Length contraction and proper length: VCE Physics Unit 4

Syllabus dot point

“Describe proper length and use the length contraction relationship with the Lorentz factor to compare lengths measured in different inertial frames at speeds approaching the speed of light”

8 min readUpdated 7 Oct 2026

Quick answer

Proper length $L_0$ is the length measured in the frame in which the object (or the distance) is at rest. An observer who sees it moving at speed $v$ along its length measures a shorter length, $L = L_0/\gamma = L_0\sqrt{1 - v^2/c^2}$ . Only the dimension along the direction of motion contracts.

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What this dot point is asking
Proper length
Length contraction
Worked examples
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What this dot point is asking

VCAA expects you to describe proper length $L_0$ and model length contraction with $L = L_0/\gamma$ , where $\gamma = 1/\sqrt{1 - v^2/c^2}$ is the Lorentz factor from the time dilation page.

Proper length

Proper length ( $L_0$ ) is the length measured in the frame of reference in which the object is at rest. Anyone moving along with the object measures this value. It is the longest length any inertial observer will measure for that object.

The same idea applies to a distance between two points that are fixed relative to each other, such as two stars, or the two ends of an accelerator tube. The proper length is the distance measured in the frame where those points are at rest.

Length contraction

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

An observer who sees the object moving past at speed $v$ measures the length $L$ , which is shorter than $L_0$ because $\gamma \ge 1$ .

The contraction is only along the direction of relative motion. Width and height are unchanged.
It is a real feature of measurement between frames, not an optical illusion and not a physical squashing felt by the object.
At everyday speeds $L \approx L_0$ to many decimal places.

How it connects to time dilation

Consider a journey between two stars. In the frame of the stars, the distance is the proper length $L_0$ and the trip takes time $t = L_0/v$ . The traveller's clock is present at both the start and the end, so it records the proper time $t_0 = t/\gamma$ . The traveller can explain the same shorter time by saying the stars are $L = L_0/\gamma$ apart and approach at speed $v$ , so $t_0 = L/v$ . Both views give the same $t_0$ . Length contraction and time dilation are two descriptions of the same physics.

Worked examples

Worked example

Example 1. A moving rocket

A rocket has a proper length of $100$ m. It passes a station at $0.80c$ . What length does the station measure?

$\gamma = 1/\sqrt{1 - 0.80^2} = 1.667$

$L = L_0/\gamma = 100/1.667 = 60.0$ m

Example 2. A journey between stars

Two stars are $4.0$ light years apart in their common rest frame. A ship travels between them at $0.90c$ . Find (a) the distance in the ship's frame and (b) the travel time measured by the ship and by the stars.

$\gamma = 2.294$

(a) $L = L_0/\gamma = 4.0/2.294 = 1.7$ light years (data to 2 significant figures; keep extra digits in the working).

(b) Ship's clock (proper time): $t_0 = L/v = 1.74/0.90 = 1.9$ years. Stars' frame: $t = L_0/v = 4.0/0.90 = 4.4$ years.

Check with time dilation: $t_0 = t/\gamma = 4.44/2.294 = 1.9$ years. The two methods agree.

Example 3. Finding the speed

An object's length measured while it moves is $80\%$ of its proper length. How fast is it moving?

$L/L_0 = \sqrt{1 - v^2/c^2} = 0.80$ , so $1 - v^2/c^2 = 0.64$ and $v^2/c^2 = 0.36$ .

$v = 0.60c$

Key fact

Proper length and proper time are measured in different frames. Proper length: the frame where the object is at rest. Proper time: the frame where both events happen at the same place. In a journey between two stars, the stars' frame has the proper length and the traveller's frame has the proper time.

Common errors

Contracting the wrong length: Divide $L_0$ by $\gamma$ , never multiply. The answer for $L$ must be less than $L_0$ . If it is longer, you have flipped the equation.
Contracting all three dimensions: Only the length parallel to the motion shrinks.
Calling the contracted length the "real" length: All inertial observers are equally entitled to their measurements. The proper length is distinguished only by being measured at rest.
Mixing frames in one calculation: Pair $L_0$ with $t$ (both measured in the stars' frame) and $L$ with $t_0$ (both measured in the traveller's frame), as in Example 2. Never combine $L$ with $t$ , or $L_0$ with $t_0$ .

Try this

Q1. Define proper length. [1 mark]

Cue. The length measured in the frame of reference in which the object is at rest.

Q2. A spacecraft has a proper length of $50.0$ m and passes Earth at $0.95c$ . Calculate its length as measured from Earth. [2 marks]

Cue. $\gamma = 3.20$ , $L = 50.0/3.20 = 15.6$ m.

Q3. A traveller moves at $0.60c$ past a $10.0$ m long rod at rest in the lab. What length does the traveller measure for the rod? [2 marks]

Cue. $\gamma = 1.25$ , $L = 10.0/1.25 = 8.0$ m.

Q4. A pilot flying at $0.80c$ says the distance between two beacons is $60$ m. What is the distance in the beacons' rest frame, and which of the two values is the proper length? [3 marks]

Cue. $L_0 = \gamma L = 1.667 \times 60 = 100$ m. The $100$ m is the proper length because it is measured in the frame where the beacons are at rest.