Special Relativity

Beginner

Throw a stone from a boat gliding across a calm lake. The ripples spread through the water at their own pace, whatever the boat is doing, so from the boat the ripples in front seem to crawl away while those behind race off. Waves in water or in air travel through a medium, and the speed you measure depends on how you move through it.

Physicists long thought light worked the same way, as a wave in an invisible “ether” filling space. The Earth circles the Sun at 30 km/s, so light should seem slightly faster in one direction than in another. In 1887 Michelson and Morley compared two light beams sent at right angles, turned their instrument around, and found no difference at all. Light needs no medium, and its speed, about 300,000 km every second, is the same for every observer.

Einstein took this seriously in 1905, and it changes time itself. In the animation a flash bounces between two mirrors, one tick per trip. In the resting clock the light goes straight up and down. The other clock rushes sideways, so, seen from the ground, its light follows longer slanted zig-zags. Light cannot speed up to make up the distance, so each tick takes longer: four ticks against five. Everything on board, heartbeats included, slows the same way, so the traveller notices nothing. Muons, fast particles made high in the air by cosmic rays, reach the ground only because their time runs slow. Moving objects are also measured shorter along their motion, and no push can bring anything with mass up to light's speed.

Check yourself: if the moving clock went faster, would its zig-zag be longer or shorter, and its ticks faster or slower? Answer: longer, so slower.

Intermediate

Frames, events and the two postulates

No experiment inside a smoothly cruising train with its blinds drawn can show whether the train is moving. An observer in relativity stands for a whole inertial frame: rulers and synchronised clocks moving at constant velocity, like a smoothly cruising train. An event happens at one place and one instant, such as a flash or the tick of a clock, so each frame gives it a position $x$ and a time $t$.

The theory rests on two postulates. First, the laws of physics take the same form in every inertial frame, so no experiment inside a closed cabin can show whether it is moving. Second, light in vacuum moves at the same speed in every inertial frame, $c\approx 3.00\times 10^{8}m/s$, however its source or the observer moves. Together they force time intervals and lengths to depend on the frame.

Reading the light-clock triangle

In a light clock a flash bounces between two mirrors a distance $d$ apart, and each crossing is one tick. In the clock's rest frame the light goes straight across, so a tick lasts $\Delta t_0=d/c$. This is a proper time: the time between two events read on a single clock present at both of them.

In a frame where the clock moves sideways at speed $v$ (the blue arrow), a tick lasts $\Delta t$ and the mirrors move on by $v\Delta t$, so the light runs along a slant. The animation draws this right triangle over the first tick: the dashed green side $c\Delta t_0$ is the mirror gap, the blue base $v\Delta t$ the shift of the clock, and the gold hypotenuse $c\Delta t$ the path of the light, still covered at speed $c$. Pythagoras gives $(c\Delta t{)}^2=(c\Delta t_0{)}^2+(v\Delta t{)}^2$, so
$\Delta t=\gamma \Delta t_0$
$\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$

The Lorentz factor $\gamma$ is 1 at rest and grows without limit as $v$ approaches $c$: it is 1.005 at $0.1c$, 1.25 at $0.6c$ and 7.09 at $0.99c$. Every process aboard, a passenger's heartbeat included, slows by the same factor; otherwise comparing them would reveal the motion.

Length contraction and simultaneity

Lengths change too. An object's length in its own rest frame is its proper length $L_0$, and a frame in which it moves measures it shorter along the motion,
$L=\frac{L_0}{\gamma }$
but not across it. So the moving clock is drawn 24 units wide instead of 30 yet just as tall, and its mirror gap $d$ is the same in both frames, as the triangle assumed.

Both effects are mutual: a passenger riding with the moving clock finds the ground clocks slow and the ground rulers short. There is no contradiction, because the two frames disagree about simultaneity. Two flashes at the same moment at the two ends of a platform are not simultaneous for a passenger on a passing train: for her, the flash ahead of the train comes first. Timing a moving clock takes two of your own clocks at the places it passes, and the moving observer says those two are not synchronised.

In every problem, first find the rest frame of the clock or object: proper times and lengths belong to it.

