Part II · Special Relativity — Chapter 2.2

The Lorentz Transformation, Derived

Two sentences go in. One transformation comes out, and there is no second candidate. Everything else in this chapter is arithmetic.

Tools you'll need  — Chapter 2.1 §7: the two postulates, which are the entire input. Chapter 0.4: linear maps, matrices, matrix multiplication, and what it means for a set of maps to be closed under composition, which is what §3 is, applied. Chapter 0.3: Taylor and binomial expansions, used every time we ask "and what does this look like at ordinary speeds?" Chapter 0.1 §3: the derivative as the coefficient of a linear approximation. The linearity argument in §1 is that idea run in reverse, and §5 differentiates a coordinate transformation.

1 · What the postulates constrain

Let's start by writing down exactly what we are given. Then we write down exactly what we are quietly assuming on top of it, which is the part usually skipped.

1.1 · The input

Now fix the geometry once and use it for the whole chapter. There are two inertial frames $S$ and $S'$, with parallel axes. $S'$ moves in the $+x$ direction of $S$ with speed $v$, and the origins coincide at the moment both clocks read zero. This is the standard configuration, and it costs nothing, because any other arrangement is this one composed with a rotation and a shift of origin. As throughout Part II, we write the relative speed in units of $c$:

Nothing is assumed yet about the range of $\beta$. That $\abs{\beta}\lt1$ will emerge in §2.3, where a larger value makes the transformation complex, and it will be proved unreachable in §5.

What we want are the four functions that take an event's coordinates $(t,x,y,z)$ in $S$ to its coordinates $(t',x',y',z')$ in $S'$. An event is a point of spacetime: a flash, a collision, a clock tick. Both observers are describing the same one. Nothing physical changes when we relabel. The question is only what the labels are.

1.2 · The three assumptions nobody states

The postulates alone do not determine the transformation. They are constraints on a class of maps, and you have to say which class. Three assumptions do that work, and it is worth being blunt: these are physical assumptions, not logical necessities. Each is testable, each has been tested, and each could in principle have come out otherwise. This is where the real content of the chapter enters, and pretending otherwise is how the derivation acquires its undeserved reputation for being a magic trick.

(H) Homogeneity of space and time

No place is special and no moment is special. Concretely: the rule relating $S$-coordinates to $S'$-coordinates is the same rule everywhere and everywhen. If you and I are both inertial observers, the dictionary between our coordinates cannot depend on where in the room we set up, or on what year it is.

This forces the transformation to be linear. Here is the argument, and it is better than the hand-wave usually offered.

Let $\Phi$ be the map from $S$-coordinates to $S'$-coordinates, acting on events. Take any two events $P$ and $Q$ and let $\Delta = Q-P$ be their coordinate separation in $S$. Homogeneity says the separation the primed observer assigns, $\Phi(Q)-\Phi(P)$, can depend on $\Delta$ but not on where the pair sits. If it did depend on that, a rod's transformed length would tell you your absolute position, and space would have a marked spot. So there is a single function $F$ with

We want to learn what kind of function $F$ can be, and the way to find out is to chain two separations together. Go from $P$ to $P+\Delta_{1}$ to $P+\Delta_{1}+\Delta_{2}$, and add the two pieces:

That is Cauchy's functional equation, and for a continuous $F$ its only solutions are the linear ones, $F(\Delta)=M\Delta$ for a constant matrix $M$. Since the origins coincide we also have $\Phi(0)=0$, and therefore $\Phi(\Delta)=F(\Delta)=M\Delta$. The transformation is a constant matrix acting on $(ct,x,y,z)$. The entries of $M$ may depend on $v$, and they had better, but they may not depend on the event.

Grind box — Cauchy's equation, and the loophole that "straight lines stay straight" leaves open

Cauchy. ⚑ The theorem is: if $F:\R^{n}\to\R^{m}$ satisfies $F(a+b)=F(a)+F(b)$ and is continuous at even one point, then $F$ is linear. Here is the sketch in one dimension. Additivity gives $F(na)=nF(a)$ for integer $n$, hence $F(a) = F(n\cdot a/n)=nF(a/n)$, so $F(qa)=qF(a)$ for every rational $q$. Continuity then upgrades rational to real. (Without continuity the theorem is false, because there exist monstrous non-measurable additive functions, built with the axiom of choice. Nobody is proposing that spacetime is coordinatised by one.)

The other argument, and why it is not quite enough. The usual alternative runs like this. A free particle moves uniformly in a straight line in $S$, by Newton's first law, and by P1 it must do so in $S'$ too. So the map takes straight worldlines to straight worldlines, and therefore it is linear. The conclusion is right, but the reasoning has a gap. The maps of $\R^{n}$ carrying straight lines to straight lines are not just the affine ones. They are the projective ones,

$$x'^{\mu} \;=\; \frac{A^{\mu}{}_{\nu}x^{\nu} + b^{\mu}}{c_{\nu}x^{\nu} + d},$$

which are perfectly good line-preserving maps with a nonconstant denominator. ⚑ (That every line-preserving bijection is of this form is the fundamental theorem of projective geometry, which we quote.) Such a map is not linear, and it does something revealing. It blows up on the hypersurface $c_{\nu}x^{\nu}+d=0$. That is a preferred set of events, a marked place and time in the universe. Homogeneity is exactly the assumption that kills the denominator, forcing $c_{\nu}=0$ and returning us to affine, hence linear once the origins coincide.

This is not an idle technicality. Some theories keep P1 and P2 but drop strict homogeneity, among them "de Sitter relativity" and the Fock–Lorentz transformations. They live precisely in that discarded denominator, and they are constrained by experiment rather than by logic. The assumption is doing work, and now you know which work.

(I) Isotropy of space

No direction is special either. Two consequences follow, and we need both.

