Part II · Special Relativity — Chapter 2.3

Minkowski Geometry

Special relativity is Euclidean geometry with one sign flipped. Everything strange about it is that sign, working.

Tools you'll need  — Chapter 2.2: the boost $ct'=\gamma(ct-\beta x)$, $x'=\gamma(x-\beta ct)$, the matrix form, and the rapidity $\phi$ with $\beta=\tanh\phi$. Nothing from 2.2 is re-derived here. Chapter 0.4 §3 and §5: linear maps as matrices, composition as matrix multiplication, and the determinant. Section 1 asks which linear maps preserve a given quadratic expression, which is exactly a 0.4 question. Chapter 0.5 §1: the inner-product axioms, and the observation that once you have a bilinear form you have lengths, angles and orthogonality. This chapter removes one of those axioms on purpose and watches which conclusions survive. Chapter 1.2 §9, worked example 2: extremising $\int\dd s$ gives straight lines in the plane and great circles on a sphere. Section 6 runs that machine one more time, in spacetime.

1 · The invariant interval

Everything in this chapter is about two events. Not two objects, and not two particles, but two events, each one a point of spacetime with coordinates $(t,x,y,z)$ in some inertial frame. The single combination of their coordinates that the whole subject turns on is this one:

Here $\Delta t = t_{2}-t_{1}$ and so on. The quantity is called the interval between the two events, or sometimes the spacetime separation.

Three warnings come with it, and all three are paid off before the chapter ends.

• The notation $\Delta s^{2}$ is a single symbol. It is not the square of something called $\Delta s$, and it can come out negative.
• It is not a distance.
• The minus signs are not a typographical accident. They are the entire subject.

1.1 · It is the same in every frame

The claim to be proved is that two observers in relative motion, handed the same two events, compute the same $\Delta s^{2}$. The proof is direct substitution and it is short. Start from the boost of Chapter 2.2,

and apply it to a separation rather than to a single event. That step is legitimate for exactly one reason, and 2.2 §1.2 worked hard to earn it. The transformation is linear, so it acts on differences the same way it acts on coordinates. Our goal is to see what happens to the two-dimensional part $c^{2}\Delta t'^{2}-\Delta x'^{2}$, so substitute the boost into it and grind:

The last line used $\gamma^{2}(1-\beta^{2})=1$, which is the definition of $\gamma$ rearranged.

Look at what became of the cross terms. They cancelled identically, and they had to. They enter the two brackets with the same coefficient and with opposite overall sign, so nothing else was available to them.

The other two directions cost us nothing, because $\Delta y'=\Delta y$ and $\Delta z'=\Delta z$. We may therefore subtract those two squares from each side without disturbing the equality, and that gives the full four-dimensional statement:

Every inertial observer, handed the same two events, computes the same $\Delta s^{2}$. They will disagree about $\Delta t$. They will disagree about $\Delta x$. They will disagree about which event came first, sometimes. They agree about this.

Chapter 2.2 reached this same equation in its §2.4 and treated it as a consistency check. We had demanded that light go at $c$, and the check confirmed that a spherical light pulse stays spherical. That reading undersells the result badly, and it is worth being precise about why.

Postulate 2 constrains only the pairs of events with $\Delta s^{2}=0$, which is one three-dimensional family out of all the pairs there are. What came back out is a statement about every pair of events, at every value of $\Delta s^{2}$. The theory handed us more invariance than we asked for. A quantity that turns out to be invariant for no reason you demanded is exactly the kind of quantity worth building a subject on.

1.2 · The reframe: this is what "length" means

Now comes the move that makes the rest of the book easier. Let's ask what a rotation is.

You probably answer "a rigid turn about an axis". That is a picture rather than a definition. The definition used in mathematics is algebraic, and it is the one Chapter 0.4 §5 was quietly preparing you for:

That is the whole content. You do not need to mention turning, or axes, or angles. Demand that a linear map leave $x^{2}+y^{2}+z^{2}$ unchanged for every vector, and rotations together with reflections are what you get. They form the group $\mathrm{O}(3)$, and requiring the determinant to be $+1$ throws away the reflections.

Notice which way the logic runs. It runs from the quadratic form to the transformations, and not the other way round.

Now notice what the preserved quantity is. It is what we call length. That is not a separate fact about the world sitting alongside the definition. It is the definition of length: the thing on which all observers with differently-oriented axes agree.

Read (2.3.4) again with that in mind. It says:

Two remarks on how much that box is claiming.

The first is that a converse is involved. What we proved above is that Lorentz transformations preserve $\Delta s^{2}$. The statement that the linear maps preserving $\Delta s^{2}$ are exactly the Lorentz transformations runs the other way, and converses need proof. The grind box below proves it in the case that matters, which is one space dimension. It also finds three extra maps hiding in the answer, which turn out to be parity, time reversal, and the two applied together.

The second is that the word "geometry" is a promise. Calling this a geometry commits us to lengths, angles, straight lines and triangles behaving like a geometry. Sections 3 to 6 make good on each of those. Section 6 finds that one of them behaves like a geometry with the inequality reversed. That is the sign flip, arriving where it hurts.

Grind box — the converse: every linear map preserving $c^{2}t^{2}-x^{2}$, classified

Work in the $(ct,x)$ plane, which is where all the content lives. Write the unknown map as a general $2\times2$ matrix and let the invariance condition tell us what its entries have to be. So let

$$M = \begin{pmatrix} a & b\\ d & e\end{pmatrix}, \qquad \begin{pmatrix}ct'\\ x'\end{pmatrix} = M\begin{pmatrix}ct\\ x\end{pmatrix},$$

and demand $c^{2}t'^{2}-x'^{2}=c^{2}t^{2}-x^{2}$ for all $(ct,x)$. The word "all" is what makes this usable, because two quadratics that agree everywhere must agree coefficient by coefficient. So expand the left side and see what the coefficients are:

$$\big(a\,ct+b\,x\big)^{2}-\big(d\,ct+e\,x\big)^{2} = \big(a^{2}-d^{2}\big)c^{2}t^{2} + 2\big(ab-de\big)ct\,x + \big(b^{2}-e^{2}\big)x^{2}.$$

Matching coefficients of the three independent monomials gives three equations:

$$a^{2}-d^{2}=1, \qquad ab-de=0, \qquad b^{2}-e^{2}=-1.$$

Solve the first. We have $a^{2}=1+d^{2}\geq1$, so $\abs a\geq1$, and any number of modulus at least one can be written as a $\cosh$. Put $a=\sigma\cosh\phi$ with $\sigma=\pm1$ and $\phi\geq0$. Then $d^{2}=\sinh^{2}\phi$, and by allowing $\phi$ to take either sign we can write $d=-\sigma\sinh\phi$ without loss of generality.

Solve the third. The same argument applies, since $e^{2}=1+b^{2}\geq1$ likewise, so $e=\tau\cosh\psi$ and $b=-\tau\sinh\psi$ with $\tau=\pm1$. Notice that we have had to allow a second angle $\psi$ here. The middle equation is what will force it to equal the first.

