D O C U M E N T 9 0 O C T O B E R 1 9 2 5 1 0 3 Further, In all cases where it is multiplied by an infinitesimal quantity, the factor may be substituted by 1. The relations then follow from your equations (7), or Then the ’s can be derived out of (10a). One thereby attains the following re- sult: 1. If the indices α, μ, ν all differ from one another, 2. If : 3. If The last two expressions follow from (3), if in it one sets , or . One can therefore also say that (3) is valid if α, μ, ν all differ from one another and if or . Now and still remain unknown. They cannot be derived from (3) the formula simply does not apply either for or for . The only thing that can still be concluded out of (10a) is The four quantities remain unknown. Through calculation one finds for the vector gαβ –g gαβ ⋅ = g – α)gαα--------( ∂ ∂xα gαv gvα) – ¦( 0= α)gαα------------ ∂ψvα ∂xα ¦( 0= ……(2) Γαv μ Γαv μ 1 2 --gμμ© - ∂gμν ∂xα ----------- ∂gαμ ∂xν ----------- - ∂gνα· ∂xμ -----------¹ –+ § = ……(3) μ α ≠ Γαα μ 1 2 -- - gμμ© ∂gμα ∂xα ----------- - ∂gαμ ∂xα ----------- - ∂gαα· ∂xμ -----------¹ -–+ § = ……(4) μ α ≠ Γαμ μ 1 2 -- - gμμ© ∂gμμ ∂xα ----------- ∂gαμ ∂xμ ----------- - ∂gμα ∂xμ -----------· -–+ ¹ § = ……(5) ν α = ν μ = ν α = ν μ = Γμα μ α μ) ≠ ( Γαα α α μ = ν α) ≠ ( α μ ν = = Γμα μ Γαμμ – Γαα α gμμ----------- ∂gμμ ∂xα 1 2 -- - gαα----------- ∂gαα ∂xα - – + + = ……(6) Γααα ϕα ϕα Γαα α – 1 2 --gαα----------- - ∂gαα ∂xα - += ……(7)