# Lorentz Transformations: Time Dilation and Length Contraction from the Light Clock
> From the constancy of the speed of light and the light-clock thought experiment we derive time dilation, Lorentz contraction and the relativity of simultaneity, then assemble them into the Lorentz transformation and the invariant spacetime interval of Minkowski spacetime.
> https://rikai.mugen-giken.com/en/physics/relativity/lorentz-transformations
## 0. Key points
- Once we accept that the speed of light is the same for everyone, the way time and space are *measured* differs from observer to observer. The Lorentz transformation is exactly what pins down that difference, no more and no less.
- A moving clock runs slow by the factor $\gamma = 1/\sqrt{1 - v^2/c^2}$ (time dilation), and a moving body is measured to be shorter by the factor $1/\gamma$ along the direction of motion (Lorentz contraction).
- The most counterintuitive consequence is the **relativity of simultaneity**. Two spatially separated events that happen at the same time in one inertial frame do not happen at the same time in another. It is this single fact that keeps time dilation and contraction from being contradictory even though each frame says the effect applies to the other.
- All three results are contained in the single pair of formulas $x' = \gamma(x - vt)$ and $t' = \gamma(t - vx/c^2)$. In the limit $v/c \to 0$ they reduce to the Galilean transformation.
- The quantity left unchanged by a Lorentz transformation is the spacetime interval $s^2 = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$. It is the "length" of four-dimensional spacetime (Minkowski spacetime), and it guarantees that causal relations do not depend on the observer.
## 1. Motivation: where the Galilean transformation breaks down
In the previous article, [Principles of Special Relativity](/en/physics/relativity/principles-of-special-relativity), we set out Einstein's two principles. Let us restate them.
- **The principle of relativity** (): the laws of physics take the same form in every inertial frame. No inertial frame is preferred over another.
- **The light postulate** (): the speed of light in vacuum, $c$, is the same regardless of the motion of the source and of the inertial frame of the observer.
The trouble is that these two collide head-on with a tacit assumption of [Newtonian mechanics](/en/physics/mechanics/newtonian-mechanics). There, the coordinates of a frame $S'$ moving with speed $v$ along the $x$ axis and those of a frame $S$ at rest are taken to be related by
$$
x' = x - vt, \qquad y' = y, \qquad z' = z, \qquad t' = t .
$$
This is the **Galilean transformation** (). The last equation $t' = t$ is the assertion that "time flows in common for all observers": the assumption of absolute time.
Under the Galilean transformation, velocities simply add. A body with speed $u$ in $S$ has speed $u' = u - v$ in $S'$. Applying this to light, with $u = c$, gives $u' = c - v$, so the speed of light in $S'$ is no longer $c$. This is incompatible with the light postulate.
At the end of the nineteenth century, the contradiction was supposed to be removed by positing a medium (the ether) that carries light, and hence a genuinely stationary frame. But the Michelson–Morley experiment of 1887 detected no ether wind whatsoever corresponding to the Earth's orbital motion at 30 km per second ().
The road Einstein chose in 1905 was to abandon whichever assumption was in conflict with experiment. What he abandoned was $t' = t$, that is, absolute time (). Once we admit that **time passes differently for different observers**, the light postulate and the principle of relativity become compatible. In this article we carry the calculation of exactly *how* it differs all the way through.
## 2. Preliminaries: events, inertial frames, notation
A single point of spacetime, that is, a specification of "when and where", is called an **event**. In an inertial frame $S$ an event is assigned four coordinates $(t, x, y, z)$.
Throughout, we assume that two inertial frames $S$ and $S'$ are in the following **standard configuration**.
1. $S'$ moves relative to $S$ with constant speed $v$ ($0 \le v < c$) in the positive $x$ direction.
2. The $x$, $y$ and $z$ axes of the two frames are respectively parallel.
3. The origins of the two frames coincide at $t = t' = 0$.
We further write $\beta = v/c$ and $\gamma = 1/\sqrt{1 - \beta^2}$, and call $\gamma$ the **Lorentz factor**.
For $0 \le \beta < 1$ we have $\gamma \ge 1$, and $\gamma \to \infty$ as $\beta \to 1$. Here are some concrete values.
| $\beta = v/c$ | $\gamma$ | $\gamma - 1$ |
|---|---|---|
| $3.0 \times 10^{-7}$ (Shinkansen) | $1.000\,000\,000\,000\,04$ | $4.4 \times 10^{-14}$ |
| $1.0 \times 10^{-4}$ (Earth's orbital motion) | $1.000\,000\,005$ | $5.0 \times 10^{-9}$ |
| $0.5$ | $1.1547$ | $0.155$ |
| $0.8$ | $1.6667$ | $0.667$ |
| $0.9$ | $2.2942$ | $1.294$ |
| $0.99$ | $7.0888$ | $6.089$ |
| $0.999$ | $22.366$ | $21.37$ |
The table shows that relativistic effects start to bite abruptly once $\beta$ exceeds about $0.1$. Even at $\beta = 0.5$ the correction is still only 15 %, while at $\beta = 0.99$ it is a factor of seven.
There is one more fact worth settling at the outset, and it follows from the two principles. If $S'$ moves with speed $v$ as seen from $S$, then $S$ moves with the same speed $v$ as seen from $S'$. This follows from the principle of relativity (the two frames are on an equal footing) together with the isotropy of space (no direction is singled out). If the speed of $S$ as seen from $S'$ differed from $v$, we could distinguish the two frames by which speed is larger, and inertial frames would cease to be equivalent, contradicting the principle of relativity. We shall use this **reciprocity** below without further comment.
## 3. The light clock and time dilation
### 3.1. Transverse lengths are unchanged
Let us first dispose of a lemma we shall need later. Discussing the light clock without checking it would leave a hole in the argument.
Let $S$ and $S'$ be inertial frames in the standard configuration. If a rod at rest in $S'$, laid along the $y$ axis (perpendicular to the direction of motion), has length $\ell_0$ as measured in $S'$, then the same rod measured in $S$ also has length $\ell_0$. The same holds in the $z$ direction.
Write the length measured in $S$ as $\ell = f(v)\,\ell_0$. By the principle of relativity, $f$ is determined by the relative velocity of the two frames alone, and by the isotropy of space it cannot depend on the direction of motion; hence $f(v) = f(-v)$, so that $f$ is a function of $|v|$ only.
Now prepare two rods $A$ and $B$ of equal length at rest. Put $A$ at rest in $S$ and $B$ at rest in $S'$, stand both parallel to the $y$ axis with their lower ends aligned at $y = 0$, and let them pass each other along the $x$ axis. Attach a brush to the upper end of $B$ so that, as it goes by, it will mark the other rod.
The whole question is this: did the brush of $B$ leave a mark on $A$, or did it sweep past above the upper end of $A$? This is a matter of whether the brush and rod $A$ touched, that is, of whether two objects were at the same place at the same time — a **local fact** — so the answer must be the same seen from any inertial frame.
