
Optional unit · institution-dependent · Interactive lesson
Special-relativity postulates
Relativity does not begin with strange formulas. It begins with two modest claims — the laws are the same in every inertial frame, and light has one speed in all of them — and with a refusal to assume that clocks in different places can be trusted to agree. Take both seriously and Galilean velocity addition has to go.
UPII-15.02 · Lesson 2 of 8 in this unit. Position is not a completion score.Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.
Build the model
Two postulates, and distant clocks stop agreeing about 'at the same time'.
The key idea
The special-relativity postulates are two claims about inertial frames: every law of physics takes the same form in all of them, and light in vacuum travels at the same speed c in all of them, whatever the source or the receiver is doing.
Example. Two laboratories, one on the ground and one aboard a spacecraft passing at 0.90c, time the same laser pulse and both get 2.998 × 10⁸ m s⁻¹ — and no experiment either can run will settle which of the two is really moving.
The complete physical model and assumptions
Special relativity is built on two claims and nothing else. The relativity principle says that every law of physics, electromagnetism included, takes the same form in every inertial frame, so no experiment done inside a uniformly moving laboratory can reveal its velocity. The light postulate says that light in vacuum travels at c in every inertial frame, whatever the source or the receiver is doing. Separately each is modest; together they are fatal to classical kinematics, because Galilean addition would give a headlight beam on a train the speed c + v in the ground frame. The only assumption left to sacrifice is the silent one buried in t' = t: that clocks in different places can be set once and then trusted by everybody. Give that up and the rest follows. Simultaneity becomes a property of a frame rather than of the universe, clocks synchronised and L₀ apart in their own rest frame are offset by vL₀/c² as seen from a frame moving at v, and the quantity every frame still agrees on is neither a time nor a distance but the interval (cΔt)² − (Δx)². The model covers inertial frames only: accelerating frames and gravity need general relativity, and at ordinary speeds the corrections fall off as β² and disappear back into the Galilean picture.
1An inertial frame is rulers and synchronised clocks
A frame is not an observer's eye; it is a coordinate system, a lattice of rulers filling space with a clock at every node, all of them reading the same time. The frame is inertial if Newton's first law holds in it: a body with no force on it moves in a straight line at constant speed. That is a testable condition, and plenty of laboratories fail it. A rotating turntable fails, because a free puck curves across it. A braking car fails, because loose objects slide forward. The Earth's surface fails only mildly: at the equator the centripetal acceleration is ω²R = (7.29 × 10⁻⁵ s⁻¹)² × 6.38 × 10⁶ m = 0.034 m s⁻², about 0.35% of g, so for a bench experiment lasting a second the laboratory counts as inertial. Notice what that definition has already smuggled in. 'All of them reading the same time' is a claim about clocks in different places, and nothing so far says how they were set. That gap is where relativity walks in.
2The relativity principle, and Maxwell's frameless speed
Galileo's principle says no experiment inside a smoothly moving ship reveals its motion, and mechanics obeys it: under x' = x − vt with t' = t, velocities shift by v but accelerations do not, so F = ma reads identically in both frames. Einstein's first postulate upgrades 'no mechanical experiment' to 'no experiment at all'. Electromagnetism immediately objects. Maxwell's equations predict waves at 1/√(ε₀μ₀) = 1/√(8.854 × 10⁻¹² × 1.257 × 10⁻⁶) = 3.00 × 10⁸ m s⁻¹, a number assembled from two bench constants with no frame attached to it. Nineteenth-century physics answered by giving light a medium, the aether, and reading Maxwell's speed as the speed in the aether's rest frame. That is a testable claim, because the Earth's 30 km s⁻¹ orbit should sweep the laboratory through the aether at β = 1.0 × 10⁻⁴, reversing every six months.
3The invariant speed is measured, not assumed
Michelson and Morley ran that test in 1887. Split a beam, send the halves along perpendicular 11 m arms, recombine them and rotate the table: the predicted fringe shift 2Lv²/(λc²), with λ = 590 nm and v = 30 km s⁻¹, is about 0.4 of a fringe, and the apparatus could resolve a fortieth of that. They found nothing, and every repetition since has found nothing. The other loophole, that light might be carried along by its source the way a ball is by a moving truck, was shut at CERN in 1964: neutral pions flying at 0.99975c decayed into photons, and those photons were clocked at 2.998 × 10⁸ m s⁻¹, not the 5.995 × 10⁸ m s⁻¹ that adding the source's speed predicts. So the second postulate is an experimental result promoted to an axiom. Read it precisely: one particular speed has the same value in every inertial frame. The ban on exceeding that speed is a consequence drawn later, not part of the postulate.
4Galilean velocity addition cannot survive both postulates
Put a headlight on a train moving at 0.60c. Galileo says the ground frame measures the beam at c + 0.60c = 1.60c; the light postulate says c. They cannot both be right, and the experiments side against Galileo. The two frames still disagree about how far the beam travelled, so if they are to agree about its speed they must also disagree about how long it took: the assumption t' = t has to go. What replaces the addition rule is u = (u' + v)/(1 + u'v/c²). Try it on a probe launched forward at 0.80c from a ship that itself passes a station at 0.60c: the station measures (0.80 + 0.60)/(1 + 0.48) = 0.946c = 2.84 × 10⁸ m s⁻¹, not the 1.40c Galileo predicts, and still under c. Try it on light, u' = c: (c + 0.60c)/(1 + 0.60) = c exactly, for any v at all. The rule is built so that c is the one speed nobody can argue about.