Energy, momentum and the speed limit

If Newton's momentum $mv$ were conserved in a collision seen from one frame, it would not be in a frame moving at a large fraction of $c$. The relativistic momentum, energy and kinetic energy all carry the factor $\gamma$:

• $p=\gamma mv$
• $E=\gamma m{c}^2$
• $K=(\gamma -1)m{c}^2$

Here $m$ is the ordinary mass, the same in every frame, and $E$ is the total energy. Even at rest a body has the rest energy $m{c}^2$; $K$ is what motion adds. An electron has $m{c}^2=511\mathrm{keV}$, so at $0.6c$ its energy is $E=1.25\cdot 511\approx 639\mathrm{keV}$ and $K\approx 128\mathrm{keV}$, while Newton's $\frac{1}{2}m{v}^2$ gives only 92 keV. Since $\gamma$ has no upper limit, neither has $K$: at $0.99c$ it is already $6.1m{c}^2$, and no finite energy brings a body with mass up to $c$.

Worked example: the clocks in the animation

On the screen light is slowed to 50 units per second and the mirrors are 76 units apart, so the clock at rest ticks every $\Delta t_0=76/50=1.52\mathrm{s}$. The other clock moves at $v=0.6c$, which is 30 units per second:
$\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{0.64}}=\frac{1}{0.8}=1.25$
$\Delta t=1.25\cdot 1.52\mathrm{s}=1.90\mathrm{s}$

These are the values on the line “1 tick: 1.52 s → 1.90 s”. The triangle checks them: in one moving tick the clock advances $30\cdot 1.90=57$ units and the light runs $50\cdot 1.90=95$ units, and $57^2+76^2=95^2$, a 3-4-5 triangle. After 7.6 s the green counter of the clock at rest reaches $7.6/1.52=5$ ticks and the red counter above the moving clock $7.6/1.90=4$: the ratio 5 : 4 is $\gamma$.

The blue arrow $v$ keeps its length because the speed is constant, and the moving box is drawn $30/1.25=24$ units wide. For the moving clock nothing is wrong: its four ticks took $4\cdot 1.52=6.08\mathrm{s}$ of its own time.

Worked example: muons from the upper air

Cosmic rays hitting the upper atmosphere produce muons, unstable particles that live on average $\Delta t_0=2.2\mu \mathrm{s}$ at rest. Do muons made 10 km up, falling at $v=0.995c$, reach the ground?

Without relativity a muon would cover only $0.995\cdot 3.00\times 10^8\cdot 2.2\times 10^{-6}\approx 660\mathrm{m}$ in one average lifetime. The 10 km trip takes $10000/(2.985\times 10^8)\approx 33.5\mu \mathrm{s}$, about 15 lifetimes, and since a fraction $e^{-n}$ is left after $n$ lifetimes, only one muon in four million would arrive.

With relativity, $\gamma =1/\sqrt{1-0.995^2}\approx 10.0$: in the Earth frame the average lifetime is about $22\mu \mathrm{s}$, the trip lasts only 1.5 lifetimes, and a fraction $e^{-1.52}\approx 0.22$, about one muon in five, survives.

In the muon's frame its clock runs normally, but the atmosphere rushes past, contracted to $10\mathrm{km}/10.0\approx 1.0\mathrm{km}$, which goes by in about $3.35\mu \mathrm{s}$: again 1.5 lifetimes. What one frame calls time dilation, the other calls length contraction.

Common mistakes

• The moving clock is not faulty: in its own frame it ticks every 1.52 s; the 1.90 s is measured from a frame in which it moves.
• Proper time is the shortest interval, so multiply it by $\gamma$, never divide: 2.0 s of proper time at $0.6c$ becomes 2.5 s, not 1.6 s.
• Do not confuse $\gamma$ with the root it contains: at $0.6c$ the root $\sqrt{1-v^2/c^2}$ is 0.8, and $\gamma$, its inverse, is 1.25.
• Contraction acts only along the motion: the mirror gap stays 76 units while the width shrinks from 30 to 24 units.
• Speeds do not simply add: a probe fired forward at $0.6c$ from a ship moving at $0.6c$ travels at $1.2/1.36\approx 0.88c$, not $1.2c$.

