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A tangent vector is just a directional derivative along a curve γ(t) through p ∈M µ µ ∂ f = ∂ f dx = dx ∂ f µ µ . (1) ∂t !γ t=t ∂x γ(t ) dt t=t dt ∂x γ(t ) 0 0 0 0 The directional derivative of a function f is obtained by applying the differential operator V to f , where
µ = µ ∂ µ = dx V V µ , where V (2) ∂x dt t=t 0 that is ∂ f = µ ∂ f = V µ V[ f ]. (3) ∂t ! = ∂x γ t t0 ∂ Thus every tangent vector at a point p can be expressed as a linear combination of the coordinate derivatives . The ∂xµ directional derivatives along the coordinate lines at p form a basis of an n-dimensional vector space whose elements ∂ are the tangent vector at p. This space is called the tangent space Tp and the basis is called a coordinate basis (∂xµ ) or holonomic frame. Thus every tangent vector at a point p can be expressed as a linear combination of the coordinate ∂ derivatives . Any tangent vectors to at p of the n-dimensional vector space Tp are referred as contravariant (∂xµ ) M vector. The commutator, [U, V], of two vector fields U and V is defined by
[U, V]( f ) = U(V( f )) V(U( f )), − ∂ ∂ for a given coordinates basis U = Uκ and V = Vλ we have that ∂xκ ∂xλ ∂ ∂ ∂ ∂ [U, V] = Uκ Vλ Vλ Uκ ∂xκ ∂xλ ! − ∂xλ ∂xκ ! ∂Vλ ∂ ∂2 ∂Uκ ∂ ∂2 = Uκ + UκVλ Vλ VλUκ ∂xκ ! ∂xλ ∂xκ∂xλ − ∂xλ ! ∂xκ − ∂xλ∂xκ ∂Vλ ∂ ∂Uκ ∂ = Uκ Vλ ∂xκ ! ∂xλ − ∂xλ ! ∂xκ ∂Vλ ∂Uλ ∂ ∂ = Uκ Vκ = Wλ , ∂xκ − ∂xκ ! ∂xλ ∂xλ
∂Vλ ∂Uλ where Wλ = Uκ Vκ . Thus, the commutator defines a new vector ∂xκ − ∂xκ [U, V] = W. (4) 2 The commutators satisfy the Jacobi identity
[U, [V, W]] + [V, [W, U]] + [W, [U, V]] = 0 (5) for arbitrary U, V and W. A one-form (1-form) or covariant vector or dual vector ω at p maps a tangent vector V into a real number, the contraction, denoted by ω, V , and this mapping is linear: h i ω, aU + bV = a ω, U + b ω, V (6) h i h i h i holds for all a, b R and U, V Tp. Linear combinations of 1-forms ω, τ are defined by the rule ∈ ∈ aω + bτ, V = a ωV + b τ, V . (7) h i h i h i An arbitrary 1-form ω is written as a linear combination,
µ ω = ωµdx , (8)
∂ where dxµ is a dual basis to the basis of tangent vectors, being called cotangent basis. The 1-forms form a { } (∂xµ ) basis of n-dimensional dual vector space called the dual or cotangent space T with basis dxµ with conditions ∗p { } µ µ ∂ ∂x µ dx , = = δ ν. (9) * ∂xν + ∂xν The contraction ω, V is the inner product between a contravariant vector V and a covariant or dual vector ω. For h i ω T and V Tp we have that inner product is ∈ ∗p ∈
µ ν ∂ ν µ ∂ ν µ µ ω, V = ωµdx , V = ωµV dx , = ωµV δ ν = ωµV . (10) h i * ∂xν + * ∂xν +
In terms of the dual basis, the differential d f on an arbitrary function f is given by ∂ f d f = dxµ. (11) ∂xµ 2.2. Tensors
A tensor of type (q, r) at a point p is a multilinear object which maps q elements of Tp and r elements of T∗p to a real number. q( ) at p denotes the set of type (q, r) tensors at p , by tensor product Tr M ∈M q ( )p = Tp Tp T∗ T∗ . (12) Tr M ⊗···⊗ ⊗ p ⊗···⊗ p q factors r factors | {z } | {z } 1 0 q ∂ ν In particular, = Tp and = T∗. An element of ( ) is written in terms of the basis and dx as T0 T1 p Tr M (∂xµ ) { }
µ1 µq ∂ ∂ ν1 νr T = T ··· ν1 νr dx dx , (13) ··· ∂x µ1 ⊗···⊗ ∂x µq ⊗ ⊗···⊗
µ1 µq where all indices run from 1 to n. The coefficients T ··· ν1 νr with contravariant indices µ1 µq and covariant ··· ∂ ··· indices ν1 νr are the components of T with respect to the basis and dxν . ··· (∂xµ ) { }
3 2.3. Differential forms, exterior product and exterior derivatives A differential form of order r, or r-form, is a totally antisymmetric tensor of type (0, r). If ρ and σ are r-form and s-form respectively, one can define a (r + s)-form ρ σ from them, where is the skew-symmetrized tensor product . The exterior product or wedge product ρ σ is the∧ tensor of type (0, r +∧s) [15]. For example, if ⊗ ∧ µ ν (i) ρ = rµdx and σ = sνdx
ρ and σ are 1-form, then
µ ν ν µ µ ν ρ σ = rµ sν(dx dx dx dx ) = rµ sν dx dx = σ ρ, ∧ ⊗ − ⊗ ∧ − ∧ ρ σ is a 2-form. ∧Now if λ µ ν (ii) ρ = rλdx , σ = sµdx and τ = tνdx then
λ µ ν ν λ µ µ ν λ λ ν µ ν µ λ µ λ ν ρ σ τ = rλ sµtν(dx dx dx +dx dx dx +dx dx dx dx dx dx dx dx dx dx dx dx ) ∧ ∧ ⊗ ⊗ ⊗ ⊗ ⊗ ⊗ − ⊗ ⊗ − ⊗ ⊗ − ⊗ ⊗ or λ µ ν ρ σ τ = rλ sµtν dx dx dx , ∧ ∧ ∧ ∧ ρ σ τ is a 3-form. ∧For∧ the skew-symmetrized tensor product , it is verified that the exterior product satisfies the fllowing axioms: ∧ 1. is linear in each dxµ 2. vanish if any two factors coincide 3. changes sign if any two factors are interchanged.
If we denote the vector space of r-forms at point p by Ωr( ) and an element ρ Ωr( ) is expaned as ∈M M ∈ M µ1 µ2 µr ρ = rµ1,µ2, ,µr dx dx dx , 1 µ1 < µ2 < ···< µr n∧, ∧···∧with r n. ≤ ··· ≤ ≤ The above axiom 2 implies that these exterior products vanish for r > n. n We have already stated that rµ1,µ2 µr are totally antisymmetric, then there are r-combinations of choices of ··· r! n set (µ1, µ2, , µr), thus the dimensional of the vector space Ωr( ) is . For example, a manifold which has ··· M r! dim( ) = 3 with dx, dy and dz. The 2-forms: dx dy, dx dz and dy dz are oriented area elements. dx dy dz is orientedM volume element. We have: ∧ ∧ ∧ ∧ ∧
1 0 1 1. dx, dy, dz Ω ( ) = T∗( ) where Ω ( ) is vector space of smooth functions. Ω ( ) has dimension {3 } ∈ M M M M = 3. 1! 3 2. dx dy, dx dz, dy dz Ω2( ) where Ω2( ) has dimension = 3. { ∧ ∧ ∧ } ∈ M M 2! 3 3. dx dy dz Ω3( ) where dim(Ω3( )) = = 1. { ∧ ∧ } ∈ M M 3!
4 Let ρ Ωr( ), σ Ωs( ) and τ Ωt( ), the exterior product is associative: ∈ M ∈ M ∈ M (ρ σ) τ = ρ (σ τ). (14) ∧ ∧ ∧ ∧ However, the commutative law is slighty changed:
ρ σ = ( 1)rsσ ρ. (15) ∧ − ∧ The definition of an r-form is extended to include ordinary function f on manifold . Functions on are called 0-form; 1-form is a covariant vector, 2-form is an antisymmetric covariant tensor of rankM 2 and so on. M If ρ is an r-form given by µ1 µ2 µr ρ = rµ1,µ2, ,µr dx dx dx , ··· ∧ ∧···∧ then the exterior derivative of ρ is written dρ and is defined by
∂rµ1,µ2, ,µr dρ = ··· dxν dxµ1 dxµ2 dxµr, (16) ∂xν ∧ ∧ ∧···∧ The exterior derivative d is a map Ωr( ) Ωr+1( ). For example: if = R3 and M → M M 1. 0-form: f = f (x, y, z) then the action of exterior derivative is ∂ f ∂ f ∂ f d f = dx + dy + dz, (17) ∂x ∂y ∂z it is identified with gradient;
2. 1-form: ρ = ρx(x, y, z) dx + ρy(x, y, z) dy + ρz(x, y, z) dz then the action of exterior derivative is
∂ρy ∂ρ ∂ρ ∂ρy ∂ρ ∂ρ dρ = x dx dy + z dy dz + x z dz dx (18) ∂x − ∂y ! ∧ ∂y − ∂z ! ∧ ∂z − ∂x ! ∧ it is identified with rotational;
3. 2-form: σ = σxy(x, y, z) dx dy + σyz(x, y, z) dy dz + σzx(x, y, z) dz dx then the action of exterior derivative is ∧ ∧ ∧ ∂σyz ∂σ ∂σxy dσ = + zx + dx dy dz (19) ∂x ∂y ∂z ! ∧ ∧ it is identified with divergence.
