Annus MirabilisEinstein in 1905
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special relativity

The equation explained

Transversale Masse=μ1−(vV)2.\text{Transversale Masse} = \frac{\mu}{1 - \left( \frac{v}{V} \right)^2} .
In wordsThe number that goes with an acceleration across the motion: the mass of the electron divided by one less the square of its speed over the velocity of light.OverviewThe transverse mass: the mass of the slow electron divided by 1−(v/V)21 - (v/V)^2. It is what a push across the motion has to overcome, with the force read in the electron's own system.Full explanationThe row across the motion is μβ2d2ydt2=εY′\mu \beta^2 \frac{d^2 y}{d t^2} = \varepsilon Y', so the sentence gives μβ2\mu \beta^2 for the mass number, and β2\beta^2 written out is 1/(1−(v/V)2)1 / (1 - (v/V)^2). At v=0.6 Vv = 0.6 \, V it is 1.5625 μ1.5625 \, \mu, where the longitudinal number is 1.953125 μ1.953125 \, \mu: the same electron is harder to push along its motion than across it. This number carries the choice § 10 made. The force in it is εY′\varepsilon Y', read in the system the electron rests in, while the acceleration is read in the resting system. The same equations give μβ\mu \beta instead if the number divided by the acceleration is taken to be ε(Y−vVN)\varepsilon \left( Y - \frac{v}{V} N \right), which is what stands in (A) before the β\beta is taken out, and which later usage reads as the force on the electron in the resting system.Every stepμβ2d2ydt2=εY′\mu \beta^2 \frac{d^2 y}{d t^2} = \varepsilon Y'The row across the motion from the display above.Transversale Masse=μβ2=μ1−(vV)2\text{Transversale Masse} = \mu \beta^2 = \frac{\mu}{1 - \left( \frac{v}{V} \right)^2}The sentence applied, with β2=1/(1−(v/V)2)\beta^2 = 1 / \left( 1 - (v/V)^2 \right) from the definition on p. 900. This is the printed form.v=0.6 V⟹μβ2=1.5625 μ,μβ3=1.953125 μv = 0.6 \, V \quad \Longrightarrow \quad \mu \beta^2 = 1.5625 \, \mu , \quad \mu \beta^3 = 1.953125 \, \muThe two numbers at six tenths of the velocity of light, for the same electron: across the motion and along it.μβd2ydt2=ε(Y−vVN)\mu \beta \frac{d^2 y}{d t^2} = \varepsilon \left( Y - \frac{v}{V} N \right)The second row of (A) again, multiplied by μβ\mu \beta instead of μβ2\mu \beta^2. Divide the right-hand side, which is β\beta times smaller than εY′\varepsilon Y', by the same acceleration and the coefficient is μβ\mu \beta. Which number is called the transverse mass depends on that choice, not on the electron.Historian's marginEinstein's transverse mass divides the force measured in the electron's momentary rest system by the acceleration measured in the resting system; dividing the force measured in the resting system by the same acceleration gives μβ\mu \beta. The two numbers answer two questions, and the electron has one mass μ\mu, which is why the paper warns that comparing different theories of the motion of the electron needs care (p. 919). Kaufmann's deflection measurements of fast β-ray electrons, made from 1901 to 1906, seemed at first to favour Abraham's rival electron theory over the Lorentz–Einstein formulas, while Bucherer's of 1908 favoured the latter, and the question stayed open for some years.Also

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