Suppose $f(v) < 1$. Seen from $S$, it is $B$ that moves, so $B$ contracts and its upper end passes below the upper end of $A$; the brush therefore leaves a mark on the side of $A$. But seen from $S'$, by reciprocity it is $A$ that moves, and since $f$ depends only on $|v|$ it contracts by the very same factor. Hence the upper end of $A$ lies below that of $B$, the brush sweeps past above the upper end of $A$, and no mark is left. We have reached opposite conclusions about the same local fact, a contradiction. Assuming $f(v) > 1$ produces the same contradiction with the roles interchanged.
Therefore $f(v) = 1$, that is, $\ell = \ell_0$.
### 3.2. The light clock
A **light clock** is a device in which a light pulse bounces back and forth between two mirrors facing each other a distance $L_0$ apart; one round trip counts as one tick. We orient it so that the line joining the mirrors is perpendicular to the direction of motion (the $x$ axis).
For two events that occur at the same place (that is, two events on the worldline of a given clock), the time the clock itself records between them is called the **proper time** between the two events and is written $\Delta \tau$. Proper time is "the elapsed time measured in the frame in which the clock is at rest", a quantity determined by the history of the clock alone.
Let $S$ and $S'$ be inertial frames in the standard configuration and consider a clock at rest in $S'$. For two events occurring on that clock, let $\Delta \tau$ be the elapsed time measured in $S'$ (the proper time) and $\Delta t$ the elapsed time measured in $S$. Then
$$
\Delta t = \gamma \, \Delta \tau = \frac{\Delta \tau}{\sqrt{1 - v^2/c^2}} .
$$
Since $\gamma \ge 1$ we have $\Delta t \ge \Delta \tau$: seen from $S$, that clock runs slow compared with the clocks of $S$.
We first prove the statement for a light clock.
**The round-trip time in $S'$.** In $S'$ the clock is at rest, the mirror separation is $L_0$, and the speed of light is $c$ by the light postulate. The light covers a distance $2L_0$, so
$$
\Delta \tau = \frac{2L_0}{c}.
$$
These two events (departure from the lower mirror, return to the lower mirror) occur at the same place in $S'$, so $\Delta \tau$ is proper time in the sense of .
**The round-trip time in $S$.** In $S$ the whole clock moves with speed $v$ in the $x$ direction. Since the mirror separation is perpendicular to the motion, it is still $L_0$ by . If the round trip takes a time $\Delta t$, the clock moves horizontally by $v\,\Delta t$ during it. As in the right-hand part of the figure above, the path of the light consists of the two equal sides of an isosceles triangle of base $v\,\Delta t$ and height $L_0$. The horizontal extent of one side is half the base, namely $v\,\Delta t/2$, so by the Pythagorean theorem the one-way path length is
$$
\sqrt{L_0^2 + \left(\frac{v\,\Delta t}{2}\right)^2}
$$
and the round-trip path length is twice this. Applying the light postulate in $S$, the speed of light in $S$ is also $c$. Hence the path length equals $c\,\Delta t$:
$$
c\,\Delta t = 2\sqrt{L_0^2 + \left(\frac{v\,\Delta t}{2}\right)^2}.
$$
Squaring both sides,
$$
c^2 \Delta t^2 = 4L_0^2 + v^2 \Delta t^2
\quad\Longrightarrow\quad
(c^2 - v^2)\,\Delta t^2 = 4L_0^2 .
$$
Since $v < c$ we have $c^2 - v^2 > 0$, so
$$
\Delta t = \frac{2L_0}{\sqrt{c^2 - v^2}} = \frac{2L_0}{c\sqrt{1 - v^2/c^2}} = \gamma \cdot \frac{2L_0}{c} = \gamma\,\Delta \tau .
$$
**Extension to an arbitrary clock.** Place next to the light clock another clock of any kind — an atomic clock, a pendulum — and synchronize them so that in $S'$ the two tick together. That "the hands of the two clocks point to the same position" is a coincidence of events at the same place, hence holds as seen from any inertial frame. Therefore the two clocks run at the same rate as seen from $S$ as well, and the relation proved for the light clock carries over unchanged.
Indeed, if the light clock and the atomic clock drifted apart within $S'$, an experiment carried out entirely inside $S'$ would detect the motion of $S'$, contradicting the principle of relativity.
When cosmic rays collide with nuclei in the upper atmosphere (at an altitude of about 15 km), large numbers of unstable particles called **muons** are produced. The mean lifetime of a muon at rest is $\tau = 2.2\ \mu\mathrm{s}$.
Classically, even travelling at the speed of light a muon could cover only
$$
c\tau = (3.0\times10^{8}\ \mathrm{m/s}) \times (2.2\times10^{-6}\ \mathrm{s}) \approx 660\ \mathrm{m}
$$
within its lifetime. That is not merely short of 15 km, it is short of 1 km, so essentially none should reach the ground. In fact about one muon per square centimetre per minute rains down on the surface.
Let us use . Taking the muon's speed to be $v = 0.995c$,
$$
\gamma = \frac{1}{\sqrt{1 - 0.995^2}} = \frac{1}{\sqrt{1 - 0.990025}} = \frac{1}{\sqrt{0.009975}} = \frac{1}{0.09987} = 10.01 .
$$
The value $2.2\ \mu\mathrm{s}$ is the lifetime in the frame in which the muon is at rest, that is, a proper time, so the lifetime measured in the ground frame is $\gamma\tau = 10.01 \times 2.2 = 22.0\ \mu\mathrm{s}$. The distance covered in that time is
$$
0.995c \times 22.0\ \mu\mathrm{s} = (2.985\times10^{8}) \times (2.20\times10^{-5}) \approx 6.6\times10^{3}\ \mathrm{m} = 6.6\ \mathrm{km},
$$
an order of magnitude more. Real muons include many that are faster still, and a sufficient number reach the surface.
This is not a thought experiment but a measured fact. In 1941 Rossi and Hall compared muon counting rates at two locations of different altitude and showed that the decay rate falls with increasing momentum. In 1963 Frisch and Smith performed the same measurement on Mount Washington (altitude about 1900 m) and at sea level, confirming the time-dilation factor quantitatively.
## 4. Contraction of space (Lorentz contraction)
Once time dilation is fixed, the way lengths are measured is fixed automatically.
The length of a body measured in the inertial frame in which it is at rest is called its **proper length** and written $L_0$. In a frame in which the body appears to move, its length is defined as the difference of the positions of its two ends, both read off **at the same time in that frame**.
The proviso "read off at the same time" cannot be dropped. If we read the position of the front end of a moving rod at 12:00 and that of the rear end at 12:01, we can obtain any length we like, however long (or short). This proviso will later be tied to .