The second exterior derivative of an r-form, it is verified that d2ρ = 0. If ρ is an r-form given by
µ1 µ2 µr ρ = rµ1,µ2, ,µr dx dx dx , ··· ∧ ∧···∧ then the d2ρ is 2 ∂ rµ1,µ2, ,µr d2ρ = ··· dxν dxλ dxµ1 dxµ2 dxµr, ∂xλ∂xν ∧ ∧ ∧ ∧···∧ 2 ∂ rµ1,µ2, ,µr since ··· is symmetric with respect ν and λ while dxν dxλ is antisymmetric. Another important property is ∂xλ∂xν ∧ d(ρ σ) = dρ σ + ( 1)r(ρ dσ) (20) ∧ ∧ − ∧ where ρ is an r-form. For example, the electromagnetic potential φ Ax (Aµ) = , Ay Az 5 ν can be assigned by 1-form A = Aνdx . It remains for us to describe the electromagnetic tensor F = dA as a 2-form,
∂A 1 ∂A ∂A dA = ν dxµ dxν = ν + ν dxµ dxν ∂xµ ∧ 2 ∂xµ ∂xµ ! ∧ 1 ∂A ∂Aµ = ν dxµ dxν + dxν dxµ 2 ∂xµ ∧ ∂xν ∧ ! 1 ∂A ∂Aµ = ν dxµ dxν 2 ∂xµ − ∂xν ! ∧ 1 µ ν = Fµν dx dx , 2 ∧
1 µ ν where F = Fµν dx dx . The components of Fµν is given by: 2 ∧
0 Ex Ey Ez − − − Ex 0 Bz By (Fµν) = − . Ey Bz 0 Bx − Ez By Bx 0 − The action of exterior derivative on F results in
1 ∂Fµν dF = dxλ dxµ dxν, 2 ∂xλ ∧ ∧ where dF = d2 A = 0, then 1 ∂ F dxλ dxµ dxν = ∂ F dxλ dxµ dxν + ∂ F dxµ dxν dxλ + ∂ F dxν dxλ dxµ = 0, λ µν 3 λ µν µ νλ ν λµ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ we can interchange dxµ dxν dxλ and dxν dxλ dxµ of above equation, ∧ ∧ ∧ ∧ dxµ dxν dxλ = dxλ dxµ dxν ∧ ∧ ∧ ∧ and dxν dxλ dxµ = dxλ dxµ dxν, ∧ ∧ ∧ ∧ such as λ µ ν ∂λFµν + ∂µFνλ + ∂νFλµ dx dx dx = 0, ∧ ∧ which is known as the Bianchi identity, that follows from this, the two homogeneous Maxwell’s equations,
∂B B = 0 and + E = 0. ∇· ∂t ∇× 2.4. The covariant derivative The exterior derivative is a limited generalization acting only on forms. So, it is necessary to introduce covariant derivative X in the direction of the vector X at p on manifold , that maps an arbitrary tensor into a tensor of same type. The∇ symbol is called affine connection. Let X, Y and MZ vector fields and f a scalar function. The covariant derivative satisfies∇ the following conditions:
X(Y + Z) = XY + X Z; (21) ∇ ∇ ∇
X+Y Z = X Z + Y Z; (22) ∇( ) ∇ ∇ f XY = f XY; (23) ∇ ∇ X( f Y) = X[ f ]Y + f XY. (24) ∇ ∇ 6 where from (3) we have that µ ∂ f X f = X[ f ] = X . ∇ ∂x µ ∂ ∂ The covariant derivative of the basis vector E = in the direction of basis vector E = can be expanded ν ∂xν µ ∂x µ in terms of basis vectors: ρ E Eν = Γ µν Eρ (25) ∇ µ where Γρ = ρ µν dx , Eµ Eν (26) D ∇ E is called connection coefficients. µ ν The covariant derivative XY, where X = X Eµ and Y = Y Eν is given by vector ∇ ν XY = Xµ E (Y Eν) ∇ ∇( µ) with conditions (23) and (24) we have
= µ ν = µ ν + ν XY X Eµ (Y Eν) X Eµ[Y ]Eν Y Eµ Eν ∇ ∇ ∇ with (25) we have ν µ ∂Y ν ρ XY = X Eν + Y Γ µν Eρ ∇ ∂xµ ! thus, this covariant derivative is given by the vector
µ ρ ν ρ XY = X ∂µY + Y Γ µν Eρ. (27) ∇ By definition, the affine connection maps two vectors X and Y to a new vector given by of the right hand side of (27), whose ρth component is ∇ µ ρ ν ρ µ ρ X ∂µY + Y Γ µν X µY . ≡ ∇ If we compute the covariant derivative E Y, we have: ∇ µ = ν + ν = ρ + νΓρ = ρ Eµ Y Eµ[Y ]Eν Y Eµ Eν ∂µY Y µν Eρ µY Eρ, ∇ ∇ ∇ ρ where µY is the ρth component of vector Eµ Y. ∇ ∇ ν The covariant derivative of 1-form ω = ωνdx is obtained from contraction of ω with a tangent vector Y,
ω, Y = f. h i ρ If we put Y = Eν and ω = dx in above equation, it becomes
ρ ρ dx , Eν = δ ν, h i then the covariant derivative of above equation with respect to Eµ is
ρ E dx , Eν = 0 ∇ µ h i ρ + ρ = Eµ dx , Eν dx , Eµ Eν 0 D∇ E D ∇ E we can use (26) ρ = Γρ Eµ dx , Eν µν, D∇ E − and as consequence of above equation we can identify
ρ ρ σ E dx = Γ µσdx . (28) ∇ µ − 7 ν Furthermore, if we calculate derivative of 1-form ω = ωνdx with respect to Eµ, we have that, = ν + ν Eµ ω Eµ[ων]dx ων Eµ dx ∇ ν ν∇ σ = ∂µωνdx ωνΓ µσdx − ν σ = ∂µωσ ωνΓ µσ dx − σ = µωσ dx , ∇ where we can identify the σth component of 1-form E ω as ∇ µ ν µωσ = ∂µωσ ωνΓ µσ. (29) ∇ − We require that the Leibnitz rule be true for any tensor products, = + Eµ (T1 T2) Eµ T1 T2 T1 Eµ T2. (30) ∇ ⊗ ∇ ⊗ ⊗ ∇ ν µ If we have a tensor of type (1,1) T = Tµ Eν dx , under the action of covariant derivative and Leibnitz rule we have that ⊗
ν µ E T = E Tµ Eν dx ∇ ρ ∇ ρ ⊗ ν µ ν µ ν µ = Eρ[Tµ ]Eν dx + Tµ E Eν dx + Tµ Eν E dx ⊗ ∇ ρ ⊗ ⊗ ∇ ρ ν µ ν σ µ ν µ σ = ∂ρTµ Eν dx + Tµ Γ ρν Eσ dx + Tµ Eν Γ ρσdx ⊗ ⊗ ⊗ − ν ν σ σ ν µ = ∂ρTµ + Γ ρσTµ Γ ρµTσ Eν dx , − ⊗ where we have that: ν ν ν σ σ ν ρTµ = ∂ρTµ + Γ ρσTµ Γ ρµTσ (31) ∇ − 2.5. The torsion and curvature Intrinsic objects defined on manifolds that measure the bends on manifolds are torsion tensor and curvature (Rie- mann) tensor. Let X, Y and Z be vectors of tangent space Tp, then the torsion tensor and curvature tensor are defined by T(X, Y) XY Y X [X, Y] torsion tensor (32) ≡ ∇ − ∇ − R(X, Y)Z X Y Z Y X Z X,Y Z curvature tensor. (33) ≡ ∇ ∇ − ∇ ∇ − ∇[ ] The torsion tensor gives an intrinsic characterization of how tangent spaces twist about a curve when they are parallel transported. The curvature or Riemann tensor describes how the tangent spaces roll along the curve. ∂ µ With respect to coordinate basis Eµ = and the dual basis dx , the components of torsion tensor is given ( ∂x µ ) { } by λ λ λ λ κ κ T µν = dx , T(Eµ, Eν) = dx , E Eν E Eµ [Eµ, Eν] = dx , Γ µν Eκ Γ νµ Eκ 0 h i h ∇ µ − ∇ ν − i h − − i where it follows that λ λ λ T µν = Γ µν Γ νµ. (34) − And the components of curvature tensor with respect to coordinate basis is given by
κ κ R λµν = dx , R(Eµ, Eν)Eλ , (35) h i where we have
R(Eµ, Eν)Eλ = E E Eλ E E Eλ E ,E Eλ ∇ µ ∇ ν − ∇ ν ∇ µ − ∇[ µ ν] = Γ ρ Γ ρ Eµ νλ Eρ Eν µλ Eρ 0, ∇ − ∇ − we recall (24) where X( f Y) = X[ f ]Y + f XY, such as, ∇ ∇ = Γ ρ + Γ ρ Γ ρ Γ ρ R(Eµ, Eν, Eλ) Eµ νλ Eρ νλ Eµ Eρ Eν µλ Eρ µλ Eν Eρ ∇ − h i − ∇ 8 ρ ρ σ ρ ρ σ = ∂µΓ νλEρ + Γ νλΓ µρEσ ∂νΓ µλ Eρ Γ µλΓ νρEσ ρ ρ σ −ρ σ −ρ = ∂µΓ νλ ∂νΓ µλ + Γ νλΓ µσ Γ µλΓ νσ Eρ, − − it follows that κ κ κ σ κ σ κ R λµν = ∂µΓ νλ ∂νΓ µλ + Γ νλΓ µσ Γ µλΓ νσ. (36) − − We readily find λ λ κ κ T µν = T νµ and R λµν = R λνµ. (37) − − 2.6. Metric tensors We have defined the contraction ω, V as the inner product between a contravariant vector V and a covariant or h i dual vector ω. For ω T and V Tp we have that inner product is ∈ ∗p ∈ µ ω, V = ωµV . h i Now, we can define a linear map gp(U, V) as the inner product between two contravariant vectors U, V Tp. The ∈ Riemannian metric gp is a type (0,2) tensor field on manifold which satisfies M (i) gp(U, V) = gp(V, U) R; ∈ (ii) gp(U, U) 0 assigns a magnitude of vector U Tp at point p. gp(U, U) = 0 is true only when U = 0. ≥ ∈ gp is called pseudo-Riemannian metric when the metric is no longer positive definite. Since gp is a type (0,2) tensor field on , it is expaned in terms of dxµ dxν as M ⊗ µ ν gp = gµν(p) dx dx . (38) ⊗ ∂ The components of gp with respect a coordinate basis Eµ = are ( ∂xµ )
gµν(p) = gp(Eµ, Eν) = gp(Eν, Eµ), (39) where we can see that (gµν) is a symmetric matrix (we can omit p in metric tensor). The matrix (gµν) has an inverse denoted by (gµν) where we have νλ λν λ gµνg = g gνµ = δµ . (40) µ The map g(U, ) : Tp R by V g(U, V) is identified with 1-form. Thus, if U are components of a M → 7→ contravariant vector U Tp, then Uµ are the components of a uniquely associated covariant vector, where ∈ ν Uµ = gµνU , (41) such as µ ν µ ν µ ν ν µ g(U, V) = g(U Eµ, V Eν) = U V g(Eµ, Eν) = gµνU V = UνV = U Vµ. (42)
Due to isomorphism between tangent space Tp and cotangent sapce T∗p we have µ µν U = g Uν. (43) In Special Relativity, two separated events with coordinates(t, x, y, z)and(t+dt, x+dx, y+dy, z+dz)in a particular inertial frame have the square of the infinitesimal interval between them given by ds2 = dt2 + dx2 + dy2 + dz2, (44) − which is known as the line element of Minkowski spacetime and is invariant under any Lorentz transformation [1, 2, 3, 4, 5, 6]. We can rewrite this line element in matrix form 1 0 0 0 dt −0 100 dx ds2 = dt dx dy dz , 0 010 dy 0 001 dz 9 or 2 µ ν ds = ηµνdx dx . (45)
It is common to regard (ηµν) as η = diag( 1, 1, 1, 1) whose signature of Minkowski spacetime metric is ( +++). The Minkowski spacetime is denoted by (R4,−η). − Any metric g in four-dimensional differentiable manifold with signatures ( +++) is called Lorentzian metric (a pseudo-Riemannian metric tensor). General Relativity isM based on the concept− of spacetime, which is a four- dimensional differentiable manifold , free of torsion with Lorentzianmetric, denoted by ( , g) or just V4. Thereare others Theories of Gravitation suchM as Einstein-Cartan(-Sciama-Kibble) Theory of GravityM concepted in a Riemann- Cartan spacetime with non-vanishing torsion and curvature. Riemann-Cartan spacetimes are denoted by U4 [16]. In accordance with the concept of line element of Minkowski spacetime (45), if we take an infinitesimal displace- µ ∂ ment dx Tp to and plug into g, we have ∂x µ ∈ M
2 µ ∂ ν ∂ µ ν ∂ ∂ µ ν ds = g dx , dx = dx dx g , = gµν dx dx (46) ∂x µ ∂x ν ! ∂x µ ∂x ν ! that represents the length of the infinitesimal arc determined by the coordinate displacement x µ x µ + dx µ. In spacetime ( , g) where the metric is Lorentzian, the vectors are divided into three classes:→ M (i) g(U, U) > 0, U is spacelike, (ii) g(U, U) < 0, U is timelike, (iii) g(U, U) = 0, U is lightlike or null. Now that a manifold can be endowed with a metric g, we can put some restrictions on the possible form of λ M connections Γ µν. We demand that the metric gµν be covariantly constant, κ κ λgµν = ∂λgµν Γ λµgκν Γ λνgκµ = 0. (47) ∇ − − Then, the affine connection is said to be a metric connection. In a coordinate basis, with (25) and (39), we obtain that ∇ Γ = = Γκ = Γκ λµν g Eλ, Eµ Eν g Eλ, µν Eκ µν gλκ. (48) ∇ In General Relativity the spacetime is V4 where the manifold is free of torsion, then with equation (47) we can obtain the Christoffel relations 1 Γ = ∂ g + ∂ g ∂ g (49) λµν 2 ν λµ µ λν λ µν − for the coordinate components of the connection. From equation (47) we have that ∂λgµν Γνλµ Γµλν = 0, (50) − − where we can rewrite above equation to 1-form language
∂gµν λ λ λ dx Γνλµ dx Γµλν dx = 0 ∂xλ − − or λ λ d(gµν) Γµλν dx Γνλµ dx = 0. − − Introducing the connection 1-forms λ Γµν Γµλν dx , (51) ≡ we have that d(gµν) Γµν Γνµ = 0. (52) − − If we have constant metric or a rigid frame, ∂λgµν = 0, then
Γµν = Γνµ. (53) − We will see in the next sections that the orthonormal and pseudo-orthonormal basis are rigid frames with the connec- tion 1-forms obeying the above propertie. 10 3. Non-coordinate basis
In General Relativity there is a possibility that to set up a system of locally inertial coordinates, valid in a suffi- ciently small region of spacetime. Strictly speaking, this must be an infinitesimal region surrounding at a point p with coordinates xµ. We can denote these local coordinates by yα. We can denote the transformation matrix, which∈M relates the two sets of coordinates - coordinate basis and locally inertial coordinates - by ∂xµ ∂yα e µ(p) = and the inverse ωα (p) = (54) α ∂yα µ ∂xµ If we set up a locally inertial frame of reference at each point of spacetime, in such a way that the directions of their µ µ axes vary smoothly from one point to another, then we obtain a set of four vector fields e0 , , e3 which specify, at each point, the directions of these axes. ··· For instance, a tangent vector at a point p can be expressed as a linear combination of a coordinate basis or ∈ M ∂ holonomic frame in accordance with equation (2). In coordinate basis, Tp is spanned by Eµ = and T∗ by { } (∂xµ ) p dx µ . Now, we can expressa new basis of vectorsin a system of locally inertial coordinates (local Lorentz spacetime) by{ linear} combinations µ eˆα = eα Eµ. (55) for a new basis of Tp and a new dual basis of T∗p,
β β ν θ˜ = ω νdx . (56)
From (55) we can see that new basis eˆα is the frame of basis vectors which is obtained by a rotation of the basis Eµ µ{ } { } preserving the orietation, where eα GL(4, R). The new basis eˆα is called non-coordinate basis of the tangent { } ∈ { } space Tp. ∂ f The directional derivative in the general spacetime coordinate E [ f ] = implies thate ˆ [ f ] also is a directional µ ∂xµ α derivative, from (55) it follows that ∂ f eˆ [ f ] = e µ , (57) α α ∂xµ or eˆα[ f ] = ∂α f, (58) the directional derivative in the local Lorentz spacetime. β Since θ˜ is a new basis of the dual space T of the tangent space Tp in local Lorentz coordinates, we require that { } ∗p β β eˆα, θ˜ = δα , h i where it follows µ β ν µ β ν µ β β eα Eµ, ω νdx = eα ω ν Eµ, dx = eα ω µ = δα , (59) h i h i α µ α µ µ µ where ω µ is the inverse of eα . Also, it is valid that ω νeα = δ ν. The matrix eα is called the tetrad or vierbein µ field. The vierbein field, eα , has two kinds of indices: µ = 1, 2, 3, 4, labels the general spacetime coordinate and ν α = 1, 2, 3, 4 labels the local spacetime. We use κ, λ, µ, ν, ρ, σ, τ, φ, etc to denote the coordinate basis Eµ and dx , β { } { } while α,β,γ,δ,ǫ, etc, to denote the non-coordinate basis eˆα and θ˜ . µ { } { } α From expression (55) wheree ˆα = eα Eµ, we can obtain a reverse expression by using ω ν,
α α µ µ ω νeˆα = ω νeα Eµ = δ ν Eµ that it gives α Eµ = ω µeˆα. (60) α α µ In the same way, the reverse expression for (56), where θ˜ = ω µdx , results that
µ µ α dx = eα θ˜ . (61) 11 ∂ f We can verify that the differential of an arbitrary scalar function f given by d f = dxµ in coordinate basis is ∂xµ given in non-coordinate basis by substituting dxµ by use of the equation (61),
∂ f d f = e µθ˜α, ∂xµ α we can use (57) and (58) where it results in α d f = θ˜ ∂α f. (62) µ ν Let a tensor field of type (1,1) given by T = T ν Eµ dx expressed in non-coordinate basis as ⊗ µ α ν β µ α ν β T = T ν(ω µeˆα) (eβ θ˜ ) = (T νω µ eβ )e ˆα θ˜ ⊗ ⊗ where we have α β T = T βeˆα θ˜ , ⊗ with a way to transit tensors from coordinate basis to non-coordinate basis by
α µ α ν T β = T ν ω µ eβ . (63)
The metric tensor g defined by (38) in terms of non-coordinate basis is
µ ν α β µ ν α µ β ν g = gµνdx dx = gαβω µω νdx dx = gαβ(ω µdx ) (ω νdx ) α ⊗ β ⊗ ⊗ g = gαβθ˜ θ˜ . (64) ⊗
The components of g with respect a coordinate basis eˆα are { }
g(ˆeα, eˆβ) = gαβ (65) where we have
µ ν µ ν µ ν g(ˆeα, eˆβ) = g(eα Eµ, eβ Eν) = eα eβ g(Eµ, Eν) = eα eβ gµν, (66) that follows that µ ν eα eβ gµν = gαβ. (67) In the inverse way it results that α β gµν = ω µω νgαβ. (68) µ The inner product is preserved by vierbein eα that maps the tangent space to local frame. Let U and V two vectors in tangent space, the inner product in coordinate basis is given by (42),
µ ν g(U, V) = gµνU V .
