1 0 8 D O C U M E N T 9 0 O C T O B E R 1 9 2 5 Substitute in (23) μ, ν, α by α, μ, ν and add the new equation to (24) resp., subtract it from (24). One obtains after dividing by 2 (6) (25) (7) (26) By cyclical permutation of μ, ν, α one obtains from these equations four others as well among these six, only three differ from one another. Instead, one may take (25), (26), and (27) Initally all the indices μ, ν, α now differ from one another. Then follows from (22), because the terms with and are of second rank, and if one adds this to (27), Therefore, if all three indices differ, one can substitute for by inserting known terms. If one now writes in this way instead of (26) and instead of (27) then addition of both these formulas to (25) yields and one obtains the value of , if one divides by or else multiplies by . Now, two of these indices in should be the same, but the third should differ. Here three cases should be distinguished. For ex[ample], one can proceed as fol- lows: Substitute in the basic equation (22) first ν by α, a second time μ by α, ν by μ and a third time, μ by α, ν by α, and α by μ. Then one obtains sequentially (always ) (28) (29) (30) If one subtracts the last equation from the sum of the two others, one obtains gμμΓαν μ gααΓμν α = gννΓαμ ν gααΓνμ α = gμμΓνα μ gννΓμα ν = ϕα ϕν gννΓμα ν gμμΓαν μ + = gμμΓνα μ gμμΓαν μ + = Γκλ π π –Γλκ gννΓαμ ν gααΓμν α + = gμμΓαν μ gννΓαμ ν = 2gμμΓαν μ = Γαν μ α μ, α ν, μ ν) ( 2gμμ 1 2 --gμμ - Γμα ν μ α gααΓμα α gμμΓαα μ + = gμμΓαα μ gααΓαμ α gααϕμ + + = gααΓαμ α gααΓμα α gααϕμ + + =
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