The big idea: Nothing can travel faster than the speed of light c. To keep that true, time and length are no longer the same for everyone — they depend on how fast you are moving. The Lorentz transformations are the rules that connect what one observer measures to what a moving observer measures.
Everything below is built from one number: the Lorentz factor γ (gamma).
Two pieces of jargon: Proper time Δt₀ — the time between two events measured by a clock that is present at both events (it is the shortest possible time).
Proper length L₀ — the length of an object measured in the frame where it is at rest (it is the longest possible length).
Every relativistic effect is scaled by the Lorentz factor γ. It depends only on the speed v as a fraction of c. At everyday speeds γ ≈ 1 (relativity hides itself); as v approaches c, γ shoots up toward infinity.
- Lorentz factor (no unit, ≥ 1)
- speed of the moving frame (m s⁻¹)
- speed of light in a vacuum (3.0 × 10⁸ m s⁻¹)
Worked example — find the Lorentz factor
A spaceship moves past Earth at v = 0.80c. Find its Lorentz factor γ.
Solution
- Write the given formula first:
- With v = 0.80c the ratio v/c = 0.80, so v²/c² = 0.80²:
- Finish the arithmetic:
Final answer
γ = 1.67 (no unit). Because v < c, γ comes out bigger than 1, as it always must.
Common slip: Keep speeds as a fraction of c. Writing v = 0.80c means v/c = 0.80, so the term is 0.80², not (0.80 × 3 × 10⁸)². Working in units of c keeps the numbers clean.
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Moving clocks tick slowly: A clock that is moving relative to you ticks slower than your own. The proper time Δt₀ (measured on the moving clock) is stretched by γ to give the longer time Δt that you measure.
- time measured by the observer who sees the clock moving (s)
- proper time — measured on the moving clock itself (s)
- Lorentz factor (no unit)
Worked example — a clock on the spaceship
The spaceship above (v = 0.80c, γ = 1.67) carries a clock that measures a proper time Δt₀ = 2.0 s for an event on board. What time does an Earth observer measure for that same event?
Solution
- Write the given formula first:
- Substitute γ = 1.67 and the proper time Δt₀ = 2.0 s:
- Work it out — keep the unit:
Final answer
Δt = 3.3 s. The Earth observer sees the moving clock take longer — the moving clock runs slow.
Lengths squeeze along the motion: An object that moves past you is measured to be shorter along its direction of motion than its proper length. Only the dimension along the motion contracts — width and height are unchanged. The proper length L₀ is divided by γ.
- length measured by the observer who sees it moving (m)
- proper length — measured in the object's rest frame (m)
- Lorentz factor (no unit)
Worked example — the length of the spaceship
The spaceship (v = 0.80c, γ = 1.67) has a proper length L₀ = 100 m. What length does an Earth observer measure as it flies past?
Solution
- Write the given formula first:
- Substitute the proper length L₀ = 100 m and γ = 1.67:
- Work it out — keep the unit:
Final answer
L = 60 m. The Earth observer measures the moving ship to be shorter than its 100 m rest length.
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You cannot just add speeds: At low speeds you simply add velocities (0.5 + 0.5 = 1.0). Near c that breaks the cosmic speed limit, so relativity uses a velocity-addition rule. The denominator keeps every result below c — you can never reach the speed of light by adding speeds.
- velocity of the object in the second frame (m s⁻¹)
- velocity of the object in the first frame (m s⁻¹)
- velocity of the second frame relative to the first (m s⁻¹)
- speed of light in a vacuum
Worked example — two ships approaching head-on
As seen from Earth, two spacecraft head straight toward each other, each moving at 0.50c. How fast does one ship measure the other to approach?
Solution
- Write the given formula first:
- Take one ship as the moving frame, v = −0.50c, and the other as u = +0.50c (approaching). Substitute in units of c:
- Finish — keep the unit (here in c):
Final answer
0.80c — NOT 1.0c. Adding 0.50c and 0.50c relativistically gives 0.80c, still safely below the speed of light.
Where it shows up: Lorentz transformations are HL only (A.5):
- Paper 1A — a one-step 'find γ', 'which is the proper time?', or 'is the result above or below c?'. - Paper 2 — determine a dilated time, a contracted length, or a relative speed, often for a muon or a fast spacecraft.
Three easy marks: (1) Always find γ first. (2) Identify the proper quantity: proper time is on the moving clock, proper length is in the rest frame. (3) Multiply by γ for time (Δt = γΔt₀); divide by γ for length (L = L₀/γ).
IB-style question — a fast-moving probe
A probe travels past a space station at v = 0.60c. A signal lamp on the probe flashes with a proper period of 5.0 s. Determine (a) the Lorentz factor and (b) the period of the flashes as measured by the station.
Solution
- (a) Start from the given Lorentz-factor formula:
- With v/c = 0.60, so v²/c² = 0.36:
- (b) The lamp is on the probe, so 5.0 s is the proper time Δt₀. Use the given time-dilation formula:
- Work it out — keep the unit:
Final answer
(a) γ = 1.25. (b) Δt ≈ 6.3 s — the station sees the flashes spaced further apart because the moving clock runs slow.