The above inner product in local frame is obtained by substituting (68) into above equation,
α β µ ν α µ β ν α β g(U, V) = ω µω νgαβU V = gαβ(ω µU )(ω νV ) = gαβ U U .
From (55) we can see that the non-coordinate basis eˆα is the frame of basis vectors which is obtained by a rotation { } of the basis Eµ preserving the orietation and inner product. The action{ of} covariant derivative in (59) results in
µ β µ β ( νeα ) ω µ + eα νω µ = 0 (69) ∇ ∇ that it follows β µ µ β ω µ νeα = eα νω µ. (70) ∇ − ∇
12 µ To lower the coordinate indice of the vierbein field, eα , we ca use
µ gµνeα = eαν
α β and when we use (68), gµν = ω µω νgαβ, on above equation, it becomes
β γ µ β µ γ β eαν = ω µω νgβγeα = (ω µeα )(gβγ ω ν) = δ αωβν from which we find eαµ = ωαµ. (71) Now, we can rewrite (68) such as α gµν = ω µωαν, (72) we can see that the vierbein fields can be interpreted as ‘square root’ of metric. From the identification of vectors with directional derivatives we can conclude that the result of the successive application of two vectors to a function depends on the order in which operators are applied. Lete ˆα ande ˆβ two vectors of a non-coordinate basis eˆα , { } ∂ ∂ eˆ = e µ ande ˆ = e ν , α α ∂xµ β β ∂xν then the commutation operation applied to a function f follows
µ ∂ ν ∂ ν ∂ µ ∂ [ˆeα, eˆβ] f = eα eβ eβ eα f ∂xµ ∂xν − ∂xν ∂xµ ! ν 2 µ 2 µ ∂eβ ∂ f µ ν ∂ f ν ∂eα ∂ f ν µ ∂ f = eα + eα eβ eβ eβ eα ∂xµ ∂xν ∂xµ∂xν − ∂xν ∂xµ − ∂xν∂xµ ! ν µ µ ∂eβ ∂ ν ∂eα ∂ = eα eβ f ∂xµ ∂xν − ∂xν ∂xµ ! ν ν µ ∂eβ ∂ µ ∂eα ∂ = eα eβ f ∂xµ ∂xν − ∂xµ ∂xν ! ν ν µ ∂eβ µ ∂eα ∂ = eα eβ f. ∂xµ − ∂xµ ! ∂xν It is util to introduce the commutator coefficients ν ν µ ∂eβ µ ∂eα γ ν eα eβ = D αβ eγ , (73) ∂xµ − ∂xµ ! where the commutator ofe ˆα ande ˆβ becomes ∂ eˆ , eˆ = Dγ e ν , α β αβ γ ∂xν h i or simply γ eˆα, eˆβ = D αβ eˆγ, (74) h i δ similar to Lie algebra. Let us now isolate the commutator coefficients from (73) by using ω ν
γ ν δ δ µ ν µ ν D αβeγ ω ν = ω ν eα ∂µeβ eβ ∂µeα − γ δ δ µ ν µ ν D αβδγ = ω ν eα ∂µeβ eβ ∂µeα − δ δ µ ν µ ν D αβ = ω ν eα ∂µeβ eβ ∂µeα . (75) − Due to the commutator from this equation, it is important to notice that
δ δ D αβ = D βα, (76) − where the commutator coefficients have antisymmetry in the index pair (αβ). 13 3.1. Conection coefficients
In the coordinate basis the connection coefficients are given by (26). Now, let eˆα be the non-coordinate basis and θ˜β the dual basis, thus the connection coefficients with respect to the non-coordinates{ } is obtained in the same way { } α that it is obtained in coordinate basis (25). Thus e eˆγ = Γ βγeˆα where it get ∇ˆβ α α Γ βγ = θ˜ , e eˆγ , (77) h ∇ˆβ i where we can use equations (55) e (56), such that the connection coefficients equation follows
α α µ ρ Γ βγ = ω µdx , e ν E eγ Eρ . (78) h ∇( β ν) i
Let us use the properties (23) of covariant derivative, f XY = f XY, where we can obtain from equation (78) the following result ∇ ∇ α α µ ν ρ Γ βγ = ω µdx , eβ E eγ Eρ , (79) h ∇( ν) i and now we can use the other propertie of covariant derivative (24), X( f Y) = X[ f ]Y + f XY, to rewrite the equation (79) as ∇ ∇ α α µ ν ρ ρ Γ βγ = ω µdx , eβ (Eν[eγ ]Eρ + eγ E Eρ) . (80) h ∇( ν) i σ From identity (25), where E Eρ = Γ νρ Eσ, we can calculate the scalar product ∇( ν) α α ν ρ µ ρ σ µ Γ βγ = ω µeβ (∂ν[eγ ]δ ρ + eγ Γ νρδ σ), (81) that results in α α ν µ ρ µ Γ βγ = ω µeβ (∂νeγ + eγ Γνρ), (82) or α α ν µ Γ βγ = ω µeβ νeγ . (83) ∇ µ In addition, it is important to observe that the νth component of covariant derivative of vierbein field eγ acts only on indice µ of coordinate basis Eµ . { } ρ ρ σ We have seen in equation (28) that E dx = Γ µσdx , similarly we have ∇ µ − γ α γ Γ βαθ˜ = e θ˜ , (84) −∇( ˜β) where the connection coefficients in term of (55) e (56) reduce to,
Γα = α βγ eγ, (eβ)θ −h µ∇ i α ρ = ν eγ Eµ, (eβ Eν)(ω ρdx ) = −h µ ∇ ν α ρ i eγ Eµ, eβ Eν (ω ρdx ) = −h µ ν∇ α ρ i+ α ρ eγ Eµ, eβ (Eν[ω ρ]dx ω ρ Eν dx ) −h µ ν α ρ α ρ∇ σ i = eγ Eµ, eβ (∂νω ρdx ω ρΓνσdx ) = −h µ ν α −α Γσ ρ i eγ Eµ, eβ (∂νω ρ ω σ νρ)dx −h µ ν α ρ − i = eγ eβ νω ρ δµ , (85) − ∇ such that α µ ν α Γ βγ = eγ eβ νω µ. (86) − ∇ It is important to note that from (83), α ν α µ Γ βγ = eβ ω µ νeγ , ∇ α µ = µ α where the underbracedterm can be replaced with aid of the| ide{zntity} (70), where we have that ω µ νeγ eγ νω µ, and consequently the connection coefficients are given by (86). ∇ − ∇
14 4. Torsion tensor in non-coordinate basis and the first Cartan equation The first structure equation of Cartan can be obtained from definition of torsion tensor (32),
T(X, Y) = XY Y X [X, Y], ∇ − ∇ − where the components of torsion tensor in non-coordinate basis follow that α α T βγ = θ˜ , T(ˆeβ, eˆγ) , h i such that α α T βγ = θ˜ , e eˆγ e eˆβ [ˆeβ, eˆγ] . h ∇ˆβ − ∇ˆγ − i With aid of δ e eˆγ = Γ βγeˆδ ∇ˆβ and from (74) where δ [ˆeβ, eˆγ] = D βγ eˆδ, the torsion tensor becomes α α δ δ δ T βγ = θ˜ , Γ βγeˆδ Γ γβeˆδ D βγ eˆδ h − − i from which we find α α α α T βγ = Γ βγ Γ γβ D βγ. (87) − − It is straightforward to notice that the difference between torsion tensor in coordinate basis of equation (34) and torsion α tensor in non-coordinate basis of equation (87) is the presence of commutator coefficient, D βγ. µ We can obtain an expression that make a transition of torsion tensor, T νρ, from coordinate basis to torsion tensor α T βγ in non-coordinate basis. From equation (87) we have α α α α T βγ = Γ βγ Γ γβ D βγ α − ν −µ α ν µ α µ ν µ ν = ω µeγ νeβ + ω µeβ νeγ ω ν eβ ∂µeγ eγ ∂µeβ − ∇ ∇ − − α ν µ ρ µ α ν µ ρ µ α µ ν µ ν = ω µeγ ∂νeβ + eβ Γ νρ + ω µeβ ∂νeγ + eγ Γ νρ ω ν eβ ∂µeγ eγ ∂µeβ − α ν ρ µ α ν ρ µ − − = ω µeγ eβ Γ νρ + ω µeβ eγ Γ νρ −α ν ρ µ µ = ω µeβ eγ Γ νρ Γ ρν , − µ µ µ we have from (34) that T νρ = Γ νρ Γ ρν which it follows − α α ν ρ µ T βγ = ω µeβ eγ T νρ. (88) If a manifold is free of torsion we have from (87) that α α α Γ βγ Γ γβ = D βγ (89) − α α α or 2Γ [βγ] = D βγ. As a consequence, in coordinate basis we have that D βγ = 0 such that α α Γ βγ = Γ γβ. (90)
If the torsion tensor vanishes and from requirement that the metric tensor must be covariantly constant, αgβγ = 0, we have that the cycle permutation of (α, β, γ) yields ∇
αgβγ = ∂αgβγ Γγαβ Γβαγ = 0 ∇ − − βgαγ = ∂βgαγ Γγβα Γαβγ = 0 ∇ − − γgαβ = ∂γgαβ Γβγα Γαγβ = 0 ∇ − − Let us compute αgβγ + βgαγ γgαβ, to obtain an expression for the connection coefficients ∇ ∇ − ∇ ∂αgβγ + ∂βgαγ ∂γgαβ Γγαβ Γβαγ Γγβα Γαβγ + Γβγα + Γαγβ = 0. − − − − − We require that the torsion tensor is vanish, then from equation (89), Γαγβ Γαβγ = Dαγβ, we have − ∂αgβγ + ∂βgαγ ∂γgαβ Γγαβ Γγαβ + Γγαβ Γγβα + Dβγα + Dαγβ = 0, − − − − with Dαγβ = Dαβγ and Dβγα = Dβαγ, the above equation results in − − 1 Γ = ∂ g + ∂ g ∂ g + D D D . (91) γαβ 2 α βγ β αγ γ αβ γαβ αβγ βαγ − − − 15 4.1. The first structure equation of Cartan Let us introduce the torsion 2-form α 1 α β γ T = T βγθ˜ θ˜ . (92) 2 ∧ From equation (87) where we have α α α α T βγ = Γ Γ D βγ, βγ − γβ − we can calculate the torsion two-form (92) as it follows 1 Tα = Γα θ˜β θ˜γ Γα θ˜β θ˜γ Dα θ˜β θ˜γ 2 βγ γβ βγ h ∧ − ∧ − ∧ i 1 α β γ α γ β α β γ = Γ βγθ˜ θ˜ + Γ γβθ˜ θ˜ D βγθ˜ θ˜ 2 ∧ ∧ − ∧ 1 h i = 2Γα θ˜β θ˜γ Dα θ˜β θ˜γ . (93) 2 βγ βγ h ∧ − ∧ i We can compute the second term in above equation by using (75) that follows
