Annus MirabilisEinstein in 1905
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special relativity

The equation explained

x=wξ+v1+vwξV2t,y=1−(vV)21+vwξV2wηt,z=0.\begin{aligned} x &= \frac{w_\xi + v}{1 + \frac{v w_\xi}{V^2}} t , \\ y &= \frac{\sqrt{1 - \left( \frac{v}{V} \right)^2}}{1 + \frac{v w_\xi}{V^2}} w_\eta t , \\ z &= 0 . \end{aligned}
In wordsIn the stationary system the point's place along the motion is its speed along that direction in the moving system added to the speed of the moving system, divided by one plus their product over the square of the speed of light, times the stationary time; and its place across the motion carries the speed across with the same denominator and a square root in the numerator.OverviewThe same motion described in the stationary system. The two speeds along the motion do not simply add: their sum is divided by a factor that grows with their product. The speed across the motion is reduced as well, although nothing pushes the point sideways.Full explanationThe transformation of § 3 is put into the equations of motion, and x,y,z,tx, y, z, t take the place of ξ,η,ζ,τ\xi, \eta, \zeta, \tau. The first equation gives the familiar composition with its denominator; the second is the one worth pausing on. The point's sideways progress is smaller in KK than wηw_\eta would suggest, by both the square root and the same denominator, and no force acts across the motion. The cause is the clock: η\eta and yy are the same length, so what has changed is the time in which that length is covered. The paper draws the conclusion under the display: the law of the parallelogram of velocities holds, on our theory, only to a first approximation.Every stepξ=β(x−vt),τ=β(t−vV2x),η=y\xi = \beta(x - vt), \quad \tau = \beta\left( t - \frac{v}{V^2} x \right), \quad \eta = yThe transformation found in § 3, which is all this section uses.β(x−vt)=wξβ(t−vV2x)\beta(x - v t) = w_\xi \beta \left( t - \frac{v}{V^2} x \right)The first equation of motion with both sides transformed. The factor β\beta stands on both sides and divides out.x(1+vwξV2)=t(wξ+v)x \left( 1 + \frac{v w_\xi}{V^2} \right) = t (w_\xi + v)The terms in xx collected on the left and those in tt on the right.x=wξ+v1+vwξV2tx = \frac{w_\xi + v}{1 + \frac{v w_\xi}{V^2}} tDivided by the bracket: the first printed equation.t−vV2x=t1−v2V21+vwξV2t - \frac{v}{V^2} x = t \frac{1 - \frac{v^2}{V^2}}{1 + \frac{v w_\xi}{V^2}}The value of xx just found, put back into the moving system's time. The terms in vwξv w_\xi cancel in the numerator.y=wηβt1−v2V21+vwξV2=1−(vV)21+vwξV2wηty = w_\eta \beta t \frac{1 - \frac{v^2}{V^2}}{1 + \frac{v w_\xi}{V^2}} = \frac{\sqrt{1 - \left( \frac{v}{V} \right)^2}}{1 + \frac{v w_\xi}{V^2}} w_\eta tThe second equation of motion with that time in it. Since β\beta is one over the square root, β\beta times 1−(v/V)21 - (v/V)^2 is the square root itself. The second printed equation.z=0z = 0The third coordinate is unchanged by the transformation, so the motion stays in the plane it started in.Also

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