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Question 28 of 34

Q.Explain length contraction.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2019Subjective· 3mImportance★★★★★
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Length contraction is the relativistic effect whereby the length of an object, measured by an observer relative to whom it is moving, is found to be smaller than its proper length, by the factor 1−v2/c2\sqrt{1-v^2/c^2}.

Let a rod be at rest in frame S′S', which moves with velocity vv along the xx-axis relative to frame SS. Let L0=x2′−x1′L_0 = x_2' - x_1' be the proper length of the rod — its length as measured by an observer at rest relative to the rod (in S′S'), where x1′x_1' and x2′x_2' are the (time-independent) positions of its two ends in S′S'.

An observer in frame SS, relative to whom the rod is moving, must measure the positions x1x_1 and x2x_2 of the two ends of the rod at the same instant of time tt (in SS) to obtain its length L=x2−x1L = x_2 - x_1 in that frame — measuring the ends at different times would give a meaningless result for a moving object.

Using the Lorentz transformation x′=γ(x−vt)x' = \gamma(x - vt), where γ=1/1−v2/c2\gamma = 1/\sqrt{1-v^2/c^2}, and applying it to both ends at the same time tt:

x1′=γ(x1−vt),x2′=γ(x2−vt)x_1' = \gamma(x_1 - vt), \qquad x_2' = \gamma(x_2 - vt)

Subtracting,

L0=x2′−x1′=γ(x2−x1)=γLL_0 = x_2' - x_1' = \gamma(x_2 - x_1) = \gamma L

so that

L=L0γ=L01−v2c2L = \dfrac{L_0}{\gamma} = L_0\sqrt{1 - \dfrac{v^2}{c^2}}

Since γ≥1\gamma \ge 1 for any v>0v>0, we have L≤L0L \le L_0: the length measured by an observer relative to whom the rod moves is always less than (or equal to) its proper length. This is length contraction (also called the Lorentz–FitzGerald contraction).

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