Use the second. Substituting the four entries into $ab-de=0$,

$$ab-de = -\sigma\tau\cosh\phi\sinh\psi + \sigma\tau\sinh\phi\cosh\psi = \sigma\tau\,\sinh(\phi-\psi) = 0,$$

so $\psi=\phi$, since $\sinh$ vanishes only at zero. Therefore

$$M = \begin{pmatrix} \sigma\cosh\phi & -\tau\sinh\phi\\ -\sigma\sinh\phi & \tau\cosh\phi\end{pmatrix} = \begin{pmatrix}\sigma&0\\0&\tau\end{pmatrix}\begin{pmatrix}\cosh\phi & -\sinh\phi\\ -\sinh\phi & \cosh\phi\end{pmatrix}.$$

Read the factorisation on the right. The second factor is exactly Chapter 2.2's boost in rapidity form. The first factor is one of four sign choices: $\sigma=\tau=+1$ is the identity, $\sigma=+1,\tau=-1$ is parity $x\to-x$, $\sigma=-1,\tau=+1$ is time reversal $t\to-t$, and $\sigma=\tau=-1$ is both. Note $\det M=\sigma\tau$.

The conclusion. The linear maps preserving $c^{2}t^{2}-x^{2}$ are precisely the boosts, possibly composed with parity and/or time reversal. So the converse holds, up to those discrete factors. That is the same structure as in Euclidean geometry, where preserving $x^{2}+y^{2}$ gives rotations possibly composed with a reflection.

The group therefore falls into four disconnected pieces. The piece containing the identity is the one with $\sigma=\tau=+1$, the pure boosts, and it is the piece physics uses without comment. It has a name, the proper orthochronous Lorentz group. Proper means $\det=+1$, and orthochronous means it does not reverse the direction of time. ⚑ That the same four-component structure survives in the full four-dimensional group, with rotations included, is quoted here and unpacked in Chapter 6.1.

A detail worth noticing. Nothing in this derivation used a postulate, a light ray, or a physical assumption. Given the quadratic form, the transformations are pure algebra. All the physics went into choosing $c^{2}t^{2}-x^{2}$ rather than $c^{2}t^{2}+x^{2}$, and Chapter 2.2 is where that choice was forced.

1.3 · The sign convention, and the one half the world uses instead

Nothing above depended on whether we write $c^{2}\Delta t^{2}-\Delta x^{2}$ or $\Delta x^{2}-c^{2}\Delta t^{2}$. The two differ by an overall factor of $-1$, they are preserved by exactly the same transformations, and every physical statement can be phrased in either. Books nevertheless choose, and then never mention it again, which is how readers get ambushed.

This book uses the timelike-positive convention, often written $(+,-,-,-)$ and sometimes called the "mostly minus" or "particle physics" signature:

The reason is Parts V to VII. In particle physics the object you write down forty times a day is the four-momentum, and its invariant square in this convention is $p\cdot p = m^{2}c^{2}$. That is mass squared: positive, with no sign to remember. In the other convention it is $-m^{2}c^{2}$, and every mass-shell condition in the book would carry a minus sign for no benefit. Since Parts V–VII are much longer than Part III, we optimise for them.

The other convention, $(-,+,+,+)$ or "mostly plus", is standard in general relativity, where the object you write forty times a day is a spatial line element and you would rather it came out positive. If you open a relativity textbook and find $\dd s^{2}=-c^{2}\dd t^{2}+\dd x^{2}$, nothing is wrong. Multiply by $-1$ and carry on.

Here is what the choice does and does not touch. What changes: the sign of $\Delta s^{2}$ classifying timelike versus spacelike, the sign in $p\cdot p$, and the sign of $\eta_{\mu\nu}$ itself. What does not change: a single physical prediction. It is a units-style choice, like measuring angles in degrees, and it carries the same amount of physics.

2 · Coordinates, the metric, and one line of index notation

We are about to need a compact way to write (2.3.1) that does not degenerate into four-fold bookkeeping every time. The notation that does this is one of the great labour-saving devices in physics. We introduce it here informally, and label it honestly as informal, because §3 onward reads a great deal better in it.

2.1 · One symbol for four coordinates

Group the four coordinates of an event into a single object with an index:

Greek indices $\mu,\nu,\rho,\sigma$ run over $0,1,2,3$; Latin indices $i,j,k$ over the spatial $1,2,3$ only. The time coordinate is $x^{0}=ct$ and not $t$, so that all four entries have the dimensions of length. That is not cosmetic: Chapter 2.2 §2.2 already showed that $ct$ and $x$ enter the boost symmetrically, and writing $t$ instead would hide the symmetry behind a units conversion.

One warning about the superscript. It is an index, not a power. So $x^{2}$ is the $y$-coordinate. Where we mean the square of something, we will write it so that the context is unmissable. This collision of notation is genuinely annoying, and it is also universal, and everyone lives with it.

2.2 · The metric, and the summation convention

Now define a $4\times4$ array of numbers

called the Minkowski metric. It carries the minus signs of (2.3.1) and does nothing else. With it, the interval is

The second form uses the Einstein summation convention: an index appearing twice in a single term, once up and once down, is summed over its whole range, and the summation sign is not written. It is a convention about typography and nothing more. The saving is nonetheless substantial. The double sum has sixteen terms, of which twelve vanish because $\eta$ is diagonal, and you never want to write that out twice.

Before leaning on the shorthand, let's check that (2.3.9) really does say what it should. Since $\eta$ is diagonal, the only surviving terms are the ones with $\mu=\nu$, so write those four out:

That is $c^{2}\Delta t^{2}-\Delta x^{2}-\Delta y^{2}-\Delta z^{2}$ ✓. So the metric is doing exactly one job at this stage: supplying a sign for each term.

In ordinary three-dimensional space the corresponding array is $\delta_{ij}=\mathrm{diag}(1,1,1)$, which supplies all plus signs. That is why nobody ever writes it, and why the metric is invisible in first-year physics. The whole of the difference between Euclidean geometry and Minkowski geometry is the difference between $\delta_{ij}$ and $\eta_{\mu\nu}$.

Grind box — the summation convention, written out once in full, and the rules for using it

The unabbreviated version of (2.3.9), all sixteen terms, with $\Delta x^{\mu}=(a,b,c,d)$ to avoid symbol collisions:

$$\begin{aligned} \eta_{\mu\nu}\Delta x^{\mu}\Delta x^{\nu} =\;& \eta_{00}aa + \eta_{01}ab + \eta_{02}ac + \eta_{03}ad\\ +\;& \eta_{10}ba + \eta_{11}bb + \eta_{12}bc + \eta_{13}bd\\ +\;& \eta_{20}ca + \eta_{21}cb + \eta_{22}cc + \eta_{23}cd\\ +\;& \eta_{30}da + \eta_{31}db + \eta_{32}dc + \eta_{33}dd\\[4pt] =\;& a^{2}-b^{2}-c^{2}-d^{2}. \end{aligned}$$

Three rules, all of which will be justified properly in Chapter 2.4 and all of which you should obey immediately.

(1) A repeated index is dead. In $\eta_{\mu\nu}\Delta x^{\mu}\Delta x^{\nu}$ the labels $\mu$ and $\nu$ have been summed away, so the result carries no index and does not depend on what you called them. Renaming $\mu\to\alpha$ throughout changes nothing. Such indices are called dummy indices, and they behave exactly like the $k$ in $\sum_{k}a_{k}$.

