1The idea in one paragraph

Two observers moving steadily past each other give the same event different coordinates. Newton's rule for relating them was simple: positions shift by vt, times are identical, velocities subtract. That is Galilean relativity, and it works perfectly at everyday speeds. It fails for light, because every observer who measures the speed of light gets the same answer, however fast they are moving. Einstein took that fact as a starting point, and the price was that time and space stop being the same for everyone. Moving clocks run slow, moving objects are shorter, and two events that are simultaneous for one observer are not for another. The Lorentz transformations are the rules that replace Galilean ones, and space-time diagrams are the picture that makes them visible. The one thing every observer agrees on is a combination of time and distance called the space-time interval.

2Reference frames and Galilean relativity

An event is something that happens at one place at one instant: a lamp flashing, a ball being caught. A reference frame is a set of rulers and synchronised clocks that an observer uses to give each event a position x and a time t. An inertial reference frame is one that is not accelerating: an observer in it sees a body with no resultant force on it stay at rest or move at constant velocity. A train running smoothly on straight track is, to a good approximation, an inertial frame; a car taking a sharp bend is not.

Newton's laws hold in every inertial frame. A ball dropped in the smooth train falls straight down, exactly as it does on the platform; no experiment done inside the train can tell you whether it is moving. That statement, that the laws of mechanics are the same in all inertial reference frames, is Galilean relativity.

Figure 1 sets up the standard arrangement you will use for the rest of this subtopic. Frame S is the platform. Frame S′ is the train, moving at velocity v in the +x direction relative to S. Their origins coincide at t = t′ = 0.

After time t, the origin of S′ has moved a distance vt. An event at position x in S is therefore at x − vt in S′. In Newton's world time is universal, so both frames give the event the same time. These are the Galilean transformation equations:

x′ = x − vt and t′ = t

Follow a moving object and they give velocities too. If the object moves at u in S, it moves at

u′ = u − v in S′

A lamp on the platform flashes at x = 200 m, t = 5.0 s. The train moves at v = 30 m s⁻¹.

A drone flies along the platform at u = 35 m s⁻¹. Its velocity measured on the train is u′ = 35 − 30 = 5 m s⁻¹. A drone flying the other way at u = −35 m s⁻¹ has u′ = −35 − 30 = −65 m s⁻¹. Signs matter: u, u′ and v are all measured along +x.

3The problem with light, and the two postulates

Now apply u′ = u − v to a beam of light moving at c along +x. Galilean relativity says an observer on a train moving at v measures the light at c − v. By the late nineteenth century there were two reasons to doubt that. Maxwell's theory of electromagnetism predicts a single speed for light, c, calculated from two constants of electricity and magnetism, with no mention of who is measuring it. And experiments designed to detect the Earth's motion through space by changes in the measured speed of light, most famously Michelson and Morley's in 1887, found no change.

Einstein, in 1905, treated these results as a law of nature and built everything else on them. The two postulates of special relativity are:

1. The laws of physics are the same in all inertial reference frames.

2. The speed of light in a vacuum is the same, c, for all inertial observers, whatever the motion of the source or the observer.

The first extends Galilean relativity from mechanics to all of physics, including electromagnetism. The second is the shock. If two observers moving relative to each other both measure the same light pulse at the same speed, then they must disagree about distances or times, or both. The rest of this page is the working-out of that disagreement.

4The Lorentz transformations

The equations that respect both postulates are the Lorentz transformations. For the same set-up as Figure 1:

x′ = γ(x − vt) and t′ = γ(t − vx/c²), where γ = 1 / √(1 − v²/c²)

γ, the Lorentz factor, is always at least 1. You are not asked to derive these equations, only to use them. The data booklet also gives them in terms of intervals, Δx′ = γ(Δx − vΔt) and Δt′ = γ(Δt − vΔx/c²), which is the form to use for the separation between two events.

Figure 2 shows how γ behaves. At everyday speeds it is 1 to many decimal places: for an airliner at 300 m s⁻¹, γ − 1 is about 5 × 10⁻¹³. Then γ is 1.25 at 0.60c, 1.67 at 0.80c, 2.0 at about 0.866c, and it grows without limit as v approaches c. Put γ = 1 and v/c² ≈ 0 into the Lorentz equations and you get x′ = x − vt and t′ = t: Galilean relativity is the low-speed limit of special relativity, which is why Newton was never caught out by a train.

Worked example. Frame S′ moves at v = 0.60c relative to S, so γ = 1 / √(1 − 0.36) = 1.25. An event happens at x = 1500 m, t = 4.0 μs in S. Find its coordinates in S′.