First: $y'=y$ and $z'=z$. Directions perpendicular to the boost are untouched. Suppose instead $y'=k(\beta)\,y$ for some factor $k$. Boosting by $-v$ afterwards must return us to $S$, so $k(\beta)k(-\beta)=1$. But isotropy says the transverse factor cannot know the sign of the boost. Reflect the $x$-axis, and a $+v$ boost becomes a $-v$ boost while $y$ is untouched, so $k(\beta)=k(-\beta)$. Put the two together and $k^{2}=1$, and since $k(0)=1$ and $k$ is continuous, $k=1$.

Second: $t'$ and $x'$ cannot depend on $y$ or $z$. By linearity, $t'=Dt+Ex+Gy+Hz$. Rotate the whole apparatus by $180^{\circ}$ about the $x$-axis. This sends $y\to-y$, $z\to-z$ and leaves the boost configuration exactly as it was, so it cannot change $t'$. That forces $G=H=0$. The same argument works for $x'$.

Grind box — the transverse argument done physically, with rods and scratches

The algebraic argument above is airtight but feels like bookkeeping. Here is the physical version, which is the one to remember, because it shows that a transverse contraction would produce a flat contradiction rather than merely a strange world.

Two identical rods, each of proper length one metre, both aligned along $y$, with their lower ends riding along the $x$-axis. One is at rest in $S$, the other at rest in $S'$, and they slide past each other. Fix a sharp scriber at the top of each rod.

Suppose $S$ finds the moving rod short, $k \lt 1$. Then as they pass, the $S$-rod's scriber passes above the tip of the $S'$-rod, and the $S'$-rod's scriber scratches the $S$-rod's side below its tip. Whether a scratch appears on a rod is a local coincidence of matter with matter. It is a single event, and every observer agrees it happened.

Now look from $S'$. By P1, $S'$'s situation is the mirror of $S$'s. $S'$ sees $S$ moving at $-v$, and by isotropy the physics cannot distinguish $+v$ from $-v$ here. So $S'$ must equally find the other rod short, and predicts the scratches the other way round. Both frames cannot be right about which rod carries a scratch, because that is one event and not two. The only escape is $k=1$. The tips meet exactly, nobody scratches anybody, and both frames agree.

Note carefully what did not happen: no such contradiction arises along $x$. There, "which rod is shorter" is settled by measurements at two separate places, and §4.3 will show that two separate places is precisely where frames stop agreeing about what "at the same time" means. Transverse: one event, no room to disagree. Longitudinal: two events, all the room in the world.

(R) Reciprocity

If $S'$ moves at $+v$ as measured in $S$, then $S$ moves at $-v$ as measured in $S'$. Equivalently: the transformation from $S'$ back to $S$ is the same transformation with $v\to-v$.

This sounds like a tautology and it is not. It is the statement that the two frames stand in a symmetric relation. Neither is privileged in a way that would let $S$ measure $S'$ receding at one rate while $S'$ measures $S$ receding at another. It follows from P1 together with isotropy, and we could derive it. We will instead assume it and flag it, because assuming it makes visible exactly how much is being asked.

2 · The derivation, in full

By §1 the transformation is linear, the transverse coordinates are untouched, and nothing depends on $y$ or $z$. So the whole problem lives in the $(t,x)$ plane, and it reads

where $A,B,D,E$ are constants, meaning functions of $v$ alone. That is four unknowns, and we have three conditions to impose. Three conditions for four unknowns sounds like one short, but it is not: condition (ii) will hand us two equations rather than one, because a light pulse can be sent in either direction along $x$ and both directions have to work. Let's impose them one at a time.

2.1 · Condition (i): the origin of $S'$ moves at $v$

This is what "$S'$ moves at $v$" means, made operational. The spatial origin of $S'$ is the locus $x'=0$, and by construction it travels along $x=vt$ in $S$. We want the relation between $A$ and $B$ that this forces, so substitute $x'=0$ and $x=vt$ into (2.2.4):

Feed that value of $B$ back into the first line of (2.2.4), and the spatial part of the transformation is already forced into the familiar shape

Notice how little that cost. And notice what it did not do. It says nothing whatsoever about $t'$. The Galilean transformation is the special case $A=1$, $D=1$, $E=0$. Every departure from Galileo is now compressed into three numbers.

2.2 · Condition (ii): a light pulse goes at $c$ in both frames

This is Postulate 2, and it is the crucial input. It is the only place where the physics of light enters. At the moment the origins coincide ($t=t'=0$, $x=x'=0$) a flash is emitted from the common origin. Consider the part of it travelling in the $+x$ direction. In $S$ it obeys $x=ct$. In $S'$, by P2, it must obey $x'=ct'$, with the same $c$ and not $c-v$.

Our aim is to turn that demand into equations relating $A$, $D$ and $E$. So put $x=ct$ into (2.2.6) and into $t'=Dt+Ex$, and see what each gives:

Both of those are proportional to $t$, so imposing $x'=ct'$ will divide the time variable out and leave a relation among the three constants alone. Do it:

That is one equation for two unknowns, so we need a second, and the flash gives us one for free. Now do the same for the part of it going in the $-x$ direction, $x=-ct$, which must obey $x'=-ct'$:

Two linear equations, two unknowns $D$ and $E$ in terms of $A$. To isolate $D$ we want the $Ec^{2}$ terms gone, so add (2.2.8) and (2.2.9). They cancel, and

That is one of the two. For the other we want the $cD$ terms gone instead, so subtract the same pair of equations rather than adding them:

Three of the four unknowns are gone, and everything is now expressed through $A$:

Let's pause on (2.2.12), because two things have already happened that no amount of staring at Galileo would have suggested.

The time equation has picked up a position term. $t'$ depends on $x$. Two events at the same $t$ but different $x$ have different $t'$. Simultaneity has already broken, before we know what $A$ is, and it broke because of the second postulate and nothing else. Everything in §4 is contained in the term $-Avx/c^{2}$.

