Annus MirabilisEinstein in 1905
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mass energy

The equation explained

Δm=−Lc2\Delta m = -\frac{L}{c^2}
In wordsThe change in the mass is minus the energy given off divided by the square of the velocity of light.OverviewWhat the paper concludes: a body that gives off energy loses mass, and the amount is the energy divided by the square of the velocity of light.Full explanationThe paper reaches this by comparing two accounting sheets for one emission, and its own way of putting the result is that the mass of a body is a measure of its energy content; if the energy changes by LL, the mass changes in the same sense by LL divided by 9⋅10209 \cdot 10^{20}, the energy being measured in ergs and the mass in grammes. The sign is worth keeping: giving energy off is a loss of mass, which is why the change is written with a minus here. The division by a very large number is why nothing about the effect is ordinary: an erg of radiation costs about 10−2110^{-21} grammes, and the paper says as much, that bodies whose energy content is variable to a high degree, radium salts for instance, are where a test might be possible.Every stepK0−K1=L(11−v2c2−1)K_0 - K_1 = L \left( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1 \right)The paper's result for the kinetic energies before and after the emission, which the display at the end of its argument prints.K0−K1≈12Lc2v2K_0 - K_1 \approx \frac{1}{2} \frac{L}{c^2} v^2At low speed the bracket is about 12v2c2\frac{1}{2} \frac{v^2}{c^2}, so the difference of the two kinetic energies is the ordinary 12mv2\frac{1}{2} m v^2 with Lc2\frac{L}{c^2} standing where the mass stands.Δm=−Lc2\Delta m = -\frac{L}{c^2}Reading that coefficient as a mass is the whole of the step: the body's kinetic energy at a given speed has fallen by exactly what a body of mass Lc2\frac{L}{c^2} would have had, so that much mass went with the light.Historian's marginThe equation is this edition's, not Einstein's. The 1905 paper prints no formula for the mass change: it states the conclusion in words and gives the divisor as the number 9⋅10209 \cdot 10^{20} in the units of the day. Writing it as Δm=−L/c2\Delta m = -L/c^2 is a rearrangement of what the paper says, and the form that became famous, E=mc2E = mc^2, appears nowhere in it. The paper's own last sentence is a proposal rather than a claim of proof: that a test may succeed with bodies whose energy content is variable to a high degree.Also

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