(2) An unrepeated index is alive, and must match on both sides. $V^{\mu}=\Lambda^{\mu}{}_{\nu}W^{\nu}$ is a legal equation: $\nu$ is summed, $\mu$ is free and appears once on each side, so the equation is four equations. By contrast $V^{\mu}=W^{\nu}$ is not an equation at all, because it does not say which component equals which.

(3) Never use the same letter three times. $\eta_{\mu\mu}\Delta x^{\mu}$ is ill-formed: the convention cannot tell which pair you meant to sum. If you need two separate sums, use two separate letters. This rule catches more algebra errors than any other.

What is not being claimed here. We have written some indices up ($x^{\mu}$) and some down ($\eta_{\mu\nu}$), and given no reason. There is a reason, it is important, and it is not that one is a row and the other a column. Chapter 2.4 §3 gives it. Until then, treat the placement as a spelling rule you are obeying on trust, and note that the trust is being tracked: every expression in this chapter has each summed index appearing once up and once down, and you may check that it does.

2.3 · What a four-vector is, provisionally

We need a working notion of "vector" in spacetime, and there is one object whose behaviour under a boost we already know completely. The displacement between two nearby events, $\dd x^{\mu}=(c\,\dd t,\dd x,\dd y,\dd z)$, transforms in exactly the way (2.3.2) says it does. We take that as our template:

Why bother now? Because the payoff arrives immediately and repeatedly. If $V^{\mu}$ and $W^{\mu}$ are both four-vectors, then the combination

is a single number that every observer agrees on. The reason is exactly the reason $\Delta s^{2}$ is agreed on, since $\Delta s^{2}$ is the special case $V=W=\Delta x$.

What we have built is a machine for manufacturing invariants, and we will keep feeding it: $\Delta x^{\mu}$ in this chapter, the four-velocity $u^{\mu}$ in §7, the four-momentum $p^{\mu}$ in Chapter 2.5 (where $p\cdot p=m^{2}c^{2}$ turns out to be the whole of relativistic kinematics), and the four-current in Chapter 2.6.

2.4 · The condition on $\Lambda$

One more line before we start using any of this, and it is the line Chapter 2.4 and Chapter 6.1 both build on. What we want is the condition a matrix has to satisfy in order to be a Lorentz transformation at all. So write the transformation in matrix form, as Chapter 2.2 §3 did, and demand invariance of the interval. In index notation:

Since the separation $\Delta x^{\rho}$ is arbitrary and both sides are symmetric in $\rho\leftrightarrow\sigma$, the coefficients must match term by term:

Let's look at what that boxed line is saying. Compare $\Lambda^{\mathsf T}\eta\Lambda=\eta$ with the condition defining a rotation matrix, $R^{\mathsf T}\,\mathbb{1}\,R=\mathbb{1}$, that is, $R^{\mathsf T}R=\mathbb{1}$. Same equation, one symbol different.

⚑ (2.3.13) is the defining equation of the Lorentz group, written $\mathrm{O}(1,3)$, where the notation records that the metric has one plus and three minuses. Chapter 6.1 will take it as the starting point rather than as the conclusion, differentiate it near the identity, and read off the six generators. Problem 4 does the counting by hand and gets six.

Two consequences drop out of that condition without any further work. Here is the first. We want to know what the condition says about the overall scale of $\Lambda$, so take determinants of (2.3.13) and use $\det(AB)=\det A\det B$ from Chapter 0.4 §5:

So Lorentz transformations preserve four-dimensional volume in spacetime, up to sign. Chapter 2.2 §3 checked $\det=+1$ for the boost explicitly.

The second consequence is the group property. The set of matrices satisfying (2.3.13) is closed under multiplication and inversion. Chapter 2.2 §3.1 verified that by brute force for boosts along one axis, and here it takes one line: $(\Lambda_{2}\Lambda_{1})^{\mathsf T}\eta(\Lambda_{2}\Lambda_{1}) = \Lambda_{1}^{\mathsf T}(\Lambda_{2}^{\mathsf T}\eta\Lambda_{2})\Lambda_{1}=\Lambda_{1}^{\mathsf T}\eta\Lambda_{1}=\eta$.

Grind box — checking $\Lambda^{\mathsf T}\eta\Lambda=\eta$ on the boost, entry by entry

Only the upper-left $2\times2$ block does anything, so drop $y$ and $z$ and take $\eta=\mathrm{diag}(1,-1)$ with

$$\Lambda = \begin{pmatrix}\gamma & -\gamma\beta\\ -\gamma\beta & \gamma\end{pmatrix} = \begin{pmatrix}\cosh\phi & -\sinh\phi\\ -\sinh\phi & \cosh\phi\end{pmatrix}.$$

The boost matrix is symmetric, so $\Lambda^{\mathsf T}=\Lambda$. Take the triple product in two stages and compute $\eta\Lambda$ first. Multiplying on the left by $\mathrm{diag}(1,-1)$ flips the sign of the second row:

$$\eta\Lambda = \begin{pmatrix}\cosh\phi & -\sinh\phi\\ \sinh\phi & -\cosh\phi\end{pmatrix}.$$

Now multiply that result on the left by $\Lambda^{\mathsf T}$, which is $\Lambda$ itself:

$$\begin{aligned} \Lambda^{\mathsf T}\eta\Lambda &= \begin{pmatrix}\cosh\phi & -\sinh\phi\\ -\sinh\phi & \cosh\phi\end{pmatrix}\begin{pmatrix}\cosh\phi & -\sinh\phi\\ \sinh\phi & -\cosh\phi\end{pmatrix}\\[6pt] &= \begin{pmatrix}\cosh^{2}\phi-\sinh^{2}\phi & -\cosh\phi\sinh\phi+\sinh\phi\cosh\phi\\ -\sinh\phi\cosh\phi+\cosh\phi\sinh\phi & \sinh^{2}\phi-\cosh^{2}\phi\end{pmatrix}\\[6pt] &= \begin{pmatrix}1&0\\0&-1\end{pmatrix} \;=\; \eta \quad\checkmark \end{aligned}$$

The diagonal entries are the hyperbolic Pythagorean identity. The off-diagonal ones cancel because the matrix is symmetric. Note which identity did the work: $\cosh^{2}-\sinh^{2}=1$. In the Euclidean case the same computation with $\eta\to\mathbb{1}$ and $\Lambda\to R(\theta)$ runs on $\cos^{2}+\sin^{2}=1$ instead. One sign, two geometries, and that is what §3 is about.

3 · Boosts are hyperbolic rotations

This is the section the chapter exists for. Everything before it was setup and everything after it is consequence.

3.1 · Two matrices, side by side

Put the Euclidean rotation and the Lorentz boost next to each other and read across.

A rotation in the $xy$ plane preserves $x^{2}+y^{2}$ and is built from a pair of functions satisfying $\cos^{2}\theta+\sin^{2}\theta=1$:

Now the boost, in the form Chapter 2.2 §6.1 put it into. A boost in the $(ct,x)$ plane preserves $c^{2}t^{2}-x^{2}$, and it is built from a pair of functions satisfying $\cosh^{2}\phi-\sinh^{2}\phi=1$:

The parallel is not loose, so let's set out the three points of contact. Each matrix is built from the two functions whose squares obey the identity matching its own quadratic form. Each preserves that form. Each is parametrised by one real number. And in both cases the parameter is additive:

The first of those follows from the circular addition formulas and the second from the hyperbolic ones, and Chapter 2.2 §6.2 multiplied the matrices out to check. That additivity deserves a name.