Where you meet it

GPS satellites move at about 3.9 km/s, so their clocks lose about 7 microseconds a day to motion, while the weaker gravity at their height makes them gain more; uncorrected, position fixes would drift by kilometres a day. Accelerator magnets are designed for $p=\gamma mv$, with protons at $\gamma$ in the thousands. Electron microscopes accelerate electrons through 300 kV to $0.78c$ ($\gamma \approx 1.59$), and hospital PET scanners detect pairs of 511 keV photons from electron–positron annihilation.

The rest energy returns in Fission, Fusion and E = mc², and the gravitational part of the GPS correction in Gravity as Curved Spacetime.

Advanced

From two postulates to the Lorentz transformation

If two inertial frames both see a light pulse move at the same speed $c$, how are the coordinates of an event in one frame related to its coordinates in the other? The answer, the Lorentz transformation, contains every effect of special relativity, so it is worth deriving step by step.

Take a frame $S^{\prime }$ that moves at velocity $v$ along the $x$ axis of a frame $S$, and set both sets of clocks to zero as the two origins pass each other, so that event has $t=t^{\prime }=0$. Galileo's rule, $x^{\prime }=x-vt$ with $t^{\prime }=t$, cannot be right. It subtracts velocities: a light pulse with $x=ct$ would have $x^{\prime }=(c-v)t$ and so move at $c-v$ in $S^{\prime }$, against the second postulate.

Three facts fix the correct rule. First, it must be linear: space and time are the same everywhere, so the rule cannot single out any place or instant, and a body that moves uniformly in one frame must move uniformly in the other.

Second, distances across the motion do not change: $y^{\prime }=y$ and $z^{\prime }=z$. Picture two identical rings flying through each other along their common axis. If motion shrank a ring, each frame would find the other ring smaller and see it pass inside its own; both cannot be true, since which ring passes inside is a plain fact.

Third, the origin of $S^{\prime }$, where $x^{\prime }=0$, sits at $x=vt$. A linear rule that gives $x^{\prime }=0$ exactly there must be proportional to $x-vt$:
$x^{\prime }=\gamma (x-vt)$
Here $\gamma$ is a factor that may depend on $v$ but not on $x$ or $t$. In $S^{\prime }$ the frame $S$ moves at $-v$, and by the first postulate neither frame is special, so the inverse has the same form, with the same factor and the opposite velocity:
$x=\gamma (x^{\prime }+v{t}^{\prime })$

The second postulate now fixes $\gamma$. Send a light pulse from the common origin at $t=t^{\prime }=0$. In $S$ it obeys $x=ct$, and in $S^{\prime }$ it obeys $x^{\prime }=c{t}^{\prime }$. Putting these into the two equations gives $c{t}^{\prime }=\gamma (c-v)t$ and $ct=\gamma (c+v)t^{\prime }$. Multiplying them and cancelling $t{t}^{\prime }$ leaves $c^2={\gamma }^2(c^2-v^2)$, so
$\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$
which is the Lorentz factor that the light clock gave for time dilation.

The rule for time follows by eliminating $x^{\prime }$. Substitute $x^{\prime }=\gamma (x-vt)$ into $x=\gamma (x^{\prime }+v{t}^{\prime })$ and solve for $t^{\prime }$; with $1-1/{\gamma }^2=v^2/c^2$ the result simplifies, and the full transformation reads
$x^{\prime }=\gamma (x-vt),t^{\prime }=\gamma \left(t-\frac{vx}{c^2}\right)$
When $v\ll c$, $\gamma \approx 1$ and the term $vx/c^2$ is negligible, so Galileo's rule returns. That new term is the heart of the matter: the time of an event in $S^{\prime }$ depends not only on when it happens in $S$ but also on where.

