Annus MirabilisEinstein in 1905
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special relativity

The equation explained

cos⁡φ′′′=cos⁡φ′′+vV1+vVcos⁡φ′′=−(1+(vV)2)cos⁡φ−2vV1−2vVcos⁡φ+(vV)2,\cos \varphi''' = \frac{\cos \varphi'' + \frac{v}{V}}{1 + \frac{v}{V} \cos \varphi''} = - \frac{\left( 1 + \left( \frac{v}{V} \right)^2 \right) \cos \varphi - 2 \frac{v}{V}}{1 - 2 \frac{v}{V} \cos \varphi + \left( \frac{v}{V} \right)^2} ,
In wordsThe cosine of the reflected light's angle in the stationary system is the cosine of its angle in the mirror's system plus the speed over the velocity of light, divided by one plus that ratio times the same cosine; in terms of the angle of incidence, it is minus one plus the square of the ratio times the cosine, less twice the ratio, over one minus twice the ratio times the cosine plus its square.OverviewThe direction of the reflected light back in the stationary system. For a moving mirror the angle of reflection is not the angle of incidence.Full explanationThe law of aberration with −v-v carries the reflected direction from kk back to KK; then the first two moves express it through the angle of incidence. For v=0v = 0 the long form is −cos⁡φ-\cos \varphi, ordinary reflection. For a moving mirror the reflected light leaves at a different angle. The light must overtake the mirror, so cos⁡φ\cos \varphi must exceed vV\frac{v}{V}; for a mirror receding at a tenth of the velocity of light from light falling at φ=π3\varphi = \frac{\pi}{3}, the numerator is 1.01⋅0.5−0.2=0.3051.01 \cdot 0.5 - 0.2 = 0.305 and the denominator 1−0.1+0.01=0.911 - 0.1 + 0.01 = 0.91, so cos⁡φ′′′≈−0.34\cos \varphi''' \approx -0.34 instead of the −0.5-0.5 of a mirror at rest.Every stepcos⁡φ′′′=cos⁡φ′′+vV1+vVcos⁡φ′′\cos \varphi''' = \frac{\cos \varphi'' + \frac{v}{V}}{1 + \frac{v}{V} \cos \varphi''}Back to KK: the law of aberration with −v-v for vv.cos⁡φ′′+vV=−(1+(vV)2)cos⁡φ+2vV1−vVcos⁡φ\cos \varphi'' + \frac{v}{V} = \frac{-\left( 1 + \left( \frac{v}{V} \right)^2 \right) \cos \varphi + 2 \frac{v}{V}}{1 - \frac{v}{V} \cos \varphi}Put in cos⁡φ′′=−cos⁡φ−vV1−vVcos⁡φ\cos \varphi'' = -\frac{\cos \varphi - \frac{v}{V}}{1 - \frac{v}{V} \cos \varphi} and bring vV\frac{v}{V} over the same denominator: −cos⁡φ+vV+vV−(vV)2cos⁡φ-\cos \varphi + \frac{v}{V} + \frac{v}{V} - \left( \frac{v}{V} \right)^2 \cos \varphi.cos⁡φ′′′=−(1+(vV)2)cos⁡φ−2vV1−2vVcos⁡φ+(vV)2\cos \varphi''' = -\frac{\left( 1 + \left( \frac{v}{V} \right)^2 \right) \cos \varphi - 2 \frac{v}{V}}{1 - 2 \frac{v}{V} \cos \varphi + \left( \frac{v}{V} \right)^2}Divide by 1+vVcos⁡φ′′1 + \frac{v}{V} \cos \varphi'', which the previous display found to be 1−2vVcos⁡φ+(vV)21−vVcos⁡φ\frac{1 - 2 \frac{v}{V} \cos \varphi + \left( \frac{v}{V} \right)^2}{1 - \frac{v}{V} \cos \varphi}; the denominators 1−vVcos⁡φ1 - \frac{v}{V} \cos \varphi cancel.Also

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