1 α β γ 1 α µ ν µ ν β γ D βγθ˜ θ˜ = ω ν eβ ∂µeγ eγ ∂µeβ θ˜ θ˜ 2 ∧ 2 − ∧ α µ ν β γ = ω νeβ ∂µeγ θ˜ θ˜ . (94) ∧ We can notice that ν α α ∂µ(eγ ω ν) = ∂µ(δγ ) = 0, where it results in α ν ν α ω ν∂µeγ = eγ ∂µω ν, − hence, the equation (94) with above identity becomes 1 Dα θ˜β θ˜γ = ∂ ωα e νe µθ˜β θ˜γ = ∂ ωα (e µθ˜β) (e νθ˜γ). 2 βγ µ ν γ β µ ν β γ ∧ − ∧ − ∧ With aid of expression (61) it follows that 1 Dα θ˜β θ˜γ = ∂ ωα (dxµ) (dxν). (95) 2 βγ µ ν ∧ − ∧ Now, if we compute an exterior derivative of expression (56) we have that
α α µ ν dθ˜ = ∂µω ν dx dx . (96) ∧ Hence we have that the equation (95) becomes
1 α β γ α D βγθ˜ θ˜ = dθ˜ , (97) 2 ∧ − where it can be returned in expression (93), that results in 1 Tα = 2Γα θ˜β θ˜γ Dα θ˜β θ˜γ = Γα θ˜β θ˜γ + dθ˜α. (98) 2 βγ βγ βγ h ∧ − ∧ i ∧ We have introduced the conection 1-form in equation (51) in coordinate basis, similarly we have the conection 1-form in non-coordinate basis α α β Γ γ Γ βγθ˜ . (99) ≡ With this conection 1-form in equation of torsion 2-form, we arrive in α α α β T = dθ˜ + Γ β θ˜ first structure equation of Cartan . (100) ∧ If the manifold is free of torsion then it follows that α α β dθ˜ = Γ β θ˜ . (101) − ∧ For non-coordinate basis, the connection coefficients can be computed from (101). 16 5. Curvature tensor in non-coordinate basis and the second Cartan equation
Let us examine the curvature tensor given by (33)
R(X, Y)Z = X Y Z Y X Z X,Y Z. ∇ ∇ − ∇ ∇ − ∇[ ]
From this definition we can obtain the components of curvature tensor in non-coordinate basis with X = eˆγ, Y = eˆδ and Z = eˆβ, such as R(ˆeγ, eˆδ)ˆeβ = e ( δeˆβ) eδ( γeˆβ) e ,e eˆβ, ∇ˆγ ∇ − ∇ˆ ∇ − ∇[ˆγ ˆδ] α with e eˆγ = Γ βγeˆα and (74) it results in ∇ˆβ α α R(ˆeγ, eˆδ)ˆeβ = e (Γ δβeˆα) e (Γ γβeˆα) Dǫ e eˆβ. ∇ˆγ − ∇ˆδ − ∇( γδ ˆǫ ) We can use covariant derivative properties (23) and (24),
f XY = f XY, ∇ ∇ for the third term and X( f Y) = X[ f ]Y + f XY ∇ ∇ for the first and second terms. Thus
α α α α ǫ R(ˆeγ, eˆδ)ˆeβ = eˆγ[Γ δβ]ˆeα + Γ δβ e eˆα eˆδ[Γ γβ]ˆeα Γ γβ e eˆα D γδ e eˆβ, ∇ˆγ − − ∇ˆδ − ∇ˆǫ that results in α α α ǫ α ǫ ǫ α R(ˆeγ, eˆδ)ˆeβ = eˆγ[Γ δβ]ˆeα eˆδ[Γ γβ]ˆeα + Γ δβΓ γαeˆǫ Γ γβΓ δαeˆǫ D γδΓ ǫβeˆα, − − − or = Γα Γα + Γǫ Γα Γǫ Γα ǫ Γα R(ˆeγ, eˆδ)ˆeβ eˆγ[ δβ] eˆδ[ γβ] δβ γǫ γβ δǫ D γδ ǫβ eˆα. n − − − o From equation (35) we have that α R(ˆeγ, eˆδ)ˆeβ = R βγδeˆα, it follows that the components of curvature tensor with respect to non-coordinate basis is given by
α α α ǫ α ǫ α ǫ α R βγδ = eˆγ[Γ δβ] eˆδ[Γ γβ] + Γ δβΓ γǫ Γ γβΓ D γδΓ ǫβ − − δǫ − α α with (58) where we havee ˆγ[Γ δβ] = ∂γΓ δβ
α α α ǫ α ǫ α ǫ α R βγδ = ∂γΓ δβ ∂δΓ γβ + Γ δβΓ γǫ Γ γβΓ δǫ D γδΓ ǫβ. (102) − − − Furthermore, similar to torsion tensor, we have to notice that the principal difference between curvature tensor in coordinate basis of equation (36) and curvature tensor in non-coordinate basis of equation (102) is the presence of ǫ commutator coefficient, D γδ.
5.1. The second structure equation of Cartan Similarly as we have worked with the torsion 2-form, we can introduce the curvature 2-form
α 1 α γ δ Θ β = R βγδθ˜ θ˜ . (103) 2 ∧ α Let us substitute R βγδ from equation (102) into curvature two-form 1 Θα = ∂ Γα θ˜γ θ˜δ ∂ Γα θ˜γ θ˜δ + Γǫ Γα θ˜γ θ˜δ Γǫ Γα θ˜γ θ˜δ Dǫ Γα θ˜γ θ˜δ , β 2 γ δβ δ γβ δβ γǫ γβ δǫ γδ ǫβ h ∧ − ∧ ∧ − ∧ − ∧ i
17 γ δ δ γ α α α with θ˜ θ˜ = θ˜ θ˜ and from equation (89) when the manifold is free of torsion, where Γ βγ Γ γβ = D βγ, we have ∧ − ∧ − 1 Θα = ∂ Γα θ˜γ θ˜δ ∂ Γα ( θ˜δ θ˜γ) + Γǫ Γα θ˜γ θ˜δ Γǫ Γα ( θ˜δ θ˜γ) (Γǫ Γǫ )Γα θ˜γ θ˜δ , β 2 γ δβ δ γβ δβ γǫ γβ δǫ γδ δγ ǫβ h ∧ − − ∧ ∧ − − ∧ − − ∧ i that it follows 1 Θα = 2 ∂ Γα θ˜γ θ˜δ + 2 Γǫ Γα θ˜γ θ˜δ Γα Γǫ θ˜γ θ˜δ + Γα Γǫ ( θ˜δ θ˜γ) , β 2 γ δβ δβ γǫ ǫβ γδ ǫβ δγ h ∧ ∧ − ∧ − ∧ i or
α α γ δ ǫ α γ δ α ǫ γ δ Θ β = ∂γΓ δβθ˜ θ˜ + Γ δβΓ γǫθ˜ θ˜ Γ ǫβΓ γδθ˜ θ˜ α γ ∧ δ α γ ∧ ǫ −δ α ǫ ∧ γ δ = ∂γΓ δβθ˜ θ˜ + (Γ γǫθ˜ ) (Γ δβθ˜ ) Γ ǫβ(Γ γδθ˜ ) θ˜ . ∧ ∧ − ∧ From definition from conection 1-form from (99), we have
α α γ δ α δ ǫ α ǫ Θ β = ∂γΓ δβθ˜ θ˜ + Γ ǫβθ˜ Γ δ +Γ ǫ Γ β. (104) ∧ ∧ ∧
The underbraced terms can be simplifyied| by use of exterior{z derivative} rule (20). If f is a 0-form and ξ is one-form we have that d( f ξ) = (d f ) ξ + f dξ, (105) ∧ ∧ ∧ then α α γ α δ γ α γ dΓ β = d(Γ γβθ˜ ) = (∂δΓ γβ)θ˜ θ˜ + Γ γβdθ˜ . ∧ γ γ β In a manifold free of torsion we have from first Cartan equation (101), dθ˜ = Γ β θ˜ , where it can be inserted into above equation that follows − ∧
α α δ γ α γ δ α γ δ α δ γ dΓ β = (∂δΓ γβ)θ˜ θ˜ + Γ γβ( Γ δ θ˜ ) = (∂γΓ δβ)θ˜ θ˜ + Γ γβθ˜ Γ δ. ∧ − ∧ ∧ ∧ Now we can change the underbraced terms from equation (104) by above result, that it results in the following curvature two-form
α α α γ Θ β = dΓ β + Γ γ Γ β second structure equation of Cartan (106) ∧ The second structure equation of Cartan is an efficient method for calculating the components of curvature tensor (102) with respect to a non-coordinate basis. The second Cartan’s equation gives an algororithm for the calculation of α curvature from the conection coefficients Γ β. 2 α We have seen that the second exterior derivative of an r-form is zero, d Γ β = 0, thus in this sense we can write the Bianchi identity for the curvature 2-form,
α α γ α γ dΘ β Θ γ Γ β + Γ γ Θ β = 0. − ∧ ∧ with aid of exterior derivative property (20).
6. Local Lorentz transformations
m(m+1) In an m-dimensional Riemannian manifold, the metric tensor gµν has 2 degrees of freedom, in 4-dimensional µ 2 spacetime it has 10 degrees of freedom. While the vierbein field eα has m degrees of freedom, in 4-dimensional spacetime it has 16 degrees of freedom. There are many non-coordinate bases which yield the same metric g, for example, the the two main non-coordinate bases are (i) orthonormal basis and (ii) pseudo-orthonormal basis. Each of them is related to the other by local rotation,
α α β θ˜′ (p) =Λ β(p)θ˜ (p) (107) 18 at each point p. From equation (56), where we have,
α α µ θ˜ = ω µdx , and where it yields α α µ α β θ˜′ (p) = ω′ µdx =Λ β(p)θ˜ (p) or α µ α β µ ω′ µdx =Λ β ω µdx where it results in α α β ω′ µ =Λ β ω µ. (108) α Under the local transformation, Λ β, the indices α, β rotated while µ, ν , from coordinate basis (world indices) are not affected. ··· ··· The basis vector of non-coordinate basis in a referential frame is given by O µ eˆα = eα Eµ (109) while = µ eˆα′ eα′ Eµ (110) α α β is given in the referential frame ′. From expression (108) where ω′ µ = Λ β ω µ and from identity (59) where µ β β O eα ω µ = δα , we can obtain µ β = β eα′ ω′ µ δα µ Λβ γ = β eα′ γ ω µ δα (111)
1 δ it can be multiplied by (Λ− ) β, such that it follows
Λ 1 δ Λβ µ γ = Λ 1 δ β ( − ) β γeα′ ω µ ( − ) βδα δ γ µ = Λ 1 δ δ γω µeα′ ( − ) α δ µ = Λ 1 δ ω µeα′ ( − ) α (112)
ν and multiplying by eδ we have that ν δ µ = ν Λ 1 δ eδ ω µeα′ eδ ( − ) α ν µ = ν Λ 1 δ δ µeα′ eδ ( − ) α that results in µ = µ Λ 1 β eα′ eβ ( − ) α. (113) Now, we can return to equation (110), where we have = µ = µ Λ 1 β eˆα′ eα′ Eµ eβ ( − ) α Eµ.
