The Special Theory of Relativity

In 1905 Albert Einstein took two simple-sounding statements seriously and followed them wherever they led. They led to moving clocks running slow, moving rods shrinking, and E = mc². Everything in this chapter follows logically from those two statements.

The principle of relativity#

Throughout this chapter, “frame” means an inertial frame: one moving with constant velocity relative to another inertial frame. Drop a stone on a station platform so it lands on your left foot; do the same in a train gliding smoothly at 120 km/h, and it again lands on your left foot. No experiment inside a smoothly moving train can tell you whether it is moving or standing at the platform, as long as it doesn’t speed up, slow down or turn.

For mechanics this follows from Newton’s laws: F = ma holds in the same form in every inertial frame, so identical experiments give identical results. There is no special “at rest” frame: you may call either frame stationary and the other moving.

A problem with light#

Maxwell’s equations predict that electromagnetic waves, including light, travel in vacuum at c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m/s. If Maxwell’s laws hold in the same form in every frame, then light must travel at c in every frame. Experiments say it does. That leads to a strange bookkeeping problem.

Suppose A, B and a light pulse W start together at t = 0. B moves right at c/2 relative to A; W moves right at c. Each observer writes down distances over time:

Common sense says A-to-W should equal A-to-B plus B-to-W, but here A-to-W equals B-to-W even though B has moved. Something about our measuring tools must differ: perhaps 1 s on A’s clock is not 1 s on B’s, or A’s metre is not B’s metre.

There are only two ways out:

1. Maxwell’s laws hold in every frame, and our ideas of time and length must change; or
2. Maxwell’s laws take different forms in different frames, so experiments with electricity, magnetism or light could reveal whether a train is moving.

Experiments rule out the second. The most famous, by Michelson and Morley (1887), tried to measure the Earth’s motion through the supposed “ether” that carried light waves. It found nothing, one of the most important null results in physics.

Einstein’s postulates#

A clock is any repeating process: a pendulum, a heartbeat, a pulse of light between two mirrors. Use the light version, insist that light goes at c for everyone, and you are forced to conclude that moving clocks tick slowly.

Sideways lengths don’t change#

First, a thought experiment. Make two identical rods, check their ends match, and paint the ends of one red, the other blue. Now slide one past the other sideways (perpendicular to their length), close enough for the paint to rub off.

If the moving rod shrank, it would leave red marks inside the blue rod’s ends. But viewed from the red rod, it is the blue rod that moves, and by the principle of relativity the moving rod should then be the one that shrinks: blue marks inside the red ends. Both can’t happen. So neither does: lengths perpendicular to the motion are unchanged.

The light clock#

Fix two mirrors a distance L apart on a rod, and bounce a light pulse between them. Each return to the bottom mirror M₁ is a “tick”. In the frame S where the mirrors are at rest, the pulse covers 2L per tick, so

Now watch the same clock from a frame S′ in which it moves sideways at speed v. Between two ticks taking Δt′, the mirrors move v Δt′ along, so the pulse travels along two slanted lines, each of length √(L² + (vΔt′/2)²). The rod length L is unchanged (it is perpendicular to the motion), and light still travels at c:

Proper time

So a moving clock runs slow by the factor γ. This holds for every kind of clock (atomic, mechanical, biological) because by postulate 1 they must all agree with the light clock in their own frame. And it is symmetric: each of two observers in relative motion finds the other’s clock running slow.

Any two events can be treated this way. If event E₁ happens at x = 0, t = 0 and event E₂ at x = L, time t, a clock moving at L/t is present at both. In its frame the events are at one place, so its reading t/γ is the proper time.

Turn the light clock so it moves along its length and a second surprise appears: moving objects are shorter. Then look at two distant events and a third: whether they happen “at the same time” depends on who is watching.

Length contraction#

Take the light clock and move it at speed v parallel to its rod. Call the rod’s length in this frame L′ (we don’t yet know if it equals L). On the way out the pulse chases the front mirror, which runs away at v: it takes L′/(c − v). On the way back the rear mirror rushes to meet it: L′/(c + v). One tick therefore lasts

But the tick must also equal γ(2L/c) by time dilation. Setting the two equal:

Which event happened first?#

A box of rest length L lies on the ground, with doors D₁ (left) and D₂ (right) and a lamp exactly in the middle. Each door opens when light reaches it. Train T₁ moves left and train T₂ moves right, both at speed v relative to the ground. The lamp flashes.