Tilted axes: the spacetime diagram

Legend

$x$	Position in $S$: the horizontal axis, the line $ct=0$.	$\mathrm{m}$
$ct$	Time in $S$ multiplied by $c$: the vertical axis, so that time and distance share one unit.	$\mathrm{m}$
$x^{\prime }$	Position axis of $S^{\prime }$: the line of events with $t^{\prime }=0$.	$\mathrm{m}$
$c{t}^{\prime }$	Time axis of $S^{\prime }$, time multiplied by $c$: the worldline of the origin of $S^{\prime }$, where $x^{\prime }=0$.	$\mathrm{m}$
$x=ct$	Worldline of a light pulse leaving $A$; it also obeys $x^{\prime }=c{t}^{\prime }$.
$A$	Event at the origin of both frames.
$B$	Event on the $x$ axis, simultaneous with $A$ in $S$.
$\Delta x$	Distance between $A$ and $B$ in $S$.	$\mathrm{m}$
$c\Delta t^{\prime }$	Time between $A$ and $B$ in $S^{\prime }$, multiplied by $c$ and read along the $c{t}^{\prime }$ axis. It is negative: $B$ comes first.	$\mathrm{m}$

A spacetime diagram draws the Lorentz transformation instead of writing it, and it shows at a glance why two frames disagree about which distant events happen at the same time. Figure 1 is drawn in the frame $S$, for $v=0.6c$.

Time is plotted upward as $ct$, the time multiplied by $c$, so that both axes of $S$ measure length in the same unit. A light pulse leaving the origin then follows the dashed 45° line $x=ct$.

The frame $S^{\prime }$ brings two axes of its own. Its time axis $c{t}^{\prime }$ is the set of events with $x^{\prime }=0$: the worldline $x=vt$ of its origin, which leans to the right of the $ct$ axis. Its position axis $x^{\prime }$ is the set of events with $t^{\prime }=0$; setting $t^{\prime }=0$ in $t^{\prime }=\gamma (t-vx/c^2)$ gives $ct=(v/c)x$, a line that leans up from the $x$ axis. Both axes of $S^{\prime }$ lean by the same angle, towards the light line from either side, so the light line still lies midway between them. The pulse therefore obeys $x^{\prime }=c{t}^{\prime }$ as well: the second postulate is built into the picture.

Now look at the events $A$ and $B$. Both lie on the $x$ axis, so in $S$ they happen at the same time, $t=0$, a distance $\Delta x$ apart. In $S^{\prime }$ the events with equal $t^{\prime }$ lie on lines parallel to the $x^{\prime }$ axis, just as events with equal $t$ lie on horizontal lines in $S$. To read the time of $B$ in $S^{\prime }$, follow the dashed line through $B$, parallel to $x^{\prime }$, until it meets the $c{t}^{\prime }$ axis. It meets that axis below $A$, at $c\Delta t^{\prime }$: by the clocks of $S^{\prime }$, the event $B$ happens first.

Putting $\Delta t=0$ into the transformation gives the size of the gap:
$\Delta t^{\prime }=-\frac{\gamma v\Delta x}{c^2}$
The minus sign says that $B$, the event further along the direction of motion, comes first. At the figure's speed $\gamma =1.25$, so $\gamma v/c=0.75$: two lab flashes at the same moment, one light-second apart, are 0.75 s apart for the moving clocks. Whether two distant events are simultaneous depends on the frame.

Time dilation, length contraction, adding speeds

The transformation now delivers the effects met earlier in a few lines each, and it shows that they hold for any clock and any rod, not only for the light clock.

Time dilation. Take a clock at rest at the origin of $S^{\prime }$. In $S$ it moves along $x=vt$, so its reading is $t^{\prime }=\gamma (t-v^{2}t/c^2)=\gamma t(1-v^2/c^2)=t/\gamma$. While a time $\Delta t$ passes in $S$, the moving clock advances only $\Delta t_0=\Delta t/\gamma$; that is, $\Delta t=\gamma \Delta t_0$.

Length contraction. A rod of proper length $L_0$ is at rest in $S^{\prime }$, so its ends are $\Delta x^{\prime }=L_0$ apart. To measure a moving rod, $S$ must mark both ends at the same time, $\Delta t=0$, and then the transformation gives $\Delta x^{\prime }=\gamma \Delta x$. The length in $S$ is therefore $L=\Delta x=L_0/\gamma$.

