25.6. DIFFERENTIATION AND CHRISTOFFEL SYMBOLS 475

gi j (z) =∂xr

∂ zi∂xs

∂ z j grs (x) , gi j (z) =∂ zi

∂xr∂ z j

∂xs grs (x) . (25.22)

Proof: We already have shown the first part of 25.21 in 25.19. Then, from 25.19,

ei (z) = ei (z) ·e j (x)ej (x) = ei (z) · ∂ zk

∂x j ek (z)ej (x)

= δik

∂ zk

∂x j ej (x) =

∂ zi

∂x j ej (x)

and this proves the second part of 25.21. Now to show 25.20,

vi (z) = v ·ei (z) = v·∂x j

∂ zie j (x) =

∂x j

∂ ziv ·e j (x) =

∂x j

∂ ziv j (x)

and

vi (z) = v ·ei (z) = v · ∂ zi

∂x j ej (x) =

∂ zi

∂x j v ·ej (x) =

∂ zi

∂x j v j (x) .

To verify 25.22,

gi j (z) = ei (z) ·e j (z) = er (x)∂xr

∂ zi ·es (x)∂xs

∂ z j = grs (x)∂xr

∂ zi∂xs

∂ z j . ■

25.6 Differentiation and Christoffel SymbolsLet F : U → Rn be differentiable. We call F a vector field and it is used to model force,velocity, acceleration, or any other vector quantity which may change from point to pointin U. Then ∂F (x)

∂x j is a vector and so there exist scalars, F i, j (x) and Fi, j (x) such that

∂F (x)

∂x j = F i, j (x)ei (x) ,

∂F (x)

∂x j = Fi, j (x)ei (x) (25.23)

We will see how these scalars transform when the coordinates are changed.

Theorem 25.6.1 If x and z are curvilinear coordinates,

Fr,s (x) = F i

, j (z)∂xr

∂ zi∂ z j

∂xs , Fr,s (x)∂xr

∂ zi∂xs

∂ z j = Fi, j (z) . (25.24)

Proof:

Fr,s (x)er (x)≡

∂F (x)

∂xs =∂F (z)

∂ z j∂ z j

∂xs ≡

F i, j (z)ei (z)

∂ z j

∂xs = F i, j (z)

∂ z j

∂xs∂xr

∂ zi er (x)

which shows the first formula of 25.23. To show the other formula,

Fi, j (z)ei (z)≡ ∂F (z)

∂ z j =∂F (x)

∂xs∂xs

∂ z j ≡

Fr,s (x)er (x)

∂xs

∂ z j = Fr,s (x)∂xs

∂ z j∂xr

∂ zi ei (z) ,