Annus MirabilisEinstein in 1905
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special relativity

The equation explained

ξ2+η2+ζ2=V2τ2.\xi^2 + \eta^2 + \zeta^2 = V^2 \tau^2 .
In wordsThe three coordinates of the moving system, each squared and added, are the square of the velocity of light times the square of that system's own time. The wave is a sphere there too, spreading with the same velocity.OverviewThe same spherical wave, written in the moving system: still a sphere, still spreading with the velocity of light. This is what makes the two principles of § 2 compatible.Full explanationThe paper transforms the previous display and says it obtains this after a simple calculation. The calculation is worth doing, because what comes out of it is the whole point of § 3. Substituting the transformation of p. 900 into ξ2+η2+ζ2−V2τ2\xi^2 + \eta^2 + \zeta^2 - V^2 \tau^2 leaves x2+y2+z2−V2t2x^2 + y^2 + z^2 - V^2 t^2, multiplied by the square of the still unknown factor. The second expression is zero because the wave is a sphere in the stationary system, so the first is zero as well. A reader who expects the moving system to see an ellipsoid, or a sphere about a displaced centre, can watch both possibilities fail here: the terms that would produce them cancel. The paper draws the conclusion in one sentence: the wave considered is, viewed from the moving system also, a spherical wave of propagation velocity VV, and with that its two fundamental principles are compatible with each other.Every stepξ2+η2+ζ2=φ(v)2[β2(x−vt)2+y2+z2],V2τ2=φ(v)2V2β2(t−vV2x)2\xi^2 + \eta^2 + \zeta^2 = \varphi(v)^2 \left[ \beta^2 (x - v t)^2 + y^2 + z^2 \right] , \qquad V^2 \tau^2 = \varphi(v)^2 V^2 \beta^2 \left( t - \frac{v}{V^2} x \right)^2The transformation of p. 900, each line squared. The unknown factor appears squared on both sides and nowhere else, so whatever it is, it cancels out of the comparison.(x−vt)2−V2(t−vV2x)2=(1−v2V2)(x2−V2t2)(x - v t)^2 - V^2 \left( t - \frac{v}{V^2} x \right)^2 = \left( 1 - \frac{v^2}{V^2} \right) \left( x^2 - V^2 t^2 \right)Expand both squares. The first is x2−2vxt+v2t2x^2 - 2 v x t + v^2 t^2 and the second is V2t2−2vxt+v2V2x2V^2 t^2 - 2 v x t + \frac{v^2}{V^2} x^2. The cross terms are equal and cancel, leaving x2(1−v2V2)−t2(V2−v2)x^2 \left( 1 - \frac{v^2}{V^2} \right) - t^2 \left( V^2 - v^2 \right), which is the right-hand side.β2(1−v2V2)=1\beta^2 \left( 1 - \frac{v^2}{V^2} \right) = 1From the definition of the factor: its square is the reciprocal of exactly this bracket.ξ2+η2+ζ2−V2τ2=φ(v)2[x2+y2+z2−V2t2]=0\xi^2 + \eta^2 + \zeta^2 - V^2 \tau^2 = \varphi(v)^2 \left[ x^2 + y^2 + z^2 - V^2 t^2 \right] = 0Put the three together. The β2\beta^2 meets the bracket and becomes 1, the transverse terms come through as y2+z2y^2 + z^2, and what is left inside is zero by the display above.Historian's marginThe calculation never uses the value of φ(v)\varphi(v), only that the same factor stands in all four lines of the transformation. So this step, which settles the compatibility of the two principles, leaves the factor exactly as unknown as it was; the paper has to determine it by a separate argument on the next two pages. It is worth separating the two questions the section is answering: whether a transformation of this shape can carry a light sphere into a light sphere, which is settled here, and which transformation of that shape is the right one, which is settled on p. 902.Also

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