5Simultaneity is fixed by a procedure, and it is frame-bound
Before you can say that two distant events happened at the same time, you must have set the clocks. Einstein's recipe leans on the second postulate: send a pulse from A at t₁, bounce it off B, receive it back at t₂, and set B's clock to t₁ + (t₂ − t₁)/2, which is legitimate only because the pulse takes the same time each way in this frame. Now run the flash experiment. A rocket 240 m long in the station frame passes at 0.40c, and a flash at its midpoint sends pulses toward the nose and the tail. In the station frame each pulse moves at c while the tail advances into one and the nose retreats from the other, so the tail is hit at 120 m ÷ 1.40c = 286 ns and the nose at 120 m ÷ 0.60c = 667 ns, a gap of 381 ns. In the rocket's frame the ends are equidistant and the arrivals coincide. The same breakdown shows up in clocks rather than in pulses: two clocks synchronised a distance L₀ apart in their own rest frame are read, from a frame moving at v along that line, as offset by vL₀/c², with the trailing clock ahead.
6Events are shared; their coordinates are not
An event is a happening with one place and one instant: a spark, a detector click, a pulse leaving a lamp. Frames give it different coordinates (t, x), and as we have just seen they can even disagree about the order of two events. What every inertial frame does agree on is the interval, (Δs)² = (cΔt)² − (Δx)². Take two laboratory clicks 200 m and 1.00 μs apart: cΔt = 300 m, so (Δs)² = 300² − 200² = +5.00 × 10⁴ m², positive. A positive interval is called timelike, and it means a signal at 200 m ÷ 1.00 μs = 2.00 × 10⁸ m s⁻¹ = 0.667c, comfortably under light speed, can run from the first click to the second. Every frame then agrees which came first, so cause can safely precede effect. The frame moving at that same 0.667c finds both clicks at one place, and there cΔt = √(5.00 × 10⁴ m²) = 224 m. Let (Δx)² be the larger term instead and (Δs)² comes out negative; the pair is then spacelike, no signal at or below c joins it, some frame calls the two events simultaneous and faster frames reverse their order, with no causal link at risk because none could exist. When (Δs)² = 0 a light pulse joins them exactly, which is the light postulate written as geometry.
Key equations
Read the units, assumptions and limits before substituting.
Galilean frame change
x' = x − v t, with t' = t
The classical answer. Its silent assumption is t' = t, and that is the one the postulates destroy.
x and x' in m, t and t' in s, v the constant relative speed of the two inertial frames in m s⁻¹The invariant speed of light
|Δr| / Δt = c for any light pulse, in every inertial frame
Neither the motion of the source nor that of the receiver changes that number.
Δr the pulse's displacement in m, Δt the elapsed time in s, c = 2.99792458 × 10⁸ m s⁻¹ exactlyVelocity combination
u = (u' + v) / (1 + u' v / c²)
Put u' = c and it returns u = c for every v, which is exactly what the light postulate demands.
u' the object's velocity in frame S', v the velocity of S' along x measured in S, u the result in S, all m s⁻¹Synchronised clocks are offset in another frame
Δt = v L₀ / c²
The trailing clock reads ahead by this much: 240 ns for L₀ = 240 m at 0.30c.
L₀ the clocks' separation in their own rest frame in m, v the other frame's speed along that line in m s⁻¹Invariant spacetime interval
(Δs)² = (c Δt)² − (Δx)²
Frames disagree about Δt and Δx and agree about Δs. Δs = 0 restates the light postulate.
Δt the time between two events and Δx their separation along x in one frame, in s and m; Δs in mSize of the departure from Galileo
β = v / c
At 8.0 km s⁻¹, β = 2.7 × 10⁻⁵ and β² = 7.1 × 10⁻¹⁰, below most laboratory precision.
β is dimensionless; classical predictions pick up fractional corrections of order β²Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set the speed to zero and both pulses land together. Raise it and the tail event slides down the backward light line while the nose event runs far up the forward one: at 0.55c the nose waits 3.44 times as long, though nothing inside the rocket has changed.