Adding speeds. The inverse transformation is the same rule with $-v$: $x=\gamma (x^{\prime }+v{t}^{\prime })$ and $t=\gamma (t^{\prime }+v{x}^{\prime }/c^2)$. A body moving at $u^{\prime }$ along $x$ in $S^{\prime }$ covers $d{x}^{\prime }=u^{\prime }d{t}^{\prime }$, so $dx=\gamma (u^{\prime }+v)d{t}^{\prime }$ and $dt=\gamma (1+u^{\prime }v/c^2)d{t}^{\prime }$. Their ratio is its speed in $S$:
$u=\frac{u^{\prime }+v}{1+u^{\prime }v/c^2}$
For $u^{\prime }=c$ it gives $u=c$, as the second postulate demands. Two speeds of $0.6c$ compose not to $1.2c$ but to $1.2/1.36=15/17\approx 0.88c$.

The size of all these effects is set by $\gamma$, and the graph in the animation plots it against $v/c$ up to 0.95, where $\gamma =3.20$. Near the origin the curve is flat, $\gamma \approx 1+\frac{1}{2}{v}^2/c^2$, only 0.5 % above 1 at $0.1c$; this is why relativity escapes everyday notice. At the dot, $0.6c$, this parabola gives only 1.18 against the true 1.25, and the slope $d\gamma /d(v/c)=(v/c){\gamma }^3$ is already 1.17. At the right edge the slope is 31, and the curve climbs towards a vertical asymptote at $v=c$.

The invariant interval and rapidity

Legend

$x$	Position in $S$: the horizontal axis.	$\mathrm{m}$
$ct$	Time in $S$ multiplied by $c$: the vertical axis, the worldline of the clock at rest.	$\mathrm{m}$
$x^{\prime }$	Position axis of $S^{\prime }$.	$\mathrm{m}$
$c{t}^{\prime }$	Time axis of $S^{\prime }$, time multiplied by $c$: the worldline of the moving clock.	$\mathrm{m}$
$x=ct$	Worldline of light from $O$: the dashed line the hyperbola approaches.
$O$	Event where the two clocks meet and start.
$Q$	Tick of the clock at rest, after the proper time $\tau$.
$P$	Tick of the moving clock, after the same proper time $\tau$.
$c\tau$	Proper time between $O$ and the tick, multiplied by $c$: the height of $Q$ and the $c{t}^{\prime }$ coordinate of $P$.	$\mathrm{m}$
$\gamma c\tau$	Time of $P$ in $S$, multiplied by $c$: longer than $c\tau$ by the factor $\gamma$.	$\mathrm{m}$
$s$	Interval from $O$, the same for every point of the hyperbola; here $s=c\tau$.	$\mathrm{m}$
$\alpha$	Tilt of the axes of $S^{\prime }$ from those of $S$: $\tan \alpha =v/c$.	$\mathrm{rad}$

Times and lengths each change from frame to frame, but one combination of them does not. This invariant, the interval, plays the part in spacetime that distance plays in ordinary space, and Figure 2 shows how to read it on a diagram.

Two identical clocks start together at the event $O$. One stays at rest in $S$ and climbs the $ct$ axis; the other moves with $S^{\prime }$ and climbs the $c{t}^{\prime }$ axis. Each ticks once after the proper time $\tau$, the time it reads itself. The clock at rest ticks at the event $Q$, at the height $c\tau$. The moving clock ticks at the event $P$, where its own axis reads $c{t}^{\prime }=c\tau$. The dashed line from $P$ across to the $ct$ axis shows when $S$ dates that tick: at $ct=\gamma c\tau$, later than $Q$. This is time dilation once more.

Notice the puzzle on paper: the segment $OP$ is drawn longer than $OQ$, yet both stand for the same $c\tau$. The curve $s$ resolves it. Both $P$ and $Q$ lie on this hyperbola, the set of all events at the same interval $s=c\tau$ from $O$. On a spacetime diagram equal intervals are read along such hyperbolas, not with a ruler.

The interval between two events is defined by
$s^2=c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2$
Put the transformation into the right-hand side, square and subtract: the cross terms cancel, the factor ${\gamma }^2(1-v^2/c^2)$ equals 1, and the same $s^2$ comes out in $S^{\prime }$. A rotation in a plane leaves $x^2+y^2$ unchanged in just the same way.

