Annus MirabilisEinstein in 1905
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special relativity

The equation explained

∂τ∂y=0∂τ∂z=0.\begin{aligned} \frac{\partial \tau}{\partial y} &= 0 \\ \frac{\partial \tau}{\partial z} &= 0 . \end{aligned}
In wordsHow fast the moving system's time changes with either coordinate across the motion is zero, for both of them.OverviewAcross the motion nothing slides: clocks set in the moving system agree with the stationary system's about what is simultaneous, so long as one moves only sideways. The sliding of the previous display happens along the direction of the motion alone.Full explanationThe paper reaches this by running the argument of the first displays along the other two axes. A ray sent along the moving system's second axis has to keep up with that axis as it is carried along at vv, so part of its velocity goes into the motion and what is left for crossing is V2−v2\sqrt{V^2 - v^2}, the figure the paper states here. The outward and the return leg then take the same stationary time, which is what makes the first-order terms cancel and leaves the rate of change across the motion at zero. The result is what allows τ\tau to be written with tt and one distance only, which the next display does.Every stepout=back=yV2−v2\text{out} = \text{back} = \frac{y}{\sqrt{V^2 - v^2}}The two legs across the motion take the same stationary time, since the ray crosses at V2−v2\sqrt{V^2 - v^2} each way and the axis is carried along with the system.12[τ(0,0,0,t)+τ(0,0,0,t+2yV2−v2)]=τ(0,y,0,t+yV2−v2)\frac{1}{2}\left[ \tau(0,0,0,t) + \tau\left( 0,0,0, t + \frac{2 y}{\sqrt{V^2 - v^2}} \right) \right] = \tau\left( 0, y, 0, t + \frac{y}{\sqrt{V^2 - v^2}} \right)The rule of § 1 again, for a ray that goes out along the second axis and comes back.yV2−v2∂τ∂t=y∂τ∂y+yV2−v2∂τ∂t\frac{y}{\sqrt{V^2 - v^2}} \frac{\partial \tau}{\partial t} = y \frac{\partial \tau}{\partial y} + \frac{y}{\sqrt{V^2 - v^2}} \frac{\partial \tau}{\partial t}Both sides to first order in a small yy, with τ(0,0,0,t)\tau(0,0,0,t) taken from each.∂τ∂y=0\frac{\partial \tau}{\partial y} = 0The terms in ∂τ∂t\frac{\partial \tau}{\partial t} are the same on both sides, so what is left is zero; divide by yy.∂τ∂z=0\frac{\partial \tau}{\partial z} = 0The third axis is the second one over again, with nothing in the argument to tell them apart.Historian's marginThe speed V2−v2\sqrt{V^2 - v^2} is not a velocity of light smaller than VV: it is how fast the ray crosses the stationary system's second axis while also keeping pace with the moving system along the first, and the paper says as much in the same sentence (p. 899). A modern account gets the same figure by adding the velocities as components of one vector of length VV.Also

Read it in the paper