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Length Contraction

Key idea: Learn length contraction using L = L0/γ, identify proper length, and solve exam-style problems with clear simultaneity conditions.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • discuss qualitatively the results of the Michelson–Morley interferometer experiment and its implications on the ether theory (knowledge of the details of the experiment is not required)
  • state the postulates of the special theory of relativity, that in all inertial frames, the laws of physics are the same and the speed of light in free space is the same regardless of the motion of the light source or observer
  • appreciate the failure of Galilean transformation equations when applied to a moving source of light
  • discuss the concept of simultaneity
  • show an understanding of the terms proper time and proper length
  • apply the Lorentz transformation equations to solve one-dimensional problems
  • Derive the time dilation formula and the length contraction formula, making use of the Lorentz factor.
  • apply the time dilation formula and the length contraction formula in related situations (e.g. the lifetime of fast-moving muons) or to solve problems
  • use the one-dimensional relativistic velocity addition formula to calculate velocities in different inertial frames or to solve problems
  • Apply the relativistic energy–momentum relation E² = (pc)² + (mc²)² to solve problems, including selecting its limiting form.
  • Show that E² = (pc)² + (mc²)² reduces to E = pc for massless particles and to E = mc² + ½mv² at low speeds.

Length contraction is the result that a moving object is measured to be shorter along the direction of motion. The key H3 formula is: L = L₀/γ where L₀ is the proper length and γ is the Lorentz factor.

Treat this as the sister result of time dilation: both come from the same Lorentz structure with different measurement conditions.

1. Definitions (Must Know)

  • Lorentz factor: γ = 1/(square root of (1-v²/c²))
  • Proper length, L₀: length measured in the frame where the object is at rest (its rest frame).
  • Contracted length, L: length measured in a frame where the object moves at speed v (measured along the direction of motion).
  • Length contraction formula: L = L₀/γ
  • Symbols used in this lesson: L₀,L (m), v (m s⁻¹), c (≈ 3.00 × 10⁸ m s⁻¹), γ (dimensionless).

Length contraction factor vs speed

The ratio L/L0 decreases with speed and drops rapidly as v approaches c.

Scroll across the graph to read all labels.

The ratio L/L0 decreases with speed and drops rapidly as v approaches c.The ratio L/L0 decreases with speed and drops rapidly as v approaches c.
Length contraction factor is 1/γ: L = L0/γ.
Open full-size graph
View figure data
Values for Length contraction factor vs speed
Speed (v/c)L/L₀
01
0.20.98
0.40.917
0.60.8
0.80.6
0.90.436
0.950.312
0.980.199
0.990.141

2. Key Ideas (What Earns Marks)

  • Length contraction only applies parallel to the relative motion. Perpendicular dimensions are unchanged in this model.
  • Proper length is the largest length: L ≤ L₀.
  • Measuring length requires recording the positions of both ends at the same time in your frame (this ties directly to relativity of simultaneity).

Quick comparison:

QuantityMeaningKey condition
Proper length L₀Object’s rest-frame lengthObject at rest in that frame
Contracted length LLength measured when object movesEndpoints measured simultaneously in the measuring frame

3. Detailed Explanations

A. Why “same time” matters for length

To measure the length of a moving rod in your frame, you need the positions of the front and back ends at the same time in your frame. Because simultaneity is frame-dependent, different frames can disagree on the length.

See: Simultaneity

B. A clean derivation using time dilation (spaceship/stars argument)

Consider two stars at rest in the Earth frame, separated by proper distance L₀ (proper length in the stars/Earth rest frame).

In Earth frame, a spaceship travels at speed v, so the travel time is: Δ t = L₀/v

The spaceship clock measures proper time for the trip (departure and arrival occur at the same place in the spaceship frame), so: Δτ = (Δ t)/γ = L₀/(γ v)

In the spaceship frame, the stars move toward the ship at speed v and the ship measures the distance between stars as L. The trip duration in the ship frame is: Δτ = L/v

Equate: L/v = L₀/(γ v) ⇒ L = L₀/γ

4. Common Mistakes

  • Using L = γ L₀ (wrong direction); moving lengths are shorter.
  • Forgetting “along the direction of motion”.
  • Mixing proper time and proper length: proper length is measured in the object’s rest frame; proper time is measured where the two events occur at the same place.
  • Measuring length using endpoints at different times (invalid in a single frame).

5. Exam Tips

  • Decide what is proper: “Which frame is the object at rest in?” That frame measures L₀.
  • Write the safe chain: L = L₀/γ, γ = 1/(square root of (1-v²/c²))
  • Quick check: if v → 0, then γ → 1 and L → L₀.

6. Worked Examples

Modelled example 1

Length at 0.80c

Core

Problem

A spacecraft has proper length L₀ = 1.0 × 10² m and passes Earth at 0.80c. Find its Earth-frame length.
Study the worked solution

  1. Identify the lengths

    Method

    The spacecraft rest frame measures L₀; Earth measures L.

    Reason

    The spacecraft moves relative to Earth and its length is measured along the motion.

