Annus MirabilisEinstein in 1905
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special relativity

The equation explained

12(1V−v+1V+v)∂τ∂t=∂τ∂x′+1V−v∂τ∂t,\frac{1}{2}\left( \frac{1}{V-v} + \frac{1}{V+v} \right) \frac{\partial \tau}{\partial t} = \frac{\partial \tau}{\partial x'} + \frac{1}{V-v} \frac{\partial \tau}{\partial t} ,
In wordsHalf the sum of one over the velocity of light less the speed of the moving system and one over their sum, times how fast the moving system's time changes with the stationary time, equals how fast it changes with the distance along the motion plus the first of those two delays times its rate of change with the stationary time.OverviewThe same condition for a point very close to the origin. Once the distance is small, only the rate at which each time changes matters, so the condition becomes an equation between rates of change.Full explanationThe paper says: choose x′x' infinitely small. The two sides of the previous display then differ only by their first-order terms in x′x', and those must agree. On the left the argument of τ\tau is moved in time alone, by the whole out-and-back delay; on the right it is moved in place by x′x' and in time by the outward delay. Dividing by x′x' leaves the rates. The step assumes τ\tau has first derivatives to work with, which the linearity assumed at the head of the section supplies. Every term here is still written with the stationary system's delays, so the unknown is the single function τ\tau and nothing about kk's own measurements has been used.Every stepτ(0,0,0,t+x′V−v+x′V+v)=τ(0,0,0,t)+(x′V−v+x′V+v)∂τ∂t\tau\left( 0,0,0, t + \frac{x'}{V-v} + \frac{x'}{V+v} \right) = \tau(0,0,0,t) + \left( \frac{x'}{V-v} + \frac{x'}{V+v} \right) \frac{\partial \tau}{\partial t}The return reading, to first order in the small delay: the value at tt plus the delay times the rate of change with tt.τ(x′,0,0,t+x′V−v)=τ(0,0,0,t)+x′∂τ∂x′+x′V−v∂τ∂t\tau\left( x',0,0, t + \frac{x'}{V-v} \right) = \tau(0,0,0,t) + x' \frac{\partial \tau}{\partial x'} + \frac{x'}{V-v} \frac{\partial \tau}{\partial t}The reflection reading, to first order in both the small distance and the small delay.τ(0,0,0,t)+12(x′V−v+x′V+v)∂τ∂t=τ(0,0,0,t)+x′∂τ∂x′+x′V−v∂τ∂t\tau(0,0,0,t) + \frac{1}{2}\left( \frac{x'}{V-v} + \frac{x'}{V+v} \right) \frac{\partial \tau}{\partial t} = \tau(0,0,0,t) + x' \frac{\partial \tau}{\partial x'} + \frac{x'}{V-v} \frac{\partial \tau}{\partial t}Both readings go into 12(τ0+τ2)=τ1\frac{1}{2}(\tau_0 + \tau_2) = \tau_1. The departure reading appears on the left halved and once more from τ0\tau_0, which together make one of it.12(1V−v+1V+v)∂τ∂t=∂τ∂x′+1V−v∂τ∂t\frac{1}{2}\left( \frac{1}{V-v} + \frac{1}{V+v} \right) \frac{\partial \tau}{\partial t} = \frac{\partial \tau}{\partial x'} + \frac{1}{V-v} \frac{\partial \tau}{\partial t}Take τ(0,0,0,t)\tau(0,0,0,t) from both sides and divide by x′x', which is the line as printed. The paper compresses these three moves into the words choose x′x' infinitely small.Also

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