Annus MirabilisEinstein in 1905
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special relativity

The equation explained

{d2xdt2=εμ1β3X,d2ydt2=εμ1β(Y−vVN),d2zdt2=εμ1β(Z+vVM).\left\{ \begin{aligned} \frac{d^2 x}{d t^2} &= \frac{\varepsilon}{\mu} \frac{1}{\beta^3} X , \\ \frac{d^2 y}{d t^2} &= \frac{\varepsilon}{\mu} \frac{1}{\beta} \left( Y - \frac{v}{V} N \right) , \\ \frac{d^2 z}{d t^2} &= \frac{\varepsilon}{\mu} \frac{1}{\beta} \left( Z + \frac{v}{V} M \right) . \end{aligned} \right.
In wordsHow fast the speed along each coordinate changes with the time: the charge divided by the mass, divided again by the cube of the factor along the motion and by the factor itself across it, times the electric force along that coordinate, with the transverse rows carrying also the speed of the electron over the velocity of light times the magnetic force across it.OverviewThe law of motion of an electron that is already moving, written in the time and coordinates of the system it moves in. Along the motion the electric force is divided by the cube of the factor; across the motion, by the factor itself.Full explanationThese are the equations the rest of § 10 draws on, and the paper labels them (A). They come from the three equations in kk by substituting the transformations of the display above, and the paper does not show that working. Two things make it come out: at the moment considered the electron moves along the XX axis with the velocity vv, so dxdt=v\frac{d x}{d t} = v and the transverse velocities vanish; and τ\tau depends on xx as well as tt, so a second derivative with respect to τ\tau is not simply a second derivative with respect to tt. Working it through gives d2ξdτ2=β3d2xdt2\frac{d^2 \xi}{d \tau^2} = \beta^3 \frac{d^2 x}{d t^2} along the motion and d2ηdτ2=β2d2ydt2\frac{d^2 \eta}{d \tau^2} = \beta^2 \frac{d^2 y}{d t^2} across it, and the transverse electric force of kk brings a further β\beta with it, which is why the transverse rows carry 1β\frac{1}{\beta} and not 1β2\frac{1}{\beta^2}.Every stepdξdτ=dξdtdτdt=β(dxdt−v)β(1−vV2dxdt)=dxdt−v1−vV2dxdt\frac{d \xi}{d \tau} = \frac{\frac{d \xi}{d t}}{\frac{d \tau}{d t}} = \frac{\beta \left( \frac{d x}{d t} - v \right)}{\beta \left( 1 - \frac{v}{V^2} \frac{d x}{d t} \right)} = \frac{\frac{d x}{d t} - v}{1 - \frac{v}{V^2} \frac{d x}{d t}}Differentiate ξ=β(x−vt)\xi = \beta (x - v t) and τ=β(t−vV2x)\tau = \beta \left( t - \frac{v}{V^2} x \right) with respect to tt and divide one by the other. The factor cancels, and what is left is the addition theorem of velocities of § 5.ddt(dξdτ)=d2xdt2⋅1−v2V2(1−vV2dxdt)2\frac{d}{d t} \left( \frac{d \xi}{d \tau} \right) = \frac{d^2 x}{d t^2} \cdot \frac{1 - \frac{v^2}{V^2}}{\left( 1 - \frac{v}{V^2} \frac{d x}{d t} \right)^2}Differentiate that quotient once more with respect to tt. The numerator of the quotient rule is d2xdt2(1−vV2dxdt)+(dxdt−v)vV2d2xdt2\frac{d^2 x}{d t^2} \left( 1 - \frac{v}{V^2} \frac{d x}{d t} \right) + \left( \frac{d x}{d t} - v \right) \frac{v}{V^2} \frac{d^2 x}{d t^2}, and the two terms in dxdt\frac{d x}{d t} cancel, leaving 1−v2V21 - \frac{v^2}{V^2}.d2ξdτ2=d2xdt2β21β=β3d2xdt2\frac{d^2 \xi}{d \tau^2} = \frac{\frac{d^2 x}{d t^2} \beta^2}{\frac{1}{\beta}} = \beta^3 \frac{d^2 x}{d t^2}At the moment considered dxdt=v\frac{d x}{d t} = v. Then 1−vV2dxdt=1−v2V2=1β21 - \frac{v}{V^2} \frac{d x}{d t} = 1 - \frac{v^2}{V^2} = \frac{1}{\beta^2}, so the fraction in the previous step is 1/β21/β4=β2\frac{1 / \beta^2}{1 / \beta^4} = \beta^2; and dividing once more by dτdt=β(1−v2V2)=1β\frac{d \tau}{d t} = \beta \left( 1 - \frac{v^2}{V^2} \right) = \frac{1}{\beta} turns it into β3\beta^3.μβ3d2xdt2=εX′=εX⟹d2xdt2=εμ1β3X\mu \beta^3 \frac{d^2 x}{d t^2} = \varepsilon X' = \varepsilon X \quad \Longrightarrow \quad \frac{d^2 x}{d t^2} = \frac{\varepsilon}{\mu} \frac{1}{\beta^3} XPut that into the first equation in kk, and use X′=XX' = X from the display above. Dividing by μβ3\mu \beta^3 gives the first row of (A).dηdτ=dydtβ(1−vV2dxdt),d2ηdτ2=β2d2ydt2\frac{d \eta}{d \tau} = \frac{\frac{d y}{d t}}{\beta \left( 1 - \frac{v}{V^2} \frac{d x}{d t} \right)} , \qquad \frac{d^2 \eta}{d \tau^2} = \beta^2 \frac{d^2 y}{d t^2}Across the motion η=y\eta = y. When this is differentiated again, the term carrying the derivative of the denominator is multiplied by dydt\frac{d y}{d t}, which is zero at the moment considered, so only d2ydt2\frac{d^2 y}{d t^2} survives; dividing by dτdt=1β\frac{d \tau}{d t} = \frac{1}{\beta} twice over the two steps leaves β2\beta^2.μβ2d2ydt2=εY′=εβ(Y−vVN)⟹d2ydt2=εμ1β(Y−vVN)\mu \beta^2 \frac{d^2 y}{d t^2} = \varepsilon Y' = \varepsilon \beta \left( Y - \frac{v}{V} N \right) \quad \Longrightarrow \quad \frac{d^2 y}{d t^2} = \frac{\varepsilon}{\mu} \frac{1}{\beta} \left( Y - \frac{v}{V} N \right)The second equation in kk, with Y′Y' from the display above. One β\beta of the β2\beta^2 is used up against the β\beta in Y′Y', which is why the transverse rows of (A) carry 1β\frac{1}{\beta}. The third row follows in the same way with ZZ and MM.Historian's margin(A) is the only equation the paper numbers, and every later step of § 10 names it: the rewriting for the two masses (p. 919), the kinetic energy (p. 920), the announcement of the properties open to experiment (p. 920), the deflectability relation (p. 920) and the radius of curvature (p. 921). The step it passes over in one sentence is the longest calculation in the section, and its shape is worth keeping: the difference between the cube and the first power of the factor is not a difference between the electric forces along and across the motion, but between the way τ\tau enters a longitudinal and a transverse second derivative.Also

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