Annus MirabilisEinstein in 1905
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special relativity

The equation explained

U=V2V−ϰ−λ2V−ϰ−λ+ϰλV<V.U = V \frac{2 V - \varkappa - \lambda}{2 V - \varkappa - \lambda + \frac{\varkappa \lambda}{V}} < V .
In wordsThe resulting speed is the speed of light times a fraction whose denominator exceeds its numerator by the product of the two amounts by which the compounded speeds fall short of light over that speed, so the result stays below it.OverviewWrite each of the two speeds as the speed of light less something positive. The composition then comes out as the speed of light times a fraction smaller than one, so two speeds below light always compound to a speed below light.Full explanationThe paper proves the claim by writing v=V−ϰv = V - \varkappa and w=V−λw = V - \lambda, with ϰ\varkappa and λ\lambda positive and smaller than VV, which any two speeds below the speed of light can be written as. The composition of the display before then takes this form, and the comparison is immediate: the denominator is the numerator plus ϰλV\frac{\varkappa \lambda}{V}, which is positive, so the fraction is less than one and UU is less than VV. The proof needs no smallness and no approximation: it holds for two speeds each as close to VV as one likes, and it is what secures the speed of light as a limit rather than merely a large number.Every stepv=V−ϰ,w=V−λv = V - \varkappa, \quad w = V - \lambdaAny two speeds below the speed of light, written by how far each falls short of it. Both shortfalls are positive and smaller than VV.v+w=2V−ϰ−λv + w = 2 V - \varkappa - \lambdaTheir sum, the numerator of the addition theorem.1+vwV2=V2+(V−ϰ)(V−λ)V2=2V2−V(ϰ+λ)+ϰλV21 + \frac{v w}{V^2} = \frac{V^2 + (V - \varkappa)(V - \lambda)}{V^2} = \frac{2 V^2 - V(\varkappa + \lambda) + \varkappa \lambda}{V^2}Its denominator, with the product multiplied out over a common denominator.U=(2V−ϰ−λ)V22V2−V(ϰ+λ)+ϰλU = \frac{(2V - \varkappa - \lambda) V^2}{2 V^2 - V(\varkappa + \lambda) + \varkappa \lambda}The quotient of the two.U=V2V−ϰ−λ2V−ϰ−λ+ϰλVU = V \frac{2 V - \varkappa - \lambda}{2 V - \varkappa - \lambda + \frac{\varkappa \lambda}{V}}Numerator and denominator divided by VV, which leaves one factor VV in front. The printed form.ϰλV>0\frac{\varkappa \lambda}{V} > 0Both shortfalls are positive, so the denominator exceeds the numerator by this amount and the fraction is smaller than one. Hence the printed conclusion that UU is less than VV.Also

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