µ γ In local frame , from (109), wheree ˆα = eα Eµ and from (60) where Eµ = ω µeˆγ, the above equation yields O = µ Λ 1 β γ eˆα′ eβ ( − ) α ω µeˆγ µ γ 1 β = eβ ω µeˆγ(Λ− ) α eˆγ = γ Λ 1 β δβ ( − ) α eˆγ so that = Λ 1 β eˆα′ ( − ) α eˆβ. (114) α β Let a tensor field of type (1,1) given by T = T βeˆα θ˜ in local frame . In any rotated local frame ′ it is given by ⊗ O O α β T′ = T ′ βeˆ′ θ˜ . α ⊗ ′ 19 α α β It is possible to obtain a rule how this tensor transforms by use of equation (107) where θ˜′ =Λ βθ˜ and the equation = Λ 1 β (114) wheree ˆα′ ( − ) α eˆβ, with respective inverses ˜α = Λ 1 α ˜ β =Λβ θ ( − ) βθ′ ande ˆα α eˆβ′ . (115)
α β Thus the tensor T = T βeˆα θ˜ in a local frame is related to the other by local rotation by ⊗ O α β T = T βeˆα θ˜ α ⊗γ 1 β δ = T β Λ α eˆ′ (Λ− ) θ˜′ γ ⊗ δ = Λγ h α Λ i1 β h ˜ δ i αT β( − ) δ eˆγ′ θ′ h i ⊗ from which we find the transformation rule,
α =Λα γ Λ 1 δ T ′ β γT δ( − ) β. (116)
This transformation rule for tensor field of type (1,1) can be written in tensor form,
1 T′ = ΛTΛ− . (117)
1 The upper non-coordinate indices are rotated by Λ while lower non-coordinate indices are rotated by Λ− . However not all objects with indices are the components of a tensor, an important example is provided by the transformation α properties of the connection coefficients Γ βγ determined by (77). The connection coefficients under the rotation is Γ α = ˜ α = Λα ˜δ Λ 1 ζ ′ βγ θ′ , eˆ eˆ′ δθ , Λ 1 ǫ ( − ) γeˆζ , h ∇ β′ γi h ∇[( − ) βeˆǫ ] i withe properties (23), (24) and (58) it follows that
Γ α = Λα ˜δ Λ 1 ǫ Λ 1 ζ ′ βγ δ θ , ( − ) β eˆǫ ( − ) γeˆζ α h 1 ǫ δ ∇ 1 ζ i 1 ζ η = Λ δ(Λ− ) θ˜ , ∂ǫ(Λ− ) eˆζ + (Λ− ) Γ ǫζ eˆη βh γ γ i α 1 ǫ h 1 ζ δ 1 ζ η δ i = Λ δ(Λ− ) β ∂ǫ(Λ− ) γδ ζ + (Λ− ) γΓ ǫζ δ η = Λα Λ 1 ǫ h Λ 1 δ +Λα Λ 1 ǫ Λ 1 ζ Γi δ δ( − ) β∂ǫ( − ) γ δ( − ) β( − ) γ ǫζ ,
Because of the first term on the right-hand side of above equation, the connection coefficients do not transform as the α α β components of a tensor. We can see how to transform the connection 1-form Γ γ by calculate Γ′ βγθ˜′ , Γ α ˜ β =Λα Λ 1 ǫ ˜ β Λ 1 δ +Λα Λ 1 ζ Γδ Λ 1 ǫ ˜ β ′ βγθ′ δ( − ) βθ′ ∂ǫ( − ) γ δ( − ) γ ǫζ ( − ) βθ′ ,
1 α β α with transformation (Λ− ) βθ˜′ = θ˜ from (115) it follows that
α =Λα ˜ǫ Λ 1 δ +Λα Λ 1 ζ Γδ ˜ǫ Γ′ γ δθ ∂ǫ( − ) γ δ( − ) γ ǫζ θ ,
˜ǫ 1 δ 1 δ we can use (62), where θ ∂ǫ(Λ− ) γ = d(Λ− ) γ, to reduce the above equation to
α α 1 δ α δ 1 ζ Γ′ γ =Λ δd(Λ− ) γ +Λ δΓ ζ (Λ− ) γ. (118)
α Also, the above transformation rule of the connection 1-form Γ β, can be obtained by rotate the torsion 2-form from first Cartan equation (100), α α α β T′ = dθ˜′ + Γ′ β θ˜′ . ∧ α α γ On the left-hand side of the above equation we have T′ =Λ γ T , while on the right-hand side we can apply (107),
α γ α γ α β γ Λ γ T = d(Λ γθ˜ ) + Γ′ β (Λ γ θ˜ ) α γ γ β α γ α ∧ γ α β γ Λ γ dθ˜ + Γ β θ˜ = dΛ γ θ˜ +Λ γdθ˜ + Γ′ β Λ γ θ˜ α γ ∧ β α ∧ γ α β γ ∧ Λ γΓ β θ˜ = dΛ γ θ˜ + Γ′ β Λ γ θ˜ . ∧ ∧ ∧ 20 α We can isolate Γ′ β to obtain α β γ α β γ α γ Γ′ βΛ γ θ˜ =Λ βΓ γ θ˜ dΛ γ θ˜ ∧ ∧ − ∧ where it reduces to α β α β α Γ′ βΛ γ =Λ βΓ γ dΛ γ, − 1 and multiplying both sides by Λ− from the right, we have α β 1 γ α β 1 γ α 1 γ Γ′ βΛ γ(Λ− ) =Λ βΓ γ(Λ− ) (dΛ γ)(Λ− ) . δ δ − δ We can use α Λβ Λ 1 γ = α β = α Γ′ β γ( − ) δ Γ′ β δ δ Γ′ δ and
α 1 γ α 1 γ α 1 γ d Λ γ)(Λ− ) δ = (dΛ γ)(Λ− ) δ +Λ γd(Λ− ) δ, h αi = Λα Λ 1 γ +Λα Λ 1 γ dδ δ (d γ)( − ) δ γd( − ) δ where it follows that α 1 γ α 1 γ (dΛ γ)(Λ− ) = Λ γd(Λ− ) . (119) δ − δ α Thus we can express Γ′ β as α α γ 1 δ α 1 γ Γ′ β =Λ γΓ δ(Λ− ) β +Λ γd(Λ− ) β. (120) Thus, the above result is the same one obtained in equation (118). It is easy to notice from expression (116), that the curvature two-form transforms as
α α γ 1 δ Θ′ β =Λ γΘ δ(Λ− ) β, (121) under a local frame transformation Λ.
7. The non-coordinate orthonormal basis General Relativity is based on the concept of spacetime, which is a 4-dimensional differentiable manifold with Lorentzian metric, denoted by ( , g) where the components of the metric tensor g of spacetime with respectM a ∂ M coordinate basis Eµ = are given by (39), ( ∂xµ )
g(Eµ, Eν) = gµν. However, for many purposes it is more convenient to use a non-coordinate orthonormal basis, or orthonormal tetrad or Lorentz frame eˆα , where we have: { } g(ˆeα, eˆβ) = ηαβ, (122) where (ηαβ) = diag( 1, 1, 1, 1) are the components of metric tensor of Minkowski reference system. The relationship between coordinate− basis with non-coordinate orthonormal basis is given by (66), where
µ ν g(ˆeα, eˆβ) = eα eβ gµν, that follows that µ ν ηαβ = eα eβ gµν. (123) In the inverse way it results that α β gµν = ω µω νηαβ. The metric tensor given by (64) is
α β g = ηαβθ˜ θ˜ = (θ˜0 ⊗θ˜0) + (θ˜1 θ˜1) + (θ˜2 θ˜2) + (θ˜3 θ˜3). (124) − ⊗ ⊗ ⊗ ⊗ The metric tensor in non-coordinate orthonormal basis is a system of locally inertial coordinates or local Lorentz spacetime. 21 7.1. The Ricci rotation coefficients In the non-coordinate orthonormal basis the Minkowskian metric tensor is constant and consequently this local reference system is a rigid frame with ∂γηαβ = 0, then we can review equation (91) where we have, 1 Γ = D D D . (125) γαβ 2 γαβ αβγ βαγ − − we can interchange γ β into above equation such as ↔ 1 Γβαγ = Dβαγ Dαγβ Dγαβ 2 − − 1 = D + D D , 2 γαβ αγβ βαγ − − where we can use equation (76) for the second term in right hand side of above equation, Dαγβ = Dαβγ, that it follows, − 1 Γ = D D D . βαγ 2 γαβ αβγ βαγ − − − Now we can compare above equation with (125) and obtain that
Γγαβ = Γβαγ. (126) − The connection coefficients in non-coordinate orthonormal basis also referred as the Ricci rotation coefficients [5, ffi 2 (n 1) 8, 12]. The antisymmetry of the Ricci rotation coe cients in the index pair (βγ) yields n −2 , in 4-dimensional 4 spacetime V , we have 24 components Γγαβ. The matrix of Ricci rotation coefficients (Γαγβ)is4 4 4 matrix and we can see this in 4 layers × × 0 Γ001 Γ002 Γ003 Γ001 0 Γ102 Γ103 Γα0β = − Γ002 Γ102 0 Γ203 − − Γ003 Γ103 Γ203 0 − − − 0 Γ011 Γ012 Γ013 Γ Γ Γ Γ = 011 0 112 113 α1β −Γ Γ 0 Γ 012 112 213 −Γ −Γ Γ 0 013 113 213 − − − 0 Γ021 Γ022 Γ023 Γ Γ Γ Γ = 021 0 122 123 α2β −Γ Γ 0 Γ 022 122 223 −Γ −Γ Γ 0 023 123 223 − − − 0 Γ031 Γ032 Γ033 Γ Γ Γ Γ = 031 0 132 133 α3β −Γ Γ Γ 032 132 0 233 −Γ −Γ Γ 033 133 233 0 − − − We can see the 24 terms, 6 in each matrix layer. µ In a coordinate basis the Ricci rotation coefficients are substituted by the Christoffel symbols Γ νρ that are sym- Γµ = Γµ 2 (n+1) 4 metric in the index pair (νρ), νρ ρν. It yields n 2 terms, where in in 4-dimensional manifold V , there are 40 α components Γ βγ. Thus, we have advantage in use of non-coordinateorthonormal basis for calculating curvature. The Cartan’s method for the calculation of curvature is more compact and efficient in many applications. The calculation of the connection 1-forms (51) are obtained from (126), γ γ Γαγβθ˜ = Γβγαθ˜ − Γαβ = Γβα, (127) − with the first Cartan equation determine Γαβ uniquely. In a non-coordinate orthonormal basis in 4-dimensional mani- fold, at most six independent connection 1-forms survive. 22 7.2. The first equation of Cartan In a non-coordinate orthonormal basis the first Cartan equation yields a system of four equations
dθ˜0 = Γ0 θ˜1 + Γ0 θ˜2 + Γ0 θ˜3 − 1 ∧ 2 ∧ 3 ∧ dθ˜1 = Γ1 θ˜0 + Γ1 θ˜2 + Γ1 θ˜3 0 2 3 (128) − 2 2 ∧ 0 2 ∧ 1 2 ∧ 3 dθ˜ = Γ 0 θ˜ + Γ 1 θ˜ + Γ 3 θ˜ − ∧ ∧ ∧ ˜3 = Γ3 ˜0 + Γ3 ˜1 + Γ3 ˜2 dθ 0 θ 1 θ 2 θ . − ∧ ∧ ∧ There are 12 Ricci rotation coefficients in above system, but with the antisymmetry of the connection 1-forms (127), Γβα = Γαβ the Ricci rotation coefficients are reduced to 6. Let α = 0 (timelike) and β = i = 1, 2, 3 (spacelike), where − we have Γ0i = Γi0, thus, − 0 00 0 Γ i = η Γ i = ( 1)Γ i or Γ i = Γi . (129) 0 − 0 0 Moreover, we have that i i j i j Γ 0 = η Γ j0 = δ Γ j0, (130) for i = 1, it yields 1 Γ 0 = Γ10, (131) 0 with respect to (129) where Γ 1 = Γ10 we have that
1 0 Γ 0 = Γ 1. (132)
The next step, let α = 1e β = 2 (both spacelike), and from (127), we have Γ = Γ , 12 − 21 1 1k 11 Γ 2 = η Γk2 = δ Γ12 = Γ12. (133)
For Γ21 we have 2 2k 22 Γ = η Γk = δ Γ = Γ = Γ , (134) 1 1 21 21 − 12 comparing the two above expression we have that
Γ1 = Γ2 . (135) 2 − 1 With these results we can write the system of equations (128) as
0 0 1 0 2 0 3 dθ˜ = Γ 1 θ˜ + Γ 2 θ˜ + Γ 3 θ˜ − 1 0 ∧ 0 1 ∧ 2 1 ∧ 3 dθ˜ = Γ 1 θ˜ + Γ 2 θ˜ + Γ 3 θ˜ − ∧ ∧ ∧ (136) dθ˜2 = Γ0 θ˜0 Γ1 θ˜1 + Γ2 θ˜3 2 2 3 − 3 0 ∧ 0 − 1 ∧ 1 2 ∧ 2 dθ˜ = Γ 3 θ˜ Γ 3 θ˜ Γ 3 θ˜ − ∧ − ∧ − ∧ 0 0 0 1 1 2 with six independent connection 1-forms: Γ 1, Γ 2, Γ 3, Γ 2, Γ 3 e Γ 3 to be determined.
7.3. The second equation of Cartan For the calculation of curvature we use the second Cartan equation from (106),
γ Θαβ = dΓαβ + Γαγ Γ β ∧ where the curvature 2-forms Θαβ are antisymmetric in the index pair αβ. From (127) we have that
γ γ γ γ Θαβ = dΓβα Γγα Γ β = dΓβα Γ α Γγβ = dΓβα + Γγβ Γ α = dΓβα Γβγ Γ α, − − ∧ − − ∧ − ∧ − − ∧ it follows that Θαβ = Θβα. (137) − 23 The antisymmetry of the curvature 2-forms, Θαβ, yields a system of six equations
Θ = dΓ + Γ Γ2 + Γ Γ3 01 01 02 ∧ 1 03 ∧ 1 Θ = dΓ + Γ Γ1 + Γ Γ3 02 02 01 ∧ 2 03 ∧ 2 = + 1 + 2 Θ03 dΓ03 Γ01 Γ 3 Γ02 Γ 3 ∧ ∧ Θ = dΓ + Γ Γ0 + Γ Γ3 12 12 10 2 13 2 ∧ 0 ∧ 2 Θ13 = dΓ13 + Γ10 Γ 3 + Γ12 Γ 3 ∧ ∧ Θ = dΓ + Γ Γ0 + Γ Γ1 23 23 20 ∧ 3 21 ∧ 3 0 0 0 1 1 2 We have seen that there are six independent connection 1-forms: Γ 1, Γ 2, Γ 3, Γ 2, Γ 3 e Γ 3 to be determined. With 1 0 1 2 conditions (132), Γ 0 = Γ 1 = Γ01, and (135), Γ 2 = Γ 1 = Γ12, the above system of equations from second Cartan equation becomes − − Θ = dΓ Γ Γ Γ Γ 01 01 − 02 ∧ 12 − 03 ∧ 13 Θ = dΓ + Γ Γ Γ Γ 02 02 01 ∧ 12 − 03 ∧ 23 Θ = Γ + Γ Γ + Γ Γ 03 d 03 01 13 02 23 ∧ ∧ (138) Θ12 = dΓ12 + Γ01 Γ02 Γ13 Γ23 ∧ − ∧ Θ = dΓ + Γ Γ + Γ Γ 13 13 01 ∧ 03 12 ∧ 23 Θ = Γ + Γ Γ Γ Γ 23 d 23 02 03 12 13. ∧ − ∧ Now let us see an example of Cartan’s formalism applied to General Relativity.
