Annus MirabilisEinstein in 1905
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special relativity

The equation explained

1V∂X∂τ=∂β(N−vVY)∂η−∂β(M+vVZ)∂ζ,1V∂β(Y−vVN)∂τ=∂L∂ζ−∂β(N−vVY)∂ξ,1V∂β(Z+vVM)∂τ=∂β(M+vVZ)∂ξ−∂L∂η,1V∂L∂τ=∂β(Y−vVN)∂ζ−∂β(Z+vVM)∂η,1V∂β(M+vVZ)∂τ=∂β(Z+vVM)∂ξ−∂X∂ζ,1V∂β(N−vVY)∂τ=∂X∂η−∂β(Y−vVN)∂ξ,\begin{aligned} \frac{1}{V} \frac{\partial X}{\partial \tau} &= \frac{\partial \beta \left( N - \frac{v}{V} Y \right)}{\partial \eta} - \frac{\partial \beta \left( M + \frac{v}{V} Z \right)}{\partial \zeta} , \\ \frac{1}{V} \frac{\partial \beta \left( Y - \frac{v}{V} N \right)}{\partial \tau} &= \frac{\partial L}{\partial \zeta} - \frac{\partial \beta \left( N - \frac{v}{V} Y \right)}{\partial \xi} , \\ \frac{1}{V} \frac{\partial \beta \left( Z + \frac{v}{V} M \right)}{\partial \tau} &= \frac{\partial \beta \left( M + \frac{v}{V} Z \right)}{\partial \xi} - \frac{\partial L}{\partial \eta} , \\ \frac{1}{V} \frac{\partial L}{\partial \tau} &= \frac{\partial \beta \left( Y - \frac{v}{V} N \right)}{\partial \zeta} - \frac{\partial \beta \left( Z + \frac{v}{V} M \right)}{\partial \eta} , \\ \frac{1}{V} \frac{\partial \beta \left( M + \frac{v}{V} Z \right)}{\partial \tau} &= \frac{\partial \beta \left( Z + \frac{v}{V} M \right)}{\partial \xi} - \frac{\partial X}{\partial \zeta} , \\ \frac{1}{V} \frac{\partial \beta \left( N - \frac{v}{V} Y \right)}{\partial \tau} &= \frac{\partial X}{\partial \eta} - \frac{\partial \beta \left( Y - \frac{v}{V} N \right)}{\partial \xi} , \end{aligned}
In wordsWritten in the moving system's time and coordinates, the same six equations keep their shape, with the electric force and the magnetic force of the stationary system mixed in pairs by the speed of the moving system over the velocity of light and multiplied by the factor.OverviewThe field equations of the first display, rewritten in the time τ\tau and coordinates ξ\xi, η\eta, ζ\zeta of the system kk that moves with velocity vv, by the transformation of § 3. The forces are still those measured in KK; only the coordinates they are written in have changed.Full explanationThe paper takes the transformation of § 3 and applies it to the equations of the first display, without showing the work. Every derivative with respect to tt and xx becomes a combination of derivatives with respect to τ\tau and ξ\xi, and when the terms are collected, the equations come back in the same shape as before. In the places where XX, YY, ZZ, LL, MM, NN stood, there now stand XX, β(Y−vVN)\beta \left( Y - \frac{v}{V} N \right), β(Z+vVM)\beta \left( Z + \frac{v}{V} M \right), LL, β(M+vVZ)\beta \left( M + \frac{v}{V} Z \right) and β(N−vVY)\beta \left( N - \frac{v}{V} Y \right). The components along the motion are left alone; each component across the motion picks up a share of the other force's perpendicular component. That regularity is what the next steps of § 6 use. One detail the paper passes over: the rows for XX and LL also need the fact that in empty space neither force has a divergence, a Maxwell equation not printed beside the six.Every stepτ=β(t−vV2x),ξ=β(x−vt),η=y,ζ=z\tau = \beta \left( t - \frac{v}{V^2} x \right), \quad \xi = \beta (x - v t), \quad \eta = y, \quad \zeta = zThe transformation of § 3, with the factor there found to be 1. It gives the time and coordinates that the moving system kk assigns to an event at xx, yy, zz and tt in KK.