Two differences are worth flagging before we exploit the parallel, because they are the places where the two geometries part company.

The signs in the off-diagonal entries. The rotation matrix has $+\sin$ above the diagonal and $-\sin$ below. The boost has $-\sinh$ in both places. So the boost matrix is symmetric and the rotation is not. That is not cosmetic. Chapter 2.2's Problem 3 used precisely this fact to prove that two boosts in different directions leave a rotation behind.

The range of the parameter. The angle $\theta$ is periodic. Rotate by $2\pi$ and you are back where you started, and the orbit of a point closes. The rapidity $\phi$ is not periodic and runs over all of $\R$, so you may boost forever and never return. In the same way $\cos$ and $\sin$ are bounded, while $\cosh$ and $\sinh$ are not.

That last contrast has a name. The rotation group is compact and the boosts are not, and this is the deepest structural difference between them. It is responsible for a great deal in Part VI. It is also the reason $\gamma=\cosh\phi$ has no upper bound while a rotation never stretches anything.

3.2 · The orbits: circles become hyperbolae

Now ask the question that turns algebra into a picture. Take one point and apply every transformation in the family. What curve does it sweep out?

Euclidean. Start at $(x_{0},y_{0})$ and rotate through every $\theta$. Since $x^{2}+y^{2}$ is preserved, every image point satisfies $x^{2}+y^{2}=r^{2}$ with $r^{2}=x_{0}^{2}+y_{0}^{2}$ fixed. The orbit is a circle. And every point of that circle is reached, since the angle covers the full range. So: the orbits of the rotation group are the circles centred on the origin, and a circle is precisely a locus of constant distance from the origin. This is so familiar that it takes an effort to notice it is a theorem.

Minkowski. Start at $(ct_{0},x_{0})$ and boost through every $\phi$. Since $c^{2}t^{2}-x^{2}$ is preserved, every image satisfies

That is the equation of a hyperbola, a rectangular one with asymptotes $ct=\pm x$. So much for the curve the image stays on. We also want to know how the point moves along it, so write the starting point in the form $(ct_{0},x_{0}) = \big(\lambda\cosh\phi_{0},\,\lambda\sinh\phi_{0}\big)$ for $k=\lambda^{2}\gt0$ and apply (2.3.16) to it:

So all the boost does is slide the hyperbolic angle: $\phi_{0}\to\phi_{0}-\phi$. Compare the Euclidean statement that a rotation slides the polar angle, $\theta_{0}\to\theta_{0}-\theta$. The structure is identical, and the parametrisation $(\cosh,\sinh)$ is doing exactly the job that $(\cos,\sin)$ does.

One consequence deserves saying out loud, because it kills a persistent confusion. The light cone $k=0$ is an orbit too. A boost slides a null point along the ray $ct=x$, multiplying both coordinates by $\ee^{-\phi}$. (Put $k=0$ into (2.3.19), or just note that $\cosh\phi-\sinh\phi=\ee^{-\phi}$.) It can shrink the coordinates toward zero or blow them up without bound, and it can never move the point off the line.

That is Postulate 2 in geometric dress. The light cone is not merely a surface that happens to be invariant. It is the degenerate member of the family of invariant hyperbolae, and no continuous boost can carry a point across it.

3.3 · Why the boosted axes look wrong, and why the ticks are wrong too

Chapter 2.2's figure showed the primed axes scissoring symmetrically toward the light line, and readers invariably ask two questions about it. First: why are the $ct'$ and $x'$ axes not perpendicular, if $S'$ is a perfectly good frame with perpendicular axes? Second: where along those axes is "one second", and why is it not where I would put it? Both have the same answer, and we can now give it.

Perpendicularity. In Euclidean geometry, two directions are perpendicular when their dot product vanishes. In this geometry the dot product is (2.3.11), with the minus signs in it. The right word here is orthogonal, and it means $\eta_{\mu\nu}V^{\mu}W^{\nu}=0$.

So let's test the two primed axes against that condition. The $ct'$ axis is the set of events with $x'=0$, so it points along $T^{\mu}=(\cosh\phi,\sinh\phi,0,0)$. The $x'$ axis is the set with $t'=0$, so it points along $X^{\mu}=(\sinh\phi,\cosh\phi,0,0)$. Their Minkowski dot product is

They are orthogonal, exactly, at every $\phi$. They do not look it, because your eye is applying the Euclidean dot product of the page, in which their product is $\cosh\phi\sinh\phi+\sinh\phi\cosh\phi=\sinh2\phi\neq0$. The axes are perpendicular in the geometry and not on the paper, and it is the paper that is wrong.

There is a pleasing detail here. The two axes are mirror images in the line $ct=x$, since swapping $\cosh\leftrightarrow\sinh$ maps one to the other. That reflection is what "scissoring symmetrically about the light line" means, and it is forced by the fact that light itself is orthogonal to itself. A null vector $N=(1,1,0,0)$ has $N\cdot N=0$.

Calibration. Where is the event "one second of $S'$ time, at $S'$'s origin"? By definition it is at $x'=0$, $ct'=1$, so its interval from the origin is $c^{2}t'^{2}-x'^{2}=1$. Intervals are invariant, so in the unprimed frame it is the point where the $ct'$ axis meets the invariant hyperbola $c^{2}t^{2}-x^{2}=1$. From the parametrisation, that point is

So the invariant hyperbolae calibrate the axes. To mark off unit ticks along any observer's time axis, intersect that axis with the hyperbolae $c^{2}t^{2}-x^{2}=1,4,9,\dots$. To mark off unit ticks along the space axis, intersect it with $x^{2}-c^{2}t^{2}=1,4,9,\dots$. The recipe works for every observer, because the hyperbolae belong to no observer.

How far off is the eye? The unit tick on the unprimed $ct$ axis sits at Euclidean distance $1$ from the origin on the page. The unit tick on the $ct'$ axis sits at page-distance

Put $\beta=0.6$ into that, where $\phi=\operatorname{artanh}0.6=\ln 2$ exactly, which is a small gift. The page-distance comes out as $\sqrt{1.25^{2}+0.75^{2}}=\sqrt{2.125}=1.4577$. So the primed observer's second is drawn 46% longer than yours. It is not longer. The page is lying, uniformly and predictably, by the factor $\gamma\sqrt{1+\beta^{2}}$:

$\beta$	$0.2$	$0.4$	$0.5$	$0.6$	$0.8$	$0.9$
$\phi$	$0.2027$	$0.4236$	$0.5493$	$0.6931$	$1.0986$	$1.4722$
page stretch	$1.041$	$1.175$	$1.291$	$1.458$	$2.134$	$3.087$

Once you know that, length contraction and time dilation become visible on a diagram rather than mysterious. Chapter 2.2 §4.3 said that two observers "slice the same four-dimensional object at different angles". The slices are the lines parallel to each observer's $x$ axis. The calibration is why the same physical rod, sliced two ways, is assigned two different lengths without anything being squeezed.