The same coefficient $A$ appears in both. That was not imposed. It fell out of adding the two light conditions. The transformation is symmetric between $ct$ and $x$, and the symmetry is easier to see if we measure time in metres, which means using $x^{0}=ct$ as the coordinate rather than $t$, as Part II does throughout. So multiply the second equation by $c$:

Now $ct$ and $x$ enter on exactly the same footing. That symmetry is the first hint that time is about to become a coordinate like the others, which is the whole thesis of Chapter 2.3.

2.3 · Condition (iii): reciprocity fixes $A$

One unknown left. Reciprocity says the inverse of the boost by $+v$ is the boost by $-v$. So apply the boost with parameter $\beta$ and then the boost with parameter $-\beta$, and demand that we get back what we started with. Writing $A_{+}=A(\beta)$ and $A_{-}=A(-\beta)$:

We started with $x$ and we must end with $x$, whatever $x$ is, so the whole prefactor in front of it has to be one:

(The same manipulation on $ct''$ gives the same condition, and the grind box does it.) That is one equation in two unknowns, $A(\beta)$ and $A(-\beta)$, unless $A$ happens to be an even function of $\beta$. Isotropy delivers exactly that, and here is how. Reflect the $x$-axis by setting $X=-x$ and $X'=-x'$. Then (2.2.13) becomes

which is a boost with parameter $-\beta$ and coefficient $A(\beta)$. Space has no handedness, so the reflected description is as good as the original, and the coefficient for parameter $-\beta$ must be what it always is: $A(-\beta) = A(\beta)$. Feed that into (2.2.15):

Take the positive root, and here is why we may. $A(0)=+1$, because no boost is no change. $A$ is continuous. And $A$ never vanishes, so it cannot cross zero to reach the negative branch. The surviving root is the one quantity this whole chapter turns on, so give it a name:

Substitute that name back into the two lines of (2.2.13), and nothing is left undetermined. The whole quarry of the chapter is now on the page.

That is the Lorentz transformation for a boost along $x$, in the form this book uses everywhere. Three things are worth noting straight away. First, $\gamma\geq1$ always. Second, $\gamma\to\infty$ as $\beta\to1$. Third, $\beta \gt 1$ would make $\gamma$ imaginary, which is a first cheap indication that $c$ is a barrier. §5 makes that indication respectable.

Grind box — the inverse, checked both ways, and the time equation of the reciprocity step

The time half of (2.2.14). Same manoeuvre:

$$\begin{aligned} ct'' &= A_{-}\big(ct' + \beta x'\big) = A_{-}A_{+}\Big[(ct-\beta x) + \beta(x - \beta ct)\Big]\\[3pt] &= A_{-}A_{+}\big(1-\beta^{2}\big)\,ct, \end{aligned}$$

which gives condition (2.2.15) again. Consistent, as it must be: two equations, one condition.

Inverting (2.2.19) directly. This time do not appeal to reciprocity at all. Just solve. Multiply the first equation by $\gamma$ and the second by $\gamma\beta$, and add:

$$\gamma\,ct' + \gamma\beta\,x' = \gamma^{2}\Big[(ct-\beta x) + \beta(x-\beta ct)\Big] = \gamma^{2}\big(1-\beta^{2}\big)\,ct = ct,$$

since $\gamma^{2}(1-\beta^{2})=1$ is (2.2.18) squared and rearranged. The same trick with the roles swapped gives $x$. Hence

$$ct = \gamma\big(ct' + \beta x'\big), \qquad x = \gamma\big(x' + \beta\,ct'\big),$$

which is (2.2.19) with $\beta\to-\beta$ and primes swapped. So reciprocity, which we assumed, is also a theorem of the result. That is a genuine consistency check rather than a circularity, because the algebra above never used it.

Numbers worth memorising. $\gamma$ is stubbornly close to $1$ until $\beta$ is large:

$\beta$	$0.01$	$0.1$	$0.5$	$0.8$	$0.9$	$0.99$	$0.999$
$\gamma$	$1.00005$	$1.005$	$1.155$	$1.667$	$2.294$	$7.089$	$22.37$

Half the speed of light buys you a 15% effect. This is exactly why nineteenth-century physics could be built without noticing. At $\beta=10^{-4}$, the Earth's orbital speed, $\gamma-1 = 5\times10^{-9}$.

2.4 · A consistency check we did not pay for

Postulate 2 was imposed only along the $x$-axis, one pulse forward and one back. But light goes in all directions, and a flash from the origin makes an expanding sphere, $x^{2}+y^{2}+z^{2}=c^{2}t^{2}$ in $S$. If the derivation is any good, that sphere had better still be a sphere in $S'$, expanding at the same $c$. We never asked for that. Let's check it.

The quantity to watch is the one that measures the radius of the sphere against the time it has had to expand. So compute the combination $c^{2}t'^{2}-x'^{2}$ from (2.2.19):

The cross terms cancel identically and $\gamma^{2}(1-\beta^{2})=1$ finishes it. Since $y'=y$ and $z'=z$, those two coordinates come along untouched, and we have proved something considerably stronger than we asked for:

So the set $c^{2}t^{2}-x^{2}-y^{2}-z^{2}=0$, which is the expanding light sphere, maps to itself. Light emitted from the common origin is a sphere of radius $ct$ in $S$ and a sphere of radius $ct'$ in $S'$, each centred on that observer's own origin, and the two origins are moving apart. Both are right. If that sounds impossible, hold the thought until §4.1. The two observers disagree about which points of the sphere are "simultaneous", and that disagreement is exactly large enough.

2.5 · The low-speed limit: Galileo was not wrong, he was first-order

A new theory that contradicted the old one everywhere would be a theory about a different universe. The correct relationship is subtler. Relativity must reduce to Galileo where Galileo was tested. So let's expand and see.

What we need first is $\gamma$ for small $\beta$, and the binomial series (Chapter 0.3) supplies it:

so $\gamma = 1 + O(\beta^{2})$. Substituting that into (2.2.19) and keeping terms through first order in $\beta$:

Now let $c\to\infty$ at fixed $v$, $x$ and $t$, which is what "ordinary speeds" means. The second term of $t'$ vanishes too, and what is left is $x'=x-vt$ and $t'=t$. That is the Galilean transformation. Not approximately, and not in spirit. Exactly, as a limit.

