,\mathrm d\beta = \frac{\delta}{1.6}\,(0.8\cdot0.6 + \arcsin0.8) = 0.880\,\delta$, compared with the $\delta$ that Earth ages meanwhile. (The trip's turning point moves a little, which changes the coasting legs by $O(\delta)$ too.) Every effect of the turn is $O(\delta)$; the age difference is $4$ years at $\delta = 0$. The proper acceleration during the turn is of order $c\cdot1.6/\delta$, unbounded as $\delta\to0$ — large acceleration, negligible time.

The figure's numbers. Coasting legs: $2\times0.6\times(5 - \delta/2) = 6 - 0.6\,\delta$. Turn: $0.880\,\delta$. Total $6 + 0.28\,\delta$: $6.28$ at $\delta = 1$, $6.56$ at $\delta = 2$. The far point moves in from $x = 4$ to $4 - 0.2\,\delta$. A long turn makes the traveller slightly older, because the slow part of the turn is spent near rest; after it, each curve rejoins the instant-turn worldline.

26 · proper time

Cauchy–Schwarz and the triangle inequality, reversed

reversed Cauchy–Schwarz inequality

If $X$ is timelike and $Y$ is any vector, $\ \eta(X, Y)^2 \ge \eta(X, X)\,\eta(Y, Y)$, with equality iff $Y$ is a multiple of $X$.

proof

Write $Y = aX + Z$ with $\eta(X, Z) = 0$. $Z$ is $0$ or spacelike (lemma of section 2). Then $\eta(X,Y)^2 = a^2\eta(X,X)^2$ and $\eta(X,X)\eta(Y,Y) = a^2\eta(X,X)^2 + \eta(X,X)\eta(Z,Z)$. The last term is $\le 0$, and $= 0$ iff $Z = 0$.

reversed triangle inequality

With $\lVert X\rVert := \sqrt{-\eta(X, X)}$, for future-directed timelike $X, Y$:

$$\lVert X + Y\rVert \ \ge\ \lVert X\rVert + \lVert Y\rVert ,$$

with equality iff $X, Y$ are parallel.

proof

$\lVert X + Y\rVert^2 = \lVert X\rVert^2 + \lVert Y\rVert^2 - 2\eta(X, Y)$. By the section 2 lemma $\eta(X, Y) < 0$, and by reversed Cauchy–Schwarz $-\eta(X, Y) \ge \lVert X\rVert\,\lVert Y\rVert$. So the right side is $\ge (\lVert X\rVert + \lVert Y\rVert)^2$.

The decomposition. Put $a = \eta(X, Y)/\eta(X, X)$, legitimate since $\eta(X, X) \ne 0$, and $Z = Y - aX$; then $\eta(X, Z) = \eta(X, Y) - a\,\eta(X, X) = 0$. The lemma that $X^\perp$ is spacelike was proved in the notes of "Two future timelike vectors pair negatively" by counting dimensions: a 2-plane on which $\eta \le 0$ would have to meet the positive-definite spatial 3-plane.

The inequality, in words. In Euclidean space $|x\cdot y| \le |x||y|$. Here, for timelike $X$, the inequality goes the other way — because $\eta(X, X) < 0$ multiplies $\eta(Z, Z) \ge 0$ and so helps rather than hurts. For two spacelike vectors neither direction holds in general. Take $X = (0, 1, 0, 0)$. With $Y = (0.5, 1, 0, 0)$, $\eta(X,Y)^2 = 1 > 0.75 = \eta(X,X)\eta(Y,Y)$, against the Euclidean direction; with $Y = (0, 0, 1, 0)$, $\eta(X,Y)^2 = 0 < 1$, against the reversed one. Indefinite forms only give inequalities on cones.

On the twins. $X = (5, 4)$, $Y = (5, -4)$: $\eta(X, Y) = -25 - 16 = -41$, $\lVert X\rVert\lVert Y\rVert = 9$, and indeed $41 \ge 9$. $\lVert X + Y\rVert^2 = 9 + 9 + 82 = 100$, so $\lVert X + Y\rVert = 10 \ge 3 + 3$.

Equality. Equality in the triangle inequality needs equality in reversed Cauchy–Schwarz, so $Y = aX$, and $a > 0$ since both are future-directed. Then the broken path is not broken.

Induction. For a polygon of future timelike steps $X_1, \dots, X_n$ from $p$ to $q$: $\lVert q - p\rVert \ge \sum\lVert X_i\rVert$, since the sum of future timelike vectors is future timelike (convexity of the cone). Proper time along a smooth curve is a limit of such sums, which gives the next slide's theorem a second proof.