∂∂t=β(∂∂τ−v∂∂ξ),∂∂x=β(∂∂ξ−vV2∂∂τ)\frac{\partial}{\partial t} = \beta \left( \frac{\partial}{\partial \tau} - v \frac{\partial}{\partial \xi} \right), \quad \frac{\partial}{\partial x} = \beta \left( \frac{\partial}{\partial \xi} - \frac{v}{V^2} \frac{\partial}{\partial \tau} \right)The chain rule. A force that depends on tt and xx through τ\tau and ξ\xi changes with tt by β\beta times its change with τ\tau, plus −βv-\beta v times its change with ξ\xi, because τ\tau grows at the rate β\beta and ξ\xi at the rate −βv-\beta v as tt advances. Along xx, τ\tau changes at the rate −βv/V2-\beta v / V^2 and ξ\xi at the rate β\beta. Along yy and zz nothing is new: ∂∂y=∂∂η\frac{\partial}{\partial y} = \frac{\partial}{\partial \eta} and ∂∂z=∂∂ζ\frac{\partial}{\partial z} = \frac{\partial}{\partial \zeta}.βV(∂Y∂τ−v∂Y∂ξ)=∂L∂ζ−β(∂N∂ξ−vV2∂N∂τ)\frac{\beta}{V} \left( \frac{\partial Y}{\partial \tau} - v \frac{\partial Y}{\partial \xi} \right) = \frac{\partial L}{\partial \zeta} - \beta \left( \frac{\partial N}{\partial \xi} - \frac{v}{V^2} \frac{\partial N}{\partial \tau} \right)Put these into the second equation of the first display, 1V∂Y∂t=∂L∂z−∂N∂x\frac{1}{V} \frac{\partial Y}{\partial t} = \frac{\partial L}{\partial z} - \frac{\partial N}{\partial x}.1V∂β(Y−vVN)∂τ=∂L∂ζ−∂β(N−vVY)∂ξ\frac{1}{V} \frac{\partial \beta \left( Y - \frac{v}{V} N \right)}{\partial \tau} = \frac{\partial L}{\partial \zeta} - \frac{\partial \beta \left( N - \frac{v}{V} Y \right)}{\partial \xi}Gather the two τ\tau-slopes on the left and the two ξ\xi-slopes on the right. On the left, βV∂Y∂τ−βvV2∂N∂τ\frac{\beta}{V} \frac{\partial Y}{\partial \tau} - \frac{\beta v}{V^2} \frac{\partial N}{\partial \tau} is 1V\frac{1}{V} times the τ\tau-slope of β(Y−vVN)\beta \left( Y - \frac{v}{V} N \right); on the right, β∂N∂ξ−βvV∂Y∂ξ\beta \frac{\partial N}{\partial \xi} - \frac{\beta v}{V} \frac{\partial Y}{\partial \xi} is the ξ\xi-slope of β(N−vVY)\beta \left( N - \frac{v}{V} Y \right). This is the second printed row. The third, fifth and sixth rows come out the same way.∂X∂x+∂Y∂y+∂Z∂z=0\frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} + \frac{\partial Z}{\partial z} = 0The first row needs one more fact, which the paper does not state: in empty space the electric force has no divergence. Written in the new coordinates it says ∂X∂ξ=vV2∂X∂τ−1β(∂Y∂η+∂Z∂ζ)\frac{\partial X}{\partial \xi} = \frac{v}{V^2} \frac{\partial X}{\partial \tau} - \frac{1}{\beta} \left( \frac{\partial Y}{\partial \eta} + \frac{\partial Z}{\partial \zeta} \right).1V∂X∂τ=∂β(N−vVY)∂η−∂β(M+vVZ)∂ζ\frac{1}{V} \frac{\partial X}{\partial \tau} = \frac{\partial \beta \left( N - \frac{v}{V} Y \right)}{\partial \eta} - \frac{\partial \beta \left( M + \frac{v}{V} Z \right)}{\partial \zeta}Substituting as before turns the first equation into βV(∂X∂τ−v∂X∂ξ)=∂N∂η−∂M∂ζ\frac{\beta}{V} \left( \frac{\partial X}{\partial \tau} - v \frac{\partial X}{\partial \xi} \right) = \frac{\partial N}{\partial \eta} - \frac{\partial M}{\partial \zeta}. Put in the divergence condition and use β(1−v2V2)=1β\beta \left( 1 - \frac{v^2}{V^2} \right) = \frac{1}{\beta}: the left side becomes 1βV∂X∂τ\frac{1}{\beta V} \frac{\partial X}{\partial \tau} and the two new terms join the right side. Multiplied by β\beta, this is the first printed row. The fourth row, for LL, comes the same way from the magnetic force's divergence.Also

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