when we remember that the contraction ω˜ α , eβ = δα β is a constant, and its derivative must be zero.

The same procedure that led to Eq. (5.63) would lead to the following:

Inspect these closely: they are very systematic. Simply throw in one term for each index; a raised index is treated like a vector and a lowered one like a one-form. The geometrical

5.4 C h r i s t o f f e l s y m b o l s a n d t h e m e t r i c

The formalism developed above has not used any properties of the metric tensor to derive covariant derivatives. But the metric must be involved somehow, because it can convert a vector into a one-form, and so it must have something to say about the relationship between their derivatives. In particular, in Cartesian coordinates the components of the one-form and its related vector are equal, and since is just differentiation of components, the components of the covariant derivatives of the one-form and vector must be equal.

But Eq. (5.67) is a tensor equation, so it must be valid in all coordinates. We conclude that

which is the component representation of Eq. (5.67).

If the above argument in words wasn’t satisfactory, let us go through it again in equations. Let unprimed indices α, β, γ , · · · denote Cartesian coordinates and primed indices α , β , γ , · · · denote arbitrary coordinates.

We begin with the statement

valid in any coordinate system. But in Cartesian coordinates

gαμ = δαμ, Vα = Vα .

Now, also in Cartesian coordinates, the Christoffel symbols vanish, so

Vα;β = Vα,β and Vα ;β = Vα ,β .

Therefore we conclude

Vα;β = Vα ;β

in Cartesian coordinates only. To convert this into an equation valid in all coordinate systems, we note that in Cartesian coordinates

Vα ;β = gαμVμ;β ,

so that again in Cartesian coordinates we have

Vα;β = gαμVμ;β .

But now this equation is a tensor equation, so its validity in one coordinate system implies its validity in all. This is just Eq. (5.68) again:

This result has far-reaching implications. If we take the β covariant derivative of Eq. (5.69), we find

Vα ;β = gα μ ;β Vμ + gα μ Vμ ;β .

Comparison of this with Eq. (5.70) shows (since V is an arbitrary vector) that we must have

in all coordinate systems. This is a consequence of Eq. (5.67). In Cartesian coordinates

gαμ;β ≡ gαμ,β = δαμ,β ≡ 0

is a trivial identity. However, in other coordinates it is not obvious, so we shall work it out as a check on the consistency of our formalism.

Using Eq. (5.64) gives (now unprimed indices are general)

In polar coordinates let us work out a few examples. Let α = r, β = r, μ = r:

grr;r = grr,r − ν rrgνr − ν rrgrν .

Since grr,r = 0 and ν rr = 0 for all ν, this is trivially zero. Not so trivial is α = θ ,

β = θ , μ = r:

gθ θ ;r = gθ θ ,r − ν θ rgνθ − ν θ rgθ ν .

With gθ θ = r2, θ θ r = 1/r and rθ r = 0, this becomes

gθ θ ;r = (r2),r − 1 (r2) − 1 (r2) = 0. r r

We use this to invert Eq. (5.72) by some advanced index gymnastics. We write three versions of Eq. (5.72) with different permutations of indices:

gαβ,μ = ν αμgνβ + ν βμgαν , gαμ,β = ν αβ gνμ + ν μβ gαν , −gβμ,α = − ν βα gνμ − ν μα gβν .

We add these up and group terms, using the symmetry of g, gβν = gνβ :

gαβ,μ + gαμ,β − gβμ,α

= ( ν αμ − ν μα )gνβ + ( ν αβ − ν βα )gνμ + ( ν βμ + ν μβ )gαν .

In this equation the first two terms on the right vanish by the symmetry of , Eq. (5.74), and we get

This is the expression of the Christoffel symbols in terms of the partial derivatives of the components of g. In polar coordinates, for example,

θ rθ = 12 gαθ (gαr,θ + gαθ ,r − grθ ,α ).

Since grθ = 0 and gθ θ = r−2, we have

This is the same value for θ rθ , as we derived earlier. This method of computing α βμ is so useful that it is well worth committing Eq. (5.75) to memory. It will be exactly the same in curved spaces.

The tensorial nature of α βμ

Since eα is a vector, eα is a 11 tensor whose components are μαβ . Here α is fixed and

μand β are the component indices: changing α changes the tensor eα , while changing

μor β changes only the component under discussion. So it is possible to regard μ and