Let $S$ and $S'$ be inertial frames in the standard configuration. Let a rigid rod at rest in $S$ and laid parallel to the $x$ axis have proper length $L_0$. Then its length $L'$ as measured in $S'$ is
$$
L' = \frac{L_0}{\gamma} = L_0\sqrt{1 - v^2/c^2} .
$$
Since $\gamma \ge 1$ we have $L' \le L_0$: lengths along the direction of motion are measured to be contracted. Lengths perpendicular to the motion are unchanged, by .
Consider an observer $O'$ at rest at the origin of $S'$. Seen from $S$, $O'$ moves along the rod with speed $v$. Take the following two events.
- Event $P$: $O'$ passes one end of the rod.
- Event $Q$: $O'$ passes the other end of the rod.
**The time difference in $S$.** In $S$ the rod is at rest with length $L_0$ and $O'$ moves with speed $v$, so
$$
\Delta t = \frac{L_0}{v}.
$$
**The time difference in $S'$.** Both $P$ and $Q$ occur at the position of $O'$, that is, at the origin of $S'$. Being two events at the same place, the elapsed time in $S'$ is the proper time of . Hence, by ,
$$
\Delta t' = \frac{\Delta t}{\gamma} = \frac{L_0}{\gamma v}.
$$
**The length in $S'$.** Seen from $S'$, the rod moves with speed $v$ (in the negative $x'$ direction). Between the moment one end passes $O'$ and the moment the other end does, the rod travels exactly its own length. Since the rod is rigid and moves uniformly, this distance is the length $L'$ of the rod measured in $S'$. Therefore
$$
L' = v\,\Delta t' = v \cdot \frac{L_0}{\gamma v} = \frac{L_0}{\gamma}.
$$
That the measurement used here — obtaining the length from the time taken to pass an observer — agrees with the measurement of , which reads both ends at the same time, will be checked in §6 using .
Let us reconsider in the frame in which the muon is at rest. In this frame the muon's lifetime is still $2.2\ \mu\mathrm{s}$ and there is no time dilation. Why, then, does it reach the ground?
As seen by the muon, it is the atmosphere and the ground that approach at $0.995c$. The atmospheric thickness of 15 km is a proper length measured in the frame at rest with the Earth, so by it appears in the muon's frame contracted to
$$
\frac{15\ \mathrm{km}}{10.01} = 1.5\ \mathrm{km} .
$$
The time it takes for this thickness to sweep past the muon is
$$
\frac{1.5\times10^{3}\ \mathrm{m}}{2.985\times10^{8}\ \mathrm{m/s}} = 5.0\ \mu\mathrm{s} .
$$
Let us check that the two computations give the same answer. In the muon frame, crossing the atmosphere takes $5.0/2.2 = 2.3$ mean lifetimes. In the ground frame, a muon covers $6.6\ \mathrm{km}$ per mean lifetime, so crossing $15\ \mathrm{km}$ takes $15/6.6 = 2.3$ mean lifetimes. The ratios agree. Either way the surviving fraction is $e^{-2.3} = 0.10$, about 10 %.
The words of the two explanations — "the lifetime is stretched" and "the distance is contracted" — could hardly be more different, yet on the single observable answer, namely how many muons reach the surface, they agree completely. In relativistic calculations this agreement always holds.
## 5. Relativity of simultaneity
Here lies the sharpest break with classical intuition. Time dilation and length contraction can be accepted with a shrug, but the following statement cannot be swallowed without careful thought.
Let $S$ and $S'$ be inertial frames in the standard configuration with $0 < v < c$. Suppose two events $P$ and $Q$ occur **at the same time** in $S'$ and are separated in the $x'$ coordinate of $S'$ by $\Delta x' = x'_Q - x'_P > 0$. Then in $S$ the two events are not simultaneous, and the time difference is
$$
\Delta t = t_Q - t_P = \frac{\gamma\, v\, \Delta x'}{c^2} > 0 .
$$
That is, as seen from $S$, the event $Q$, the one ahead in the direction of motion, happens later.
To make the argument concrete, let us realize $P$ and $Q$ with a definite apparatus. Consider a carriage of length $\Delta x'$ at rest in $S'$, with a lamp at its centre. The lamp flashes once; let $P$ be the arrival of the light at the rear end and $Q$ its arrival at the front end.
**Simultaneous in $S'$.** In $S'$ the speed of light is $c$ in both directions (the light postulate), and the distance from the centre to either end is $\Delta x'/2$. The light therefore reaches the two ends in equal times, so $P$ and $Q$ are simultaneous in $S'$.
**The time difference in $S$.** In $S$ the carriage is contracted to $L = \Delta x'/\gamma$ by , and the whole carriage moves with speed $v$ in the positive $x$ direction. Take the instant of the flash to be $t = 0$ and the lamp's position at that instant to be $x = 0$, and write $x_-(t)$ for the position of the rear end and $x_+(t)$ for that of the front end:
$$
x_-(t) = -\frac{L}{2} + vt, \qquad x_+(t) = \frac{L}{2} + vt .
$$
The speed of light in $S$ is also $c$, so the backward-going light travels along $x = -ct$ and the forward-going light along $x = ct$. The times $t_P$ and $t_Q$ at which each catches its end are
$$
-c\,t_P = -\frac{L}{2} + v\,t_P \ \Longrightarrow\ t_P = \frac{L}{2(c+v)},
\qquad
c\,t_Q = \frac{L}{2} + v\,t_Q \ \Longrightarrow\ t_Q = \frac{L}{2(c-v)} .
$$
The rear end advances towards the light while the front end runs away from it, so $t_P < t_Q$. Computing the difference,
$$
\Delta t = t_Q - t_P = \frac{L}{2}\left(\frac{1}{c-v} - \frac{1}{c+v}\right)
= \frac{L}{2}\cdot\frac{(c+v)-(c-v)}{c^2 - v^2}
= \frac{L\,v}{c^2 - v^2} .
$$
Since $c^2 - v^2 = c^2(1 - \beta^2) = c^2/\gamma^2$,
$$
\Delta t = \frac{\gamma^2 L v}{c^2} = \frac{\gamma^2 v}{c^2}\cdot\frac{\Delta x'}{\gamma} = \frac{\gamma\, v\, \Delta x'}{c^2}.
$$
From $v > 0$ and $\Delta x' > 0$ we get $\Delta t > 0$.
We have proved this for one particular pair of events produced by a particular apparatus. That the same formula holds for any two events simultaneous in $S'$ follows at once from . Since is derived independently, without using the present result, there is no circularity.
Let us first estimate it on an everyday scale. For a $400\ \mathrm{m}$ Shinkansen train travelling at $320\ \mathrm{km/h}$ ($v = 88.9\ \mathrm{m/s}$), the time difference in the ground frame between two events simultaneous inside the train is, taking $\gamma \approx 1$,
$$
\Delta t \approx \frac{v L_0}{c^2} = \frac{88.9 \times 400}{(3.0\times10^{8})^2} = \frac{3.56\times10^{4}}{9.0\times10^{16}} = 4.0\times10^{-13}\ \mathrm{s} .
$$
That is 0.4 picoseconds. No human could possibly notice it. It is hardly surprising that we believed simultaneity to be absolute.
