Annus MirabilisEinstein in 1905
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special relativity

The equation explained

ξ=aV(t−vV2−v2x′).\xi = a V \left( t - \frac{v}{V^2 - v^2} x' \right) .
In wordsThe distance along the motion in the moving system is the factor still to be found times the velocity of light times the stationary time less the speed of the moving system over the square of the velocity of light less the square of that speed, times the distance along the motion.OverviewThe same ray, with the moving system's time written out in the stationary system's terms. It gives the moving distance in terms of a stationary time and a stationary distance, which is half of what the section is after.Full explanationNothing new is assumed here: the expression for τ\tau that the previous displays reached is put into ξ=Vτ\xi = V \tau. What the line still contains is a stationary time tt as well as a distance, so it does not yet say where the ray is in terms of position alone. The paper removes tt in the next two displays by asking, of this same ray, how far it has got in the stationary system at the stationary time tt, and putting that in. The factor aa is still the unknown of the previous display, and it rides along untouched.Every stepξ=Vτ\xi = V \tauThe ray, measured in the moving system, travels at the velocity of light.τ=a(t−vV2−v2x′)\tau = a \left( t - \frac{v}{V^2 - v^2} x' \right)The moving system's time, as the section has just found it.ξ=aV(t−vV2−v2x′)\xi = a V \left( t - \frac{v}{V^2 - v^2} x' \right)Put the second into the first. The line as printed.Also

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