3.4 · Seeing it

rapidity φ

switch to Euclidean

reset angle

timelike event

spacelike event

null event

MINKOWSKI invariant (ct)² − x² rapidity φ = 0.0000 β = tanh φ = 0.00000 γ = cosh φ = 1.00000

in S: ct = +1.50000 x = +0.50000 in S′: ct′ = +1.50000 x′ = +0.50000

Δs² in S = +2.0000000000 | Δs² in S′ = +2.0000000000 | difference = 0.00e+0

TIMELIKE, absolute future — proper time from the origin τ = 1.414214 yr

Grind box — why $\phi$ deserves to be called an angle: it is twice a sector area, exactly as $\theta$ is

Calling $\phi$ an "angle" is so far only an analogy of algebraic form. Here is the fact that makes it more than that, and the computation is one line in each geometry.

Chapter 0.7 §5 derived Green's theorem, one corollary of which is that the area swept by the radius vector from the origin along a curve $\big(u(s),v(s)\big)$ is

$$A = \frac12\int\big(u\,\dd v - v\,\dd u\big) = \frac12\int\big(u\dot v - v\dot u\big)\,\dd s.$$

Circle. Parametrise the unit circle as $u=\cos s$, $v=\sin s$. Then

$$u\dot v - v\dot u = \cos^{2}s + \sin^{2}s = 1 \qquad\Longrightarrow\qquad A = \frac{\theta}{2}.$$

Hyperbola. Parametrise the unit hyperbola $u^{2}-v^{2}=1$ as $u=\cosh s$, $v=\sinh s$. Then

$$u\dot v - v\dot u = \cosh^{2}s - \sinh^{2}s = 1 \qquad\Longrightarrow\qquad A = \frac{\phi}{2}.$$

The same integrand, the same value, the same conclusion: the parameter is twice the area of the sector it sweeps out, in both geometries. That is the honest sense in which $\phi$ is an angle. And it explains additivity without any trigonometry: applying two transformations sweeps two adjacent sectors, and areas add. Angles add because areas add, in the circular case and the hyperbolic case alike.

It also explains why the analogy could not have been avoided. "Angle" was never fundamentally about turning. It was about the invariant measure along the orbit of a one-parameter group, and both groups have one. Chapter 6.1 will call this measure the group parameter and will not care which geometry it came from.

A numerical check, since the claim is easy to test. Take $\phi=1$. The sector bounded by the $u$-axis, the hyperbola, and the ray to $(\cosh1,\sinh1)=(1.5431,1.1752)$ has area $\tfrac12(1.5431)(1.1752) - \int_{1}^{1.5431}\sqrt{u^{2}-1}\,\dd u = 0.9068 - 0.4068 = 0.5000$, which is $\phi/2$ ✓. (The integral evaluates to $\tfrac12 u\sqrt{u^{2}-1}-\tfrac12\operatorname{arcosh}u$ between the limits.)

4 · Causal structure

In Euclidean space, $x^{2}+y^{2}+z^{2}$ is positive unless both points coincide. Every pair of distinct points is at a positive distance, and there is nothing more to say. Flip the sign and a new phenomenon appears immediately: $\Delta s^{2}$ can be positive, negative, or zero, and which it is turns out to be the most important thing about a pair of events.

4.1 · Three kinds of separation

The classification is Lorentz-invariant, and the proof is two words: $\Delta s^{2}$ is invariant, so the sign of $\Delta s^{2}$ is invariant. That is the entire argument, and its brevity is the point. Everything else in this section is unpacking what that one-line theorem means.

It is worth being clear about how strong that is. Whether two events are timelike separated is not a matter of who is looking, how fast they are moving, or which coordinates they chose. It is a fact about the pair, of exactly the same standing as "these two points are three metres apart" in Euclidean geometry. It is a frame-independent property, and it is the first one we have met that is not just a number but a structure.

4.2 · The light cone

Fix an event $O$ and ask which events are null-separated from it. With $O$ at the origin, the condition $\Delta s^{2}=0$ reads

At each time $t\gt0$ this is a sphere of radius $ct$, which is the expanding flash of light emitted at $O$. Stack those spheres up the time axis and you get a cone. Do the same for $t\lt0$ and you get a second cone opening downward, made of the light that could have converged on $O$. Together they are the light cone of $O$.

The name comes from what it looks like drawn. Suppress $y$ and $z$ and it is the pair of $45^{\circ}$ lines in the figure of §3.4. Suppress only $z$ and it is an honest cone.

The cone partitions spacetime into three regions, and every one of them is invariant, because the cone is invariant and the sign of $\Delta t$ inside it is invariant (§4.3):

Region	Condition	Meaning
Absolute future	$\Delta s^{2}\geq0$ and $\Delta t\gt0$	Events $O$ can influence. Every observer agrees they happen after $O$.
Absolute past	$\Delta s^{2}\geq0$ and $\Delta t\lt0$	Events that can influence $O$. Every observer agrees they happen before $O$.
Elsewhere	$\Delta s^{2}\lt0$	No causal contact in either direction. Time order is frame-dependent.

The word absolute is doing real work there and is not decoration. It marks the parts of "before" and "after" that survive the loss of absolute simultaneity. Chapter 2.2 destroyed the universal "now", and it is easy to conclude from that wreckage that time order is entirely conventional. It is not. Precisely the causally connectable part of it is absolute. That is the minimum required for physics to make sense, and, as we are about to see, it is also exactly the maximum available.

"Elsewhere" is the genuinely new region, and it has no Newtonian counterpart. In Galilean physics every event is either before, after, or simultaneous with $O$, and the three cases exhaust everything. Here the third of those has swollen from a three-dimensional slice into a four-dimensional region, and it contains, at this instant, everything in the universe more than a few light-nanoseconds away from you. The Andromeda galaxy has no "now". It has a two-million-year-thick band of events about which your question "what is happening there right now?" has no frame-independent answer.

4.3 · The causality theorem

Claim. If two events are timelike or null separated, every inertial observer agrees on which one happened first.

Proof. Suppose $\Delta s^{2}\geq0$. The first thing to extract from that hypothesis is a bound on how far apart in space the two events are, relative to how far apart in time, so write the condition out and then drop the two transverse terms:

The division there is legitimate because $\Delta t\neq0$. (If $\Delta t=0$ then $\Delta s^{2}\leq0$, with equality only for coincident events.) Now choose axes so that the boost is along $x$. That costs nothing, since a rotation changes neither $\Delta t$ nor the classification. What we want is a comparison between $\Delta t'$ and $\Delta t$, so take the time equation of (2.3.2) and factor $c\Delta t$ out of it:

The sign of $\Delta t'$ now rests entirely on the sign of that bracket, because $\gamma$ is positive and $c\Delta t$ carries the sign we are testing. So bound the bracket below, using the inequality we have just derived:

The last step used $\abs\beta\lt1$, which is not an extra assumption but a property of every Lorentz transformation. Since $\gamma\gt0$ and the bracket is strictly positive, $\Delta t'$ has the same sign as $\Delta t$ in every frame. $\blacksquare$

Notice exactly where the proof lives: in the collision of two inequalities, $\abs{\Delta x/c\Delta t}\leq1$ from the separation being timelike, and $\abs\beta\lt1$ from the observer being an observer. Neither alone is enough. If observers could exceed $c$, or if influences could be spacelike, the bracket could reach zero and change sign. Causality is protected by the product of two things being less than one.