That is the honest statement of the relationship, and it is worth saying what it means. Galilean relativity is not a mistake that got corrected. It is the leading term of an expansion, and the reason it looked like the whole truth for three centuries is that nobody could measure the next term. The same pattern will recur. Newtonian gravity is the leading term of general relativity (Chapter 3.6), and classical mechanics is a limit of quantum mechanics (Chapter 4.10). Old theories almost never die. They get demoted to leading order.

But look closely at (2.2.23), because the two corrections do not enter at the same order, and that asymmetry explains the entire history of the subject:

• Lengths and time rates are wrong by a factor $\gamma$, that is by $\tfrac12\beta^{2}$, which is second order. At $\beta=10^{-4}$ that is $5\times10^{-9}$, a part in two hundred million.
• Simultaneity is wrong by $-vx/c^{2}$, which is first order in $\beta$, but multiplied by a distance. At $v=30\ \mathrm{km\,s^{-1}}$ and $x=1\ \mathrm{m}$ this is $3.3\times10^{-13}\ \mathrm{s}$, which is small. But scale $x$ up to a satellite orbit, $2\times10^{7}\ \mathrm{m}$, and it becomes $6.7\ \mathrm{\mu s}$. That is $2.0\ \mathrm{km}$ of light travel, and it would wreck a navigation system in an afternoon.

So the largest relativistic effect at low speed is the one about clocks at different places, and it is the one nobody notices, because noticing it requires two clocks far apart and a way to compare them. That is exactly the experiment nineteenth-century physics could not do and twenty-first-century engineering does continuously.

3 · Matrix form and the group property

Chapter 0.4 taught you to see a linear map as a matrix, and composition as matrix multiplication. Let's do that here. Write the four coordinates as $x^{\mu}=(ct,x,y,z)$, with time measured in metres so that all four carry the same units. Then (2.2.19) is

The index placement $\Lambda^{\mu}{}_{\nu}$ puts the upper index on the row and the lower one on the column, and it is not decoration. Chapter 2.4 explains exactly what promise it encodes, and derives this same matrix as a Jacobian $\partial x'^{\mu}/\partial x^{\nu}$. For now it is a matrix.

Two features are worth naming immediately. It is symmetric, which we will use in Problem 3. And its determinant is $\gamma^{2}-\gamma^{2}\beta^{2}=\gamma^{2}(1-\beta^{2})=1$, so boosts preserve four-dimensional volume in spacetime. That fact matters in statistical mechanics and in Chapter 5.8.

3.1 · Two boosts make a boost

Now the question that decides whether we have a theory or a formula. Boost from $S$ to $S'$ at $\beta_{1}$, then boost from $S'$ to $S''$ at $\beta_{2}$, along the same axis. Is the composite a boost? To find out we multiply the two matrices, and only the upper-left $2\times2$ block does anything:

We want that product to look like a single boost matrix, which has ones on the diagonal. So pull out the common factor $1+\beta_{1}\beta_{2}$, and give the ratio that survives on the off-diagonal a name of its own:

Then the product is $\gamma_{1}\gamma_{2}(1+\beta_{1}\beta_{2})$ times the matrix $\left(\begin{smallmatrix}1 & -\beta_{3}\\ -\beta_{3} & 1\end{smallmatrix}\right)$, and it is a boost provided that prefactor equals $\gamma_{3}=(1-\beta_{3}^{2})^{-1/2}$. It does equal that, and the grind box does the algebra, so

Grind box — the prefactor really is $\gamma_{3}$, and one identity that keeps reappearing

Compute $1-\beta_{3}^{2}$ from (2.2.26):

$$1-\beta_{3}^{2} = \frac{(1+\beta_{1}\beta_{2})^{2}-(\beta_{1}+\beta_{2})^{2}}{(1+\beta_{1}\beta_{2})^{2}}.$$

Expand the numerator and watch the cross terms die:

$$\begin{aligned} (1+ab)^{2}-(a+b)^{2} &= 1 + 2ab + a^{2}b^{2} - a^{2} - 2ab - b^{2}\\[3pt] &= 1 - a^{2} - b^{2} + a^{2}b^{2} \;=\; \big(1-a^{2}\big)\big(1-b^{2}\big). \end{aligned}$$

So

$$\sqrt{1-\beta_{3}^{2}} = \frac{\sqrt{1-\beta_{1}^{2}}\,\sqrt{1-\beta_{2}^{2}}}{1+\beta_{1}\beta_{2}} = \frac{1}{\gamma_{1}\gamma_{2}\big(1+\beta_{1}\beta_{2}\big)},$$

hence $\gamma_{3}=\gamma_{1}\gamma_{2}(1+\beta_{1}\beta_{2})$, which is the prefactor. ✓

Keep that identity. The factorisation $(1+ab)^{2}-(a+b)^{2}=(1-a^{2})(1-b^{2})$, along with its sibling $(1-ab)^{2}-(a-b)^{2}=(1-a^{2})(1-b^{2})$, will do the work again in §5, where it proves that you cannot reach $c$ by combining sub-$c$ speeds. It appears once more in §6, where it turns out to be the hyperbolic addition formula in disguise. Three appearances of one factorisation is usually a sign that something structural is going on, and it is.

The other group axioms. $\Lambda(0)=I$ is the identity. $\Lambda(-\beta)$ is the inverse, since (2.2.26) with $\beta_{2}=-\beta_{1}$ gives $\beta_{3}=0$. Associativity is free, because matrix multiplication is associative. So the boosts along a fixed axis form a one-parameter group, and §6 finds the parameter that makes it look like one.

This matters more than it looks. Closure is not automatic. It is a statement that the set of transformations is a self-consistent family, that "the frames of physics" is a coherent notion, and that no chain of boosts can ever take you outside the theory. Had the composition of two boosts failed to be a boost, the whole scheme would have been incoherent.