    Working

    L₀ = 100 m, L < L₀

  2. Calculate gamma

    Method

    Evaluate the Lorentz factor.

    Reason

    v/c = 0.80.

    Working

    γ = 1/(square root of (1-0.80²)) = 1.67

  3. Contract

    Method

    Divide proper length by γ.

    Reason

    The moving length is shorter.

    Working

    L = 100/1.67 = 60 m

Guided practice 2

Find speed from contraction factor

About 5 min

Problem

A rod is measured to be half its proper length: L = 0.50L₀. Find v/c.

Try this before viewing the solution

Hints

Hint 1: extract gamma
Use L/L₀ = 1/γ before rearranging the Lorentz factor.
Hint 2: use the positive magnitude
Once v²/c² = 3/4, speed is the positive square root.
View solution step by step

  1. Find gamma

    Method

    Invert the length fraction.

    Reason

    L/L₀ = 1/γ.

    Working

    0.50 = 1/γ ⇒ γ = 2

  2. Find speed

    Method

    Invert and square the Lorentz-factor relation.

    Reason

    This isolates the squared speed ratio.

    Working

    1-v²/c² = 1/4 ⇒ v/c = square root of (3/4) = 0.866

Common misconception 3

Length at 0.60c

Find and correct the mistake

Learner claim

A 5.0 m proper-length rod moves along its length at 0.60c. A learner calculates L = γ L₀ = 6.25 m. Explain the error and correct the method.

Try this before viewing the solution

Required relation

View solution step by step

  1. Apply the ordering check

    Method

    The moving length must be smaller than the proper length.

    Reason

    L₀ is measured in the rod rest frame and is the maximum length.

    Working

    L < 5.0 m

  2. Calculate gamma

    Method

    At 0.60c, γ = 1.25.

    Reason

    1/square root of (1-0.60²) = 1.25.

    Working

    γ = 1.25

  3. Correct the relation

    Method

    Divide by γ.

    Reason

    Multiplication would incorrectly lengthen the moving rod.

    Working

    L = 5.0/1.25 = 4.0 m

Examiner practice 4

Contraction to 80% of proper length

4 marks

Examination question

An object is measured to have L = 0.80L₀. Find v/c. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Relate the fraction

    1 mark

    Method

    Use the contraction ratio.

    Reason

    L/L₀ = 1/γ.

    Working

    0.80 = 1/γ

  2. Find gamma

    1 mark

    Method

    Invert the ratio.

    Reason

    The Lorentz factor exceeds one.

    Working

    γ = 1.25

  3. Rearrange

    1 mark

    Method

    Square the reciprocal relation.

    Reason

    This isolates v²/c².

    Working

    1-v²/c² = 1/1.25² = 0.64

  4. Calculate

    1 mark

    Method

    Take the positive speed magnitude.

    Reason

    Speed is non-negative.

    Working

    v/c = square root of 0.36 = 0.60

Challenge 5

Which length is “proper”?

Minimal support

Independent transfer

You measure a passing spaceship as 60 m long; its pilot measures 100 m. Classify L₀ and L, state the relevant rest frame, and explain why the values differ.

Try this before viewing the solution

Hints

Hint 1: start from rest
Orient by asking which observer is at rest relative to the object whose length is measured.
View solution step by step

  1. Identify proper length

    Method

    The pilot’s 100 m is L₀.

    Reason

    The spacecraft is at rest in the pilot frame.

    Working

    L₀ = 100 m

  2. Identify moving length

    Method

    Your 60 m is L.

    Reason

    The spacecraft moves relative to your frame, and its endpoints are recorded simultaneously in that frame.

    Working

    L = 60 m

  3. Explain the relation

    Method

    The moving length is contracted along the motion.

    Reason

    Lorentz transformations give L = L₀/γ.

    Working

    60 = 100/γ

7. Mind Stretchers

Mind stretcher 1: Why contraction and time dilation are consistentExtension

In the spaceship/stars scenario, Earth says “distance is L₀ and time is Δ t”. The spaceship says “time is smaller (Δτ)”.

What must the spaceship conclude about the distance, and why?

Answer

It must conclude the distance is smaller: L = L₀/γ. Otherwise v = L/Δτ would exceed the relative speed v measured in both frames, creating an inconsistency with Lorentz transformations and the invariance of c.

Mind stretcher 2: The “garage and pole” paradox (qualitative)Extension

A moving pole is length-contracted and can “fit” inside a garage in the garage frame. In the pole’s rest frame, the garage is length-contracted and seems too short. How can both frames be consistent?

Answer

The key is relativity of simultaneity. “Both ends inside at the same time” is frame-dependent. The garage frame’s “door closes when the back end enters” and the pole frame’s “door close events” do not occur simultaneously in both frames, so there is no contradiction.

8. Optional/Enrichment: Low-Speed Approximation

For v≪ c, γ ≈ 1 + (1/2)v²/c², so: L ≈ L₀(1-(1/2)v²/c²) This explains why length contraction is negligible at everyday speeds.

Next in the maintained sequence: Relativistic Velocity Addition.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027