A sharper version comes free. We would like a lower bound on the size of $\Delta t'$ and not just on its sign, so rearrange the invariance relation into $c^{2}\Delta t'^{2}=\Delta s^{2}+\abs{\Delta\vv x'}^{2}\geq\Delta s^{2}$ and take square roots:

with equality exactly in the frame where $\Delta\vv x'=0$. So for a timelike pair, $\Delta t'$ does not merely fail to vanish. It is bounded away from zero by a frame-independent amount, and no observer can even make the two events nearly simultaneous. Worked example 1 puts numbers on this.

4.4 · Spacelike: the order is genuinely up for grabs

Now the other case, and it is not a near miss. Claim. If two events are spacelike separated, there is a frame in which they are simultaneous, a frame in which $A$ precedes $B$, and a frame in which $B$ precedes $A$.

Proof. Take $\Delta s^{2}\lt0$ and rotate the axes so that the spatial separation lies along $x$. Then $\Delta y=\Delta z=0$ and $c^{2}\Delta t^{2}\lt\Delta x^{2}$, so $\Delta x\neq0$ and we may divide by it:

Now go looking for the frame in which the two events are simultaneous. Reading (2.3.25), the bracket there vanishes, and with it $\Delta t'$, at the single velocity

By (2.3.28) that value satisfies $\abs{\beta_{0}}\lt1$, so it is an allowed velocity and the frame exists. Now for the two orders. The quantity $c\Delta t'=\gamma(c\Delta t-\beta\Delta x)$ is a continuous, strictly monotonic function of $\beta$ that passes through zero at $\beta_{0}$, its derivative in $\beta$ never vanishing for spacelike separations. So it is positive on one side of $\beta_{0}$ and negative on the other. Boosts with $\beta$ slightly less than $\beta_{0}$ give one order, and boosts with $\beta$ slightly more give the other. $\blacksquare$

In the simultaneous frame the invariance of the interval gives $-\Delta x'^{2}=\Delta s^{2}$, so the two events are separated by

That quantity is the proper distance. It is the spacelike counterpart of proper time, and it is the distance measured by the one observer for whom the two events are "at the same moment". The two cases mirror each other exactly. Just as $\sqrt{\Delta s^{2}}$ is the smallest $c\abs{\Delta t'}$ any observer can report for a timelike pair, $\sqrt{-\Delta s^{2}}$ is the smallest $\abs{\Delta\vv x'}$ any observer can report for a spacelike pair, by the identical algebra.

Two events being reversible in order sounds alarming until you notice what it costs to arrange, and what it does not permit. Reversal requires $\abs\beta\gt\abs{\beta_{0}}=c\abs{\Delta t}/\abs{\Delta x}$, and that is possible exactly when the separation is spacelike. The condition for the reordering to be achievable and the condition for the events to be causally disconnected are the same condition. That is not luck. It is the theorem below.

4.5 · Why $c$ is a causal speed limit and not merely light's speed

Assemble the two results. Time order is absolute for causally connectable pairs, and reversible for causally disconnected ones. Nothing so far forbids a faster-than-light influence. We have just not allowed one yet. Let's allow one and watch what happens.

Suppose some influence travels from event $A$ to event $B$ at speed $u\gt c$ in some frame. It could be a signal, a particle, or a "quantum of whatever". Then $\abs{\Delta \vv x}=u\abs{\Delta t}\gt c\abs{\Delta t}$, so $A$ and $B$ are spacelike separated. By §4.4 there is then an inertial frame in which $B$ precedes $A$. But $A$ was the cause and $B$ the effect. So:

An effect preceding its cause is uncomfortable, but it is not yet a contradiction. The contradiction takes one more step, and that step is constructive. With two superluminal signals you can send a message into your own past, and here is how.

The construction

Work with $c=1$, in seconds and light-seconds, and take the idealised case of a signal that is instantaneous in the frame of whoever sends it. (The finite-$u$ version is in the grind box. It needs $\beta\gt2u/(u^{2}+1)$ and is otherwise identical.)

Step 1. Alice is at rest in frame $S$ at $x=0$. At $t=0$ she sends Bob a signal that is instantaneous in $S$, meaning that it travels along the line $t=0$. Bob sits at $x=L=1$ light-second, so the reception event is

Step 2. Bob is not at rest in $S$. He moves at $\beta=0.6$, so he is at rest in $S'$. We are going to need the reception event in his coordinates, so transform $P$ with the boost:

Step 3. On receiving the message, Bob immediately replies with a signal that is instantaneous in his own frame, so it travels along the line $t'=t'_{P}=-0.75$. The question is where that line crosses Alice's worldline $x=0$. On $x=0$ we have $ct'=\gamma\,ct$, so

Alice receives Bob's reply at $t=-0.6\ \mathrm{s}$: six tenths of a second before she sent the original message. In general the return time is $t=-\beta L/c$, so the effect is available at will and can be made as large as you like by increasing $L$.

Step 4, the contradiction. Alice now writes a program: if a reply arrives before $t=0$, do not send the message. The premise of the loop is that the message was sent. The conclusion is that it was not. This is not a paradox of interpretation, a puzzle about free will, or a subtlety about what "cause" means. It is a straightforward logical contradiction, and it is produced by three ingredients: the existence of superluminal signalling, the relativity of simultaneity, and Postulate 1, which grants Bob the same right as Alice to call his own frame's instantaneous signal instantaneous. Since the last two are established, the first must go.

Grind box — the loop with a finite superluminal speed, and the exact threshold

The instantaneous case above is clean, but it looks like a limiting trick. It is not one. Any $u\gt c$ works, provided the relative velocity of the two parties is large enough, and here is that condition derived.

Units $c=1$. Alice at $x=0$ sends at $t=0$ a signal of speed $u\gt1$ in $S$ toward $+x$, and Bob receives it at

$$P = \big(t_{P},\,x_{P}\big) = \big(L/u,\ L\big).$$

Bob is at rest in $S'$ with velocity $\beta$, so transform that reception event into his coordinates:

$$t'_{P} = \gamma L\left(\frac1u-\beta\right), \qquad x'_{P} = \gamma L\left(1-\frac{\beta}{u}\right).$$

Bob replies with a signal of speed $u$ in his own frame, travelling in the $-x'$ direction: $x'(t') = x'_{P} - u\big(t'-t'_{P}\big)$. Alice's worldline $x=0$ is, in $S'$, the line $x'=-\beta t'$. Setting the two equal and solving,

$$t'\big(u-\beta\big) = x'_{P}+u\,t'_{P} \qquad\Longrightarrow\qquad t' = \frac{x'_{P}+u\,t'_{P}}{u-\beta}.$$

Substitute the two transformed coordinates and factor out $\gamma L$:

$$x'_{P}+u\,t'_{P} = \gamma L\left[\left(1-\frac\beta u\right)+\left(1-\beta u\right)\right] = \gamma L\left[2-\beta\left(u+\frac1u\right)\right].$$

Alice's own clock reads $t=t'/\gamma$ on her worldline, so the reply arrives at

$$t \;=\; \frac{L}{u-\beta}\left[2-\beta\,\frac{u^{2}+1}{u}\right].$$

Since $u\gt\beta$, the sign is the sign of the bracket, and the reply beats the message ($t\lt0$) exactly when

$$\boxed{\;\beta \;\gt\; \frac{2u}{u^{2}+1}.\;}$$

Sanity checks. The threshold is less than $1$ for every $u\gt1$, since $2u/(u^{2}+1)\lt1 \Leftrightarrow (u-1)^{2}\gt0$ ✓, so an allowed $\beta$ always exists. As $u\to\infty$ the threshold goes to $0$: with truly instantaneous signals any relative motion suffices, matching the main text. As $u\to1^{+}$ the threshold goes to $1$: at exactly light speed you would need an observer at exactly light speed, of which there are none, and causality is safe. The barrier is not a convention. It is precisely where the construction stops working.