But look hard at (2.2.26). The composed parameter is not $\beta_{1}+\beta_{2}$. Boost by $0.5c$ and then by $0.5c$ again and you get

not $1$. Two half-light-speed boosts give eight tenths, and no finite chain of sub-$c$ boosts will ever reach $c$. §5 proves that properly and §6 explains it. Velocity is the wrong variable to be adding.

4 · The three consequences

Now we cash it in. Three results, and every one of them is a two-line substitution into (2.2.19). The difficulty in this material is never the algebra. It is knowing which two events you are transforming. So each subsection begins by stating the operational question before any symbol moves: who measures what, with which apparatus, and using which pair of events.

4.1 · Relativity of simultaneity

The operational question. Two firecrackers explode. An observer in $S$, using clocks that have been synchronised throughout $S$, records both explosions at the same reading. What do the clocks of $S'$ record, given that they have been synchronised throughout $S'$?

Take two events with $\Delta t=0$ and $\Delta x\neq0$ in $S$. We want their time separation in $S'$, so apply (2.2.19) to the separation itself. That is legitimate because the transformation is linear, which is what §1 bought us:

Events simultaneous in $S$ are not simultaneous in $S'$, unless they happen at the same place. The sign matters and is worth reading off. Suppose $\Delta x \gt 0$, so that the second event is further along the direction of $S'$'s motion. Then $\Delta t' \lt 0$, so in $S'$ the leading event happened first. The frame that is moving forward sees the forward event as earlier.

A concrete case

A train of proper length $300\ \mathrm{m}$ passes a platform at $\beta=0.6$, so $\gamma=1.25$. Lightning strikes both ends of the train, simultaneously as judged on the platform.

In the platform frame $S$ the train is contracted (§4.3) to $300/1.25=240\ \mathrm{m}$, so the two strike events have $\Delta x=240\ \mathrm{m}$ and $\Delta t=0$. Putting those numbers into (2.2.29) gives, on the train,

and $\Delta x' = \gamma\Delta x = 1.25\times240 = 300\ \mathrm{m}$ ✓, which is the train's own length, as it must be. So on the train, the front is struck $600\ \mathrm{ns}$ before the rear. Not "appears to be". It is, in every sense that the train's own clocks can express.

Check it without the transformation, as a sanity test. Both flashes travel at $c$ in the platform frame and meet at the midpoint of the two strike locations. Meanwhile the train's midpoint is moving toward the front strike, so it runs into the front flash first. Now argue in the train frame. The train observer sits midway between the two scorch marks on the train, both flashes travel to him at $c$ over equal distances, and the front flash arrives first. Equal distances, equal speeds, different arrival times, and therefore different emission times. The front strike happened first. Two independent routes, one answer. ✓

4.2 · Time dilation

The operational question. A single clock rides along with $S'$ and ticks twice. The elapsed time it reads between the ticks is $\Delta\tau$. That quantity is called the clock's proper time, and it is what that particular clock actually displays. What is the elapsed time between those same two ticks according to $S$, where the reading must be taken off two different synchronised clocks, one at each tick's location?

The clock is at rest in $S'$, so the two ticks have $\Delta x'=0$ and $\Delta t'=\Delta\tau$. We want $\Delta t$, so use the inverse transformation from the grind box in §2.3:

Since $\gamma\geq1$, more time elapses in $S$ than the moving clock records. A moving clock runs slow, by exactly $\gamma$. Note the asymmetry in the apparatus, which is where all the content is: one clock on one side, two clocks on the other.

The symmetric version, and why it is not a contradiction

Run the identical argument the other way. A clock at rest in $S$ ticks twice, $\Delta x=0$, $\Delta t=\Delta\tau_{S}$. Then (2.2.19) gives directly $c\Delta t'=\gamma(c\Delta\tau_{S}-0)$, so $\Delta t'=\gamma\Delta\tau_{S}$. Each observer finds the other's clock slow, by the same factor. Both statements are true. Both are derived from the same transformation. There is no contradiction, and here is the accounting that shows it.

Set $\beta=0.6$, $\gamma=1.25$. In $S$ there are two clocks, $C_{1}$ at $x=0$ and $C_{2}$ at $x=12$ light-seconds, synchronised in $S$. A traveller carrying clock $C'$ flies from $C_{1}$ to $C_{2}$.

Told by $S$: the trip takes $\Delta t = 12/0.6 = 20\ \mathrm{s}$. The traveller's clock, moving, runs slow, and reads $20/1.25 = 16\ \mathrm{s}$ on arrival. $C_{2}$ reads $20\ \mathrm{s}$.

Told by $S'$: the traveller is at rest and the whole $S$ apparatus streams past at $0.6c$. The distance from $C_{1}$ to $C_{2}$ is contracted to $12/1.25=9.6$ light-seconds, so it takes $9.6/0.6 = 16\ \mathrm{s}$. That matches the traveller's own clock, since it is his own clock. During those $16$ seconds the $S$ clocks, being the moving ones now, tick only $16/1.25 = 12.8\ \mathrm{s}$.

Contradiction? $C_{2}$ must read $20\ \mathrm{s}$ when the traveller arrives. That is a local coincidence, two objects at the same place, and no one can disagree about it. Yet $S'$ says the $S$ clocks advanced only $12.8\ \mathrm{s}$. The missing $7.2\ \mathrm{s}$ is exactly the simultaneity offset. At the moment the traveller passes $C_{1}$, which we may call $t'=0$, the $S$ clocks that $S'$ regards as simultaneous satisfy $t = \beta x/c$. So $C_{2}$, sitting at $x=12$ light-seconds, already reads

In the traveller's reckoning, $C_{2}$ was never synchronised with $C_{1}$ in the first place. It was $7.2\ \mathrm{s}$ ahead before the trip began. Add that head start to what the clock ticked during the trip, and the books balance:

Both frames agree on every local coincidence, meaning on which clock reads what when it is next to which other clock. They disagree only about the synchronisation of distant clocks, which is not an observable but a convention imposed frame by frame. The symmetry of time dilation is consistent precisely because simultaneity fails, and the failure is exactly the right size. It could not have been any other size, because (2.2.29) and (2.2.31) came out of the same matrix.

