31.6. DIFFERENTIATION AND CHRISTOFFEL SYMBOLS 623

31.6 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) (31.23)

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

Theorem 31.6.1 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) . (31.24)

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 31.23. 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) ,

and this shows the second formula for transforming these scalars.Now F (x) = F i (x)ei (x) and so by the product rule,

∂F

∂x j =∂F i

∂x j ei (x)+F i (x)∂ei (x)

∂x j . (31.25)

Now ∂ei(x)∂x j is a vector and so there exist scalars,

{ki j

}such that

∂ei (x)

∂x j =

{ki j

}ek (x) .

Thus {ki j

}ek (x) =

∂ 2y

∂x j∂xi

and so {ki j

}ek (x) ·er (x) =

{ki j

}δ

rk =

{ri j

}=

∂ 2y

∂x j∂xi ·er (x) (31.26)

Therefore, from 31.25, ∂F∂x j =

∂Fk

∂x j ek (x)+F i (x)

{ri j

}ek (x) which shows

Fk, j (x) =

∂Fk

∂x j +F i (x)

{ki j

}. (31.27)

This is sometimes called the covariant derivative.