Notice that t′ is not t. In special relativity the time of an event depends on the frame, and it depends on where the event is: the vx/c² term.

A tidy way to work is in units where c is simple. Light travels 300 m in 1 μs, so if you measure distances in metres and times in microseconds, c = 300 m μs⁻¹ and 0.60c = 180 m μs⁻¹.

5Relativistic velocity addition

The Lorentz transformations replace u′ = u − v with:

u′ = (u − v) / (1 − uv/c²)

Here u is an object's velocity measured in S, v is the velocity of S′ relative to S, and u′ is the object's velocity measured in S′. All three are along the x axis, with signs.

Two spacecraft approach the Earth from opposite sides, each at 0.80c relative to the Earth. What is the velocity of craft B measured by craft A? Put A's rest frame as S′, moving at v = +0.80c. Craft B moves at u = −0.80c in the Earth frame S.

Galilean addition would give −1.60c, faster than light. The relativistic answer is less than c, as it must be. Now test the formula on light itself: put u = c.

Every observer gets c, whatever v is. The second postulate is built into the formula. Watch the sign in the denominator: when u and v point opposite ways, uv is negative and the denominator becomes 1 + |uv|/c².

6The space-time interval, proper time and proper length

Two observers disagree about Δx and about Δt between two events. They agree on one combination of them, the space-time interval Δs:

(Δs)² = (cΔt)² − (Δx)², the same in every inertial frame

It is an invariant, like the length of a rod that looks different from different angles but is the same rod. Check it with the event of section 4 (separated from the origin event x = 0, t = 0):

Same number. That is also the fastest way to check a Lorentz calculation.

The interval gives two special measurements their meaning.

The proper time interval Δt₀ between two events is the time measured in the frame in which the two events happen at the same place. It is the time read by a single clock that is present at both events, such as the wristwatch of a traveller who is there when each thing happens. In that frame Δx = 0, so (Δs)² = (cΔt₀)²: the interval is just the proper time multiplied by c.

The proper length L₀ of an object is its length measured in the frame in which the object is at rest.

Say both definitions in these words. "Proper" does not mean correct; every frame's measurement is equally correct. It means "belonging to": the proper time belongs to the clock that is at both events, the proper length to the frame the object sits in.

7Time dilation and length contraction

Δt = γΔt₀ and L = L₀/γ

Time dilation. The time between two events is shortest in the frame where they happen at the same place (the proper time). In every other frame it is longer, by the factor γ. A spacecraft travels at 0.80c (γ = 5/3) and the astronaut's watch records 3.0 years between leaving one star and reaching the next. Both events happen at the astronaut, so the watch reads the proper time.

Length contraction. An object is longest in its own rest frame. Measured from a frame in which it moves along its length, it is shorter by the factor γ. The spacecraft is 120 m long at rest, so observers who watch it pass at 0.80c measure L = 120 ÷ (5/3) = 72 m. Only the length along the direction of motion contracts; its width does not.

The two effects are symmetric. The astronaut sees the Earth's clocks run slow and the Earth's rulers contracted, by the same γ. There is no contradiction, because the two observers also disagree about which events are simultaneous (section 8), and "whose clock is slow" is a comparison that needs simultaneity to be defined.

Two checks save more marks than any formula. Identify the proper quantity before you calculate, and check the direction: the proper time is the shortest time and the proper length the longest length.

These are real effects, not illusions caused by light taking time to arrive. They are what the measurements give after light travel time has been allowed for.

8The relativity of simultaneity

Two events that happen at the same time in one frame do not, in general, happen at the same time in another. Look at the interval form of the time transformation:

Δt′ = γ(Δt − vΔx/c²)

If two events are simultaneous in S, Δt = 0, but they are at different places, Δx ≠ 0, then Δt′ = −γvΔx/c², which is not zero.

Two lamps stand 600 m apart beside a track and flash at the same instant in the ground frame S. A train passes at 0.60c.

In the train's frame the flashes are 1.5 μs apart. The negative sign says that the lamp at larger x, the one the train is travelling towards, flashes first. A passenger is not misreading their clocks: in the train frame, that lamp really does flash first.

Two points keep this from turning into nonsense. Events that are simultaneous and at the same place (Δx = 0) are simultaneous for everyone. And the order of two events can only be reversed if no signal could travel between them, which is why cause always comes before effect for every observer.

