Covariant differentiation in curvilinear coordinates

Covariant vectors and metric

A set of four quantities Aa, where a=0,1,2,3, is called a contravariant vector if under a transformation of coordinates x→ x′(x) it transforms as coordinate differentials dxa, while a set of four quantities Aa is called a covariant vector if it transforms1 as derivatives of a scalar ∂φ/∂xa,
Aa = ∂xa
∂x′b
 A′b,  Aa = ∂x′b
∂xa
 A′b.
(1)
The contraction AaBa is clearly invariant under coordinate transformation, AaBa = A′aB′a.
The metric tensor gab defines the invariant element of curvilinear coordinates, ds2 = gabdxadxb. Since dxa is an arbitrary contra-variant vector one can conclude that the construction gabdxb transforms as a co-variant vector and thus that the metric tensor connects contra- and co-variant vectors, Aa=gabAb.

Covariant differential

In curvilinear coordinates the differential of a vector dAa is not a covariant quantity, since generally dAa=d(gabAb) ≠ gabdAb. Indeed in curvilinear coordinates the unit vectors are generally not orthogonal and not normalized, therefore a covariant differential, denoted DAa, has to contain an additional contribution (see the exercise) which is customarily written through the so-called Christoffel symbols
DAa = dAa + ΓabcAbdxc  .
(2)
Differential of a scalar d(AaBb) is already a covariant quantity, therefore D(AaBa)=d(AaBb) for an arbitrary Ba, which leads to
DAa = dAa − ΓbacAbdxc  .
(3)
For DAa to be a covariant quantity one needs
DAa=gabDAb=DgabAb,
(4)
i.e. the covariant derivative of the metric tensor2 has to vanish, Dgab=0, which defines the Christoffel symbols,
Γabc= 1
2
dgab
dxc
 − dgbc
dxa
 + dgac
dxb

 .
(5)

Geodesic as a constant velocity trajectory

A particle in curvilinear coordinates moves in such a way that the covariant derivative of its velocity ua vanishes, Dua=0. This leads to the so called geodesic equation,
d2xa
ds2
 +Γabc dxb
ds
 dxc
ds
 = 0.
(6)

Exercises

  1. Prove (5) e.g. by rewriting Dgab=0 through Christoffel symbols and making a linear combination of three such equations with appropriately renamed indexes.
  2. Let x,y be Cartesian coordinates in a flat two-dimensional space with metric dl2=dx2+dy2. Consider polar coordinates x=rcosθ, y=rsinθ (another notation xr ≡ r, xθ ≡ θ)
    1. From3 d[(r)\vec]=dr[(e)\vec]r+dθ[(e)\vec]θ find [(e)\vec]a, a=r,θ.
    2. Find gab=[(e)\vec]a·[(e)\vec]b (a,b=r,θ) and check, that it is identical to the tensor from the metric dl2=dxadxbgab.
    3. Find gab=gab−1 and [(e)\vec]a.
    4. Consider a vector [(A)\vec]=Aa[(e)\vec]a with a differential
      d
      →
      A
       
      		 =dAa
      →
      e
       
      a 
      		 +Aa
      ∂
      →
      e
       
      a 
      ∂xb
      		 dxb
      and a covariant differential
      DAa ≡
      →
      e
       
      		 a
       
      		 ·d
      →
      A
       
      		 =dAa+
      →
      e
       
      		 a
       
      		 ·
      ∂
      →
      e
       
      b 
      ∂xc


      		 Abdxc.
      Prove that
      DAa ≡
      →
      e
       
      a 
      		 ·d
      →
      A
       
      		 =dAa−
      →
      e
       
      		 b
       
      		 ·
      ∂
      →
      e
       
      a 
      ∂xc


      		 Abdxc.
      The expression in parentheses is apparently the Christoffel symbol Γabc = ([(e)\vec]a·[(∂[(e)\vec]b)/(∂xc)]). Calculate it and compare with (5).
    5. Consider the parametric equations for a straight line in x, y coordinates,
      d2x
      ds2
      		 =0  ,     d2y
      ds2
      		 =0.
      Make a variable substitution and obtain the corresponding equations in the r,θ coordinates. Prove that they are identical to (6).

Footnotes:

1 in the following AaBa ≡ ∑aAaBa
2  considering D(AaBb) one can derive the covariant differential of a tensor, DFab=dFab+ΓacdFcbdxd+ΓbcdFacdxd.
3 d[(r)\vec] ≡ dx[(e)\vec]x+dy[(e)\vec]y, where [(e)\vec]x·[(e)\vec]x=[(e)\vec]y·[(e)\vec]y=1, [(e)\vec]x·[(e)\vec]y=0.

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On 5 Oct 2007, 17:52.