Derivatives of one forms and tensors of higher types

Since a scalarφdepends on no basis vectors, its derivatived˜φis the same as its covariant derivative∇φ. We shall almost always use the symbol∇φ. To compute the derivative of a one-form (which as for a vector won’t be simply the derivatives of its components), we use

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the property that a one-form and a vector give a scalar. Thus, ifp˜is a one-form andVis an arbitrary vector, then for fixedβ,∇_βp˜is also a one-form,∇_βV is a vector, and˜p,V ≡φ is a scalar. In any (arbitrary) coordinate system this scalar is just

φ=p_αVα. (5.59)

Therefore∇_βφis, by the product rule for derivatives,

∇βφ=φ,β =∂pα

∂xβ Vα+pα ∂Vα

∂xβ. (5.60)

But we can use Eq. (5.50) to replace∂Vα/∂xβin favor ofVα;β, which are the components of∇_βV:

∇βφ=∂p_∂xβα Vα+p_αVα;_β −p_αVμα_μβ. (5.61) Rearranging terms, and relabeling dummy indices in the term that contains the Christoffel symbol, gives ∇βφ= _∂p α ∂xβ −pμμαβ Vα+p_αVα;β. (5.62) Now, every term in this equation except the one in parentheses isknownto be the compo- nent of a tensor, for an arbitrary vector fieldV. Therefore, since multiplication and addition of components always gives new tensors, it must be true that the term in parentheses is also the component of a tensor. This is, of course, the covariant derivative ofp˜:

(∇_βp˜)_α :=(∇ ˜p)_αβ:=p_α;β =pα,β−pμμαβ. (5.63) Then Eq. (5.62) reads

∇β(pαVα)=pα;βVα+pαVα;β. (5.64) Thus covariant differentiation obeys the same sort of product rule as Eq. (5.60). Itmust do this, since in Cartesian coordinates∇ is just partial differentiation of components, so Eq. (5.64) reduces to Eq. (5.60).

Let us compare the two formulae we have, Eq. (5.50) and Eq. (5.63): Vα;β =Vα,β+Vμαμβ,

p_α;β =pα,β−pμμαβ.

There are certain similarities and certain differences. If we remember that the derivative indexβ is thelastone on, then the other indices are the only ones they can be without raising and lowering with the metric. The only thing to watch is the sign difference. It may help to remember thatα_μβ was related to derivatives of the basis vectors, for then it is reasonable that−μ_αβbe related to derivatives of the basis one-forms. The change in sign means that the basis one-forms change ‘oppositely’ to basis vectors, which makes sense

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when we remember that the contraction ˜ωα,e_β =δα_β is aconstant, and its derivative must be zero.

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

∇βT_μν=T_μν,β −Tαναμβ−Tμαανβ; (5.64)

∇βAμν=Aμν,β+Aανμαβ+Aμαναβ; (5.65)

∇βBμ_ν =Bμ_ν,β +Bανμαβ−Bμαανβ. (5.66) Inspect these closely: they areverysystematic. Simply throw in oneterm for each index; a raised index is treated like a vector and a lowered one like a one-form. The geometrical meaning of Eq. (5.64) is that∇_βT_μνis a component of the0₃tensor∇T, whereTis a0₂ tensor. Similarly, in Eq. (5.65),Ais a2₀tensor and∇Ais a2₁tensor with components

∇βAμν.

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 areequal, and since∇is just differentiation of components, the components of the covariant derivatives of the one-form and vector must be equal. This means that ifV is an arbitrary vector andV˜ =g(V, ) is its related one-form, then in Cartesian coordinates

∇βV˜ =g(∇_βV, ). (5.67) But Eq. (5.67) is a tensor equation, so it must be valid inallcoordinates. We conclude that

V_α;β =gαμVμ;β, (5.68)

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 equa- tions. Let unprimed indicesα,β,γ,· · · denote Cartesian coordinates and primed indices α_{,β,γ,· · · denotearbitrarycoordinates.}

We begin with the statement

V_α =g_αμVμ, (5.69)

valid in any coordinate system. But in Cartesian coordinates g_αμ=δ_αμ, V_α =Vα.

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Now, also in Cartesian coordinates, the Christoffel symbols vanish, so V_α;β =Vα,β and Vα;β =Vα,β. Therefore we conclude

V_α;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

V_α;β =gαμVμ;β.

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

V_α;β =gαμVμ

;β (5.70)

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

V_α;β =gαμ;β Vμ

+g_αμVμ;β.

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

g_αμ;β ≡0 (5.71)

in all coordinate systems. This is a consequence of Eq. (5.67). In Cartesian coordinates g_αμ;_β ≡g_αμ,_β =δ_αμ,_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)

g_αβ;μ=gαβ,μ−ναμgνβ−νβμgαν. (5.72) In polar coordinates let us work out a few examples. Letα=r,β=r,μ=r:

grr;r=grr,r−νrrgνr−νrrgrν.

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

β=θ,μ=r:

gθθ;r =gθθ,r−νθrgνθ−νθrgθν.

Withg_θθ =r2,θ_θr =1/randrθr =0, this becomes

g_θθ;r =(r2),r− 1 r(r 2 )−1 r(r 2 )=0.

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So it works, almost magically. But it is important to realize that it is not magic: it fol- lows directly from the facts thatg_αβ,μ=0 in Cartesian coordinates and thatgαβ;μare the components of thesametensor∇gin arbitrary coordinates.

Perhaps it is useful to pause here to get some perspective on what we have just done. We introduced covariant differentiation in arbitrary coordinates by using our understand- ing of parallelism in Euclidean space. We then showed that the metric of Euclidean space is covariantly constant: Eq. (5.71). When we go on to curved (Riemannian) spaces we will have to discuss parallelism much more carefully, but Eq. (5.71) will still be true, and therefore so will all its consequences, such as those we now go on to describe.