Now at relativistic speed. Suppose a spacecraft of proper length $100\ \mathrm{m}$ flies at $v = 0.8c$. Then $\gamma = 1/\sqrt{1-0.64} = 1/0.6 = 5/3$, so
$$
\Delta t = \frac{\gamma v L_0}{c^2} = \frac{(5/3)\times(0.8c)\times 100}{c^2} = \frac{(5/3)\times 0.8 \times 100}{3.0\times10^{8}} = \frac{133.3}{3.0\times10^{8}} = 4.4\times10^{-7}\ \mathrm{s},
$$
roughly $440$ nanoseconds. When the lights at the bow and the stern come on "simultaneously" aboard the ship, an observer on the ground records the bow light as coming on $440$ nanoseconds later. Both records are correct.
says that "seen from $S$, the clocks of $S'$ run slow", but by the principle of relativity it must be equally true that "seen from $S'$, the clocks of $S$ run slow". Is this not a contradiction?
It is not. The claim "the clocks of $S$ are running slow" acquires meaning only once the readings of several clocks at different places are compared **at the same time in some frame**. By , that "same time" differs between $S$ and $S'$, so the two parties are not making the same comparison. The observer in $S$ compares one clock of $S'$ against a row of clocks laid out in $S$; the observer in $S'$ does the reverse. Since the comparisons differ, no clash arises when both conclude that the other's clocks are slow.
Time dilation, length contraction and the relativity of simultaneity are not three separate phenomena but three faces of a single transformation. In the next section we write that transformation down.
## 6. The Lorentz transformation
Let $S$ and $S'$ be inertial frames in the standard configuration. Under the principle of relativity, the light postulate, and the assumption — which follows from the homogeneity of spacetime — that the coordinate transformation is linear, the coordinates of one and the same event are related by
$$
\begin{aligned}
x' &= \gamma\,(x - v t), \\
y' &= y, \\
z' &= z, \\
t' &= \gamma\left(t - \frac{v x}{c^2}\right)
\end{aligned}
$$
where $\gamma = 1/\sqrt{1 - v^2/c^2}$. The inverse transformation is obtained by replacing $v$ with $-v$:
$$
x = \gamma\,(x' + v t'), \qquad t = \gamma\left(t' + \frac{v x'}{c^2}\right) .
$$
In the limit $v/c \to 0$ it reduces to the Galilean transformation $x' = x - vt$, $t' = t$.
**Step 1 (the $y$ and $z$ components).** By , lengths perpendicular to the motion are unchanged, so $y' = y$ and $z' = z$.
**Step 2 (the form of the $x$ transformation).** From the assumption of linearity together with rotational symmetry about the $x$ axis (if $x'$ depended on $y$ or $z$, we could distinguish the configuration from the one obtained by rotating it through $180$ degrees about the $y$ axis), we may write $x' = a x + b t$, where $a$ and $b$ are constants determined by $v$ alone.
The origin of $S'$ is the point defined by $x' = 0$, and seen from $S$ it moves along $x = vt$. Substituting $x = vt$ must therefore give $x' = 0$ identically, so
$$
a\,(vt) + b\,t = (av + b)\,t = 0 \quad(\text{for all } t) \quad\Longrightarrow\quad b = -av .
$$
Rewriting $A = a$, we get
$$
x' = A\,(x - vt) .
$$
**Step 3 (the transformation in the reverse direction).** By the principle of relativity, the same form of law holds for $S$ as seen from $S'$. By reciprocity, the velocity of $S$ as seen from $S'$ is $-v$, so
$$
x = A(-v)\,\bigl(x' - (-v)t'\bigr) = A(-v)\,(x' + v t').
$$
By the isotropy of space, $A$ does not depend on the direction of the velocity, that is, $A(-v) = A(v) = A$. Hence
$$
x = A\,(x' + v t').
$$
**Step 4 (fixing $A$ by the light postulate).** At $t = t' = 0$ emit light from the origin in the positive $x$ direction. By the light postulate, $x = ct$ holds in $S$ and $x' = ct'$ in $S'$. Substituting these into the formulas of Steps 2 and 3,
$$
c t' = A\,(ct - vt) = A t\,(c - v), \qquad c t = A\,(ct' + v t') = A t'\,(c + v).
$$
Taking a point some time after the light has left the origin, we have $t \ne 0$. Then the left-hand side of the second equation $ct = A t'(c+v)$ is nonzero, so $t' \ne 0$ as well. Multiplying the two equations side by side,
$$
c^2 t t' = A^2 t t' (c-v)(c+v) = A^2 t t' (c^2 - v^2),
$$
and dividing by $t t' \ne 0$,
$$
A^2 = \frac{c^2}{c^2 - v^2} = \frac{1}{1 - v^2/c^2} = \gamma^2 .
$$
Since we consider transformations preserving the orientation of the $x$ axis, $A > 0$, and therefore $A = \gamma$.
**Step 5 (the time transformation).** Substitute $x' = \gamma(x - vt)$ from Step 2 into $x = \gamma(x' + vt')$ from Step 3:
$$
x = \gamma\bigl(\gamma(x - vt) + v t'\bigr) = \gamma^2 x - \gamma^2 v t + \gamma v t'.
$$
Solving for $\gamma v t'$,
$$
\gamma v t' = x - \gamma^2 x + \gamma^2 v t = x\,(1 - \gamma^2) + \gamma^2 v t .
$$
Using
$$
1 - \gamma^2 = 1 - \frac{1}{1-\beta^2} = \frac{(1-\beta^2) - 1}{1-\beta^2} = \frac{-\beta^2}{1-\beta^2} = -\gamma^2\beta^2 ,
$$
we get
$$
\gamma v t' = -\gamma^2 \beta^2 x + \gamma^2 v t
\quad\Longrightarrow\quad
t' = \gamma\left(t - \frac{\beta^2}{v}x\right).
$$
Since $\beta^2/v = (v^2/c^2)/v = v/c^2$, this gives
$$
t' = \gamma\left(t - \frac{v x}{c^2}\right).
$$
The inverse transformation follows from the same argument as in Step 3 ($v \to -v$).
**The limit.** As $v/c \to 0$ we have $\gamma \to 1$ and $vx/c^2 \to 0$, so $x' \to x - vt$ and $t' \to t$: the Galilean transformation is recovered.
The assumption used in the proof, that the transformation is linear, can be derived from the homogeneity of spacetime (). In an inertial frame a free particle moves uniformly in a straight line, which on a spacetime diagram is a straight line. A straight line in $S$ must again be a straight line in $S'$, so the transformation must map straight lines to straight lines, that is, it must be an affine transformation. The standard-configuration condition that the origins coincide at $t = t' = 0$ then removes the constant term, leaving a linear transformation.
The fact that a bijection mapping straight lines to straight lines must be affine is known as the fundamental theorem of affine geometry.