4.3 · Length contraction

The operational question, and it is the whole difficulty. A rod lies at rest in $S'$ along the $x'$-axis. Its ends are at $x'_{L}$ and $x'_{R}$, always, so its length in $S'$ is unambiguous: $L_{0}=x'_{R}-x'_{L}$, called the proper length. Now ask $S$ for the rod's length. In $S$ the rod is moving, so "where its ends are" is a function of time, and to get a length you must locate both ends at the same moment. At the same moment according to whom? According to $S$, because that is what it means for $S$ to measure a length.

So a length measurement is a pair of events (locate left end, locate right end) constrained by $\Delta t=0$ in the measuring frame. The moment you write that down, the answer is forced, because the constraint is a simultaneity constraint and §4.1 already told you those are frame-dependent.

Apply (2.2.19) to the pair. In $S$ we have $\Delta x = L$, which is what we want, and $\Delta t=0$, which is the definition of measuring. In $S'$ we have $\Delta x' = L_{0}$, since the rod is at rest there and it does not matter when you look. Then

which is the answer, but written the wrong way round: it gives the proper length in terms of the measured one. Divide through by $\gamma$ to put the measured length on the left, where we want it:

A moving rod is short, by $\gamma$, along its direction of motion only, since $y'=y$ and $z'=z$ mean nothing happens across it. And Chapter 2.1's loose end is now tied. This is precisely the $\sqrt{1-\beta^{2}}$ that FitzGerald and Lorentz had to postulate to explain Michelson–Morley. Here it is a theorem.

It follows from simultaneity — watch

The derivation used $\Delta t=0$, which by (2.2.29) means the two measurement events are not simultaneous in $S'$:

From the rod's own frame, then, $S$ located the right end first and the left end $\beta L_{0}/c$ later. Since the rod is not going anywhere in $S'$, that sloppiness costs $S$ nothing. The ends are where they always are, which is why $S$ still gets a definite answer. But it is exactly why the answer is not $L_{0}$. To see that the choice of event pair is doing all the work, let's redo the calculation with a pair that is simultaneous in $S'$ instead:

Same rod, same frames, different pair of events, and the answer flips from $L_{0}/\gamma$ to $\gamma L_{0}$. The contraction is not a property of the rod. It is a property of the measurement, and specifically of whose simultaneity slices the measurement uses. Nothing was squeezed. Two observers sliced the same four-dimensional object at different angles and got different cross-sections, exactly as two people slicing the same sausage at different angles get ellipses of different lengths. Chapter 2.3 makes that picture precise. Here the algebra already contains it.

boost β

β = 0.375

β = 0.8

twin worldlines

β = 0.300 γ = 1.04828 φ = 0.30952 (light lines fixed at 45°)

S: Δt = +1.2000 yr (A first) S′: Δt′ = +0.2516 yr (A first)

twin worldlines: off

5 · Velocity addition

An object moves at velocity $u$ along $x$ in frame $S$. What velocity $u'$ does $S'$ assign it? Galileo says $u'=u-v$, which is the velocity-composition rule of Chapter 2.1 §1.2, and it is exactly what Postulate 2 refuses to accept for light. So let's derive the replacement.

Velocity is a ratio of displacements rather than of coordinates, so what we want is the transformation applied to small displacements. Take differentials of (2.2.19). That is legitimate because the coefficients are constants, so the differentials obey the same linear relations as the coordinates do:

The velocity in $S'$ is $\dd x'/\dd t'$, so divide the first of these by the second. The factors of $\gamma$ cancel, which is a small mercy, and a hint that the result depends on $\beta$ less violently than one might fear:

and the last step is only relabelling. Write $u=\dd x/\dd t$ for the velocity in $S$, and the rule stands in the form this book quotes everywhere:

The dimensionless form is (2.2.26) with a sign flipped, as it should be: composing two boosts and transforming a velocity are the same operation seen from two sides.

5.1 · The postulate reproduces itself

The first thing to ask of a new rule is whether it respects the postulate that produced it. So set $u=c$ and turn the crank:

That holds for any $v$ with $\abs{v}\lt c$. This is a real check, not a tautology dressed up. We imposed the invariance of $c$ only for a pulse emitted from the origin at $t=0$ along $\pm x$. The transformation that resulted now certifies that any object moving at $c$ in $S$ moves at $c$ in every frame, anywhere, at any time, in either direction. A theory that failed this would have been inconsistent with its own founding assumption.

5.2 · You cannot get to $c$ by adding speeds

Claim. If $\abs{\beta_{u}}\lt1$ and $\abs{\beta_{v}}\lt1$ then $\abs{\beta_{u'}}\lt1$.

Proof. First, the denominator never vanishes, since $\abs{\beta_{u}\beta_{v}}\lt1$ gives $1-\beta_{u}\beta_{v}\gt0$. What we want is the sign of $1-\beta_{u'}^{2}$, so compute it, using the identity from the §3 grind box in its second form, $(1-ab)^{2}-(a-b)^{2}=(1-a^{2})(1-b^{2})$:

Every factor on the right is strictly positive, so $1-\beta_{u'}^{2}\gt0$, so $\abs{\beta_{u'}}\lt1$. $\blacksquare$

Read (2.2.42) once more, because it says more than the claim. It says the "distance from $c$", measured by $1-\beta^{2}$, is multiplicative under composition. Combine two sub-light speeds and the product of their deficits is what survives. You can multiply small positive numbers together forever and never reach zero. The light barrier is not a wall someone installed. It is the statement that a product of positives is positive.

