D O C . 1 5 8 O N R I E M A N N C U R V A T U R E T E N S O R 2 9 1 righthand side trace-free by restricting the total energy-momentum tensor to coincide with that of the electromagnetic field, i.e. by positing eq. (2a) below outright. In the present document, Einstein instead first writes down field equations where both the left and the righthand side are trace-free, without demanding a particular energy-momentum tensor. One can recover the Einstein equations with cosmological constant (eq. [2]) by taking the covariant derivative of the trace-free field equa- tions as introduced here, then use the contracted Bianchi identity for the Einstein tensor, the conser- vation condition for the energy-momentum tensor and the condition of metric compatibility to conclude that is a constant, which one identifies with . Inserting into the trace-free field equations gives eq. (2), corrected so that the third term on the right is + , as in Einstein 1919a, p. 353. [7]The reference here should be to the unnumbered equation immediately following eq. (2). [8]The standard field eq. (1) together with eq. (3) imply that everywhere, whereas the trace- free field equations (2a), together with eq. (3), only imply that everywhere. See Einstein 1919a (Vol. 7, Doc. 17), p. 351 and p. 353 for details. [9]In Einstein 1919a (Vol. 7, Doc. 17, p. 352), Einstein had argued that the difference between the Ricci scalar R and the constant , which he would quickly relate to the cosmological constant, acts as a negative pressure inside charged elementary particles, and thus ensures their stability. [10]Einstein had disavowed the path of Weyl and Eddington before see especially Doc. 71, Doc. 94, and Einstein 1925w (Doc. 92). [11]Rainich 1925b. The short note consists of two parts. Rainich’s account of Riemann curvature tensors is Einstein’s starting point in the present paper. Rainich writes: “What a Riemann tensor does is to assign to every two-dimensional direction, or orientation, a certain number—its curvature the first of the two parts mentioned above is characterized by the property that it assigns to two (abso- lutely) perpendicular orientations equal curvatures, while the second part assigns to such orientations opposite curvatures. We may mention that of the twenty constants which are needed to give the com- plete Riemann tensor, the first part involves 11 and the second 9.” The second part of Rainich’s note then associates the first (symmetric) part of the Riemann tensor thus defined with gravity, via relating it to the left-hand side of the Einstein equations with cosmological constant, while the second (anti- symmetric) part of the Riemann tensor is associated with electromagnetism. Rainich states that these results follow easily from his much longer articles Rainich 1924a, 1924b, and 1925a. [12]The Riemann curvature tensor can be uniquely decomposed into three parts that are irreducible representations of the full Lorentz group. Einstein’s is one of these irreducible parts, whereas his can be split further into the trace-free Weyl tensor and a term dominated by the Ricci scalar, i.e., the trace of the Ricci tensor. From this decomposition, one sees that Einstein’s theorem below is correct: vanishes if and only if vanishes. See Stephani et al. 2009, p. 37, for details. [13]Rainich 1925b had pointed out that the symmetric has eleven independent components, whereas the antisymmetric has nine independent components. Rainich wanted to associate the former tensor with the gravitational and the latter with the electromagnetic field. In the present paper, Einstein, contrary to Rainich, associates with the pure gravitational field. The reason lies in the fact that he wanted to represent gravity not by the Einstein tensor but by the trace-free tensor that he had pioneered in Einstein 1919a (Vol. 7, Doc. 17). He had already pointed out to Besso (Doc. 138) that the equations (2a) and (3) of the present document, based on the trace-free ten- sor, constitute 9 equations for the 14 quantities (gravity) and (electromagnetism). [14]Here the text should say “antisymmetrisch” rather than “asymmetrisch.” R kT – 4λ R 4λ kT += λgim R 0= R const. = R0 Aik,lm Sik,lm Aik,lm Rik 1 4 --gikR - – Sik,lm Aik,lm Aik,lm Rik 1 4 --gikR - – gμν ϕμ