Annus MirabilisEinstein in 1905
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special relativity

The equation explained

(βξ−aβvVξ)2+(η−bβvVξ)2+(ζ−cβvVξ)2=R2.\left( \beta \xi - a \beta \frac{v}{V} \xi \right)^2 + \left( \eta - b \beta \frac{v}{V} \xi \right)^2 + \left( \zeta - c \beta \frac{v}{V} \xi \right)^2 = R^2 .
In wordsSeen from the moving system at its time zero, the travelling sphere is the surface where the square of the factor times the first coordinate, less the first direction cosine times that factor times the speed over the velocity of light times the same coordinate, plus two like squares for the other coordinates, equals the square of the radius: an ellipsoid.OverviewThe travelling sphere as the moving system sees it at one instant of its own time, τ=0\tau = 0: not a sphere but an ellipsoid.Full explanationThe paper asks what surface the sphere of the previous display is when regarded in kk. It takes the instant τ=0\tau = 0 of kk's time and rewrites the sphere's equation in kk's coordinates with the transformation of § 3. Because kk's instant is not a single instant of KK, points at different ξ\xi are caught at different times tt, and the sphere comes out stretched and slanted: an ellipsoid. Its volume is compared with the sphere's in the next display.Every stepτ=β(t−vV2x)=0\tau = \beta \left( t - \frac{v}{V^2} x \right) = 0The instant in kk at which the paper looks. For each point it fixes the time in KK at which that point is caught.x=βξ,t=βvV2ξx = \beta \xi, \quad t = \beta \frac{v}{V^2} \xiAt τ=0\tau = 0, § 3's transformation gives x=β(ξ+vτ)=βξx = \beta (\xi + v \tau) = \beta \xi and t=β(τ+vV2ξ)=βvV2ξt = \beta \left( \tau + \frac{v}{V^2} \xi \right) = \beta \frac{v}{V^2} \xi; and y=ηy = \eta, z=ζz = \zeta.x−Vat=βξ−aβvVξ,y−Vbt=η−bβvVξx - V a t = \beta \xi - a \beta \frac{v}{V} \xi, \quad y - V b t = \eta - b \beta \frac{v}{V} \xiPut these into the brackets of the sphere's equation. The third bracket, with zz and the third cosine, becomes ζ\zeta less the third cosine times βvVξ\beta \frac{v}{V} \xi in the same way.(βξ−aβvVξ)2+(η−bβvVξ)2+(ζ−cβvVξ)2=R2\left( \beta \xi - a \beta \frac{v}{V} \xi \right)^2 + \left( \eta - b \beta \frac{v}{V} \xi \right)^2 + \left( \zeta - c \beta \frac{v}{V} \xi \right)^2 = R^2The sphere's equation written in kk's coordinates at τ=0\tau = 0: the display.Also

Read it in the paper