The analogy with rotation is exact. Define the rapidity $\phi$ by $v/c=\tanh \phi$. Then $\gamma =\cosh \phi$ and $\gamma v/c=\sinh \phi$, and the transformation becomes $c{t}^{\prime }=ct\cosh \phi -x\sinh \phi$ and $x^{\prime }=x\cosh \phi -ct\sinh \phi$. This is a rotation in the $(x,ct)$ plane by the angle $\phi$, with hyperbolic functions in place of circular ones. Figure 2 shows the same boost with a ruler: the axes of $S^{\prime }$ tilt by the angle $\alpha$, with $\tan \alpha =v/c=\tanh \phi$. The rapidities of successive boosts along one line simply add, which is the velocity-addition law written in terms of rapidity.

The animation's speed gives a check of these relations. In the animation $\phi =\mathrm{artanh}0.6=\ln 2$, so $\gamma =\cosh (\ln 2)=(2+\frac{1}{2})/2=1.25$, and two such boosts give $\tanh (2\ln 2)=\tanh (\ln 4)=15/17$, the speed found above.

The light clock measures an interval. One tick of the moving clock spans $\Delta t=1.90\mathrm{s}$ and $\Delta x=57$ units in $S$, with light at 50 units per second, so $c\Delta t=95$ units and $(c\Delta t{)}^2-\Delta x^2=95^2-57^2=76^2$. The interval is the mirror gap, 76 units, the same in every frame; divided by $c$, it is the proper time of one tick, 1.52 s.

Along a worldline that bends, add up the proper time of each short straight piece: $\tau =\int dt/\gamma$. Because of the minus signs in $s^2$, motion through space reduces the proper time instead of adding to it. Among all worldlines between two events, the straight, inertial one therefore has the longest proper time, the reverse of Euclid's geometry, where the straight line is the shortest. A twin who travels away and turns back ages less, as the worked problem below shows.

Energy and momentum as one four-vector

Legend

$m{c}^2$	Rest energy: one leg of the triangle, copied onto the hypotenuse by the arc.	$\mathrm{J}$
$pc$	Momentum multiplied by $c$: the other leg.	$\mathrm{J}$
$E$	Total energy, $\gamma m{c}^2$: the hypotenuse.	$\mathrm{J}$
$K$	Kinetic energy, $E-m{c}^2$: the part of the hypotenuse beyond the swung leg.	$\mathrm{J}$
$\theta$	Angle between the $m{c}^2$ leg and the hypotenuse: $\sin \theta =pc/E=v/c$.	$\mathrm{rad}$

If momentum is to be conserved in every inertial frame, energy and momentum must transform together, just as $t$ and $x$ do. Built into one four-vector, that pair leads to the relation that Figure 3 draws as a right triangle.

First read the triangle. Its legs are the rest energy $m{c}^2$ and $pc$, the momentum times $c$; its hypotenuse is the total energy $E$, and the right angle sits at the top, so $E^2=(pc{)}^2+(m{c}^2{)}^2$. Swing the leg $m{c}^2$ about the left vertex down onto the hypotenuse, along the dashed arc: the piece of the hypotenuse left over at the right-hand end is $K=E-m{c}^2$, the kinetic energy. At the left vertex the angle $\theta$ has $\sin \theta =pc/E$, which equals $v/c$, as shown below.

Now the derivation. The step $(cdt,d\vec{x})$ between two nearby events on a body's worldline transforms like $(ct,\vec{x})$. Dividing it by the proper time $d\tau =dt/\gamma$, which is the same in all frames, keeps that property; the result is the four-velocity $\gamma (c,\vec{v})$. Multiplying by the mass $m$ gives the four-momentum
$(E/c,\vec{p})=(\gamma mc,\gamma m\vec{v})$
Its space part, $\vec{p}=\gamma m\vec{v}$, is the momentum conserved in collisions. Its time part is $E/c$, and a boost mixes the two, just as it mixes $ct$ and $x$: the momentum seen in $S^{\prime }$ contains energy from $S$. So momentum can be conserved in every frame only if energy is conserved too, and the two laws become one.