Live results
- TAIL HIT
- 286 ns
- NOSE HIT
- 667 ns
- GAP, NOSE − TAIL
- 381 ns
- RATIO (1+β)/(1−β)
- 2.33
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Rocket speed v/c: 0.40 c; Rocket length in station frame: 240 m
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The usual error is to explain the relativity of simultaneity as a delay in seeing. Students say the station observer only thinks the tail was hit first because the tail happened to be nearer their eye. That is not it. A frame's verdict on simultaneity comes from a lattice of clocks already synchronised throughout that frame, each event stamped by the clock standing right beside it, so every light-travel time has been subtracted before any comparison is made. The 381 ns gap in the station frame is what survives that subtraction; it is a fact about the frame's coordinates, not about anyone's eyesight. The opposite error is to conclude that everything is relative. Three things are not: the speed of light, the interval (Δs)², and the time order of any two events close enough for a signal at or below c to connect them.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A probe receding from a station at 0.60c switches on a beacon when the station-frame gap between them is 4.50 × 10¹⁰ m, and the pulse travels back to the station. Find the travel time the station should record (a) if Galilean addition of velocities were correct, and (b) from the second postulate. Take c = 3.00 × 10⁸ m s⁻¹.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Galilean picture: the light leaves a source receding at 0.60c and heads back toward the station, so the station should measure it closing at c − 0.60c = 0.40c = 1.20 × 10⁸ m s⁻¹.
- That gives t = 4.50 × 10¹⁰ m ÷ 1.20 × 10⁸ m s⁻¹ = 375 s.
- Second postulate: the motion of the source is irrelevant to the pulse's speed in the station frame, which is c = 3.00 × 10⁸ m s⁻¹ whatever the probe was doing at emission.
- That gives t = 4.50 × 10¹⁰ m ÷ 3.00 × 10⁸ m s⁻¹ = 150 s.
- The two predictions differ by 225 s, a factor of 2.5, far outside any plausible timing error. This is a decisive test, not a fine correction.
Answer. Galilean: 375 s. Relativistic: 150 s. Spacecraft ranging agrees with 150 s, so the source's velocity adds nothing to the light's, and here Galileo is wrong by a factor of 2.5.
MediumExample 2
A platform 240 m long carries a clock at each end, and the two are Einstein-synchronised in the station frame. A train passes at 0.30c, running from the west clock toward the east clock. In the train's frame, do the platform clocks read the same? If not, which one is ahead, and by how much? Take c = 3.00 × 10⁸ m s⁻¹.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- The synchronisation was carried out with light signals timed in the station frame. Nothing in that procedure makes the same settings look synchronised from a frame moving along the platform.
- For two clocks a rest-frame distance L₀ apart along the line of relative motion, a frame moving at v finds them offset by Δt = vL₀/c². The platform is at rest in the station frame, so L₀ = 240 m.
- Δt = (0.30 × 3.00 × 10⁸ m s⁻¹)(240 m) ÷ (3.00 × 10⁸ m s⁻¹)² = 2.16 × 10¹⁰ ÷ 9.00 × 10¹⁶ = 2.4 × 10⁻⁷ s.
- Which one is ahead: in the train's frame the platform slides west, so the east clock is the trailing clock, and the trailing clock reads ahead. The train is running toward the clock that is ahead of the other.
- Checks: the offset equals the time light needs to cross 72 m, it is proportional to v, and it vanishes at v = 0, recovering the Galilean answer that synchronisation is absolute.
Answer. No. In the train's frame the east clock, the one the train is heading toward, reads 240 ns (0.24 μs) ahead of the west clock.
HardExample 3
In a laboratory frame, event 1 is a spark at x = 0 at t = 0 and event 2 is a spark at x = 600 m at t = 1.00 μs. (a) Classify the pair using the invariant interval. (b) Find the speed of the inertial frame in which the two sparks are simultaneous, and their separation in that frame. (c) Decide whether the first spark could have caused the second. Take c = 3.00 × 10⁸ m s⁻¹.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Put both pieces in metres: cΔt = 3.00 × 10⁸ m s⁻¹ × 1.00 × 10⁻⁶ s = 300 m, and Δx = 600 m.
- (Δs)² = (cΔt)² − (Δx)² = (300 m)² − (600 m)² = 9.00 × 10⁴ − 3.60 × 10⁵ = −2.70 × 10⁵ m². It is negative, so the pair is spacelike.
- A frame moving at v along the +x direction finds Δt' = 0 when its clock offset exactly cancels the laboratory's time gap, which needs v = c²Δt/Δx = (3.00 × 10⁸)² × 1.00 × 10⁻⁶ ÷ 600 = 1.50 × 10⁸ m s⁻¹ = 0.500c.
- In that frame Δt' = 0, so (Δs)² = −(Δx')², giving Δx' = √(2.70 × 10⁵ m²) = 520 m. Because the interval is the same number in every frame, that separation came out without transforming a single coordinate.
- Causality: a signal from event 1 would have to cover 600 m in 1.00 μs, a speed of 6.00 × 10⁸ m s⁻¹ = 2.00c. Impossible, and consistently, frames faster than 0.500c put event 2 first, so the pair has no frame-independent order to protect.
Answer. (a) Spacelike, (Δs)² = −2.70 × 10⁵ m². (b) v = 1.50 × 10⁸ m s⁻¹ = 0.500c, with Δx' = 520 m. (c) No: it would need a 2.00c signal, and frames disagree about which spark came first.
Quick check
Select an answer and explain it before checking.
Review & continue
The special-relativity postulates are two claims about inertial frames: every law of physics takes the same form in all of them, and light in vacuum travels at the same speed c in all of them, whatever the source or the receiver is doing.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.