26 · proper time

The straight worldline is the longest

theorem: the twin paradox

Let $q - p$ be future timelike. Among all future-directed timelike piecewise-$C^1$ curves from $p$ to $q$, the straight segment has the greatest proper time, $\lVert q - p\rVert/c$, and it is the only one that does.

proof

Boost and translate so $p = 0$ and $q = (cT, \vec 0)$. Parametrise by $t\in[0, T]$. Then $\tau = \int_0^T\sqrt{1 - |\dot{\vec x}|^2/c^2}\,\mathrm dt \le T$, with equality iff $\dot{\vec x} \equiv 0$.

and there is no shortest

Zig-zag at speed $\beta c$ between $p$ and $q$: $\tau = T\sqrt{1 - \beta^2} \to 0$ as $\beta \to 1$. The infimum is $0$ and is not attained.

From $p$ to $q$, $10$ years apart on the straight clock. The twins' path at $0.8c$: $6$. A zig-zag at $0.95c$, never more than $1.2$ light-years from the straight line: $3.1$. Faster zig-zags push it towards $0$.

Full proof

The frame. $q - p$ is future timelike, $= (c\Delta t, \Delta\vec x)$ with $|\Delta\vec x| < c\Delta t$. Rotate $\Delta\vec x$ onto the $x$-axis, then boost with $\beta = |\Delta\vec x|/c\Delta t < 1$: in the new frame $q - p = (cT, \vec 0)$ with $cT = \lVert q - p\rVert$ (the interval is invariant). Proper time is invariant, so it may be computed in this frame.

The parameter. Along a future timelike curve $t$ is strictly increasing, so it runs from $0$ to $T$ and parametrises the curve; $\vec x(t)$ is piecewise $C^1$ with $\vec x(0) = \vec x(T) = 0$.

The inequality. $0 < \sqrt{1 - |\dot{\vec x}|^2/c^2} \le 1$ pointwise, with equality iff $\dot{\vec x}(t) = 0$. Integrating gives $\tau \le T$; if $\tau = T$, the continuous non-negative function $1 - \sqrt{\cdots}$ integrates to zero on each piece, so vanishes, so $\dot{\vec x} \equiv 0$, $\vec x \equiv 0$, and the curve is the straight segment.

The figure. Four out-and-back zig-zags at $\beta = 0.95$ over $T = 10$: each leg lasts $10/8$ years and reaches $0.95\times10/8 = 1.19$ light-years out, and $\tau = 10\sqrt{1 - 0.9025} = 3.12$. The twins' single zig at $0.8$ gives $10\times0.6 = 6$.

The zig-zag. Go out at $+\beta c$ for time $T/2n$, back at $-\beta c$ for $T/2n$, $n$ times. Each piece has $\sqrt{1 - \beta^2}$ times its coordinate time, so $\tau = T\sqrt{1-\beta^2}$, arbitrarily small, while the curve stays within $\beta cT/2n$ of the straight one: arbitrarily close in space, arbitrarily far in proper time.

Checks. Dimensions: $\lVert q - p\rVert/c$ is a time ✓. Limit $c\to\infty$: the integrand tends to $1$ for every curve, and every path takes the same time — Newton's absolute time, with no twin effect ✓.

The twins, as a corollary. Earth's worldline is the straight segment from departure to reunion; the traveller's is not; so the traveller's proper time is strictly less. Whose clock "really" ran slow is not a well-posed question; which worldline is longer is.

26 · proper time

A maximum principle — worth ten uncomfortable minutes

Unit 11 asked whether a stationary curve is a minimum. For $\tau$ the Lagrangian in $\dot x$ is $F = \sqrt{1 - \dot x^2/c^2}$:

$$F_{\dot x\dot x} = -\frac{1}{c^2\,(1 - \dot x^2/c^2)^{3/2}} < 0 .$$

Unit 11's Legendre condition for a minimum, $F_{\dot x\dot x} \ge 0$, fails everywhere; the one for a maximum holds everywhere. The sign of $\eta$ has turned the variational problem upside down.

The two integrands against $\dot x$ ($c = 1$). Arc length's is convex: a zig-zag at $\pm0.8$ averages $1.28 > 1$, so straight is shortest. Proper time's is concave: the zig-zag averages $0.6 < 1$, so straight is longest.