Like the interval, this four-vector has a length that every frame agrees on: $(E/c{)}^2-p^2={\gamma }^2{m}^2{c}^2(1-v^2/c^2)=m^2{c}^2$. Multiplied by $c^2$, it reads
$E^2=(pc{)}^2+(m{c}^2{)}^2$
Dividing the space part by the time part gives $pc/E=v/c$, so $v=p{c}^2/E$, and this is the $\sin \theta$ of Figure 3. A massless particle has $E=pc$ and therefore moves at $c$.

For small speeds, expand $\gamma$ in powers of $v^2/c^2$:
$E=\gamma m{c}^2\approx m{c}^2+\frac{1}{2}m{v}^2+\frac{3}{8}m\frac{v^4}{c^2}$
Newton's kinetic energy is the first correction, and $K=(\gamma -1)m{c}^2$ is the exact form. The rest energy $m{c}^2$ that remains at $v=0$ is real: a hot body is heavier by $\Delta E/c^2$, and a nucleus is lighter than its separate parts. The mass $m$ is the same in every frame; what grows with speed is $\gamma$.

At the animation's speed, $0.6c$, the triangle of Figure 3 is a 3-4-5 triangle: $pc=\gamma mvc=0.75m{c}^2$ and $E=1.25m{c}^2$, and indeed $0.75^2+1^2=1.25^2$. The kinetic energy is $K=0.25m{c}^2$, while Newton's $\frac{1}{2}m{v}^2=0.18m{c}^2$ is 28 % too low.

Doppler shift, fields and acceleration

Three short applications show how far the transformation reaches: to light from moving sources, to electric and magnetic fields, and to bodies that speed up.

Doppler shift. Light needs no medium, so its Doppler shift can depend only on the relative velocity of source and observer. Let a source recede at $v$ and emit a crest every $T$ in its own frame. In our frame the crests leave every $\gamma T$, because of time dilation, and between two crests the source moves away by $v\gamma T$, which adds $v\gamma T/c$ to the travel time of the next crest. The received period is therefore
$\gamma T(1+v/c)=T\sqrt{\frac{1+v/c}{1-v/c}}$

At $0.6c$ the factor is exactly 2, which is $e^{\phi }$ for the rapidity $\phi =\ln 2$: a 500 nm spectral line arrives at 1000 nm from a receding source and at 250 nm from an approaching one. For light received at right angles to the motion the distance does not change, so only time dilation remains, lowering the frequency by the factor $\gamma$.

Fields. A charge at rest has only an electric field. In a frame where it moves, it is a current and has a magnetic field too, and the magnetic force on a moving charge is an electric force in the charge's own rest frame: a boost mixes electric and magnetic fields as it mixes $t$ and $x$. Maxwell's equations already obey the Lorentz transformation, so it was mechanics, not electromagnetism, that had to change.

Acceleration. Accelerated bodies need no new theory while spacetime is flat: at each instant, use the inertial frame moving with the body. Suppose a ship keeps a constant proper acceleration $a$, the one the passengers feel. In each short stretch $d\tau$ of ship time it gains the same small rapidity $ad\tau /c$, and rapidities add, so $\phi =a\tau /c$ grows uniformly and $v=c\tanh (a\tau /c)$ never reaches $c$. One year of ship time at $9.81m/{\mathrm{s}}^2$ gives $a\tau /c\approx 1.03$ and $v\approx 0.77c$.

Worked problem: a round trip to a nearby star

Legend

$t$	Time in the Earth frame, drawn in years: the horizontal axis, which is also Earth's own worldline.	$\mathrm{s}$
$x$	Distance from Earth along the trip, drawn in light-years.	$\mathrm{m}$
$L_0$	Distance to the star in the Earth frame: 3.00 light-years.	$\mathrm{m}$

A traveller flies to a star and back and returns younger than her twin on Earth. This is the twin paradox, and this problem applies the whole lesson to it. Figure 4 draws the trip in the Earth frame, with the time $t$ in years across and the distance $x$ in light-years up, so that light runs at 45°. Earth stays on the $t$ axis, and the star, at the distance $L_0$, is the dashed horizontal line. The bent line is the traveller's worldline, climbing to the star and falling back. The numbers are clock readings: when she turns, Earth's clock shows 5 and hers 4; when she lands, 10 and 8.

