Year 12 Physics Module 7 ⏱ ~45 min 5 MC · 2 Short Answer Lesson 10 of 14

Length Contraction

In 1940, Bruno Rossi and colleagues measured cosmic-ray muons at mountain altitude and sea level. Classical physics predicts muons travelling at $v = 0.998c$ with a lifetime of $\tau_0 = 2.2\,\mu$s should travel only 659 m before decaying, far short of the 10 km from their creation point. Yet they arrive at sea level. In the muon's rest frame the reason is length contraction: the atmosphere is only $10\,000/\gamma \approx 625$ m thick, well within their classical range.

Today's hook: Cosmic ray muons are created 10 km above Earth at $v = 0.998c$. Their rest-frame lifetime is $2.2\,\mu$s, classically giving a range of only 659 m. Yet particle detectors at sea level detect them easily, they travel 10,000 m. In the muon's own rest frame, where did those 10 km go? And if a 100 m spaceship passes Earth at 0.8c, what length does Earth measure, and why?
0/5TASKS
Your guided path0 of 8 done · keep going
1
You’re here

Orient and predict

Predict what Earth measures for a passing spaceship, then set your goals and vocabulary for length contraction.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you read, predict

A 100 m spaceship flies past Earth at 0.8$c$. Observers on Earth measure its length as it passes.

  1. Will Earth observers measure the ship's length to be more than, less than, or exactly 100 m?
  2. The ship's captain measures the ship's length while flying. What does she measure?
  3. A tunnel on Earth is 80 m long. At what speed would the ship need to travel so that Earth observers measure it as exactly 80 m, fitting entirely inside the tunnel?

Write your predictions before reading on, you will revisit them at the end.

Warm-up, which statement about length contraction is correct?

Learning Intentions
goals

Know, Length Contraction

  • Moving objects contract along direction of motion
  • $L = L_0/\gamma$
  • Only lengths parallel to motion contract

Understand, Proper Length

  • $L_0$ measured in object's rest frame
  • $L$ is always less than or equal to $L_0$
  • Contraction and time dilation are inseparable

Can Do, Solve Problems

  • Calculate contracted lengths from proper length and velocity
  • Relate length contraction to time dilation
  • Analyse muon and barn paradox scenarios
Scan these before reading
vocab
Proper length ($L_0$)The length of an object measured in the frame where the object is at rest, always the maximum measured length.
Contracted length ($L$)The shorter length measured in any frame where the object is moving; $L = L_0/\gamma \leq L_0$.
Length contractionThe relativistic reduction in measured length of an object moving relative to the observer, in the direction of motion only.
Lorentz factor ($\gamma$)$\gamma = 1/\sqrt{1-v^2/c^2} \geq 1$; quantifies both time dilation and length contraction at speed $v$.
Barn–ladder paradoxA thought experiment where a contracted ladder fits inside a barn in one frame but not another; resolved by the relativity of simultaneity.
Cross-lesson links: L12 showed time dilation, clocks in motion tick slower. L13 shows the complementary result: distances shrink in the direction of motion. The 1940 Rossi cosmic-ray muon experiment provides the clearest real-world evidence, muons at $0.998c$ with $\tau_0 = 2.2\,\mu$s classically survive only 659 m, yet reach sea level from 10 km altitude because in their frame the atmosphere is contracted to just $\sim 625$ m. L14 uses these ideas to derive $E = mc^2$, and L15 unifies length contraction and time dilation through the relativity of simultaneity.

Core Content

2

Measure a contracted length

See how simultaneous endpoint measurements give $L = L_0/\gamma$, and which dimensions do not contract.

1
The Mathematics of Length Contraction
+5 XP

Space shrinks in the direction of motion

A spaceship built to be 100 m long flies past Earth at $0.8c$. Two Earth observers standing 100 m apart simultaneously record the position of the ship's nose and tail as it passes. They find the gap is only 60 m, the ship appears shorter in the direction of its travel. Yet the captain aboard measures a perfectly normal 100 m ship. To understand why both measurements are correct, consider a rod of proper length $L_0$ at rest in the spaceship: the ship passes Earth at speed $v$, and the two events "front of rod at a fixed Earth point" and "back of rod at the same Earth point" occur at the same location in Earth's frame but at different positions in the ship's frame.

