Annus MirabilisEinstein in 1905
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special relativity

The equation explained

Longitudinale Masse=μ(1−(vV)2)3,\text{Longitudinale Masse} = \frac{\mu}{\left( \sqrt{1 - \left( \frac{v}{V} \right)^2} \right)^3} ,
In wordsThe number that goes with an acceleration along the motion: the mass of the electron divided by the cube of the square root of one less the square of its speed over the velocity of light.OverviewThe longitudinal mass: the mass of the slow electron divided by the cube of 1−(v/V)2\sqrt{1 - (v/V)^2}. It is what a push along the motion has to overcome.Full explanationTake the row of the display above that belongs to the motion, μβ3d2xdt2=εX′\mu \beta^3 \frac{d^2 x}{d t^2} = \varepsilon X', and apply the sentence: the number in front of the acceleration is the mass number. So the longitudinal mass is μβ3\mu \beta^3, and writing β\beta out gives the printed form. It is μ\mu when the electron is at rest and grows without limit as vv approaches VV: at v=0.6 Vv = 0.6 \, V, where β=1.25\beta = 1.25, it is 1.953125 μ1.953125 \, \mu. Along the motion the force is the same number in both systems, since X′=XX' = X, so this coefficient is the ratio of force to acceleration whichever of the two systems the force is read in. The transverse coefficient of the next display is not like that.Every stepμβ3d2xdt2=εX′\mu \beta^3 \frac{d^2 x}{d t^2} = \varepsilon X'The row along the motion from the display above.Longitudinale Masse=μβ3\text{Longitudinale Masse} = \mu \beta^3The sentence applied: the number in front of the acceleration is the mass number.β=11−(vV)2⟹μβ3=μ(1−(vV)2)3\beta = \frac{1}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} \quad \Longrightarrow \quad \mu \beta^3 = \frac{\mu}{\left( \sqrt{1 - \left( \frac{v}{V} \right)^2} \right)^3}The factor as the paper defines it on p. 900. Cubing the reciprocal is dividing by the cube, which is the printed form.v=0.6 V⟹β=1.25,μβ3=1.953125 μv = 0.6 \, V \quad \Longrightarrow \quad \beta = 1.25 , \quad \mu \beta^3 = 1.953125 \, \muA worked value: at six tenths of the velocity of light the square root is 1−0.36=0.8\sqrt{1 - 0.36} = 0.8, so β=1.25\beta = 1.25 and its cube is 1.9531251.953125.Historian's marginThe longitudinal number does not depend on where the force is read. A force along the motion has the same value in the system the electron rests in and in the resting system, which the display above shows as εX=εX′\varepsilon X = \varepsilon X', so μβ3\mu \beta^3 is the ratio of force to acceleration under either reading. That is why the two mass formulas of this page are not alike in status: the next one carries the choice of § 10's sentence, and this one does not.Also

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