Let us check that the transformation just derived reproduces all three earlier results.
- **Time dilation.** For a clock at rest in $S'$ ($\Delta x' = 0$), the inverse transformation $t = \gamma(t' + vx'/c^2)$ gives $\Delta t = \gamma\,\Delta t'$, in agreement with .
- **Relativity of simultaneity.** For two events with $\Delta t' = 0$, we get $\Delta t = \gamma v \Delta x'/c^2$, in agreement with .
- **Lorentz contraction.** Measure both ends of a rod at rest in $S$ at the same time in $S'$, that is, take $\Delta t' = 0$. Eliminating $t$ from $x' = \gamma(x - vt)$ and $t' = \gamma(t - vx/c^2)$ gives $x = \gamma(x' + vt')$, so with $\Delta t' = 0$ we have $\Delta x = \gamma\,\Delta x'$, that is, $\Delta x' = \Delta x/\gamma = L_0/\gamma$. This agrees with , and it confirms that the measurement of (reading both ends at the same time) and the measurement used in the proof of (obtaining the length from a passage time) give the same answer.
Let $S$ and $S'$ be inertial frames in the standard configuration. If a particle moves in the $x'$ direction with constant velocity $u'$ in $S'$, then its velocity $u$ in $S$ is
$$
u = \frac{u' + v}{1 + \dfrac{u' v}{c^2}} .
$$
Conversely, the velocity in $S'$ is given in terms of the velocity $u$ in $S$ by $u' = (u - v)/(1 - uv/c^2)$. In particular, if $|u'| < c$ and $|v| < c$ then $|u| < c$, and if $u' = c$ then $u = c$.
Writing the inverse transformation of for infinitesimal changes,
$$
dx = \gamma\,(dx' + v\,dt'), \qquad dt = \gamma\left(dt' + \frac{v\,dx'}{c^2}\right).
$$
Taking the ratio and dividing numerator and denominator by $\gamma\,dt'$,
$$
u = \frac{dx}{dt} = \frac{dx' + v\,dt'}{dt' + v\,dx'/c^2}
= \frac{\dfrac{dx'}{dt'} + v}{1 + \dfrac{v}{c^2}\dfrac{dx'}{dt'}}
= \frac{u' + v}{1 + \dfrac{u'v}{c^2}} .
$$
Substituting $u' = c$,
$$
u = \frac{c + v}{1 + \dfrac{cv}{c^2}} = \frac{c+v}{1 + v/c} = \frac{c(c+v)}{c+v} = c ,
$$
so the speed of light is $c$ in every inertial frame. One sees that the light postulate is built into the transformation.
That $|u'| < c$ and $|v| < c$ imply $|u| < c$ follows from two identities. First,
$$
c - u = \frac{c\left(1 + \dfrac{u'v}{c^2}\right) - (u' + v)}{1 + \dfrac{u'v}{c^2}}
= \frac{c + \dfrac{u'v}{c} - u' - v}{1 + \dfrac{u'v}{c^2}}
= \frac{\dfrac{(c - u')(c - v)}{c}}{1 + \dfrac{u'v}{c^2}} .
$$
Indeed, expanding the numerator gives $(c-u')(c-v)/c = (c^2 - cv - cu' + u'v)/c = c - v - u' + u'v/c$, which matches. The same computation gives
$$
c + u = \frac{c\left(1 + \dfrac{u'v}{c^2}\right) + (u' + v)}{1 + \dfrac{u'v}{c^2}}
= \frac{\dfrac{(c + u')(c + v)}{c}}{1 + \dfrac{u'v}{c^2}} .
$$
From $|u'| < c$ and $|v| < c$ we have $c - u' > 0$, $c - v > 0$, $c + u' > 0$ and $c + v > 0$, so both numerators are positive. Also $|u'v|/c^2 < 1$, so the denominator $1 + u'v/c^2$ is positive. Hence $c - u > 0$ and $c + u > 0$, that is, $|u| < c$.
Suppose a rocket $A$ recedes from the Earth at $0.5c$ and launches forward from itself a rocket $B$ at a further $0.5c$ as seen from $A$. Galilean reasoning would make the speed of $B$ relative to the Earth equal to $c$. By ,
$$
u = \frac{0.5c + 0.5c}{1 + \dfrac{(0.5c)(0.5c)}{c^2}} = \frac{c}{1 + 0.25} = \frac{c}{1.25} = 0.8c .
$$
More extremely, taking $u' = v = 0.9c$,
$$
u = \frac{1.8c}{1 + 0.81} = \frac{1.8c}{1.81} = 0.9945c ,
$$
which again falls short of $c$. The reason no amount of addition ever exceeds the speed of light is the denominator in the velocity addition law. This structure, in which $c$ is an upper bound, is what fixes the form of momentum and energy in [Relativistic Mechanics (E=mc²)](/en/physics/relativity/relativistic-mechanics) ().
## 7. Minkowski spacetime and the spacetime interval
If both $t$ and $x$ change under a Lorentz transformation, what is the "real" quantity? Just as $x^2 + y^2$ is unchanged when the coordinate axes are rotated in Euclidean geometry, there is a quantity unchanged by Lorentz transformations.
For two events $E_1 = (t_1, x_1, y_1, z_1)$ and $E_2 = (t_2, x_2, y_2, z_2)$, writing $\Delta t = t_2 - t_1$ and so on, the quantity
$$
s^2 = c^2 \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2
$$
is called the square of the **spacetime interval** between the two events. Since $s^2$ can be negative, the symbol $s^2$ is a conventional notation for the squared quantity; $s$ itself need not be real.
Let $S$ and $S'$ be inertial frames in the standard configuration, and let the coordinate differences of the same pair of events be $(\Delta t, \Delta x, \Delta y, \Delta z)$ in $S$ and $(\Delta t', \Delta x', \Delta y', \Delta z')$ in $S'$. Then
$$
c^2\Delta t'^2 - \Delta x'^2 - \Delta y'^2 - \Delta z'^2 = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2 .
$$
That is, the spacetime interval is invariant under Lorentz transformations.
Since is a linear transformation, it applies to coordinate differences as it stands. The relations $\Delta y' = \Delta y$ and $\Delta z' = \Delta z$ come from taking differences in $y' = y$ and $z' = z$, so we need only compute the $t$ and $x$ part.
$$
\begin{aligned}
c^2\Delta t'^2 - \Delta x'^2
&= c^2\gamma^2\left(\Delta t - \frac{v\Delta x}{c^2}\right)^2 - \gamma^2\left(\Delta x - v\Delta t\right)^2 \\
&= \gamma^2\left[c^2\Delta t^2 - 2v\,\Delta t\,\Delta x + \frac{v^2}{c^2}\Delta x^2 - \Delta x^2 + 2v\,\Delta x\,\Delta t - v^2\Delta t^2\right].
