Annus MirabilisEinstein in 1905
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special relativity

The equation explained

s2=Δx2+Δy2+Δz2−c2Δt2=Δx′2+Δy′2+Δz′2−c2Δt′2\begin{aligned}s^2 &= \Delta x^2 + \Delta y^2 + \Delta z^2 - c^2 \Delta t^2 \\ &= \Delta x'^2 + \Delta y'^2 + \Delta z'^2 - c^2 \Delta t'^2\end{aligned}
In wordsThe squared interval between two events: the squared distance between them along the three coordinates, less the square of the velocity of light times the time between them in the stationary system. The same number comes out from the coordinates and the time of any other system.OverviewOne number, built from the distance and the time between two events, that every system computes alike. It is a later way of stating what the transformation does.Full explanationBoth systems disagree about the distance between two events and about the time between them, and this combination of the two is what they agree on. Checking it is arithmetic with the transformation, and the same cancellation that carried the light sphere of § 3 into the moving system carries this: the cross terms in Δx Δt\Delta x \, \Delta t cancel and the factor meets the bracket it is built from. A reader who has met the contraction and the slowing of a clock as separate results can take this as the single statement both come from.Every stepΔx′=γ(Δx−vΔt),Δt′=γ(Δt−vc2Δx)\Delta x' = \gamma (\Delta x - v \Delta t), \qquad \Delta t' = \gamma \left( \Delta t - \frac{v}{c^2} \Delta x \right)The transformation, applied to the differences between two events.Δx′2−c2Δt′2=γ2[(Δx−vΔt)2−c2(Δt−vc2Δx)2]\Delta x'^2 - c^2 \Delta t'^2 = \gamma^2 \left[ (\Delta x - v \Delta t)^2 - c^2 \left( \Delta t - \frac{v}{c^2} \Delta x \right)^2 \right]Square both and subtract. Everything is inside one factor of γ2\gamma^2.(Δx−vΔt)2−c2(Δt−vc2Δx)2=(1−v2c2)(Δx2−c2Δt2)(\Delta x - v \Delta t)^2 - c^2 \left( \Delta t - \frac{v}{c^2} \Delta x \right)^2 = \left( 1 - \frac{v^2}{c^2} \right) \left( \Delta x^2 - c^2 \Delta t^2 \right)Expand. The cross terms are −2vΔxΔt-2 v \Delta x \Delta t in both and cancel, leaving Δx2(1−v2c2)−Δt2(c2−v2)\Delta x^2 \left( 1 - \frac{v^2}{c^2} \right) - \Delta t^2 \left( c^2 - v^2 \right), which is the right-hand side.γ2(1−v2c2)=1⟹Δx′2−c2Δt′2=Δx2−c2Δt2\gamma^2 \left( 1 - \frac{v^2}{c^2} \right) = 1 \quad \Longrightarrow \quad \Delta x'^2 - c^2 \Delta t'^2 = \Delta x^2 - c^2 \Delta t^2The factor meets the bracket and the two sides are equal. The transverse terms are unchanged by the transformation, so adding them to both sides completes the display.Historian's marginThis is not in the paper. It is Minkowski's, from his lecture of 1908, and the lab's own sentence says so before the display. Nothing in § 3 or § 4 requires it, and the site's account of the 1905 argument never uses it as a premise: the transformation is derived from the two principles, and this invariance is a consequence of the transformation rather than an axiom it rests on. It is offered here as a later way of stating the same facts, which is what it is.Also

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