The resolution of the apparent algebra is careful application of which frame carries proper time. The events "front of rod passes the point" and "back of rod passes the point" occur at the same Earth position, so Earth clocks measure proper time $\Delta t_0$ for this interval. The ship frame measures $\Delta t = \gamma \Delta t_0$. Since $L_0 = v \Delta t_{\text{ship}}$ and $L = v \Delta t_0$:

Length contraction equation

$$L = \dfrac{L_0}{\gamma}$$

The moving rod is measured to be shorter by factor $\gamma$. Like time dilation, it is reciprocal: each observer sees the other's lengths contracted. And like time dilation, the key properties are:

  • Only lengths parallel to the direction of motion contract. Perpendicular dimensions are unchanged.
  • $L \leq L_0$ always, proper length is the maximum measured length.
  • Length contraction and time dilation are two views of the same underlying spacetime geometry.
At rest (L₀) Rod L₀ Moving (L = L₀/γ) Rod L < L₀ Height unchanged

Figure 1, Length contraction: the rod contracts along its direction of motion but its height (perpendicular) stays the same

Stop & Check

A rod has proper length 2.0 m. It moves past you at $0.6c$. What length do you measure? If the rod is rotated 90° so it moves perpendicular to its length, what length do you measure now?

Length contraction: $L = L_0/\gamma$ where $\gamma \geq 1$, so $L \leq L_0$ always. Proper length $L_0$ is measured in the object's rest frame, always the longest. Contracted length $L$ is shorter. Only the dimension parallel to motion contracts; perpendicular dimensions are unchanged. The effect is real and reciprocal.

Write $L = L_0/\gamma$ and note which dimensions do and don't contract.

A rod of proper length 5.0 m moves at $0.6c$ past an observer. The observer measures its length as:

3

Explain the muon evidence in two frames

Match the Earth-frame time dilation account to the muon-frame length contraction account of the same arrival.

2
The Muon Revisited, Two Explanations
+5 XP

Time dilation and length contraction are two sides of the same coin

We just saw that length contraction is described by $L = L_0/\gamma$ with contraction only along the direction of motion. That raises a question: if the same muon experiment was explained in L12 by time dilation, how does length contraction give the same answer from a different frame? This card answers it → both explanations are perspectives on one spacetime geometry.

In Lesson 12, we explained how muons reach Earth's surface using time dilation: in Earth's frame, the muon's lifetime is extended by $\gamma \approx 10$ at $v \approx 0.995c$, allowing it to travel ~6.6 km rather than the classical 660 m.

But we can also explain it entirely from the muon's frame: the muon is at rest, and Earth's atmosphere rushes toward it at $0.995c$. The atmosphere's thickness is length-contracted from 15 km to $15/\gamma \approx 15/10 = 1.5$ km (or ~650 m for $\gamma \approx 22$ at $0.9995c$). The muon, living only 2.2 $\mu$s in its own frame, travels $0.995c \times 2.2\ \mu\text{s} \approx 650$ m, just enough to reach the contracted surface.

Both explanations are correct. They are two perspectives on the same relativistic reality. Time dilation and length contraction are inseparable: you cannot have one without the other. They are two manifestations of the same spacetime geometry.

Summary, Length Contraction Equations

$L = \dfrac{L_0}{\gamma}$   (contracted length from proper length)

$L_0 = \gamma L$   (proper length from contracted measurement)

$L_{\perp} = L_{0\perp}$   (no contraction perpendicular to motion)

Earth's Frame Atmosphere 15 km (proper) Earth surface muon v ≈ 0.995c Time dilation: τ = γτ₀ ≈ 22 μs Muon's Frame Atmosphere ~1.5 km (contracted) Earth surface μ Earth rushing up Length contraction: L = L₀/γ ≈ 1.5 km

Figure 2, Same physical event (muon reaches Earth's surface) explained two ways: Earth's frame uses time dilation ($\tau = \gamma\tau_0$); muon's frame uses length contraction ($L = L_0/\gamma$). Both give the same outcome

Stop & Check

A spaceship has proper length 300 m. Earth observers measure it as 180 m as it flies past. Calculate its speed. The ship passes a space station 900 m long (proper length). How long does the ship measure the station to be?