\end{aligned}
$$
The second term $-2v\Delta t\Delta x$ and the fifth term $+2v\Delta x\Delta t$ cancel. Collecting the rest in $\Delta t^2$ and $\Delta x^2$,
$$
\begin{aligned}
c^2\Delta t'^2 - \Delta x'^2
&= \gamma^2\left[c^2\Delta t^2\left(1 - \frac{v^2}{c^2}\right) - \Delta x^2\left(1 - \frac{v^2}{c^2}\right)\right] \\
&= \gamma^2\left(1 - \frac{v^2}{c^2}\right)\left(c^2\Delta t^2 - \Delta x^2\right).
\end{aligned}
$$
Since $\gamma^2 = 1/(1 - v^2/c^2)$ we have $\gamma^2(1 - v^2/c^2) = 1$, so
$$
c^2\Delta t'^2 - \Delta x'^2 = c^2\Delta t^2 - \Delta x^2 .
$$
Subtracting $\Delta y^2 + \Delta z^2$ from both sides gives the claim.
The four-dimensional space carrying this invariant is called **Minkowski spacetime**. Rather than treating three dimensions of space and one of time as separate things, one regards $(ct, x, y, z)$ as the coordinates of a single point. In the words of Minkowski's own lecture of 1908, space and time by themselves sink into shadow, and only a union of the two preserves an independent reality.
The difference from Euclidean space is nothing but a sign. In Euclidean space the invariant is $\Delta x^2 + \Delta y^2$; in Minkowski spacetime it is $c^2\Delta t^2 - \Delta x^2$. This single minus sign turns rotations into hyperbolic transformations (boosts) and endows spacetime with a causal structure.
A few words on how to read the diagram. Taking the vertical axis to be $ct$ (time multiplied by $c$ so as to have the dimension of length), the worldlines of light become exactly the $45$-degree lines $ct = \pm x$.
The time axis $ct'$ of $S'$ is "the set of events with $x' = 0$", namely the line $x = vt$, which in the $(x, ct)$ plane reads $x = \beta\,(ct)$. The space axis $x'$ of $S'$ is "the set of events with $t' = 0$"; setting $t' = \gamma(t - vx/c^2)$ of equal to $0$ gives $ct = \beta x$. Thus the $ct'$ axis tilts away from the $ct$ axis, and the $x'$ axis away from the $x$ axis, each by an angle whose tangent equals $\beta$, in both cases towards the light worldline. The two axes close in symmetrically about the light worldline, and as $\beta \to 1$ both come to coincide with it.
That the $x$ axis (the line of simultaneity of $S$) and the $x'$ axis (that of $S'$) are different lines is the geometric meaning of .
For two distinct events, the sign of $s^2$, and — when $s^2 \ge 0$ — the temporal order, are the same in every inertial frame. In detail:
1. When $s^2 > 0$ (the events are **timelike** separated) or $s^2 = 0$ (**lightlike**), the sign of $\Delta t$ is the same in every inertial frame.
2. When $s^2 < 0$ (**spacelike** separated), there exist inertial frames with $\Delta t > 0$, with $\Delta t = 0$, and with $\Delta t < 0$.
That the sign of $s^2$ is frame independent is exactly . Reorienting the coordinate axes, we may assume the two events lie on the $x$ axis, so in what follows $\Delta y = \Delta z = 0$.
**Proof of (1).** The condition $s^2 \ge 0$ means $c^2\Delta t^2 \ge \Delta x^2$, that is, $|\Delta x| \le c\,|\Delta t|$. If $\Delta t = 0$ then $|\Delta x| \le 0$, so $\Delta x = 0$ and the two events coincide; hence for two distinct events $\Delta t \ne 0$. By ,
$$
\Delta t' = \gamma\left(\Delta t - \frac{v\,\Delta x}{c^2}\right)
$$
and since $\gamma > 0$, the sign of $\Delta t'$ is determined by the sign of the bracket. Estimating the magnitude of the second term,
$$
\left|\frac{v\,\Delta x}{c^2}\right| \le \frac{|v|}{c^2}\cdot c\,|\Delta t| = \frac{|v|}{c}\,|\Delta t| < |\Delta t|
$$
(the last inequality uses $|v| < c$). Since the second term is strictly smaller in magnitude than the first, the sign of the bracket agrees with the sign of $\Delta t$. Hence $\Delta t'$ and $\Delta t$ have the same sign.
**Proof of (2).** The condition $s^2 < 0$ means $|\Delta x| > c\,|\Delta t|$, and in particular $\Delta x \ne 0$. Set
$$
v_0 = \frac{c^2\,\Delta t}{\Delta x} .
$$
Then $|v_0| = c^2|\Delta t|/|\Delta x| < c^2 \cdot |\Delta x| / (c\,|\Delta x|) = c$, so $v_0$ is an admissible relative velocity. In the frame moving with this $v_0$,
$$
\Delta t' = \gamma\left(\Delta t - \frac{v_0 \Delta x}{c^2}\right) = \gamma\left(\Delta t - \Delta t\right) = 0 ,
$$
so the two events are simultaneous. Taking $v$ slightly larger or slightly smaller than $v_0$ makes $\Delta t'$ change sign continuously. Hence there exist frames in which the temporal order is reversed.
This proposition is what guarantees causality in relativity. Cause and effect are always connected at a speed no greater than that of light (that is, timelike or lightlike separated), so their order is the same for every observer. The order can be reversed only for spacelike separated events, that is, events that not even light can bridge and which therefore cannot influence each other. In short: simultaneity is relative, but causality is absolute.
A pole of proper length $10\ \mathrm{m}$ passes through a barn of proper length $8\ \mathrm{m}$ at $\beta = \sqrt{3}/2 \approx 0.866$ ($\gamma = 2$). The barn has a door at each end.
**In the barn frame $S$.** The pole contracts to $10/\gamma = 5\ \mathrm{m}$, so it fits comfortably inside the $8\ \mathrm{m}$ barn. If both doors are shut at a certain instant, the pole is entirely inside the barn.
**In the pole frame $S'$.** The barn contracts to $8/\gamma = 4\ \mathrm{m}$ while the pole stays $10\ \mathrm{m}$ long. The pole can never fit inside the barn.
This looks contradictory, but is at work. Let us check with coordinates. Measure lengths in metres and times as $ct$ (also in metres). In $S$, place the entrance door at $x = 0$ and the exit door at $x = 8$, and let the rear end of the pole be at $x = 0$ and its front end at $x = 5$ when $ct = 0$. The pole is entirely inside for $0 \le ct \le 3/0.866 = 3.46$, so let us shut both doors at $ct = 2$. The two events are
- Event $F$ (shutting the entrance door): $(ct, x) = (2, 0)$
- Event $R$ (shutting the exit door): $(ct, x) = (2, 8)$
and they are simultaneous in $S$. Apply in the form $ct' = \gamma(ct - \beta x)$, $x' = \gamma(x - \beta\, ct)$.
$$
\begin{aligned}
F:\quad & ct' = 2\,(2 - 0.866\times 0) = 4, \qquad x' = 2\,(0 - 0.866\times 2) = -3.46, \\
R:\quad & ct' = 2\,(2 - 0.866\times 8) = -9.86, \qquad x' = 2\,(8 - 0.866\times 2) = 12.5 .