7.4. Example of Schwarzschild spacetime When the non-null components of metric tensor g are only diagonal components, it is practical to work with the non-coordinate orthonormal basis, as the case may be Schwarzschild spacetime. In coordinate basis, the metric of the Schwarzschild spacetime is given by
2M 1 ds2 = 1 dt2 + dr2 + r2(dθ2 + sin2 θdφ), (139) − − r ! 1 2M − r where the parameters run over the range r > 2M, 0 θ<π and 0 φ < 2π. The metric of the Schwarzschild spacetime is already in diagonal form, thus it is easy to≤ rewrite it in non-coordinate≤ orthonormal basis in accordance with (124) 2 α β 0 0 1 1 2 2 3 3 ds = ηαβθ˜ θ˜ = θ˜ θ˜ + θ˜ θ˜ + θ˜ θ˜ + θ˜ θ˜ . ⊗ − ⊗ ⊗ ⊗ ⊗ From (56) we have that
0 0 µ 1 1 µ 2 2 µ 3 3 µ θ˜ = ω µdx , θ˜ = ω µdx , θ˜ = ω µdx and θ˜ = ω µdx . (140)
α β and from (68) where gµν = ω µω νηαβ, it follows that
0 0 1 1 2 2 3 3 gµν = ω µω νη00 + ω µω νη11 + ω µω νη22 + ω µω νη33 0 0 1 1 2 2 3 3 = ω µω ν + ω µω ν + ω µω ν + ω µω ν, − the diagonal terms of gµν
0 2 1 2 2 2 3 2 gtt = (ω t) , grr = (ω r) , gθθ = (ω θ) and gφφ = (ω φ) where the vierbein fields are interpreted as square root of metric and it results in
0 2M 1 1 2 3 ω t = 1 , ω r = , ω θ = r and ω φ = r sin θ. (141) r − r 2M 1 r q − 24 Thus the 1-form terms from (140) become
2M dr θ˜0 = dt 1 , θ˜1 = , θ˜2 = rdθ and θ˜3 = r sin θdφ. (142) r − r 2M 1 r q − The inverse terms are
θ˜0 2M θ˜2 θ˜3 dt = , dr = θ˜1 1 , dθ = and dφ = . (143) 2M r − r r r sin θ 1 r q − With (142) and (143) we can calculate 2-form dθ˜α, dθ˜0 results in • M/r2 M/r2 dθ˜0 = (dr dt) = θ˜1 θ˜0, 2M ∧ 2M ∧ 1 r 1 r q − q − dθ˜1 results in • M/r2 dθ˜1 = − (dr dr) = 0, 2M 3/2 ∧ 1 r − dθ˜2 results in • 2M 1 r dθ˜2 = dr dθ = q − θ˜1 θ˜2, ∧ r ∧ dθ˜3 results in • 2M 1 r cot θ dθ˜3 = sin θdr dφ + r cos θdθ dφ = q − θ˜1 θ˜3 + θ˜2 θ˜3. ∧ ∧ r ∧ r ∧ With these results we can solve the first structure equation of Cartan as the system of equations (136),
2 M/r 1 0 0 1 θ˜ θ˜ = Γ 1 θ˜ , − 2M ∧ ∧ 1 r q − 0 = Γ0 θ˜0 + Γ1 θ˜2 + Γ1 θ˜3, 1 ∧ 2 ∧ 3 ∧ 2M 1 r q − θ˜1 θ˜2 = Γ1 θ˜1, − r ∧ − 2 ∧ 2M 1 r cot θ q − θ˜1 θ˜3 θ˜2 θ˜3 = Γ1 θ˜1 Γ2 θ˜2. − r ∧ − r ∧ − 3 ∧ − 3 ∧ Thus, we have four independent connection 1-forms
0 M 0 Γ 1 = θ˜ = Γ01, 2 2M − r 1 r q − 2M 1 r Γ1 = q − θ˜2 = Γ , 2 − r 12 2M 1 r Γ1 = q − θ˜3 = Γ , 3 − r 13 cot θ Γ2 = θ˜3 = Γ , 3 − r 23 25 0 0 α with Γ 2 = Γ 3 = 0. Now we can calculate dΓ β with objective to solve the second structure equation of Cartan. The 0 calculation of 2-form dΓ 1 follows that
2 0 2M M 2M/r 0 M 0 dΓ 1 = · dr θ˜ + dθ˜ 2M 3/2 −r3 1 2M − 2r2 1 ∧ r2 1 2M r r r q − − q − M M 2M M M = + 2 1 θ˜1 θ˜0 + θ˜1 θ˜0 − 2M 2M r − r ∧ 2M · 2M ∧ r3 1 r 1 r r2 1 r2 1 r − r r q − q − q − M M M2 = + 2 + θ˜1 θ˜0 3 M − r r 2M r4 1 2 ∧ − − r 2M = θ˜0 θ˜1 r3 ∧ or 2M dΓ = θ˜0 θ˜1 (144) 01 − r3 ∧ 1 The calculation of 2-form dΓ 2 follows that
1 2M 1 2M 1 − r 2 − r 2 dΓ 2 = d q θ˜ q dθ˜ − r ∧ − r 2M 2M 2M 2M/r2 1 r 1 r = 1 r2 dr θ˜2 q − q − θ˜1 θ˜2 r − r − ∧ − r · r ∧ 2r 1 2M − r q 2M 2M 1 r M 2M 1 = q − 1 − r θ˜1 θ˜2 r2 − r − r − r2 ∧ r3 1 2M − r M q = θ˜1 θ˜2 − r3 ∧ or M dΓ = θ˜1 θ˜2. (145) 12 − r3 ∧ 1 The calculation of 2-form dΓ 3 follows that
2M 2M 1 M 1 r dΓ1 = − r θ˜1 θ˜3 q − dθ˜3 3 r2 − r3 ∧ − r 2M 2M 2M 1 M 1 r 1 r cot θ = − r θ˜1 θ˜3 q − q − θ˜1 θ˜3 + θ˜2 θ˜3 , r2 − r3 ∧ − r r ∧ r ∧ where it reduces to 2M M 1 r dΓ = θ˜1 θ˜3 q − cot θ θ˜2 θ˜3. (146) 13 − r3 ∧ − r2 ∧ 2 The calculation of 2-form dΓ 3 follows that
2 cot θ 3 cot θ 3 dΓ 3 = d θ˜ dθ˜ , − r ∧ − r
26 where it results is 1 dΓ = θ˜2 θ˜3. (147) 23 −r2 ∧ The curvature 2-forms are founf from the second structure Cartan equation (138). 2M Θ = dΓ Γ Γ Γ Γ = θ˜0 θ˜1, 01 01 − 02 ∧ 12 − 03 ∧ 13 − r3 ∧ M Θ = dΓ + Γ Γ Γ Γ = θ˜0 θ˜2, 02 02 01 ∧ 12 − 03 ∧ 23 r3 ∧ M Θ = dΓ + Γ Γ + Γ Γ = θ˜0 θ˜3, 03 03 01 ∧ 13 02 ∧ 23 r3 ∧ M Θ = dΓ + Γ Γ Γ Γ = θ˜1 θ˜2, 12 12 01 ∧ 02 − 13 ∧ 23 − r3 ∧ M Θ = dΓ + Γ Γ + Γ Γ = θ˜1 θ˜3, 13 13 01 ∧ 03 12 ∧ 23 − r3 ∧ 2M Θ = dΓ + Γ Γ Γ Γ = θ˜2 θ˜3. 23 23 02 ∧ 03 − 12 ∧ 13 r3 ∧ We have introduced the curvature two-form given by equation (103),
1 γ δ Θαβ = Rαβγδθ˜ θ˜ . 2 ∧ Thus we can obtain the Riemann curvature tensor in non-coordinate orthonormal basis. For 2M Θ = θ˜0 θ˜1, 01 − r3 ∧ we have 1 1 2M R θ˜0 θ˜1 + R θ˜1 θ˜0 = θ˜0 θ˜1, 2 0101 ∧ 2 0110 ∧ − r3 ∧ that results in 2M R = . 0101 − r3 In the same way we have for others components of Riemann tensor, M R = , 0202 r3 M R = , 0303 r3 M R = , 1212 − r3 M R = , 1313 − r3 2M R = . 2323 r3 Now we can compute the components of Ricci tensor
γ αγ Rβδ = R βγδ = η Rαβγδ.
The components of above Riemann tensor show that we have readily Rβδ = 0 with β , δ, however it is possible to compute R00, R11, R22 and R33, but they are null as it follows 2M M M R = R1 + R2 + R3 = + + = 0, 00 010 020 030 − r3 r3 r3 2M M M R = R0 + R2 + R3 =+ = 0, 11 101 121 131 r3 − r3 − r3 27 M M 2M R = R0 + R1 + R3 = + = 0, 22 202 212 232 − r3 − r3 r3 M M 2M R = R0 + R1 + R2 = + = 0. 33 303 313 323 − r3 − r3 r3 = = 1 = Thus Rαβ 0 and consequently we have that the Einstein tensor Gαβ Rαβ 2 gαβR 0, that is a vacuum solution. In coordinate basis or world indices we can obtain that −
α β Rµν = ω µω ν Rαβ (148) where clearly Rµν = 0. However we can se that the non null components of Riemann tensor in coordinate basis are obtained from α β γ δ Rµνρσ = ω µω νω ρω σ Rαβρσ, (149) where with aid of (141). It follows that
2 2 2M 2M R = ω0 ω1 ω0 ω1 R = ω0 ω1 = , trtr t r t r 0101 t r r3 r3 − ! −
2 2 M 2M M M 2M R = ω0 ω2 ω0 ω2 R = ω0 ω2 = 1 r2 = 1 , tθtθ t θ t θ 0202 t θ r3 r r3 r r − ! − ! 2 2 M 2M M M 2M R = ω0 ω3 ω0 ω3 R = ω0 ω2 = 1 r2 sin2 θ = 1 sin2 θ, tφtφ t φ t φ 0303 t φ r3 r r3 r r − ! − ! 2 1 1 2 1 2 1 2 2 2 M r M M 2M − Rrθrθ = ω rω θω rω θ R1212 = ω r ω θ − = = 1 , r3 − 2M r3 − r − r ! 1 r − 2 2 1 1 3 1 3 1 2 3 2 M r sin θ M M 2M − 2 Rrφrφ = ω rω φω rω φ R1313 = ω r ω φ − = = 1 sin θ, r3 − 1 2M r3 − r − r ! − r 2 2 2M 2M R = ω2 ω3 ω2 ω3 R = ω2 ω3 = r4 sin2 θ = 2Mr sin2 θ. θφθφ θ φ θ φ 2323 θ φ r3 r3 ! After some practices, we can see that the non-coordinate orthonormal basis to advance in an easy method to calculate the Riemann tensor of spacetimes that have the diagonal shape in the metric tensor g in coordinate basis.