Numbers. Take $u=2c$, so the threshold is $2\cdot2/5=0.8$. With $L=1$ light-second:

$\beta$	$0.80$	$0.85$	$0.90$	$0.95$
reply arrives at $t$ (s)	$0.000$	$-0.109$	$-0.227$	$-0.357$

At threshold the reply arrives exactly as the message departs; beyond it, before.

One honest caveat. What the argument rules out is superluminal signalling, meaning the transmission of information or influence, which is inconsistent with relativity plus logic. It does not forbid superluminal coordinate speeds, which Chapter 2.2 §5 already showed are harmless (the spotlight, the scissors), and it does not by itself forbid a particle that always moves faster than $c$ and cannot be used to signal. ⚑ Whether such a particle is consistent quantum-mechanically is a separate question with a well-known answer, which is that it is not, and the argument there is Chapter 5.1's rather than ours.

5 · Proper time is the length of a worldline

So far the interval has connected two events by a straight line. Now let's bend it.

A particle's history in spacetime is a curve, called its worldline. Along that curve, take two neighbouring events separated by $\dd x^{\mu}=(c\,\dd t,\dd x,\dd y,\dd z)$. Suppose the particle moves slower than light, which by §4 is exactly the statement that its worldline is timelike at every point. Then $\dd s^{2}\gt0$, and we may define

What we want is $\dd\tau$ expressed through the coordinate time $\dd t$ and the particle's ordinary speed, since those are the things an observer measures. So divide the right-hand side by $c^{2}\dd t^{2}$ and pull $\dd t$ back out in front:

Here $\beta(t)$ is the particle's instantaneous speed in units of $c$. It is a function of $t$, because the particle may do whatever it likes along the way. Now add the contributions up by integrating along the path:

This quantity is the proper time along the worldline. Three things about it, in increasing order of importance.

It is invariant. Each $\dd s$ is invariant, and a sum of invariants is invariant. Two observers computing (2.3.36) for the same path will use different $\beta(t)$, different $t_{1},t_{2}$, and different intermediate values, and will get the same number.

It is a functional of the path, not a function of the endpoints. This is the language of Chapter 1.2. Here $\tau$ takes an entire curve as input and returns one number, in precisely the way that the arc-length functional $\mathcal{A}[y]=\int\sqrt{1+y'^{2}}\,\dd x$ of 1.2 §1 does. Set the two of them side by side:

There is the sign again, in the only place it could possibly be. Proper time is arc length. It is the length of the worldline, measured with the geometry that spacetime actually has. Everything §6 does follows from taking that sentence at face value.

It is what a clock reads. This is the part that keeps proper time from being an abstraction. Take the interval between two neighbouring events on a clock's own worldline. In the clock's instantaneous rest frame it is not moving, so $\dd x'=\dd y'=\dd z'=0$ and $\dd s^{2}=c^{2}\dd t'^{2}$. That is to say, the interval is the time elapsed on that clock. Since $\dd s^{2}$ is invariant, the same holds computed in any frame at all.

So $\tau$ is the accumulated reading of a clock carried along the path. And (2.3.36) says that a moving clock accumulates less of it than the coordinate time, by exactly $\gamma$ at each instant. That is Chapter 2.2's time dilation, now derived as a statement about the length of a curve rather than about clocks being defective.

Grind box — the assumption hidden in "a clock measures proper time"

The argument above computed $\dd\tau$ in the clock's instantaneous rest frame, and then added the pieces up. There is an assumption buried in "instantaneous", and it deserves naming because it is a physical input rather than a mathematical one.

The clock hypothesis. An ideal clock's rate depends only on its instantaneous velocity, not on its acceleration. Equivalently: an accelerating clock reads the same as a momentarily co-moving inertial clock, at every moment.

This is not a theorem of special relativity. It is an extra assumption, and it is easy to see that it could fail: a pendulum clock in an accelerating rocket manifestly does not obey it, and neither does any device whose mechanism is disturbed by being shaken. What the hypothesis really asserts is that some physical processes are ideal in this sense, so that "proper time" is realisable and not merely definable.

⚑ Experimentally it holds to an astonishing degree. Muons circulating in a storage ring at $\gamma\approx29$ experience a proper acceleration of order $10^{18}g$, and their decay rate is slowed by exactly the factor $\gamma$ predicted from their speed alone, with no acceleration-dependent correction at the $10^{-3}$ level. Atomic clocks flown on aircraft and the clocks aboard GPS satellites agree with (2.3.36) integrated along their actual trajectories. We quote those results. Deriving them would require the internal dynamics of the clock.

Why it matters twice. First, without it, §6's twin calculation would be incomplete, because you could always object that the turnaround does something unmodelled to the traveller's clock. Second, in general relativity the clock hypothesis is promoted to a postulate. There, proper time along a worldline is defined to be what a clock carried along it reads, and that definition is the bridge between the geometry and any measurement at all. Chapter 3.3 leans on it in its first paragraph.

6 · The twin paradox is a theorem about triangles

Now the payoff. In the plane, the straight line is the shortest path between two points, and everyone has known that since childhood. In spacetime the corresponding theorem is true with the inequality turned around, and the reversal is the minus sign, doing its most consequential work.

6.1 · The theorem

Proof. Let the two events be $E_{1}$ and $E_{2}$, timelike separated. The whole proof rests on picking the right frame first, so let's pick it. By §4.3 and the invariance of the interval there is a frame in which the two events occur at the same place. Take $\beta_{\text{rest}}=\Delta x/(c\Delta t)$, which satisfies $\abs{\beta_{\text{rest}}}\lt1$ precisely because the separation is timelike, exactly mirroring the construction in §4.4.

Work in that frame from here on. There $\Delta\vv x=0$, and the coordinate time between the events is $T=\Delta t$, which by (2.3.27) is $\sqrt{\Delta s^{2}}/c$.

The straight worldline joining them is the one at rest in this frame. Along it $\beta\equiv0$, so (2.3.36) gives $\tau_{\text{straight}}=T$.

Now take any timelike path between the same two events. It has the same endpoints, so it runs over the same coordinate-time interval, and that lets us compare the two proper times integral by integral:

The inequality holds because $\sqrt{1-\beta^{2}}\leq1$ for every $\beta$, with equality only at $\beta=0$. So every path loses, and it loses strictly unless it is at rest for the entire journey, which is to say unless it is the straight one. $\blacksquare$

That is the whole proof. It is four lines because we chose the right frame first, and we were allowed to choose it because the answer is frame-independent. The integrand $\sqrt{1-\beta^{2}}$ is a penalty for moving, charged continuously, and the straight path is the one that never pays it.

6.2 · The same result from Chapter 1.2's machinery

The direct argument is airtight but it uses a special frame. The variational argument does not, and it is the one that survives into Part III, so it is worth doing even though we already have the answer.

