Annus MirabilisEinstein in 1905
Reading preferences

Reading preferences

Keep every explanation and worked example on the page. Nothing moves or plays by itself, and the larger instruments load only when you ask.

Line length
Type size
Contrast
Paragraph spacing

mass energy

The equation explained

H0=H1+[L21−vVcos⁡φ1−(vV)2+L21+vVcos⁡φ1−(vV)2]=H1+L1−(vV)2.\begin{aligned} H_0 &= H_1 + \left[ \frac{L}{2} \frac{1 - \frac{v}{V} \cos \varphi}{\sqrt{1 - \left(\frac{v}{V}\right)^2}} + \frac{L}{2} \frac{1 + \frac{v}{V} \cos \varphi}{\sqrt{1 - \left(\frac{v}{V}\right)^2}} \right] \\ &= H_1 + \frac{L}{\sqrt{1 - \left(\frac{v}{V}\right)^2}}. \end{aligned}
In wordsThe body's energy before it emits, measured in the moving system, equals its energy after plus the two light complexes as that system measures them, which together come to the energy carried off by the light divided by the square root set by the speed of the moving system against the speed of light, with no trace left of the angle the light was sent at.OverviewThe same emission, entered in the traveller's books. Each half of the light transforms by the rule of the first display, one gaining what the other loses, so the direction of the emission drops out and the total is larger than in the body's own system.Full explanationBy the principle of relativity, the energy principle holds in the moving system (ξ,η,ζ)(\xi, \eta, \zeta) as well. Call the body's energies there H0H_0 before the emission and H1H_1 after. Each of the two light complexes must now be measured in that system, and the relation of page 639 does it. The first, sent at the angle φ\varphi, carries the factor 1−vVcos⁡φ1 - \frac{v}{V} \cos \varphi; the second, sent the opposite way, carries 1+vVcos⁡φ1 + \frac{v}{V} \cos \varphi, because reversing the ray direction reverses the sign of the cosine. Added, the two cosine terms cancel and what is left is L1−(vV)2\frac{L}{\sqrt{1 - \left(\frac{v}{V}\right)^2}}. The angle has gone, and the traveller records the light as carrying more energy than LL. This is the whole asymmetry the paper needs: the same emission costs the body LL in one system and more than LL in the other.Every stepL21−vVcos⁡φ1−(vV)2\frac{L}{2} \frac{1 - \frac{v}{V} \cos \varphi}{\sqrt{1 - \left(\frac{v}{V}\right)^2}}The first light complex, of energy L2\frac{L}{2} in the system where the body rests, measured in the moving system by the relation of page 639.L21+vVcos⁡φ1−(vV)2\frac{L}{2} \frac{1 + \frac{v}{V} \cos \varphi}{\sqrt{1 - \left(\frac{v}{V}\right)^2}}The second complex travels the opposite way, so its ray direction makes the supplementary angle with the xx-axis and its cosine changes sign. Nothing else in the factor changes.L2(1−vVcos⁡φ)+(1+vVcos⁡φ)1−(vV)2=L1−(vV)2\frac{L}{2} \frac{\left(1 - \frac{v}{V} \cos \varphi\right) + \left(1 + \frac{v}{V} \cos \varphi\right)}{\sqrt{1 - \left(\frac{v}{V}\right)^2}} = \frac{L}{\sqrt{1 - \left(\frac{v}{V}\right)^2}}Adding the two, the terms in cos⁡φ\cos \varphi cancel and the numerator is 22, which the L2\frac{L}{2} in front turns into LL. The result no longer depends on the direction of the emission.H0=H1+L1−(vV)2H_0 = H_1 + \frac{L}{\sqrt{1 - \left(\frac{v}{V}\right)^2}}The energy principle in the moving system, with the light's total energy there written out. Since the square root is smaller than one, the traveller's books show the body losing more than LL.Historian's marginThe two emissions are equal in the system where the body rests, and unequal in the moving one; only their sum is free of the angle. Einstein keeps the angle φ\varphi general and lets it cancel, rather than choosing a convenient direction, so the conclusion does not depend on how the body was oriented. From this paragraph on, φ\varphi is the angle the light is emitted at, where on page 639 it was the direction of a light complex already present.Also

Read it in the paper