Send the animation's clock at $0.6c$ to a star 3.00 light-years away in the Earth frame and straight back, ignoring the brief turnaround. On Earth each leg takes $3.00/0.6=5.00$ years, 10 in all. The traveller's clock runs slow by $\gamma =1.25$, so it shows 4.00 years per leg, 8 in all.

Her own view agrees. In her frame the distance to the star is contracted to $L_0/\gamma =3.00/1.25=2.40$ light-years, which she covers at $0.6c$ in the same 4.00 years. The interval agrees too: in years and light-years, each leg has $(c\Delta t{)}^2-\Delta x^2=5^2-3^2=4^2$, a proper time of 4 years. The ratio 10 : 8 is the 5 : 4 of the tick counters in the animation.

Why is the trip not symmetric, when on the outward leg each twin finds the other's clock slow? To see it, let each twin send a light signal every year by her own clock, and count the signals that arrive. The Doppler factor at $0.6c$ is 2: while the twins move apart, the signals arrive every 2 years; while they approach, every half year.

The thin lines of Figure 4 are Earth's yearly signals, rising at 45° from the $t$ axis until they meet her worldline, and the dots mark where they arrive. On the way out she receives one every 2 years of her time, 2 signals in her 4 years; the signal of year 2 reaches her just as she turns. On the way back she receives one every half year, 8 more; the signal of year 10 arrives as she lands. So the dots crowd together after the turn, and her count of 10 signals says that Earth has aged 10 years.

Earth sees the other side. It receives her signals every 2 years until the light from her turn reaches it. She turned 3 light-years away in year 5, so Earth sees the turn only in year 8, three years after the event: 4 signals in 8 years. In the last 2 years her signals come every half year, 4 more: 8 in all, so she has aged 8 years.

The asymmetry is the change of frame. The signals she receives speed up halfway through her trip, at the turn that she alone feels, while those Earth receives speed up only 2 years before she is back. In the language of the interval, her bent worldline has the shorter proper time.

Limits of validity

Special relativity assumes flat spacetime, with inertial frames that extend everywhere. This section shows where that assumption, and with it the theory, stops.

Gravity breaks it first. A clock lower in a gravitational field runs slow, by a fraction equal to the difference in gravitational potential divided by $c^2$: about $7\times 10^{-10}$ between the Earth's surface and far space. GPS clocks lose about 7 µs a day to their motion and gain about 46 µs a day to their height, so they are tuned to tick slightly slow before launch. Where gravity varies across the region studied, spacetime is curved and general relativity takes over; special relativity still holds in any small, freely falling laboratory.

At everyday speeds the corrections, of order $\frac{1}{2}{v}^2/c^2$, are tiny: $5\times 10^{-15}$ for a car at 30 m/s and $3.5\times 10^{-13}$ for an airliner at 250 m/s. Yet atomic clocks flown around the world in 1971 gained or lost tens to hundreds of nanoseconds, as predicted once gravity was included.

At the other extreme nothing has failed: 6.8 TeV protons in the Large Hadron Collider have $\gamma \approx 7250$ and trail light by less than 3 m/s, and muons circling a storage ring at $\gamma \approx 29$ live 29 times longer than at rest.

Quantum mechanics sets the last limit. Above a collision energy of $2m{c}^2$ in the centre-of-mass frame, 1.02 MeV for electrons, a particle and its antiparticle can appear. Only the union of relativity and quantum mechanics can describe this, and that union predicted antimatter before its discovery in 1932.

Where this leads

Gravity as Curved Spacetime keeps the interval but lets spacetime curve, so that free fall follows the straightest worldlines and clocks depend on height. Fission, Fusion and E = mc² turns rest energy into the power of reactors and stars: fusing hydrogen into helium releases 0.7 % of the mass. Photoelectric Effect and Wave–Particle Duality give light the energy $E=hf$ and the momentum $p=E/c=h/\lambda$, the massless case of the four-momentum. Particles and Forces: the Standard Model uses the conservation of four-momentum to find particles that leave no track. Looking back, Magnetic Force on Moving Charges and Electromagnetic Waves hold the fields that mix and the speed $c$ that started it all.