The second derivative. $F = (1 - u^2/c^2)^{1/2}$ with $u = \dot x$; $F' = -(u/c^2)(1 - u^2/c^2)^{-1/2}$; $F'' = -\frac{1}{c^2}(1 - u^2/c^2)^{-1/2} - \frac{u^2}{c^4}(1 - u^2/c^2)^{-3/2} = -\frac{1}{c^2}(1 - u^2/c^2)^{-3/2}$. The Euler–Lagrange equation of $\int F\,\mathrm dt$ is $\frac{\mathrm d}{\mathrm dt}F' = 0$, so $F'$ and hence $\dot x$ are constant: stationary curves are straight, as the theorem already said.

Reading the figure. For a path from $p$ to $q$ parametrised by $t$, $\dot x$ averages to $0$ (the path returns). A concave $F$ lies below its tangent line at $0$: $F(\dot x) \le F(0) + F'(0)\,\dot x$. Average over the trip; the last term averages to zero, so $\frac1T\int F(\dot x)\,\mathrm dt \le F(0)$: the time-average of the integrand is at most its value on the straight path, with equality only for $\dot x \equiv 0$ when $F$ is strictly concave. For a convex $F$ every inequality turns round. (Here $F'(0) = 0$ anyway; the tangent argument is what survives in curved spacetime, where only the local, second-order version holds.) The Legendre condition is the infinitesimal version of the same convexity: the sign of $F_{\dot x\dot x}$. The numbers: $\sqrt{1 + 0.64} = 1.28$, $\sqrt{1 - 0.64} = 0.6$.

26 · proper time

Why the maximum matters later

why it matters later

Unit 32 says free fall follows geodesics, and unit 27 takes $-mc^2\tau$ as the free action. In Lorentzian signature a timelike geodesic is a local maximum of proper time: a freely falling clock ages the most.

the discomfort

Every intuition about "shortest paths" is backwards here, and only for timelike curves. For spacelike curves $\eta$ is indefinite on the variations and a straight line is neither maximum nor minimum. In curved spacetime the maximum is only local, up to unit 11's conjugate points.

The flat-space facts of this section, side by side with the Euclidean ones they reverse:

	Euclidean length	proper time
straight line	shortest	longest
other extreme	no longest	no shortest
polygon form	$\|x + y\| \le \|x\| + \|y\|$	$\lVert X + Y\rVert \ge \lVert X\rVert + \lVert Y\rVert$

Why ten minutes. In a Riemannian space, "geodesic" means "locally shortest", and every intuition about geodesics is shaped by that. The same word in relativity means "locally longest in proper time" for timelike curves. The equations — unit 32's geodesic equation — are identical in form, because they only express stationarity; the nature of the extremum is what changes. Section 5's theorem is the flat-space version: straight is longest, globally, and no shortest exists.

Conjugate points. In curved spacetime a timelike geodesic maximises proper time only until the first conjugate point along it, exactly as a great-circle arc on a sphere minimises length only until the antipode (unit 11's problem). A thrown ball and a clock in circular orbit can both leave and return to the same satellite; both are geodesics, they age differently, and only one is the maximum. Unit 37 meets this.

The sign of the action. With $S = -mc^2\tau$ the free particle minimises $S$ locally, and in the Newtonian limit $-mc^2\sqrt{1 - v^2/c^2} \approx -mc^2 + \tfrac12mv^2$, recovering unit 12's $L = T$ up to a constant. The minus sign is there so that the non-relativistic limit is the familiar one, not because anything is minimised.

26 · proper time

The twins, closed

method	structure it uses	Earth	traveller	where the asymmetry shows
proper time	length of a curve under $\eta$	$10$	$6$	the traveller's worldline is bent; bent is shorter
radio counts	light cones and the Doppler factor $e^{\varphi} = 3$	$10$	$6$	the flip: traveller's year $3$ of $6$, Earth's year $9$ of $10$
simultaneity	hyperplanes $\eta$-orthogonal to the worldline	$10$	$6$	a $6.4$-year jump at the turn

the whole unit in one sentence

A group fixed by structure up to one speed, the quadratic form it preserves, and every effect — dilation, contraction, Doppler, causal order, the twins — is a property of that form.

Checks, all passed. $\gamma = 5/3$; rapidity $\ln 3$; Doppler factor $e^{\ln 3} = 3$; $1 + 9 = 10$, $3 + 3 = 6$; $1.8 + 6.4 + 1.8 = 10$; $\lVert(10,0)\rVert = 10 \ge 3 + 3$. As $c \to \infty$ every column reads $10$.

Why they had to agree. The three methods look like three physical stories — clocks, radios, and changing ideas of "now" — but each is a computation with $\eta$ on the same two worldlines. Proper time integrates $\sqrt{-\eta}$; radio signals travel along null vectors of $\eta$; simultaneity is $\eta$-orthogonality. A disagreement would be an arithmetic error.

