Annus MirabilisEinstein in 1905
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special relativity

The equation explained

W=∫εXdx=∫0vβ3vdv=μV2{11−(vV)2−1}.W = \int \varepsilon X d x = \int_0^v \beta^3 v d v = \mu V^2 \left\{ \frac{1}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} - 1 \right\} .
In wordsThe energy of motion of the electron: the work its charge and the electric force do along the path, which comes to the mass of the electron times the square of the velocity of light, times the factor at the velocity reached less one.OverviewThe kinetic energy of the electron: the energy the electrostatic field gives up while the electron is driven from rest, which is μV2\mu V^2 times the factor less one.Full explanationAn electron starts from rest at the origin and is driven along the XX axis by an electrostatic force. The energy withdrawn from the field is ∫εX dx\int \varepsilon X \, d x, and because the electron is to be slowly accelerated and so to give off no energy as radiation, the paper sets that equal to the energy of motion WW. The first row of (A) holds throughout the motion, so εX\varepsilon X can be replaced by μβ3\mu \beta^3 times the rate at which the velocity changes, and the integral over the path becomes an integral over the velocity from 0 to the value reached. Carrying it out gives μV2\mu V^2 times the bracket. The energy grows without limit as vv approaches VV: the paper adds that WW becomes infinitely great for v=Vv = V, and that velocities greater than that of light have no possibility of existence, as in its earlier results. The middle member is printed without the mass factor μ\mu (err-typo-p920-1), while the two outer members carry it.Every stepεX=μβ3d2xdt2=μβ3dvdt\varepsilon X = \mu \beta^3 \frac{d^2 x}{d t^2} = \mu \beta^3 \frac{d v}{d t}The first row of (A), multiplied by μβ3\mu \beta^3. The paper says this equation holds throughout the whole motion considered, and the electron's velocity along the XX axis is the vv the factor is built from.∫εX dx=μ∫β3dvdt dx=μ∫0vβ3v dv\int \varepsilon X \, d x = \mu \int \beta^3 \frac{d v}{d t} \, d x = \mu \int_0^v \beta^3 v \, d vOn the path dx=v dtd x = v \, d t, so the time drops out and dvdt dx\frac{d v}{d t} \, d x becomes v dvv \, d v. The limits are the velocity at the start, which is 0, and the velocity reached. The plate prints this middle member without the μ\mu.ddv(V21−(vV)2)=v(1−(vV)2)3/2=β3v\frac{d}{d v} \left( \frac{V^2}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} \right) = \frac{v}{\left( 1 - \left( \frac{v}{V} \right)^2 \right)^{3/2}} = \beta^3 vThe integral needs no substitution: differentiating V2βV^2 \beta with respect to vv gives back β3v\beta^3 v, so V2βV^2 \beta is what the integral accumulates.μ∫0vβ3v dv=μV2{11−(vV)2−1}\mu \int_0^v \beta^3 v \, d v = \mu V^2 \left\{ \frac{1}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} - 1 \right\}The accumulated value at the velocity reached, less its value at v=0v = 0, where the factor is 1. This is the printed right-hand member.μV2{11−(vV)2−1}≈12μv2\mu V^2 \left\{ \frac{1}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} - 1 \right\} \approx \frac{1}{2} \mu v^2A check at low speed: for small vV\frac{v}{V} the square root is about 1−12(vV)21 - \frac{1}{2} \left( \frac{v}{V} \right)^2 and its reciprocal about 1+12(vV)21 + \frac{1}{2} \left( \frac{v}{V} \right)^2, so the brace is about 12(vV)2\frac{1}{2} \left( \frac{v}{V} \right)^2 and the whole is the energy of motion of ordinary mechanics.Historian's marginThe middle member prints ∫0vβ3v dv\int_0^v \beta^3 v \, d v with no μ\mu in front (err-typo-p920-1), so as printed it does not equal the members on either side of it; the omission is confined to that member and the stated result is the one the arithmetic gives. Two things are worth noticing about what is printed. The condition that the electron give off no energy as radiation is an assumption of this display, not a result: it is what lets the energy taken from the field be set equal to the energy of motion. And the expression is a difference. μV2\mu V^2 appears here only as a coefficient, and the paper assigns no energy to the electron at rest; that question is the subject of the paper Einstein sent to the Annalen in September 1905.Also

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