Annus MirabilisEinstein in 1905
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special relativity

The equation explained

Δx=Δx′1−v2c2=L0γ\Delta x = \Delta x' \sqrt{1 - \frac{v^2}{c^2}} = \frac{L_0}{\gamma}
In wordsThe distance between the ends in the stationary system, both read at one time, is the distance between them in the system the rod rests in times the square root of one less the square of the speed over the velocity of light: that is, the rod's own length divided by the factor.OverviewA rod is measured shorter along its motion, by the factor, when both ends are read at one time of the system doing the measuring.Full explanationThe condition is the whole of it: both ends are read at one time of the stationary system, which is what the sentence above the display means by Δt=0\Delta t = 0. Put that into the transformation of the display above and the time term drops out, leaving the two lengths in this ratio. The measurement is not about what a camera would record, and nothing has happened to the rod: the two systems disagree about which pairs of readings are simultaneous, and a length is defined by readings taken at one time. Read the same rod at one time of the system it rests in and the ratio goes the other way, which is why neither system is the one that is right.Every stepΔx′=γ(Δx−vΔt)\Delta x' = \gamma (\Delta x - v \Delta t)The transformation of the display above, applied to the difference between the two ends rather than to one position.Δt=0⟹Δx′=γ Δx\Delta t = 0 \quad \Longrightarrow \quad \Delta x' = \gamma \, \Delta xBoth ends read at one time of the stationary system. The time term drops and the two lengths differ by the factor alone.Δx=Δx′γ=Δx′1−v2c2\Delta x = \frac{\Delta x'}{\gamma} = \Delta x' \sqrt{1 - \frac{v^2}{c^2}}Turn it over. Since the factor is never less than 1, the length measured in the stationary system is never the greater of the two.v=0.6 c⟹Δx=0.8 Δx′v = 0.6 \, c \quad \Longrightarrow \quad \Delta x = 0.8 \, \Delta x'A worked value: at six tenths of the velocity of light the factor is 1.25 and the measured length is four fifths of the rod's own.Historian's marginThe paper does not state the result as a ratio of lengths. § 4 asks what shape a sphere carried along has, and answers with the axes of an ellipsoid, R1−(v/V)2R \sqrt{1 - (v/V)^2}, RR, RR, which is the display two below this one. Writing it as L0/γL_0 / \gamma is this edition's arrangement, and L0L_0 is a modern name for the length in the system the rod rests in; the paper has no symbol for it here.Also

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