\end{aligned}
$$
In the pole frame the shutting of the exit door, event $R$, happens $13.9$ **earlier** than the entrance event $F$. This agrees with $\gamma \beta L_0 = 2 \times 0.866 \times 8 = 13.9$ from .
So the sequence of events in the pole frame is this. First, far away, the exit door closes and opens again (at that moment the front end of the pole is still at $x' = 10$ and has not reached the door at $x' = 12.5$). Much later, the entrance door, having already been passed by the rear end of the pole, closes. At no moment is the pole trapped in the barn between two closed doors.
In both frames the local facts — "when each door closed, the pole was not at that door's position" — are the same. The only point of disagreement was the answer to the question "were both doors closed at the same time?", a question that is frame dependent to begin with.
## 8. Exercises
(1) Compute the Lorentz factor $\gamma$ for $\beta = 0.6$.
(2) Aboard a spacecraft flying at $0.6c$ as seen from the Earth, the onboard clock advances by 1 year. How many years pass on the Earth?
(3) Find the speed $\beta$ for which $\gamma = 2$.
(1) $\gamma = 1/\sqrt{1 - 0.6^2} = 1/\sqrt{1 - 0.36} = 1/\sqrt{0.64} = 1/0.8 = 1.25$.
(2) The one year ticked off by the onboard clock is the time between two events at the same place (aboard the ship), hence proper time in the sense of . By , the elapsed time measured on the Earth is $\Delta t = \gamma\,\Delta\tau = 1.25 \times 1 = 1.25$ years.
(3) From $\gamma = 1/\sqrt{1-\beta^2} = 2$ we get $\sqrt{1-\beta^2} = 1/2$; squaring both sides, $1 - \beta^2 = 1/4$, so $\beta^2 = 3/4$ and $\beta = \sqrt{3}/2 \approx 0.866$.
Suppose a muon is created at an altitude of $10\ \mathrm{km}$ and flies straight down at $\beta = 0.999$. Take the proper lifetime of the muon to be $2.2\ \mu\mathrm{s}$ and $c = 3.0\times10^{8}\ \mathrm{m/s}$.
(1) Find the mean lifetime of this muon as measured in the ground frame, and the distance it covers in that time.
(2) In the frame in which the muon is at rest, how thick does the $10\ \mathrm{km}$ atmospheric layer appear? How long does it take for that thickness to go by?
(3) Verify that (1) and (2) give the same conclusion as to whether the muon reaches the ground.
First compute $\gamma$. From $\beta^2 = 0.999^2 = 0.998001$,
$$
\gamma = \frac{1}{\sqrt{1 - 0.998001}} = \frac{1}{\sqrt{0.001999}} = \frac{1}{0.04471} = 22.4 .
$$
(1) Since $2.2\ \mu\mathrm{s}$ is a proper time, by the mean lifetime in the ground frame is $22.4 \times 2.2 = 49.2\ \mu\mathrm{s}$. The distance covered in that time is
$$
0.999 \times (3.0\times10^{8}) \times (49.2\times10^{-6}) = (2.997\times10^{8})\times(4.92\times10^{-5}) \approx 1.47\times10^{4}\ \mathrm{m} = 14.7\ \mathrm{km}.
$$
(2) The atmospheric thickness of $10\ \mathrm{km}$ is a proper length measured in the ground frame, so by it contracts in the muon frame to
$$
\frac{10\ \mathrm{km}}{22.4} = 0.446\ \mathrm{km} = 446\ \mathrm{m} .
$$
The time for this to go by at speed $0.999c$ is
$$
\frac{446}{2.997\times10^{8}} = 1.49\times10^{-6}\ \mathrm{s} = 1.49\ \mu\mathrm{s}.
$$
(3) In (1): "the muon can cover $14.7\ \mathrm{km}$ per mean lifetime, so traversing $10\ \mathrm{km}$ takes only $10/14.7 = 0.68$ lifetimes." In (2): "crossing $446\ \mathrm{m}$ takes only $1.49\ \mu\mathrm{s}$, which is $0.68$ of the lifetime $2.2\ \mu\mathrm{s}$." The ratios agree. Either computation gives a surviving fraction of $e^{-0.68} = 0.51$, about half. The observable quantity (how many arrive) is frame independent — the same structure as in .
A spacecraft of proper length $300\ \mathrm{m}$ flies at $v = 0.6c$ relative to the ground. Aboard the ship, the clocks at the bow and at the stern are properly synchronized ($\gamma = 1.25$, $c = 3.0\times10^{8}\ \mathrm{m/s}$).
(1) By how much do the events "the stern clock reads 0" and "the bow clock reads 0" differ in time in the ground frame?
(2) If the two clocks are photographed at a single instant of the ground frame, which one appears behind, and by how much?
(3) Explain why the answers to (1) and (2) differ.
Let $S'$ be the ship frame and $S$ the ground frame, with the stern at $x' = 0$ and the bow at $x' = 300$.
(1) The two events are simultaneous in $S'$ (both at $t' = 0$) and $\Delta x' = 300\ \mathrm{m}$. By ,
$$
\Delta t = \frac{\gamma v \Delta x'}{c^2} = \frac{1.25 \times 0.6c \times 300}{c^2} = \frac{1.25 \times 0.6 \times 300}{3.0\times10^{8}} = \frac{225}{3.0\times10^{8}} = 7.5\times10^{-7}\ \mathrm{s},
$$
that is, the bow event happens $750\ \mathrm{ns}$ later.
(2) This is the converse question. Write $t'_{+}$ for the reading of the bow clock, $t'_{-}$ for that of the stern clock, and $x_{+}$, $x_{-}$ for their respective positions in $S$. Applying $t' = \gamma(t - vx/c^2)$ from to the two clocks at a fixed time $t$ of $S$ and taking the difference, the $t$ term drops out:
$$
t'_{+} - t'_{-} = -\frac{\gamma v}{c^2}\,(x_{+} - x_{-}) .
$$
The distance $x_{+} - x_{-}$ between bow and stern measured at a single time in $S$ is $300/1.25 = 240\ \mathrm{m}$ by . Hence
$$
t'_{+} - t'_{-} = -\frac{1.25 \times 0.6c \times 240}{c^2} = -\frac{180}{3.0\times10^{8}} = -6.0\times10^{-7}\ \mathrm{s}.
$$
The bow clock appears $600\ \mathrm{ns}$ behind the stern clock.