9Space-time diagrams

A space-time diagram plots events on axes of position x (horizontal) and ct (vertical). Using ct rather than t puts both axes in metres, and makes the path of light a line at 45°. Figure 3 is one.

Each event is a point. The history of a particle is a line of events, its world line. The guide limits you to constant velocity, so world lines are straight.

• A particle at rest in S (A in Figure 3) has a vertical world line: x stays the same while ct increases.
• A particle moving at v (B) has a world line tilted from the ct axis by an angle θ, with

tan θ = v/c

• Light has v = c, so θ = 45°. No world line can be tilted further from the ct axis than 45°, because nothing travels faster than light.

B moves at 0.5c, so θ = tan⁻¹ 0.5 = 26.6°. Event E, where B passes A, is a point both world lines share.

Adding the axes of a second frame. The ct′ axis of frame S′ is the world line of the origin of S′, so it is tilted from the ct axis by θ, where tan θ = v/c. The x′ axis is the set of events with ct′ = 0, and it is tilted up from the x axis by the same angle θ. The two primed axes close towards the 45° light line like a pair of scissors; the faster S′ moves, the more they close. Figure 4 draws them for v = 0.60c, θ = 31°.

To read an event's coordinates in either frame, draw lines through it parallel to the other axis:

• For S: drop a line parallel to the ct axis to read x, and a line parallel to the x axis to read ct. This is what you always do on a graph.
• For S′: draw a line parallel to the x′ axis to meet the ct′ axis and read ct′, and a line parallel to the ct′ axis to meet the x′ axis and read x′.

Figure 4 uses the event from section 4 in units of 300 m (so ct = 4 means 1200 m, or 4.0 μs). The two dashed amber lines reach the primed axes at ct′ = 1.25 and x′ = 3.25, which are 375 m and 975 m, the Lorentz results.

Simultaneity on the diagram. In S, events with the same ct lie on a horizontal line. In S′, events with the same ct′ lie on a line parallel to the x′ axis, which is tilted. Figure 5 shows events A and B that are simultaneous in S. The lines through them parallel to x′ meet the ct′ axis at different heights, and B's is lower, so in S′ event B happens first. This is section 8 drawn rather than calculated.

The scales are not the same. One unit on the ct′ axis is longer on the page than one unit on the ct axis. The guide says the scales are set by lines of constant space-time interval: because (ct)² − x² is the same in both frames, the curve (ct)² − x² = 1 passes through ct = 1 on the ct axis and through ct′ = 1 on the ct′ axis. Those curves are hyperbolae, drawn in Figure 6. The curve x² − (ct)² = 1 does the same job for the x and x′ axes.

Figure 6 also shows time dilation and length contraction as geometry. The tick ct′ = 1 on the S′ clock's world line sits at ct = 1.25 in S: S sees one unit of S′ time take γ = 1.25 units of its own. Length contraction can be read the same way, by comparing where the two ends of a moving rod are at one instant in each frame. You will not be asked to construct hyperbolae, only to know that they are why the scales differ.

10The evidence: muon decay

Muons are unstable particles made when cosmic rays hit the upper atmosphere. At rest, a population of muons decays with a half-life of about 1.5 μs. They arrive at the Earth's surface at speeds close to c, in numbers far larger than a Newtonian calculation allows. Measurements of this, comparing muon counts on a mountain with counts at sea level, were among the first direct tests of time dilation.

Work it through. A detector on a mountain top counts muons moving straight down at 0.995c (γ = 10.0). A second detector is 3.0 km lower, at sea level. What fraction of the muons should reach the lower detector?

Without relativity:

In the Earth frame, with time dilation. The muon's decay runs on its own clock, which is present at both events (passing the top detector, passing the bottom one), so it measures the proper time.

In the muon's frame, with length contraction. The muon is at rest, and the mountain rushes up at 0.995c. The 3.0 km height is a proper length in the Earth frame, so the muon measures it contracted: 3000 ÷ 10.0 = 300 m. That takes 300 ÷ (0.995 × 3.00 × 10⁸) = 1.0 μs to go past, the same 0.67 half-lives, the same 63%.

Figure 7 puts the two stories side by side.

The Earth frame explains the survival by time dilation; the muon frame explains it by length contraction. Both give the same prediction, and the prediction matches the counts. That is what the guide means when it says muon experiments are evidence for both effects. The mistake to avoid is mixing the frames: using the contracted 300 m with the dilated 10 μs gives nonsense.

c is a limit built into the theory, not an engineering problem, much as absolute zero is in thermal physics.