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Special relativity

From Einstein's postulates derive time dilation, length contraction, the Lorentz transformations, relativistic momentum and E = mc², and apply them to problems.

Check yourself
In this lesson you will learn
  • State the two postulates of relativity and their consequences
  • Calculate time dilation and length contraction with the Lorentz factor
  • Add velocities relativistically
  • Find relativistic momentum, rest energy and kinetic energy

If the atomic clocks on GPS satellites ran without relativistic corrections, navigation would become useless within hours. Muons created high in the atmosphere live only a couple of microseconds, yet they reach the ground. Both facts are explained by the special theory of relativity, created by Albert Einstein in 1905.

Two postulates

  1. Principle of relativity: the laws of physics are the same in all inertial frames; no experiment can detect “absolute rest”.
  2. Constancy of the speed of light: light in vacuum travels at the same speed c ≈ 3 · 10⁸ m/s in every inertial frame, whatever the motion of the source.

The second postulate contradicts everyday experience: a ball thrown forward in a train moves faster relative to the ground, but light does not. As a result, time and distance are not absolute: two events simultaneous in one frame need not be simultaneous in another — the relativity of simultaneity.

Time dilation and length contraction

Consider a “light clock”: light bounces between two mirrors a distance L apart. For an observer moving with the clock one tick lasts Δt₀ = 2L/c. If the clock moves at speed v, an observer on the ground sees the light travel along a slanted path: (cΔt/2)² = L² + (vΔt/2)². This gives Δt = Δt₀ / √(1 − v²/c²).

γ = 1 / √(1 − v²/c²), Δt = γ · Δt₀, L = L₀ / γγ = 1 / √(1 − v²/c²), Δt = γ · Δt₀, L = L₀ / γ
where:
  • γLorentz factor (γ ≥ 1)
  • Δt₀proper time — the interval measured in the clock's own rest frame, in s
  • L₀proper length — measured in the frame where the object is at rest, in m

Lengths contract only along the direction of motion; perpendicular dimensions are unchanged.

v / cγ
0.11.005
0.61.25
0.81.667
0.997.09
0.99922.4
At low speeds γ ≈ 1 and relativistic effects are unnoticeable.
Example 1: how do muons reach the ground?

A muon's mean proper lifetime is 2.2 μs. It moves at 0.995c. Find its lifetime in the Earth's frame and the mean distance it covers in that time. What would the answer be without relativity?

Show solution
γ = 1 / √(1 − 0.995²) ≈ 10.0.
Δt = γ · Δt₀ ≈ 10 · 2.2 = 22 μs.
Distance: 0.995 · 3 · 10⁸ · 22 · 10⁻⁶ ≈ 6.6 km.
Without relativity: 0.995 · 3 · 10⁸ · 2.2 · 10⁻⁶ ≈ 660 m — the muons would not get far.
In the muon's own frame the atmosphere is contracted by γ instead: same conclusion.

Lorentz transformations and velocity addition

x′ = γ(x − vt), t′ = γ(t − vx/c²), u = (u′ + v) / (1 + u′v/c²)x′ = γ(x − vt), t′ = γ(t − vx/c²), u = (u′ + v) / (1 + u′v/c²)
where:
  • x′, t′coordinates of an event in frame S′, which moves along x at speed v relative to S
  • u′, uvelocity of a body in S′ and in S

For v ≪ c they reduce to the Galilean ones: x′ = x − vt, t′ = t, u = u′ + v.

Length contraction also follows from the Lorentz transformations. A rod is at rest in S′ with its ends at x′₁ and x′₂, so L₀ = x′₂ − x′₁. To measure its length in S we record both ends at the same time t: x′₂ − x′₁ = γ(x₂ − vt) − γ(x₁ − vt) = γ(x₂ − x₁). Hence L = x₂ − x₁ = L₀/γ.

Example 2: ships flying towards each other

a) Two spaceships each fly towards each other at 0.6c relative to the Earth. What speed does the pilot of the first ship measure for the second? b) A ship 100 m long flies at 0.8c. What length is measured from the Earth?

Show solution
a) u = (0.6c + 0.6c) / (1 + 0.6 · 0.6) = 1.2c / 1.36 ≈ 0.88c — below c, not 1.2c!
b) γ = 5/3, L = L₀ / γ = 100 · 3/5 = 60 m.

Relativistic momentum and energy

p = γ · m · v, E = γ · m · c², E₀ = m · c², Ek = (γ − 1) · m · c², E² = (pc)² + (mc²)²
where:
  • mmass of the body (rest mass — invariant), in kg
  • Etotal energy, in J
  • E₀rest energy, in J

At low speeds the binomial expansion gives γ = (1 − β²)^(−1/2) ≈ 1 + β²/2. Then Ek = (γ − 1)mc² ≈ mc² · v²/(2c²) = mv²/2 — the classical formula is recovered. As v → c, γ → ∞: an infinite energy would be needed to bring a massive body to the speed of light.

Example 3: a fast electron

An electron (m = 9.11 · 10⁻³¹ kg) moves at 0.8c. Find its rest energy, kinetic energy and momentum. Check that E² = (pc)² + (mc²)².

Show solution
E₀ = mc² = 9.11 · 10⁻³¹ · 9 · 10¹⁶ ≈ 8.2 · 10⁻¹⁴ J ≈ 0.51 MeV.
γ = 5/3: Ek = (5/3 − 1) · 0.51 ≈ 0.34 MeV.
p = γmv = (5/3) · 9.11 · 10⁻³¹ · 2.4 · 10⁸ ≈ 3.64 · 10⁻²² kg · m/s.
Check: pc ≈ 1.093 · 10⁻¹³ J, mc² ≈ 0.820 · 10⁻¹³ J, √(1.093² + 0.820²) · 10⁻¹³ ≈ 1.366 · 10⁻¹³ J = γmc². ✓
The classical mv²/2 would give only ≈ 0.16 MeV.
Example 4: the energy of one gram

Find the rest energy equivalent to 1 g of mass and express it in kilowatt-hours.

Show solution
E₀ = mc² = 10⁻³ · (3 · 10⁸)² = 9 · 10¹³ J.
Since 1 kW · h = 3.6 · 10⁶ J, E₀ = 2.5 · 10⁷ kW · h = 25 million kW · h.
Nuclear reactions convert only a small fraction of the mass, yet that is still millions of times more than chemical burning.
Interactive
Loading simulation…
x = v/c. The first curve is γ, the second its low-speed approximation 1 + β²/2. Up to about v ≈ 0.3c the curves nearly coincide; as v → c, γ grows without limit.

Key points

  • Postulates: the laws of physics are the same in all inertial frames; c is the same for all observers.
  • γ = 1/√(1 − v²/c²); Δt = γΔt₀ (time dilates), L = L₀/γ (length contracts).
  • Velocity addition: u = (u′ + v)/(1 + u′v/c²) — the result never exceeds c.
  • p = γmv, E = γmc², E₀ = mc², Ek = (γ − 1)mc², E² = (pc)² + (mc²)².

Check yourself

10 questions. Every correct answer earns XP.

1 / 10
What is the Lorentz factor at v = 0.6c?