5.3 · And at ordinary speeds, Galileo

To see what the new rule costs at everyday speeds, take $uv/c^{2}\ll1$ and expand the denominator (Chapter 0.3):

Now put numbers in. Two cars approach each other at $30\ \mathrm{m\,s^{-1}}$ each, so $u=+30$ and $v=-30$. The Galilean answer is $60\ \mathrm{m\,s^{-1}}$ and the correction is

a relative error of one part in $10^{14}$. This is why the Galilean rule survived so long unchallenged: on a motorway it is wrong in the fourteenth decimal place.

6 · Rapidity — the parameter that behaves

Velocities do not add. Something must, because §3 showed that boosts along an axis form a one-parameter group, and a one-parameter group has an additive parameter almost by definition. Composing two elements should add their labels, the way rotating by $30^{\circ}$ and then $40^{\circ}$ gives $70^{\circ}$. We chose a bad label, that is all. Let's find the good one.

The place to look is the composition rule (2.2.26) itself,

which is not a formula anyone would invent. But it is a formula everyone has seen. It is the addition theorem for the hyperbolic tangent. So let's define the rapidity $\phi$ to be the thing whose hyperbolic tangent the velocity is:

This is a legitimate change of variable. $\tanh$ maps $\R$ one-to-one onto $(-1,1)$, which is exactly the allowed range of $\beta$, and it is smooth with a smooth inverse. Every subluminal velocity has exactly one rapidity, and every real number is somebody's rapidity.

6.1 · What $\gamma$ becomes

Before we can rewrite the boost we need $\gamma$ in the new variable. Divide the identity $\cosh^{2}\phi-\sinh^{2}\phi=1$ by $\cosh^{2}\phi$ to get $1-\tanh^{2}\phi=1/\cosh^{2}\phi$. Hence

Both entries of the boost matrix are now single hyperbolic functions, so substitute them into (2.2.24) and keep only the block that does anything:

and the transformation reads $ct'=ct\cosh\phi - x\sinh\phi$, $x'=x\cosh\phi-ct\sinh\phi$.

6.2 · Rapidities add

The claim to test is that composing two boosts adds their rapidities, so multiply two of these matrices together and see what comes out:

writing $c_{i}=\cosh\phi_{i}$ and $s_{i}=\sinh\phi_{i}$. The hyperbolic addition formulas are $\cosh(a+b)=\cosh a\cosh b+\sinh a\sinh b$ and $\sinh(a+b)=\sinh a\cosh b+\cosh a\sinh b$, and they turn that matrix into

Rapidities add. Plainly, exactly, with no correction term. And taking $\tanh$ of both sides recovers (2.2.26), since $\tanh(\phi_{1}+\phi_{2})=(\tanh\phi_{1}+\tanh\phi_{2})/(1+\tanh\phi_{1}\tanh\phi_{2})$. So §5 and §6 are the same statement written in two variables.

Check the arithmetic of (2.2.28) in the new currency: $\operatorname{artanh}(0.5)=0.549306$, twice that is $1.098612=\ln3$, and $\tanh(\ln3)= (3-\tfrac13)/(3+\tfrac13)=0.8$ ✓. Two boosts of $0.5c$ do not give $c$. They give rapidity $2\times0.5493$, which is velocity $0.8c$.

Grind box — hyperbolic functions from scratch, in case they are rusty

Definitions, and everything else follows:

$$\cosh\phi = \frac{\ee^{\phi}+\ee^{-\phi}}{2}, \qquad \sinh\phi = \frac{\ee^{\phi}-\ee^{-\phi}}{2}, \qquad \tanh\phi=\frac{\sinh\phi}{\cosh\phi}.$$

The Pythagorean identity. Multiply out:

$$\cosh^{2}\phi-\sinh^{2}\phi = \frac{\big(\ee^{\phi}+\ee^{-\phi}\big)^{2}-\big(\ee^{\phi}-\ee^{-\phi}\big)^{2}}{4} = \frac{4\,\ee^{\phi}\ee^{-\phi}}{4} = 1.$$

Compare $\cos^{2}+\sin^{2}=1$. The point $(\cos\theta,\sin\theta)$ traces a circle, and the point $(\cosh\phi,\sinh\phi)$ traces a hyperbola. That hyperbola is Chapter 2.3's central object.

Addition. Multiply exponentials:

$$\begin{aligned} \cosh a\cosh b + \sinh a \sinh b &= \tfrac14\Big[\big(\ee^{a}+\ee^{-a}\big)\big(\ee^{b}+\ee^{-b}\big) + \big(\ee^{a}-\ee^{-a}\big)\big(\ee^{b}-\ee^{-b}\big)\Big]\\[3pt] &= \tfrac14\Big[2\ee^{a+b}+2\ee^{-a-b}\Big] \;=\; \cosh(a+b). \end{aligned}$$

The $\sinh$ formula goes the same way. Dividing them gives the $\tanh$ addition theorem, which is (2.2.45).

The inverse. Solve $\beta=\tanh\phi$ for $\phi$. Writing $w=\ee^{2\phi}$,

$$\beta = \frac{\ee^{\phi}-\ee^{-\phi}}{\ee^{\phi}+\ee^{-\phi}} = \frac{w-1}{w+1} \;\Longrightarrow\; w = \frac{1+\beta}{1-\beta} \;\Longrightarrow\; \phi = \tfrac12\ln\frac{1+\beta}{1-\beta}.$$

Useful values: $\operatorname{artanh}(0.5)=0.5493$, $\operatorname{artanh}(0.8)=1.0986=\ln3$, $\operatorname{artanh}(0.9)=1.4722$, $\operatorname{artanh}(0.99)=2.6467$. Note how slowly rapidity grows once $\beta$ is close to $1$. Going from $0.9c$ to $0.99c$ costs about $1.17$ of rapidity, roughly the same as going from rest to $0.8c$. In Chapter 2.5 that observation acquires teeth, because the energy needed is what grows without bound.

Derivatives. $\cosh'=\sinh$, $\sinh'=\cosh$, $\tanh'=1/\cosh^{2}=1-\tanh^{2}$, all immediate from the exponential definitions. Problem 4 uses the last one.