• Ground frame: the pulses travel L/2 each at c, so both doors open at t = L/2c: simultaneously.
• Frame of T₁: the box (length L/γ) moves right at v. D₁ comes to meet its pulse, D₂ runs away from its pulse: Δt₁′ = (L/γ)/2(c + v), Δt₂′ = (L/γ)/2(c − v). D₁ opens first.
• Frame of T₂: by the same reasoning, D₂ opens first.

Moving clocks are out of sync#

To synchronize two clocks, put a lamp midway and set each to zero when its flash arrives. Do that for clocks on the ground at D₁ and D₂ and, separately, for a row of clocks along train T₁. When D₁ opens, a train clock C₁ is opposite it; when D₂ opens, clock C₂ is opposite. On the ground the openings are simultaneous, but the train clocks say D₂ opened Lvγ/c² later. So, judged from the ground, the train clocks disagree at a single instant.

Again this is symmetric: from the train, the ground clocks are the ones out of step.

If space and time behave differently at high speed, so must momentum and energy. The repaired laws contain a speed limit, a mass that grows with speed, and the most famous equation in physics.

Relativistic momentum#

At speeds close to c, F = ma no longer describes motion correctly. We will state the correct laws without deriving them. The momentum of a particle is

A steady push never reaches c

Start a particle from rest and apply a constant force F. Then p = Ft, so

Mass is energy#

In relativity, matter is a concentrated form of energy. A mass m is equivalent to an energy

At rest a particle has rest-mass energy E₀ = m₀c². Moving, its total energy is E = γm₀c². Expanding γ for small v:

Energy and momentum together

Eliminating v between p = γm₀v and E = γm₀c² gives a relation that holds in every frame:

The ultimate speed#

No material particle can be pushed to c, and no signal or information can travel faster than c. If it could, there would be frames in which an effect came before its cause: a bird hit before the gun was fired.

Twins say goodbye; one flies to a star and back at high speed. Each sees the other’s clock running slow, so who ends up younger? Working it through carefully pulls together everything in this chapter.

The trip#

Twins Asha and Bina live on Earth. Bina boards spaceship S₁ heading for planet P, which is 6 light-years away (in the Earth frame) and at rest relative to Earth. At P she jumps across to ship S₂ heading home. Both ships move at 0.6c relative to Earth, so γ = 1.25.

Asha’s view (Earth)#

Each leg takes 6 ly / 0.6c = 10 years of Earth time, 20 years in all. Bina is moving the whole time, so her clocks, including her heartbeat and her ageing, run slow by γ: each leg lasts 10/1.25 = 8 years for her. Asha predicts Bina returns having aged 16 years while Asha aged 20. Bina will be 4 years younger.

This is a real effect: fast-moving unstable particles held in circular orbits by magnets live measurably longer than identical particles at rest.

Bina’s view, and the paradox#

On S₁, Earth and P are moving, so their 6 ly separation is contracted to 6/1.25 = 4.8 ly. The planet approaches at 0.6c and arrives in 4.8/0.6 = 8 years by her clock. The return is the same: 16 years total. Bina agrees with Asha about her own age.

But Bina also sees Asha’s clock as moving, so running slow: in Bina’s 16 years Asha should age only 16/1.25 = 12.8 years. So Bina expects Asha to be the younger one. Each thinks the other is younger. Who is right?

The resolution#

The situation isn’t symmetric. Asha stays in one inertial frame. Bina changes frames at the planet, from S₁ to S₂, and when you change frames you change which distant events count as “now” (the previous lesson). Track the clocks carefully:

The time-dilation argument covered only the two coasting legs (6.4 + 6.4 = 12.8 years). It missed the 7.2 years by which “Earth now” jumps forward when Bina switches ships. Add it in and both twins agree: Asha aged 20 years, Bina 16.

Why the turning twin is younger

On a spacetime diagram the stay-at-home twin follows a straight line and the traveller a bent one. In relativity a straight (inertial) path between two events always has the longest proper time; any detour is shorter. That makes Bina the younger.