Muon experiment, two frames, same outcome: Earth frame uses time dilation ($\tau = \gamma\tau_0 \approx 22\,\mu$s); muon frame uses length contraction (atmosphere $= 15/\gamma \approx 1.5$ km). Both give the same physical result. To find speed from length: $\gamma = L_0/L$, then $v = c\sqrt{1-1/\gamma^2}$. Contraction is reciprocal.

Write both muon-experiment explanations side by side and the formula for finding $v$ from a measured length.

A spaceship of proper length 300 m is measured as 150 m by Earth observers. Its speed is:

4

Resolve the barn and ladder

Calculate both contracted lengths at $0.8c$, then settle the apparent contradiction with the relativity of simultaneity.

3
Worked Example, The Barn and Ladder Paradox
+5 XP

When contraction creates apparent contradictions

We just saw that the same muon event looks different in different frames but both frames agree on the physical outcome. That raises a question: what happens when length contraction in one frame seems to give a contradictory result in another, can a ladder "fit" in a barn in one frame but not another? This card answers it → the barn-ladder paradox, resolved by the relativity of simultaneity.

Problem

A barn has proper length 10 m. A ladder has proper length 15 m. The ladder moves at $0.8c$ toward the barn.

  1. How long is the ladder in the barn's frame?
  2. How long is the barn in the ladder's frame?
  3. Can the ladder fit entirely inside the barn in the barn's frame?
  4. Can the ladder fit inside the barn in the ladder's frame?
  5. Explain why there is no true paradox.
Step 1, Ladder in barn frame

$\gamma = 1/\sqrt{1-0.8^2} = 1/\sqrt{1-0.64} = 1/0.6 = 1.667$

$L_{\text{ladder}} = 15/1.667 = $ 9.0 m

Step 2, Barn in ladder frame

$L_{\text{barn}} = 10/1.667 = $ 6.0 m

Step 3, In barn frame

The contracted ladder (9.0 m) is shorter than the 10 m barn. Yes, both doors can be closed simultaneously with the ladder inside, for a moment the ladder fits entirely within the barn.

Step 4, In ladder frame

The contracted barn (6.0 m) cannot contain the 15 m ladder. No, the ladder extends well beyond either end of the barn at any instant.

Step 5, Resolving the paradox

The key is the relativity of simultaneity. In the barn frame, the events "front of ladder reaches back door" and "back of ladder reaches front door" are simultaneous: both doors close at the same instant. In the ladder frame, these events are not simultaneous. The front door closes first (when the front reaches it), then the back door closes later. The ladder is never fully enclosed at any single instant in its own frame. Both frames are correct; they simply disagree about simultaneity. There is no physical contradiction because "fitting inside" depends on which events you call simultaneous.

Barn frame (barn at rest) Barn 10 m Ladder 9 m (contracted) Ladder fits! Both doors close simultaneously Ladder frame (ladder at rest) Ladder 15 m (proper, at rest) Barn 6 m (contracted) Ladder does not fit, doors can't both close simultaneously Resolution: simultaneity is frame-dependent

Figure 3, Barn–ladder paradox: in the barn frame the contracted ladder (9 m) fits; in the ladder frame the contracted barn (6 m) cannot contain the ladder. Both are correct, simultaneity is relative

Stop & Check

A spaceship of proper length 200 m flies past a space station at $0.6c$. The station measures the ship as it passes. How long does the ship take to completely pass a fixed point on the station, according to station clocks?

Barn–ladder at $v = 0.8c$: ladder = 9 m in barn frame (fits); barn = 6 m in ladder frame (does not fit). No paradox: in barn frame both doors close simultaneously; in ladder frame they close at different times. "Fitting inside" requires simultaneous events, simultaneity is frame-dependent, so both frames are correct.

Write the two contracted lengths and the one-sentence resolution using simultaneity.

5

Clear the misconceptions

Correct the four classic length-contraction traps and lock in the five-step exam method.