7 · The paradoxes, dismantled

There are perhaps a dozen famous special-relativity paradoxes and they are all the same paradox. Here are the two that matter, worked to the bottom, and then the general statement.

7.1 · The twin paradox

The setup. Two twins, both aged $20$. One stays on Earth. The other flies to a star $4$ light-years away at $\beta=0.8$ ($\gamma=5/3$), turns round, and comes back. Here is the Earth-frame arithmetic. The outbound leg takes $4/0.8=5$ years, so the round trip takes $10$ years. The traveller's clock is the moving one, so by (2.2.31) it records $10/\gamma = 10\times0.6 = 6$ years. The traveller returns aged $26$, and the stay-at-home is $30$.

The apparent paradox, stated as sharply as it deserves. Motion is relative. From the traveller's point of view, Earth receded at $0.8c$ and came back. By §4.2, each observer finds the other's clock slow, symmetrically and by the same $\gamma$. So the traveller should equally conclude that Earth's clock ran slow throughout. His own $6$ years, divided by $\gamma=5/3$, gives $3.6$ years for the Earth twin, who should therefore be the younger one on his return. Both cannot be true. When the two stand next to each other at the end, comparing wrinkles is a local coincidence, and there is one fact of the matter.

Where the symmetry fails. It fails at the turnaround, and it fails physically, not verbally. The stay-at-home twin occupies a single inertial frame for the whole story. The traveller does not. He is at rest in one inertial frame going out, and in a different one coming back. The switch is detectable from inside the ship without looking out of the window. An accelerometer registers it, coffee spills, and the traveller is pressed into his seat. There is no symmetry to appeal to, because the two histories are physically different, and no amount of "motion is relative" makes them the same. One worldline is straight. The other has a kink.

That is the resolution in one sentence, and people rightly find it unsatisfying. It explains why the twins may differ without showing where the missing $6.4$ years went, that being the gap between the traveller's naive $3.6$ and the actual $10$. So let's track them.

Following the traveller's own accounting, year by year

Work in Earth-frame coordinates $(ct,x)$ measured in light-years and years, with the departure at the origin. The turnaround is the event $T=(ct,x)=(5,4)$.

Outbound leg. The traveller is at rest in the frame $S'$ moving at $\beta=+0.8$. His own elapsed time is, from (2.2.19) applied to the event $T$,

The next thing we need is his verdict on Earth. Which of Earth's events does he call simultaneous with his turnaround? Those with the same $t'$, that is those with $ct-\beta x = 1.8$. On Earth $x=0$, so

Consistent with time dilation as the traveller sees it: his $3$ years, divided by $\gamma=5/3$, gives $1.8$ years on the Earth clock. From out there, Earth is the moving one and Earth is running slow. ✓

Inbound leg. After the turn he is at rest in $S''$ moving at $\beta=-0.8$. Its simultaneity slices are $ct+\beta x = \text{const}$, so run the same question through the same turnaround event $T=(5,4)$ and see which Earth event he now calls "now":

The jump. An instant before the turn, the traveller's "now on Earth" is year $1.8$. An instant after, it is year $8.2$. Nothing happened to Earth. The traveller changed which slice of spacetime he calls "now", and the slice swung through Earth's worldline:

That jump is the piece the naive argument left out, so add the three contributions to the Earth twin's clock and watch the books balance exactly:

while the traveller's own clock reads $3+3=6$ years. Every year is accounted for. The traveller's claim "Earth's clock ran slow" was correct on each leg. What he cannot do is stitch the two legs' notions of "now" together and pretend they were one. The naive symmetric argument gave Earth $1.8+1.8=3.6$ years, and the $6.4$ it was missing is exactly the span his simultaneity slice swept across Earth's worldline during the turn. That turn was the one moment at which he was not an inertial observer, and therefore the one moment at which his slices were not entitled to line up.

Turn on the twin toggle in the figure above and watch this happen. The traveller's two simultaneity slices are the green dashed lines through the turnaround, and where they cross $x=0$ they bracket exactly the jump. Drag $\beta$ and the numbers move, but the three pieces always sum to the stay-at-home twin's total.

7.2 · The pole and the barn

The setup. A runner carries a pole of proper length $15\ \mathrm{m}$ through a barn of proper length $10\ \mathrm{m}$ with a door at each end, at $\beta=0.8$ so $\gamma=5/3$. In the barn's frame the pole is contracted to $15\times0.6=9\ \mathrm{m}$, so it fits, with a metre to spare. The farmer shuts both doors simultaneously at the moment it is inside, and the pole is briefly wholly enclosed. That last clause is the claim to be tested.

The apparent paradox. In the runner's frame the pole is $15\ \mathrm{m}$ and the barn is contracted to $10\times0.6=6\ \mathrm{m}$. The pole is two and a half times the barn's length. It cannot possibly be enclosed. But "the pole was inside the barn with both doors shut" ought to be a fact, not an opinion.

The resolution. It is not a fact, because it is not a local coincidence. It is a statement that two events at different places, front door shuts and back door shuts, are simultaneous. By §4.1 that is frame-dependent, and here the disagreement is enormous. With $\Delta x = 10\ \mathrm{m}$,

so in the runner's frame the two door-closings are $13.33\ \mathrm{m}$ of light travel apart, which is $44.5\ \mathrm{ns}$, with the far door going first. They do not shut together at all. The far door shuts and reopens long before the pole's nose gets there. Much later, the near door shuts behind the pole's tail. The pole passes through a barn that is opening and closing around it like a slow shutter, and it is never enclosed for an instant.

Both frames are right, and here is the decisive point: they agree on everything that is not a convention. Was the pole ever hit by a door? No, in both frames. Did the nose emerge from the far door before the far door shut? Problem 2 tabulates every event in both frames, and the answer is the same in both. What they disagree on is only "did the two doors shut at the same time", which is not an observable but a labelling.

7.3 · The general lesson

8 · Worked examples

9 · Your turn