(3) The two questions ask different things. (1) asks for "the time difference in $S$ between two events simultaneous in $S'$", (2) for "the difference in the readings of two clocks viewed at a single time in $S$". The two differ by a factor $\gamma$, and indeed $750/1.25 = 600$. To check consistency: at time $t = 0$ in $S$ the stern clock reads 0, and at $t = 750\ \mathrm{ns}$ the bow clock reads 0. At that instant $t = 750\ \mathrm{ns}$, the stern clock reads $750/1.25 = 600\ \mathrm{ns}$ by . So the bow (0 ns) is $600\ \mathrm{ns}$ behind the stern (600 ns), in agreement with (2).
One of a pair of twins travels in a spacecraft at $\beta = 0.8$ ($\gamma = 5/3$) to a star $4$ light years from the Earth, turns around immediately and comes back. The other stays on the Earth.
(1) Find the round-trip time in the Earth frame and the time elapsed aboard the spacecraft.
(2) In the spacecraft frame the Earth's clock ought to run slow, and yet on return it is the Earth twin who has aged more. In the outbound spacecraft frame, what does the Earth clock read just before the turnaround? In the inbound spacecraft frame, what does it read just after?
(3) Using the result of (2), explain why the Earth twin ages more.
(1) In the Earth frame each leg of $4$ light years at $0.8c$ takes $4/0.8 = 5$ years, so $10$ years for the round trip. The onboard clock ticks proper time, so by it records $10/\gamma = 10 \times 0.6 = 6$ years. Seen in the spacecraft frame this agrees: the distance contracts by to $4 \times 0.6 = 2.4$ light years, so each leg takes $2.4/0.8 = 3$ years and the round trip $6$ years.
(2) In the outbound spacecraft frame, at the instant when $3$ years of the ship's proper time have elapsed (just before the turnaround), the Earth clock reads $3/\gamma = 3 \times 0.6 = 1.8$ years by . In the inbound spacecraft frame the same argument shows that the Earth clock advances by only $1.8$ years during the $3$ years until the return. Since the Earth clock reads $10$ years on return, in the inbound frame just after the turnaround the Earth clock must read $10 - 1.8 = 8.2$ years.
(3) Across the turnaround, "what the Earth clock reads" jumped from $1.8$ years to $8.2$ years, a leap of $6.4$ years. The clock did not break. Because the spacecraft switched inertial frames, **the line of simultaneity of the spacecraft swung round**.
Let us verify this with . Measure distances in light years and times in years, so that $c = 1$. Let $T$ be the turnaround event, $A$ the event at which the Earth clock reads $1.8$ years, and $B$ the event at which it reads $8.2$ years.
- $A$ and $T$ are simultaneous in the outbound spacecraft frame. In that frame the distance between the two events is the contracted distance $\Delta x' = 2.4$ light years. By , the time difference in the Earth frame is $\gamma v \Delta x'/c^2 = (5/3)\times 0.8 \times 2.4 = 3.2$ years. Indeed $5 - 1.8 = 3.2$.
- $B$ and $T$ are simultaneous in the inbound spacecraft frame. The direction of the velocity is reversed, so the offset reverses too, and the time difference in the Earth frame is again $3.2$ years. Indeed $8.2 - 5 = 3.2$.
Together, $3.2 + 3.2 = 6.4$ years, exactly the size of the jump.
The Earth twin stays in a single inertial frame from beginning to end, whereas the travelling twin changes inertial frames on the way. is a statement made under the proviso "as seen from a single inertial frame", so it cannot be applied unchanged to the twin who switches frames. This asymmetry produces the asymmetry of the answer. Broken down from the spacecraft's point of view, the time elapsed on the Earth is $1.8 + 6.4 + 1.8 = 10$ years, while the spacecraft itself records $3 + 3 = 6$ years, in agreement with the Earth-frame computation.
## References
- A. Einstein, "Zur Elektrodynamik bewegter Körper", *Annalen der Physik* 17 (1905), 891–921. The original paper on special relativity; Part I (the kinematical part) proceeds from the definition of simultaneity to the Lorentz transformation.
- E. F. Taylor and J. A. Wheeler, *Spacetime Physics*, 2nd ed., W. H. Freeman, 1992 — Chapters 1–3. Its arrangement, taking the spacetime interval as the starting point, corresponds to §7 of this article.
- Katsuhiko Sato, *Sōtaisei Riron*, Iwanami Kiso Butsuri Series 9, Iwanami Shoten, 1996 (in Japanese) — the chapters on special relativity.
- R. Resnick, *Introduction to Special Relativity*, Wiley, 1968 — Chapter 2 (relativistic kinematics). Careful treatment of the relativity of simultaneity and of the various paradoxes.
- B. Rossi and D. B. Hall, "Variation of the Rate of Decay of Mesotrons with Momentum", *Physical Review* 59 (1941), 223. An early quantitative test of time dilation.
- D. H. Frisch and J. H. Smith, "Measurement of the Relativistic Time Dilation Using μ-Mesons", *American Journal of Physics* 31 (1963), 342. A test by muon counting on Mount Washington and at sea level.
## Appendix: Rapidity (the velocity parameter)
Using hyperbolic functions, the Lorentz transformation can be written in a form remarkably like a rotation. The quantity $\theta$ defined by $\beta = \tanh\theta$ is called the **rapidity**. The ranges $-1 < \beta < 1$ and $-\infty < \theta < \infty$ correspond one to one.
When $\tanh\theta = \beta$,
$$
\cosh\theta = \frac{1}{\sqrt{1 - \tanh^2\theta}} = \frac{1}{\sqrt{1-\beta^2}} = \gamma,
\qquad
\sinh\theta = \cosh\theta \cdot \tanh\theta = \gamma\beta ,
$$
so can be written
$$
\begin{pmatrix} ct' \\ x' \end{pmatrix}
=
\begin{pmatrix} \cosh\theta & -\sinh\theta \\ -\sinh\theta & \cosh\theta \end{pmatrix}
\begin{pmatrix} ct \\ x \end{pmatrix} .
$$
This is the usual rotation matrix with $\cos$ and $\sin$ replaced by $\cosh$ and $\sinh$, and it is called a **hyperbolic rotation**. The identity $\cosh^2\theta - \sinh^2\theta = 1$ is what guarantees that this transformation preserves $c^2t^2 - x^2$ ().
The advantage of this formulation is that becomes plain addition. The addition formula for hyperbolic functions,
$$
\tanh(\theta_1 + \theta_2) = \frac{\tanh\theta_1 + \tanh\theta_2}{1 + \tanh\theta_1 \tanh\theta_2},
$$
is, on setting $\beta_i = \tanh\theta_i$, precisely the velocity addition law. In other words, **rapidity is additive**: composing two boosts corresponds to adding their rapidities. Speeds never exceed $c$ because no finite sum of finite $\theta$ ever makes $\tanh\theta$ reach $1$. This viewpoint is useful when treating motion with constant acceleration (the relativistic version of uniformly accelerated motion), and when patching together local inertial frames () in [An Invitation to General Relativity (the Equivalence Principle)](/physics/relativity/equivalence-principle) and [Curved Spacetime and Gravity (How GPS Works)](/physics/relativity/curved-spacetime).