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Topic 4 of 10

Length Contraction

A rod moving relative to an observer appears shorter than when at rest — its length is $L=L_0/\gamma$ along the direction of motion. Like time dilation, this is a real physical effect, not an illusion. Transverse dimensions are unaffected. The two effects — time dilation and length contraction — are two sides of the same Lorentz transformation.

Key Concepts

Proper Length

L_0

is the length of an object measured in its rest frame — the longest measured length. Any other frame measures a shorter value.

Lorentz Contraction

L=L_0/\gamma

. Only the dimension parallel to the velocity is contracted; dimensions perpendicular to

v

are unchanged. For

v\ll c

L\approx L_0

Ladder Paradox

A ladder of proper length

L_0

fits inside a barn of proper length

L_b < L_0

. If the ladder moves fast enough,

L=L_0/\gamma<L_b

— the ladder fits. Resolved by relativity of simultaneity: the two ends of the ladder enter/exit simultaneously in neither frame.

Penrose-Terrell Rotation

Although a moving sphere is length-contracted, the visual appearance is of a rotated sphere (not a squashed one) due to light travel-time effects. Lorentz contraction is real but visual appearance involves additional optics.

Key Equations

Length contraction

L = \frac{L_0}{\gamma} = L_0\sqrt{1-\beta^2}

Length measured in lab frame; L₀ = proper length (rest frame value).

Transverse dimensions

y' = y, \quad z' = z

Perpendicular dimensions are unchanged by a boost along x.

Proper length from interval

L_0 = \sqrt{-(\Delta s)^2} = \sqrt{\Delta x^2 - c^2\Delta t^2}

Spacelike interval between the ends (at the same rest-frame time) gives proper length.

Worked Example

Contracted Length of a Moving Spacecraft

Problem

A spacecraft has proper length $L_0=500$ m. It flies past a space station at $v=0.80c$ . What length does the station measure?

Solution

$\gamma=1/\sqrt{1-0.64}=1/0.6=1.667$ .

L = \frac{L_0}{\gamma} = \frac{500}{1.667} = 300\text{ m}

The spacecraft appears only 300 m long to the station — 40% shorter than its rest length.

Answer L = 300 m.

Practice

Exercises

7 problems

1 of 7

Compare the two frames: the blue ship (rest frame, top) always shows L₀; the amber ship (lab frame, bottom) is shorter. A rod with proper length L₀ = 100 m moves at β = 0.60. Use L = L₀/γ to find the lab-frame length.

length contraction

Lab speed β 0.00c → L = 100.0 m

Compare the two frames: the rest-frame ship (blue, top) always has length L₀; the lab-frame ship (amber, bottom) is shorter by 1/γ.

Contracted length

L = m

Given: L₀ = 100 m, β = 0.6

2 of 7

Watch the spacecraft fly past the station — the green arrow shows how short it appears. A spacecraft with proper length L₀ = 500 m passes at β = 0.80. What does the station measure for its length?

flyby

Spacecraft speed β 0.80c → L = 300 m

The station (blue box) is always L₀ wide. Watch the spacecraft shrink as you increase β — the green arrow shows how short it appears.

Lab-frame length

L = m

Given: L₀ = 500 m, β = 0.8

3 of 7

A moving object appears to have length $L=120$ m. Its proper length is $L_0=200$ m. Find $\gamma$ .

Unlock Exercise 3