5bTime Dilation and Length Contraction
Explore how moving clocks run slow and moving objects shrink — two of the most astonishing and experimentally confirmed predictions of Einstein's special relativity.
Time dilation and length contraction are not science fiction — they are measured every day in particle accelerators and GPS satellites. Understanding these effects is essential for modern technology and fundamental physics.
Lesson Overview
Two of the most striking predictions of special relativity are time dilation and length contraction. Time dilation means that a moving clock ticks more slowly than a stationary one: t = t₀ / √(1 − v²/c²). Length contraction means that a moving object is shorter along its direction of motion: L = L₀√(1 − v²/c²). The Lorentz factor γ = 1/√(1 − v²/c²) appears in both equations. These effects are real and have been confirmed experimentally, most famously by the extended lifetimes of cosmic-ray muons.
Key Concepts
Lorentz Factor γ
γ = 1/√(1 − v²/c²); always ≥ 1; approaches ∞ as v → c
Time Dilation
t = γt₀; moving clocks run slow; t > t₀
Proper Time (t₀)
Time measured by a clock at rest relative to the event
Length Contraction
L = L₀/γ; moving objects are shorter; L < L₀
Proper Length (L₀)
Length measured in the frame where the object is at rest
Twin Paradox
Traveling twin ages less due to time dilation; resolved by asymmetry of frames
A spaceship travels at v = 0.6c. Calculate the Lorentz factor γ.
An astronaut travels at 0.6c for a proper time of t₀ = 10 years (measured on the ship). How much time passes on Earth?
A spaceship is 100 m long at rest. It travels at 0.6c. What is its length as measured by a stationary observer?
Cosmic-ray muons are created 10 km above Earth and travel at 0.998c. Their proper lifetime is 2.2 μs. How long do they live as measured from Earth?
In the twin paradox, twin A stays on Earth and twin B travels at 0.8c for 6 years (Earth time) and returns. How much does twin B age? (γ at 0.8c = 5/3)
Calculate γ for a spaceship traveling at v = 0.8c.
Hint: Use γ = 1/√(1 − v²/c²). Substitute v/c = 0.8, so v²/c² = 0.64.
A clock on a spaceship moving at 0.8c (γ = 5/3) measures a proper time of 9 years. How much time passes on Earth?
Hint: Use t = γt₀. Multiply the proper time by the Lorentz factor.
A rocket is 200 m long at rest. At 0.8c (γ = 5/3), what length does a stationary observer measure?
Hint: Use L = L₀/γ. Divide the proper length by the Lorentz factor.
Why do muons created in the upper atmosphere reach Earth's surface, even though their proper lifetime seems too short for the journey?
Hint: Consider time dilation from Earth's frame and length contraction from the muon's frame.
In the twin paradox, why does the traveling twin age less? Why is this not a true paradox?
Hint: The situation is not symmetric — only one twin actually accelerates (changes direction). The asymmetry resolves the apparent paradox.
Key Vocabulary
Time Dilation
The phenomenon where a moving clock ticks more slowly than a stationary clock, as predicted by special relativity: t = γt₀.
Example: An astronaut traveling at 0.9c for 10 years (ship time) would find that about 23 years have passed on Earth.
Length Contraction
The shortening of an object's length along its direction of motion as measured by a stationary observer: L = L₀/γ.
Example: A 100 m spaceship traveling at 0.866c (γ = 2) appears only 50 m long to a stationary observer.
Lorentz Factor (γ)
The factor γ = 1/√(1 − v²/c²) that quantifies relativistic effects; equals 1 at rest and increases toward infinity as v approaches c.
Example: At v = 0.6c, γ = 1.25; at v = 0.866c, γ = 2; at v = 0.995c, γ ≈ 10.
Proper Time (t₀)
The time interval measured by a clock that is at rest relative to the events being timed; the shortest possible time between two events at the same location.
Example: The muon's own internal clock measures its proper lifetime of 2.2 μs; Earth observers measure a longer dilated time.
Interactive Practice — 5 Questions
A spaceship travels at 0.6c (γ = 1.25). A clock on the ship measures 8 years. How much time passes on Earth?
A 500 m long spaceship travels at 0.6c (γ = 1.25). What length does a stationary observer measure?
What is the Lorentz factor γ for an object at rest (v = 0)?
The muon experiment confirms time dilation because:
In the twin paradox, the traveling twin ages less because:
Independent Practice
Calculate γ for v = 0.5c, v = 0.866c, and v = 0.99c. What trend do you notice as v approaches c?
An astronaut travels at 0.866c (γ = 2) for 20 years of Earth time. How much does the astronaut age? How far has the ship traveled (Earth frame)?
A spaceship is 300 m long at rest. At v = 0.866c (γ = 2), what length does a ground observer measure? What length does the pilot measure?
Explain the muon experiment: why do muons created 10 km up reach Earth's surface despite their short proper lifetime? Address both the Earth frame (time dilation) and the muon frame (length contraction).
★ Twin A stays on Earth. Twin B travels at 0.995c (γ ≈ 10) to a star 10 light-years away (Earth frame) and returns. (a) How long does the trip take in Earth's frame? (b) How much does twin B age? (c) What is the distance to the star in twin B's frame?
ChallengeCommon Mistakes
Thinking time dilation means time passes faster for the moving observer.
The moving observer's clock runs slower — they age less. Earth observers measure more elapsed time than the traveler experiences.
Applying length contraction to all dimensions of an object.
Length contraction only occurs along the direction of motion. Dimensions perpendicular to motion are unchanged.
Confusing proper time and dilated time — using t₀ when you need t or vice versa.
Proper time t₀ is always the smaller value (measured by the moving clock). Dilated time t = γt₀ is always larger (measured by the stationary observer).
Math Tips
Always calculate γ first: γ = 1/√(1 − v²/c²). Then: dilated time t = γt₀ (larger than proper time). Contracted length L = L₀/γ (smaller than proper length).
Common γ values to memorize: v = 0.6c → γ = 1.25; v = 0.8c → γ = 5/3 ≈ 1.667; v = 0.866c → γ = 2; v = 0.995c → γ ≈ 10.