8. Newman-Penrose null tetrad - pseudo-orthonormal basis
A pseudo-orthonormalbasis or null tetrad formalism is due to Newman and Penrose [14]. It has provedvery useful in the construction of exact solutions on General Relativity. This formalism is adapted to treatment of the propagation of radiation in spacetime as will be seen at the end of this section with the example of Brinkmann metric. We can start with metric of Minkowski spacetime in orthonormal basis where the line element is
ds2 = dt2 + dx2 + dy2 + dz2. − An alternative coordinate system for Minkowski spacetime can be through the introduction of advanced and retarded null coordinates, u and v with the following coordinate transformations 1 1 u = (t x) and v = (t + x) , (150) √2 − √2 so that the line element of Minkowski spacetime can be
ds2 = 2dudv + dy2 + dz2. −
28 = = dx = With these null coordinates, we can see that if u is constant, we have du 0, then dx dt or dt 1, the speed of light. The null coordinate u may be thought of as light ray at the speed of light in direction +x, while the null coordinate v may be thought of as light ray at the speed of light in direction x. These null coordinates, v and u are orthogonal to the yz-plane. In the similar way, it is convenient to parametrise t−he yz-plane in terms of the complex coordinate 1 1 ζ = (y + iz) , ζ¯ = (y iz) , (151) √2 √2 − thus we can see that in terms of these coordinate transformations the line element ds2 of Minkowski spacetime is
ds2 = 2dudv + 2dζdζ.¯ (152) − Now we can express the 1-form terms of ds2 as
θ˜0 = dv, θ˜1 = du, θ˜2 = dζ, θ˜3 = dζ,¯ where we have the metric tensor g in pseudo-orthonormal coordinate basis given by
g = 2θ˜0 θ˜1 + 2θ˜2 θ˜3 (153) − ⊗ ⊗ that in matrix form is 0 1 0 0 θ˜0 − ˜1 = ˜0 ˜1 ˜2 ˜3 1 0 00 θ g θ θ θ θ − ˜2 0 0 01 θ ˜3 0 0 10 θ or α β g = γαβθ˜ θ˜ (154) ⊗ since 0 1 0 0 1− 0 00 (γ ) = (155) αβ −0 0 01 0 0 10 γαβ is defined the components of metric tensor with respect to the non-coordinate pseudo-orthonormal basis. Again, from a tangent space Tp spanned by Eµ in coordinate basis, we can obtain the non-coordinate pseudo- orthonormal basis by a rotation. From equation (55){ } we have
µ eˆα = eα Eµ,
Thus we have a tetrad,
µ µ eˆ0 = e0 Eµ = k Eµ = k, eˆ = e µ E = lµ E = l, 1 1 µ µ (156) µ µ eˆ2 = e2 Eµ = m Eµ = m, eˆ = e µ E = m¯ µ E = m¯ . 3 3 µ µ The above set of complex null vectors eˆα = k, l, m, m¯ is called a Newman-Penrose null tetrad. This is usually abbreviated to NP null tetrad [13]. The reason{ } for{ it designa} tion is because the condition (67), where we have,
µ ν eα eβ gµν = γαβ.
It is straightforward to notice that the diagonal terms of metric tensor γαβ from (155) are zero. For Example
µ ν e0 e0 gµν = γ00 = 0, 29 where µ ν µ k k gµν = k kµ = 0. And the same way the other three contractions are null,
µ µ µ l lµ = 0, m mµ = 0 andm ¯ m¯ µ = 0
The non-null contractions are given by µ ν e0 e1 gµν = γ01, µ that results in k lµ = 1 and − µ ν e2 e3 gµν = γ23, µ that results m m¯ µ = 1. The relationships (150) and (151), in flat spacetime implies that, ∂ ∂ ∂ ∂ k = , l = , m = andm ¯ = , (157) ∂v ∂u ∂ζ ∂ζ¯ where we see that m andm ¯ are complex conjugates. In coordinate basis we can see from (150) and (151) that,
1 1 0 0 1 1 1 1 1 0 1 0 kµ = , lµ = , mµ = , andm ¯ µ = , (158) 0 −0 1 1 √2 √2 √2 √2 0 0 i i − From equation (59) β µ β β eˆα, θ˜ = eα ω µ = δα , h i we can identify that 0 µ 0 µ 0 eˆ , θ˜ = e ω µ = k ω µ = 1, h 0 i 0 µ 0 where from condition k lµ = 1, the above equation results for ω µ = lµ. In the same way we have − − µ 1 µ 1 e1 ω µ = l ω µ = 1,
µ 1 that from kµl = 1, it results in ω µ = kµ. For the others terms we have − − µ 2 µ 2 2 µ e ω µ = m ω µ = 1 ω µ = m¯ , 2 ⇒ and µ 3 µ 3 3 µ e ω µ = m¯ ω µ = 1 ω µ = m . 3 ⇒ α α µ So, as θ˜ = ω µdx , we have that the differential 1-forms, the basis for cotangent space in pseudo-orthonormal coordinate, are 0 0 µ µ θ˜ = ω µdx = lµdx − θ˜1 = ω1 dxµ = k dxµ µ µ (159) 2 2 µ − µ θ˜ = ω µdx = m¯ µdx θ˜3 = ω3 dxµ = m dxµ µ µ In coordinate basis, from (158) we have 1 1 0 0 1 −1 1 −1 1 0 1 0 k = , l = , m = , andm ¯ = , (160) µ 0 µ −0 µ 1 µ 1 √2 √2 √2 √2 0 0 i i − µ µ where we have the only nonzero contractions k lµ = 1 and m m¯ µ = 1. − 30 Now we can see the way to write the components of metric tensor in coordinate basis from the terms lµ, kµ, mµ andm ¯ µ. Recall equation (68), α β gµν = ω µω νγαβ, it results in
0 1 1 0 2 3 3 2 gµν = ω µω νγ01 + ω µω νγ10 + ω µω νγ23 + ω µω νγ32 = ( lµ)( kν)( 1) + ( kµ)( lν)( 1) + (¯mµ)(mν)(1) + (mµ)(¯mν)(1) − − − − − − = (lµkν + lνkµ) + (¯mµmν + m¯ νmµ), − where we can simplify the symmetrization of index pairs by round brackets
gµν = 2k µlν + 2m µm¯ ν . (161) − ( ) ( ) 8.1. Ricci rotation coefficients We have seen that in orthonormal basis there are 24 Ricci rotation coefficients. But in pseudo-orthonormal basis, because m andm ¯ are complex conjugates, the number of Ricci rotation coefficients reduces to 16 coefficients. The Ricci rotation coefficients are given by Γαβγ = g(ˆeα, e eˆγ). ∇ˆβ In terms of NP null tetrad we have for Γ100 the following calculation
µ ρ µ ν ρ µ ν ρ µ ν ρ Γ = g(ˆe , e eˆ ) = g(l Eµ, kν E k Eρ) = l k g(Eµ, νk Eρ) = l k νk g(Eµ, Eρ) = l k νk gµρ, 100 1 ∇ˆ0 0 ∇( ν) ∇ ∇ ∇ it follows that µ ν Γ = l k νkµ. 100 ∇ µ µ Notice that Γ100 = Γ100 because l and k are real numbers. Similarly to orthonormal basis, the pseudo-orthonormal basis is a rigid frame with ∂δγαβ = 0, so the Ricci rotation coefficients are given by (125) and results in (126), µ ν Γγαβ = Γβαγ, such as Γ = Γ = k k νlµ. − 100 − 001 − ∇ The calculation for Γ200 is
µ ρ µ ν Γ = g(ˆe , e eˆ ) = g(m Eµ, kν E k Eρ) = m k νkµ. 200 2 ∇ˆ0 0 ∇( ν) ∇ µ ν and for Γ = m¯ k νkµ, where we ca see that 300 ∇ µ ν Γ = m k νkµ = Γ . 300 ∇ 200
The calculation for Γ101 is
µ ρ µ ν Γ = g(ˆe , e eˆ ) = g(l Eµ, kν E l Eρ) = l k νlµ. 101 1 ∇ˆ0 1 ∇( ν) ∇ where we observe that Γ101 = Γ101 . µ ν µ ν We have Ricci rotation coefficient Γ = l k νmµ and Γ = l k νm¯ µ, where we can see Γ = Γ . With 102 ∇ 103 ∇ 103 102 these calculations we can recognize the 24 Ricci rotation coefficient Γαβγ as
µ ν µ ν µ ν µ ν Γ100 = l k νkµ Γ200 = m k νkµ Γ300 = m¯ k νkµ Γ201 = m k νlµ µ ν∇ µ ν ∇ µ ν ∇ µ ν ∇ Γ301 = m¯ k νlµ Γ302 = m¯ k νmµ Γ110 = l l νkµ Γ210 = m l νkµ µ ν ∇ µ ν∇ µ ν∇ µ ν ∇ Γ310 = m¯ l νkµ Γ211 = m l νlµ Γ311 = m¯ l νlµ Γ312 = m¯ l νmµ µ ν∇ µ ν∇ µ ν∇ µ ∇ν (162) Γ120 = l m νkµ Γ220 = m m νkµ Γ320 = m¯ m νkµ Γ221 = m m νlµ µ ν∇ µ ν ∇ µ ν ∇ µ ν ∇ Γ321 = m¯ m νlµ Γ322 = m¯ m νmµ Γ130 = l m¯ νkµ Γ230 = m m¯ νkµ µ ν ∇ µ ν∇ µ ν∇ µ ν ∇ Γ = m¯ m¯ νkµ Γ = m m¯ νlµ Γ = m¯ m¯ νlµ Γ = m¯ m¯ νmµ 330 ∇ 231 ∇ 331 ∇ 332 ∇ Newman and Penrose classify 12 independent complex linear combination of the above Ricci rotation coefficients and called spin coefficients as follows
µ ν κ = Γ = Γ¯ = m k νkµ, − 200 300 ∇ 31 µ ν ρ = Γ = Γ¯ = m¯ m νkµ, − 320 230 ∇ µ ν σ = Γ = Γ¯ = m m νkµ, − 220 330 ∇ µ ν τ = Γ = Γ¯ = m l νkµ, − 210 310 ∇ µ ν ν = Γ = Γ¯ = m¯ l νlµ, 311 211 ∇ µ ν µ = Γ = Γ¯ = m¯ m νlµ, 321 231 ∇ µ ν λ = Γ = Γ¯ = m¯ m¯ νlµ, 331 211 ∇ µ ν π = Γ = Γ¯ = m¯ k νlµ, 301 201 ∇ 1 1 ǫ = (Γ Γ ) = lµkν k m¯ µkν m , 2 100 302 2 ν µ ν µ − − ∇ − ∇ 1 1 β = (Γ Γ ) = lµmν k m¯ µmν m , 2 120 322 2 ν µ ν µ − − ∇ − ∇ 1 1 γ = (Γ Γ ) = kµlν l mµlν m¯ , 2 011 213 2 ν µ ν µ − ∇ − ∇ 1 1 α = (Γ Γ ) = kµm¯ ν l mµm¯ ν m¯ . (163) 2 031 233 2 ν µ ν µ − ∇ − ∇ From expression (52), where we have to connection 1-forms in pseudo-orthonormal coordinate that
d(γαβ) Γαβ Γβα = 0, − − and with a constant metric, a rigid frame, d(γαβ) = 0, it results to connection 1-forms that
Γαβ = Γβα. − In this frame we have α αδ Γ β = γ Γδβ, (164) and we have that only non-null components of metric tensor from (155) are γ01 = γ10 = 1 and γ23 = γ32 = 1 where α − 0 we can use above equation (164) to find the connection 1-forms Γ β from Γδβ. The calculation for Γ 0 follows
0 00 01 02 03 Γ 0 = γ Γ00 + γ Γ10 + γ Γ20 + γ Γ30, which is simplified in Γ0 = Γ . 0 − 10 The calculations for the others connection 1-forms are displayed below
0 00 01 02 03 Γ 1 = γ Γ01 + γ Γ11 + γ Γ21 + γ Γ31 = 0; 0 00 01 02 03 Γ 2 = γ Γ02 + γ Γ12 + γ Γ22 + γ Γ32 = Γ12; 0 00 01 02 03 − Γ 3 = γ Γ03 + γ Γ13 + γ Γ23 + γ Γ33 = Γ13; 1 10 11 12 13 − Γ 0 = γ Γ00 + γ Γ10 + γ Γ20 + γ Γ30 = 0; 1 10 11 12 13 Γ 1 = γ Γ01 + γ Γ11 + γ Γ21 + γ Γ31 = Γ01; 1 10 11 12 13 − Γ 2 = γ Γ02 + γ Γ12 + γ Γ22 + γ Γ32 = Γ02; 1 10 11 12 13 − Γ 3 = γ Γ03 + γ Γ13 + γ Γ23 + γ Γ33 = Γ03; 2 20 21 22 23 − Γ 0 = γ Γ00 + γ Γ10 + γ Γ20 + γ Γ30 = Γ03; Γ2 = γ20Γ + γ21Γ + γ22Γ + γ23Γ = −Γ ; 1 01 11 21 31 − 13 32 2 20 21 22 23 Γ 2 = γ Γ02 + γ Γ12 + γ Γ22 + γ Γ32 = Γ23; 2 20 21 22 23 − Γ 3 = γ Γ03 + γ Γ13 + γ Γ23 + γ Γ33 = 0; 3 30 31 32 33 Γ 0 = γ Γ00 + γ Γ10 + γ Γ20 + γ Γ30 = Γ02; 3 30 31 32 33 − Γ 1 = γ Γ01 + γ Γ11 + γ Γ21 + γ Γ31 = Γ12; 3 30 31 32 33 − Γ 2 = γ Γ02 + γ Γ12 + γ Γ22 + γ Γ32 = 0; 3 30 31 32 33 Γ 3 = γ Γ03 + γ Γ13 + γ Γ23 + γ Γ33 = Γ23. (165) The above terms can be rearranged in a matrix form Γ 0 Γ Γ − 01 − 12 − 13 α 0 Γ01 Γ02 Γ03 (Γ β) = − − − . (166) Γ03 Γ13 Γ23 0 − − − Γ02 Γ12 0 Γ23 − − Thus with these connection 1-forms we can see the first structure equation of Cartan (101),
α α β dθ˜ = Γ β θ˜ , − ∧ in pseudo-orthonormal coordinate basis that results the four below equations
0 0 2 3 dθ˜ = Γ01 θ˜ + Γ12 θ˜ + Γ13 θ˜ 1 ∧ 1 ∧ 2 ∧ 3 dθ˜ = Γ01 θ˜ + Γ02 θ˜ + Γ03 θ˜ ∧ ∧ ∧ (167) dθ˜2 = Γ θ˜0 + Γ θ˜1 + Γ θ˜2 03 13 23 3 ∧ 0 ∧ 1 ∧ 3 dθ˜ = Γ02 θ˜ + Γ12 θ˜ Γ23 θ˜ ∧ ∧ − ∧ The dual basis for complex null tetrad from (159),
0 µ 1 µ 2 µ 3 µ θ˜ = lµdx , θ˜ = kµdx , θ˜ = m¯ µdx and θ˜ = mµdx − − has for each component the following complex conjugate
0 µ 0 1 µ 1 2 µ µ 3 3 µ 2 θ˜ = lµdx = θ˜ , θ˜ = kµdx = θ˜ , θ˜ = m¯ µdx = mµdx = θ˜ and θ˜ = m¯ µdx = θ˜ . − − The complex conjugate for first equation of system (167), results in
d(θ˜0) = Γ θ˜0 + Γ θ˜2 + Γ θ˜3 01 ∧ 12 ∧ 13 ∧ where it reduces to
0 0 3 2 dθ˜ = Γ01 θ˜ + Γ12 θ˜ + Γ13 θ˜ dθ˜0 = Γ ∧ θ˜0 + Γ ∧ θ˜2 + Γ ∧ θ˜3 01 ∧ 12 ∧ 13 ∧ and it yields Γ01 = Γ01, Γ12 = Γ13. The complex conjugate for second equation of system (167), results in
d(θ˜1) = Γ θ˜1 + Γ θ˜2 + Γ θ˜3 01 ∧ 02 ∧ 03 ∧ where it reduces to
1 1 3 2 dθ˜ = Γ01 θ˜ + Γ02 θ˜ + Γ03 θ˜ dθ˜1 = Γ ∧ θ˜1 + Γ ∧ θ˜2 + Γ ∧ θ˜3 01 ∧ 02 ∧ 03 ∧ 2 3 and it yields Γ02 = Γ03. The complex conjugate for third equation of system (167), results in dθ˜ = dθ˜ , that is
Γ θ˜0 + Γ θ˜1 + Γ θ˜2 = Γ θ˜0 + Γ θ˜1 Γ θ˜3, 03 ∧ 13 ∧ 23 ∧ 02 ∧ 12 ∧ − 23 ∧ 33 and it results that Γ23 = Γ32. We verify that exchanging the indices 2 and 3 implies complex conjugation. Thus we can write the system of equations (167) with three equations,
dθ˜0 = Γ θ˜0 + Γ θ˜2 + Γ θ˜3 01 ∧ 12 ∧ 12 ∧ dθ˜1 = Γ θ˜1 + Γ θ˜2 + Γ θ˜3 , (168) 01 02 02 2 ∧ 0 ∧ 1 ∧ 2 dθ˜ = Γ02 θ˜ + Γ12 θ˜ + Γ23 θ˜ ∧ ∧ ∧ where the fourth equation of system (167) is obtained from complex conjugation of third equation of system (168). We can see that in pseudo-orthonormal basis - complex null tetrad - the number of independent connection 1-forms Γαβ reduces from six to four.
