Math 2.0

Relativity and gravitational theory Special relativity: kinematics Proposition

Length contraction

Builds on Lorentz transformationLorentz factor

Statement

Let SS and S′S' be inertial frames with coordinates (t,x,y,z)(t, x, y, z) and (t′,x′,y′,z′)(t', x', y', z'). Let S′S' move with velocity vv along the xx axis of SS, where ∣v∣<c\abs{v} < c. Let a rod be at rest in S′S', lying along the x′x' axis, with length L0L_0 in S′S' (its proper length). Define its length LL in SS as the distance between the positions of its two ends at one and the same time tt of SS. Then L=L0γ,γ=11−v2/c2,L = \frac{L_0}{\gamma}, \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}, where γ\gamma is the Lorentz factor. Lengths perpendicular to the direction of motion are the same in both frames.

Remarks

Proof idea: the Lorentz transformation gives x′=γ(x−vt)x' = \gamma (x - v t). For the two ends at the same time tt, subtracting gives L0=Δx′=γΔx=γLL_0 = \Delta x' = \gamma \Delta x = \gamma L. Since y′=yy' = y and z′=zz' = z, transverse lengths do not change. Example: a rod of proper length 11 metre moving at v=0.6cv = 0.6c has length 0.80.8 metre in SS. The measurement at equal times in SS is essential: the two measuring events are not simultaneous in S′S'. This is why each observer can find the other's rods contracted without contradiction.

Going deeper

For a student

Length contraction and time dilation are two faces of one Lorentz transformation, and in many problems either one gives the answer. For a muon flying down through the atmosphere, the Earth observer uses time dilation, while the muon uses length contraction of the atmosphere. Paradoxes such as the ladder paradox are resolved by keeping track of which events are simultaneous in which frame.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Length_contraction
About this page
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Kindstatement, proposition
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MSC83A05
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AboutLorentz transformation
UsesLorentz transformation, Lorentz factor
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Changed3 November 2026

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