What the unit has not done. It has said nothing about momentum, energy, mass or forces. The action $-mc^2\tau$ of the last slide is where unit 27 begins. It has also said nothing about accelerated observers' own coordinates — how the world looks from the traveller's cabin during a turn of finite length — which is unit 30's Rindler coordinates, and the first place a non-constant metric appears on a flat spacetime.

Reading. Taylor and Wheeler, Spacetime Physics (2nd ed.), chapters 1–5, for the physics with numbers. Naber, The Geometry of Minkowski Spacetime, chapter 1, for the mathematics at this unit's level: it proves the reversed inequalities and the structure of the Lorentz group, and its section on causality relations is Zeeman's theorem in full. Stop at the end of that chapter: what follows is the electromagnetic field tensor, which is unit 28, and then spinors, which this course does not use. Bondi, Relativity and Common Sense, for the Doppler route.

26 · checkpoint

Done when

checkpoint

On paper and without notes, you can:

derive the Lorentz transformation from homogeneity, isotropy, the group axioms and one invariant speed, showing where each assumption is used and why $K < 0$ is excluded;
compute proper time along any specified worldline, piecewise or smooth;
prove the twin paradox is the reversed triangle inequality for timelike vectors in $\eta$, including the reversed Cauchy–Schwarz inequality it rests on.

Failing it is unambiguous

A derivation that assumes $a = \gamma$ without reciprocity, or never shows $K$ is constant; a proper time with the wrong sign under the root or missing its $1/c$; a triangle inequality with no proof that $\eta(X, Y) < 0$.

27 relativistic dynamics, which needs 16 as well · 28 the field tensor, after 27 and 25 · 30 the equivalence principle, where the traveller's turn gets its own coordinates.

Try the checkpoint on a trip that is in neither the deck nor the sheet: a traveller goes to a star $3$ light-years away at $0.6c$ and returns at $0.6c$. Earth time: $2\times3/0.6 = 10$ years. $\gamma = 1.25$, so the traveller's time is $8$ years. Rapidity $\operatorname{artanh}0.6 = \tfrac12\ln4 = \ln 2$, Doppler factor $2$. Radio: the traveller receives $4\times\tfrac12 = 2$ pulses out and $4\times2 = 8$ back, total $10$ ✓; light of the turn at $(5, 3)$ reaches Earth at $t = 8$, so Earth receives $8/2 = 4$ then $2\times2 = 4$, total $8$ ✓. Simultaneity: $5 \mp 0.6\cdot3$ gives $3.2$ and $6.8$, a jump of $3.6$; and $3.2 + 3.6 + 3.2 = 10$ ✓. Vectors: $\lVert(5,3)\rVert = 4$ each, $4 + 4 = 8 \le 10$ ✓.

Then, without looking, rewrite the proof that $K$ is a constant: it is the equal-diagonal-entries argument on the product of two boosts, and it is the step most often skipped.

Minkowski space: relativity as the geometry of a quadratic form

One trip, three ways to see the same six years

The group, from structure alone

Events, frames, and the question

Homogeneity gives linearity

Isotropy reduces it to two dimensions

Only one constant is left free

Three worlds, one per sign of $K$

The invariant speed, and the work each postulate does

The form, and the group of the form

The interval is invariant

Minkowski space, defined

Timelike, null, spacelike: the light cone

Two future timelike vectors pair negatively

The Lorentz group and its four pieces

Boosts are hyperbolic rotations

Velocity addition is tanh addition

The traveller's $0.8c$ is two half-light-speed boosts

Causality alone forces the group

Simultaneity, and what follows from it

The relativity of simultaneity

Einstein synchronisation: simultaneity by radar

Time dilation and length contraction, as corollaries

Minkowski diagrams that do not lie

Length contraction is a slice of a strip

The twins on the diagram: simultaneity jumps at the turn

The pole and the barn

The pole and the barn: when the doors shut

Doppler: the rapidity, exponentiated

Doppler factors multiply

The twins by radio: when does the shift flip?

The twins by radio: what Earth receives

Aberration and the headlight effect

The headlight cone, drawn

Causal structure

What survives of "before"

The turn's elsewhere, on Earth's worldline

No superluminal signalling

What the antitelephone proves, and what it does not

Proper time, and the twin paradox as a theorem

Proper time: the length of a worldline

The twins by proper time

It is not the acceleration

Cauchy–Schwarz and the triangle inequality, reversed

The straight worldline is the longest

A maximum principle — worth ten uncomfortable minutes

Why the maximum matters later

The twins, closed

Done when

Failing it is unambiguous

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