Extremise the functional (2.3.36). In one space dimension, with $\dot x=\dd x/\dd t$,

This is a Chapter 1.2 problem of exactly the type solved in its worked example 2. The Euler–Lagrange equation is

Now notice that $F$ contains no $x$ at all, which is the statement that spacetime is homogeneous and has no preferred place in it. So the second term vanishes, and what is left says that the first bracket is a constant of the motion:

Solve for $\dot x$ exactly as 1.2 did for the plane geodesic: square, clear the denominator, and collect,

Constant velocity, which is to say $x=vt+x_{0}$, a straight worldline. We assumed nothing about the answer, only stationarity.

There is a free bonus sitting in that last expression. Whatever real constant $K$ you pick, the resulting $\abs{\dot x}$ comes out strictly less than $c$. The extremal paths are automatically timelike, and the variational principle cannot produce a superluminal answer even if you ask it to.

The variational calculation identifies the straight line as stationary. Which kind of stationary point it is has to come from somewhere else, and §6.1 supplies the answer: a maximum. Chapter 1.2 §5.1 made the same distinction for the action and warned that stationary need not mean minimal. Here is the cleanest possible example of that warning being necessary.

6.3 · The reverse triangle inequality

The theorem of §6.1 has a clean algebraic form worth stating on its own, because it is the exact mirror image of the most familiar inequality in geometry.

First the vocabulary. Call a four-vector $V^{\mu}$ timelike if $V\cdot V\gt0$, and future-pointing if $V^{0}\gt0$. For such vectors write $\abs V\equiv\sqrt{V\cdot V}$, which is a real number. With those two words in place the theorem reads:

with equality only when $U$ and $V$ are parallel. Compare the Euclidean statement $\norm{u+v}\leq\norm u+\norm v$, and notice that the inequality has flipped. In the plane, going via a third point makes the journey longer. In spacetime it makes the elapsed time shorter, and the two-leg twin path is precisely $U+V$, with $U$ the outbound leg and $V$ the inbound one. Problem 1 asks you to prove (2.3.43). The proof turns on a reversed Cauchy–Schwarz inequality $U\cdot V\geq\abs U\abs V$, which is also flipped, and flipped for the same reason.

6.4 · Closing the loop with Chapter 2.2

Chapter 2.2 §7.1 worked the twin paradox in detail. The traveller goes $4$ light-years at $\beta=0.8$, turns, and returns. Earth ages $10$ years and the traveller ages $6$. The resolution offered there was that the traveller changes inertial frames while the stay-at-home does not, followed by a careful accounting of the $6.4$ years that the traveller's simultaneity slice sweeps across Earth's worldline during the turnaround. All correct, and all now unnecessary.

Here is the same result in one sentence. The travelling twin ages less because a bent timelike path is shorter, and shorter means less proper time.

Two paths, same endpoints, different lengths. Nobody is scandalised that a detour adds mileage. The only unfamiliar thing is that in this geometry a detour subtracts time.

Notice what the geometric statement does not require. It never mentions acceleration, it never uses a non-inertial frame, and it never needs the traveller's point of view at all. Acceleration enters only as the answer to a different question, which is why is one path bent? And a kink in a worldline is detectable from the inside, by an accelerometer, which is why there was never a symmetry between the twins to appeal to.

The often-heard slogan "the twin paradox is really about acceleration" is therefore half wrong. Acceleration is what makes a path bent. Bentness is what costs proper time. And the amount of time lost is not a function of the acceleration but of the whole shape of the path. Worked example 2 makes that concrete by comparing three paths between the same two events, one of them smoothly accelerated throughout with no kink anywhere.

Grind box — the second variation, and why the extremum is a maximum rather than a minimum

Section 6.1 proved the maximum by a global argument. Here is the local version, which is what Chapter 1.2 §5.1's second-variation machinery would say, and it exposes the sign in a different place.

Work in the frame where the endpoints coincide spatially, and perturb the straight path $x(t)=0$ by a small $\varepsilon\,\xi(t)$ with $\xi(t_{1})=\xi(t_{2})=0$. Then $\beta = \varepsilon\dot\xi/c$ and

$$\tau[\varepsilon] = \int\sqrt{1-\frac{\varepsilon^{2}\dot\xi^{2}}{c^{2}}}\ \dd t = \int\left[1-\frac{\varepsilon^{2}\dot\xi^{2}}{2c^{2}} - \frac{\varepsilon^{4}\dot\xi^{4}}{8c^{4}}-\cdots\right]\dd t,$$

by the binomial series of Chapter 0.3. Read off the two variations:

$$\delta\tau = 0 \quad\text{(no term linear in }\varepsilon\text{)}, \qquad \delta^{2}\tau = -\frac{\varepsilon^{2}}{c^{2}}\int\dot\xi^{2}\,\dd t \;\lt\; 0 .$$

The first variation vanishes for every perturbation, confirming stationarity without solving anything. The second variation is negative definite, meaning strictly negative unless $\dot\xi\equiv0$, which is to say unless the perturbation is nothing at all. Hence a maximum, and a genuine one rather than a saddle.

Where the sign came from. Compare the Euclidean calculation for a straight line in the plane, perturbed the same way: $\int\sqrt{1+\varepsilon^{2}\xi'^{2}}\,\dd x = \int[1+\tfrac12\varepsilon^{2}\xi'^{2}+\cdots]$, giving $\delta^{2}\mathcal{A}=+\varepsilon^{2}\int\xi'^{2}\gt0$: a minimum. The two calculations are identical except for the sign inside the square root, which is the sign in the metric, which flips the sign of the second variation, which flips minimum into maximum. One symbol, all the way through.

And a warning that will matter in Part III. This is a local statement. It says the straight path beats all nearby paths. In flat spacetime the global statement is true as well, and that is what §6.1 proved. In curved spacetime it can fail. There can be several geodesics between the same two events, and only one of them maximises. That is the conjugate-point phenomenon of Chapter 1.2 §5.1, and in Chapter 3.8 the several geodesics become the several images of a gravitationally lensed quasar.

7 · Four-velocity, briefly

One construction before we stop. It is short, it is forced, and Chapter 2.5 is built entirely on it.

Ordinary velocity is $\dd\vv x/\dd t$, and Chapter 2.2's Problem 1 exposed its defect. The numerator is a piece of a four-vector, but the denominator is a coordinate, so the quotient transforms in a mess. Both the top and the bottom change when you change frames, and they do not change compatibly.

The repair is to divide by something invariant instead, and §5 has just supplied the only natural candidate. Define the four-velocity

Since $\dd x^{\mu}$ transforms as a four-vector and $\dd\tau$ is a number every observer agrees on, $u^{\mu}$ is a four-vector. The numerator picks up a factor of $\Lambda$ and the denominator picks up nothing. That is the entire reason for the definition.

We will want its components in terms of ordinary velocity, and they follow at once from $\dd\tau=\dd t/\gamma$ together with the chain rule:

And now the property that makes this the right object to have defined. Feed $u^{\mu}$ into the invariant-manufacturing machine of §2.3 and compute its own square:

That holds identically, for every particle, at every moment, in every frame, whatever it is doing. Check it against the components: $\gamma^{2}c^{2}-\gamma^{2}v^{2}=\gamma^{2}c^{2}(1-\beta^{2})=c^{2}$ ✓.

8 · Worked examples

9 · Your turn