Annus MirabilisEinstein in 1905
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special relativity

The equation explained

∂τ∂x′+vV2−v2∂τ∂t=0.\frac{\partial \tau}{\partial x'} + \frac{v}{V^2 - v^2} \frac{\partial \tau}{\partial t} = 0 .
In wordsHow fast the moving system's time changes with the distance along the motion, plus the speed of the moving system over the square of the velocity of light less the square of that speed, times how fast it changes with the stationary time, is zero.OverviewThe same equation, tidied. It says the moving system's time cannot depend on the stationary time alone: move along the direction of the motion and the setting of the clocks slides, by an amount this equation fixes.Full explanationThe arithmetic is only the combining of the two fractions, but the result is the whole of the relativity of simultaneity in one line. If ∂τ∂x′\frac{\partial \tau}{\partial x'} were zero, clocks that the moving system calls synchronous would be synchronous in the stationary system too, and the two systems would share a single time. The equation says instead that the two rates of change stand in a fixed ratio, vV2−v2\frac{v}{V^2 - v^2}, which vanishes only when v=0v = 0. The paragraph that follows notes that the origin was not special: any other point could have been the ray's starting point, so the equation holds for all values of x′x', yy and zz.Every step12(1V−v+1V+v)=12⋅(V+v)+(V−v)(V−v)(V+v)=VV2−v2\frac{1}{2}\left( \frac{1}{V-v} + \frac{1}{V+v} \right) = \frac{1}{2} \cdot \frac{(V+v) + (V-v)}{(V-v)(V+v)} = \frac{V}{V^2 - v^2}Add the two fractions over the common denominator V2−v2V^2 - v^2; the two vv terms cancel and 2V2V is halved.VV2−v2∂τ∂t−1V−v∂τ∂t=∂τ∂x′\frac{V}{V^2 - v^2} \frac{\partial \tau}{\partial t} - \frac{1}{V-v} \frac{\partial \tau}{\partial t} = \frac{\partial \tau}{\partial x'}Take the term in ∂τ∂t\frac{\partial \tau}{\partial t} on the right over to the left, and the term in ∂τ∂x′\frac{\partial \tau}{\partial x'} over to the right.1V−v=V+vV2−v2\frac{1}{V-v} = \frac{V+v}{V^2 - v^2}Put the second fraction over the same denominator, so the two can be subtracted.V−(V+v)V2−v2∂τ∂t=−vV2−v2∂τ∂t=∂τ∂x′\frac{V - (V+v)}{V^2 - v^2} \frac{\partial \tau}{\partial t} = - \frac{v}{V^2 - v^2} \frac{\partial \tau}{\partial t} = \frac{\partial \tau}{\partial x'}Subtract: V−V−v=−vV - V - v = -v.∂τ∂x′+vV2−v2∂τ∂t=0\frac{\partial \tau}{\partial x'} + \frac{v}{V^2 - v^2} \frac{\partial \tau}{\partial t} = 0Bring both terms to one side, which is the line as printed.Also

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