3 6 6 D O C U M E N T 4 2 5 O N G E N E R A L R E L A T I V I T Y The ’s, for any choice of , which are subject only to the symmetric con- dition (4) do not form any symmetric tensor. If the tensor is broken down into a sym- metric and antisymmetric one according to the equation , (5) whereby , (6) then it suggests itself to set the tensor equal to the metric tensor but to regard the tensor , which satisfies the relation , (7) as the tensor for the electromagnetic field.[8] First a comment in support of the limiting symmetry condition (4). From (2) fol- lows the displacement law for the covariant vector through the natural stipulation that the scalar product of a contravariant and a covariant vector does not change upon parallel displacement. From this follows the law[9] . From this follows, in the familiar manner, the tensorial character of . From this and from the tensorial character of , the tensorial character of can then be concluded. From this and also from the foregoing, it then follows that also has a tensorial nature. Therefore, the symmetry condition (4) is necessary if the unequivocal character of the vector’s covariant extension is to be preserved. In Eddington’s theory the 40 quantities appear as unknown functions of the ’s, similar to the 14 quantities and in the original theory of relativity. Now, the unsolved problem by Eddington involves finding the equations necessary Rkl Γμν α Γμν α Γνμ α = Rkl) ( Rkl gkl φkl += φkl 1 2 -- - ∂Γkα α ∂xl ------------ ∂Γlα α ∂xk ----------- – = gkl) ( gkl φkl) ( ∂φkl ∂xm --------- - ∂φlm ∂xk ---------- - ∂φmk ∂xl ----------- - + + 0 = δBμ Γμβ α Bαdxβ = ∂Bμ ∂xν --------- Γμβ α Bα – ∂Bμ ∂xν --------- ∂Bν ∂xμ -------- -– [p. 34] Γμν α Γνμ α – ∂Bμ ∂xν --------- Γνμ α Bα – Γμν α xν gμν φμ