Annus MirabilisEinstein in 1905
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special relativity

The equation explained

τ=β(t−vV2x),ξ=β(x−vt),X′=X,η=y,Y′=β(Y−vVN),ζ=z,Z′=β(Z+vVM).\begin{aligned} \tau &= \beta \left( t - \frac{v}{V^2} x \right) , \\ \xi &= \beta (x - v t) , &\qquad X' &= X , \\ \eta &= y , &\qquad Y' &= \beta \left( Y - \frac{v}{V} N \right) , \\ \zeta &= z , &\qquad Z' &= \beta \left( Z + \frac{v}{V} M \right) . \end{aligned}
In wordsThe moving system's time and its coordinates, in terms of the time and the coordinates of the stationary one, through the factor, the speed of the moving system and the velocity of light; and the moving system's electric force in terms of the electric force and the magnetic force of the stationary one.OverviewThe transformations § 10 needs, gathered in one place: the time and coordinates from § 3, the electric force from § 6.Full explanationNothing is derived here. The left column is § 3's transformation of the time and the coordinates, with β\beta for the factor 1/1−(v/V)21 / \sqrt{1 - (v/V)^2} defined on p. 900, and the right column is § 6's transformation of the electric force, with the factor φ(v)\varphi(v) already fixed at 1 there. The paper has just laid down that τ=ξ=η=ζ=0\tau = \xi = \eta = \zeta = 0 when t=x=y=z=0t = x = y = z = 0, so the two origins meet and no further shift is needed. Across the motion the coordinates are unchanged, η=y\eta = y and ζ=z\zeta = z, while the electric force across the motion is not: each transverse component picks up a component of the stationary system's magnetic force, with the sign set by which axis it is.Every stepτ=β(t−vV2x),ξ=β(x−vt),η=y,ζ=z\tau = \beta \left( t - \frac{v}{V^2} x \right) , \quad \xi = \beta (x - v t) , \quad \eta = y , \quad \zeta = z§ 3's transformation, with the origins made to meet at t=x=y=z=0t = x = y = z = 0.X′=X,Y′=β(Y−vVN),Z′=β(Z+vVM)X' = X , \quad Y' = \beta \left( Y - \frac{v}{V} N \right) , \quad Z' = \beta \left( Z + \frac{v}{V} M \right)§ 6's transformation of the electric force, with its factor φ(v)\varphi(v) already fixed at 1 (p. 909).η=y,Y′=β(Y−vVN)\eta = y , \quad Y' = \beta \left( Y - \frac{v}{V} N \right)The pair worth reading together: across the motion the coordinate is unchanged while the electric force is not. That difference is what makes the transverse mass of p. 919 what it is.Historian's marginThe two columns are the two results § 10 rests on, and the paper reprints them rather than referring back. The right column is the one § 6 reads in words (pp. 909-910): a charge at rest in kk feels (X′,Y′,Z′)(X', Y', Z'), so a charge carried with kk through a purely magnetic force in the stationary system feels an electric force in its own system, which the older mode of expression called the electromotive force.Also

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