4
Common Misconceptions and HSC Strategies
+5 XP

Avoiding the traps that cost marks

We just saw how the barn-ladder paradox is resolved by simultaneity being frame-dependent. That raises a question: what misconceptions about length contraction do students most often bring into the HSC exam? This card answers it → four misconceptions corrected and a five-step exam strategy.

Misconception 1: Proper length is Earth-frame length or the "correct" length.
Wrong. Proper length is the length measured in the frame where the object is at rest. If a spaceship's dimensions are measured while the ship is in your lab, that is the proper length. If the ship is moving, you measure a contracted length.

Misconception 2: Using $L = \gamma L_0$ instead of $L = L_0/\gamma$.
Wrong. Since $\gamma \geq 1$ always, the contracted length must be shorter than proper length. If your answer gives $L > L_0$, you have the equation inverted. A quick check: "moving lengths shrink, so $L < L_0$."

Misconception 3: All dimensions contract.
Wrong. Only the dimension parallel to the direction of motion contracts. A cube moving left–right contracts to a thin slab in that direction, but its height and depth remain the same. This is why perpendicular measurements are absolute in SR.

Misconception 4: Length contraction is just a measurement effect, the object is "really" its proper length.
Wrong. Length contraction is just as physically real as time dilation. Both result from the structure of spacetime. "Real" length depends on the reference frame, just as "real" time does.

HSC Exam Strategy

Step-by-step for length contraction problems:

  1. Identify proper length $L_0$ (measured in the object's rest frame)
  2. Calculate $\gamma = 1/\sqrt{1-v^2/c^2}$
  3. Apply $L = L_0/\gamma$ (contracted length is shorter)
  4. Check: $L < L_0$? If not, the formula is inverted
  5. Remember: perpendicular dimensions do not contract
Stop & Check

A square spaceship of side 10 m (proper) travels at $0.8c$ in the direction of one side. An observer on a space station watches it pass. What shape and dimensions does the observer measure? Show your working and explain your answer for each dimension.

Four traps: (1) "Proper" = rest frame, not Earth frame. (2) Use $L = L_0/\gamma$ (divide), never $L = \gamma L_0$, contracted must be shorter. (3) Only the dimension parallel to motion contracts; perpendicular sides unchanged. (4) Contraction is physically real, not a measurement illusion. HSC strategy: identify $L_0$ → $\gamma$ → $L = L_0/\gamma$ → check $L < L_0$.

List the four misconceptions and the five-step exam strategy.

In the barn–ladder paradox, the resolution depends on:

6

Apply and consolidate

Run your own contraction calculations, then argue the barn and ladder out in full.

Activity 1, Length Contraction Calculations
ApplyBand 4

Apply $L = L_0/\gamma$ to a range of scenarios

  1. A rod of proper length 2.0 m moves at $0.6c$. Calculate $\gamma$ and the contracted length. Verify by finding $v$ such that the rod contracts to 1.0 m.
  2. A muon at $v \approx 0.995c$ sees Earth's atmosphere contracted from 15 km. Calculate $\gamma$ and the contracted thickness the muon measures. Show how this explains the muon reaching sea level without using time dilation.
  3. A spaceship of proper length 200 m is measured as 160 m by a space station. Calculate the ship's speed and express it as a fraction of $c$.
  4. A cube (proper side 5.0 m) moves at $0.8c$ parallel to one face. State the dimensions an observer measures and explain why two sides are unchanged.
Activity 2, Analysing the Barn–Ladder Paradox
AnalyseBand 6

Resolve the apparent contradiction using the relativity of simultaneity

  1. A barn has proper length 12 m. A ladder has proper length 20 m, moving at $0.8c$. Calculate the ladder's length in the barn frame and the barn's length in the ladder frame.
  2. In the barn frame, both doors close simultaneously with the ladder inside. In the ladder frame, the ladder extends beyond both ends at the same instant. Explain why there is no physical contradiction, referencing simultaneity.
  3. A student argues "the ladder must really fit or really not fit, they can't both be right." Identify the flaw in this argument and explain the correct interpretation.
  4. Sketch a spacetime diagram showing the worldlines of the front door, back door, and both ends of the ladder to illustrate the simultaneity issue. (You may draw this freehand and describe it.)