8.2. The second equation of Cartan in pseudo-orthonormal basis
We have seen that in pseudo-orthonormal basis there are four connection 1-forms Γ01, Γ02, Γ12 and e Γ23, with complex conjugate Γ02 = Γ03, Γ12 = Γ13 and Γ23 = Γ32. We can observe that from second equation of Cartan (106),
γ Θαβ = dΓαβ + Γαγ Γ β, ∧ it can be solved reducing from six to four the number of independent curvature 2-forms Θ01, Θ02, Θ13 e Θ32. Let us calculate the curvature 2-form Θ02
γ Θ02 = dΓ02 + Γ0γ Γ 2 = dΓ + Γ ∧ Γ1 + Γ Γ2 + Γ Γ3 , 02 01 ∧ 2 02 ∧ 2 03 ∧ 2 1 2 3 where we can see the values of Γ 2, Γ 2 and Γ 2 on matrix (166), then it results in
Θ = dΓ + Γ ( Γ ) + Γ ( Γ ) + Γ (0), 02 02 01 ∧ − 02 02 ∧ − 23 03 ∧ being simplified in Θ = dΓ + Γ (Γ + Γ ). (169) 02 02 02 ∧ 01 23 The calculation for curvature 2-form Θ13 is
γ Θ13 = dΓ13 + Γ1γ Γ 3 = dΓ + Γ ∧ Γ0 + Γ Γ2 + Γ Γ3 13 10 ∧ 3 12 ∧ 3 13 ∧ 3 = dΓ13 + Γ10 ( Γ13) + Γ12 (0) + Γ13 Γ23 = dΓ + Γ ∧ (Γ− + Γ ) ∧ ∧ 13 13 ∧ 10 23 where we obtain Θ = dΓ Γ (Γ + Γ ) (170) 13 13 − 13 ∧ 01 32 The calculation for curvature 2-form Θ01 is
Θ = dΓ + Γ Γ1 + Γ Γ2 + Γ Γ3 01 01 01 ∧ 1 02 ∧ 1 03 ∧ 1 = dΓ01 + Γ01 ( Γ01) + Γ02 ( Γ13) + Γ03 ( Γ12) = dΓ Γ ∧ Γ− Γ Γ∧ ,− ∧ − 01 − 02 ∧ 13 − 03 ∧ 12 and finally for Θ32, we have
Θ = dΓ + Γ Γ0 + Γ Γ1 + Γ Γ2 32 32 30 ∧ 2 31 ∧ 2 32 ∧ 2 = dΓ32 + Γ30 ( Γ12) + Γ31 ( Γ02) + Γ32 Γ23 = dΓ + Γ ∧ Γ− Γ Γ∧ ,− ∧ 32 03 ∧ 12 − 02 ∧ 13 here we can sum the above equations
Θ + Θ = d(Γ + Γ ) + 2Γ Γ . (171) 01 32 01 32 13 ∧ 02 34 Thus we can rearrange the three complex equations (169), (170) and (171) in a system of 2-forms equations of the second equation of Cartan, Θ = dΓ + Γ (Γ + Γ ) 02 02 02 ∧ 01 23 Θ = Γ Γ Γ + Γ 13 d 13 13 ( 01 32) . (172) − ∧ Θ01 + Θ32 = d(Γ01 + Γ32) + 2Γ13 Γ02 ∧ The above system of 2-forms curvature equations has proved very useful in Newman-Penrose formalism [12, 17], where exact solutions of Einstein equations are obtained.
8.3. Example of Brinkmann spacetime An example most important which represents gravitational waves is the Brinkmann spacetime. This spacetime can be used to explain the non-expanding waves, known as plane-fronted waves with parallel rays, or simply pp- waves. The pp-waves may represent gravitational waves, electromagnetic waves, massless radiation associated with Weyl fermions, some other forms of matter moving at the speed of light, or any combination of these. The metric for pp-waves can be written in the form ds2 = 2dudv 2Hdu2 + 2dζ dζ,¯ (173) − − where H = H(ζ, ζ,¯ u). It is straightforward to notice that the shape of the Brinkmann metric is almost the same shape of the pseudo- orthonormal metric of Minkowski spacetime (152)
ds2 = 2dudv + 2dζdζ,¯ − where the aditional term 2Hdu2 is responsible for curvature of spacetime. When H = 0, the metric (173) reduces to Minkowski spacetime.− We can intuit that the the easiest way to work with Brinkmann metric is with complex null tetrad. In non-coordinate basis pseudo-orthonormal the metric tensor is given by (153),
g = 2θ˜0 θ˜1 + 2θ˜2 θ˜3, − ⊗ ⊗ where we can immediately identify θ˜2 = dζ e θ˜3 = dζ¯. While the other sector of metric can be can be identified by
2θ˜0 θ˜1 = 2dudv 2Hdu2, − ⊗ − − α α µ with θ˜ = ω µ dx , we have that
0 1 0 µ 1 ν du dv + Hdu du = θ˜ θ˜ = (ω µ dx ) (ω ν dx ) ⊗ ⊗ 0⊗ 0 ⊗ 1 1 = (ω u du + ω v dv) (ω u du + ω v dv) 0 1 ⊗ 0 1 0 1 = ω uω u du du + ω uω vdu dv + ω vω udv du 0 1 ⊗ ⊗ ⊗ +ω vω vdv dv, ⊗ where it yields the following equations
0 1 ω uω u = H; 0 1 0 1 ω uω v + ω vω u = 1; 0 1 ω vω v = 0.
The solution for the above system of equations is displayed in a matrix,
0 0 0 0 ω u ω v ω ζ ω ζ¯ H 1 0 0 1 1 1 1 α = ω u ω v ω ζ ω ζ¯ = 1 000 (ω µ) 2 2 2 2 , (174) ω u ω v ω ζ ω ζ¯ 0 010 3 3 3 3 ω u ω v ω ζ ω ¯ 0 001 ζ
35 thus we obtain the dual basis as θ˜0 = Hdu + dv θ˜1 = du . (175) θ˜2 = dζ θ˜3 = dζ¯ α The matrix of vierbein is obtained from inversion of matrix (ω µ), u v ζ ζ¯ e0 e0 e0 e0 0 1 00 u v ζ ζ¯ µ e1 e1 e1 e1 1 H 0 0 (eα ) = ¯ = − , (176) e u e v e ζ e ζ 0 0 10 2 2 2 2 e u e v e ζ e ζ¯ 0 0 01 3 3 3 3 Then the tetrad of vectors of the pseudo-orthonormalbasis is
eˆ0 = k = ∂v = = eˆ1 l ∂u H∂v − , (177) eˆ2 = m = ∂ζ eˆ = m¯ = ∂ 3 ζ¯ the above tetrad can be expressed as = α ∂ k k ∂xα α ∂ l = l α ∂x , (178) m = mα ∂ ∂xα = α ∂ m¯ m¯ ∂xα ∂ = where ∂xα (∂u,∂v,∂ζ,∂ζ¯) and 0 1 0 0 1 H 0 0 kα = , lα = , mα = , m¯ α = . (179) 0 −0 1 0 0 0 0 1 The exterior derivatives of 1-forms of (175) result that the only non-null is dθ˜0 ∂H ∂H ∂H dθ˜0 = dζ du + dζ¯ du + du du, ∂ζ ∧ ∂ζ¯ ∧ ∂u ∧ ∂H = here we denote ∂ζ H,ζ and the above equation reduces in
0 dθ˜ = H,ζ dζ du + H,¯ dζ¯ du, ∧ ζ ∧ or 0 2 1 3 1 dθ˜ = H,ζ θ˜ θ˜ + H,¯ θ˜ θ˜ . (180) ∧ ζ ∧ Let us put this result in the first equation of Cartan (168),
dθ˜0 = Γ θ˜0 + Γ θ˜2 + Γ θ˜3 01 ∧ 12 ∧ 12 ∧ dθ˜1 = Γ θ˜1 + Γ θ˜2 + Γ θ˜3 , 01 02 02 2 ∧ 0 ∧ 1 ∧ 2 dθ˜ = Γ02 θ˜ + Γ12 θ˜ + Γ23 θ˜ ∧ ∧ ∧ where we obtain 1 2 1 3 0 2 3 H,ζ θ˜ θ˜ H,¯ θ˜ θ˜ = Γ θ˜ + Γ θ˜ + Γ θ˜ − ∧ − ζ ∧ − 01 ∧ 12 ∧ 13 ∧ 0 = Γ θ˜1 + Γ θ˜2 + Γ θ˜3 , 01 02 02 ∧ 0 ∧ 1 ∧ 2 0 = Γ02 θ˜ + Γ12 θ˜ + Γ23 θ˜ ∧ ∧ ∧ 36 the solution of this system results in
1 1 Γ = H,ζ θ˜ and Γ = H,¯ θ˜ . (181) 12 − 13 − ζ
The only independent non-null Ricci rotation coefficient is Γ12 = Γ13. We can solve the second equation of Cartan for Brinkmann spacetime with (172)
Θ = dΓ + Γ (Γ + Γ ) 02 02 02 ∧ 01 23 Θ = Γ Γ Γ + Γ 13 d 13 13 ( 01 32) . − ∧ Θ01 + Θ32 = d(Γ01 + Γ32) + 2Γ13 Γ02 ∧ Here we can see that just the second equation of the above system has the nonzero 1-form Γ13. Then the only nonzero curvature 2-form is
1 ∂H,ζ¯ 1 ∂H,ζ¯ 1 ∂H,ζ¯ 1 Θ13 = dΓ13 = d( H,¯ θ˜ ) = du θ˜ dζ θ˜ dζ¯ θ˜ − ζ − ∂u ∧ − ∂ζ ∧ − ∂ζ¯ ∧ 2 1 3 1 = H, ¯ θ˜ θ˜ H,¯ ¯ θ˜ θ˜ (182) − ζζ ∧ − ζζ ∧ From definition of curvature 2-form (103), we have
α 1 α γ δ Θ β = R βγδθ˜ θ˜ , 2 ∧ such that
0 1 0 2 0 3 1 2 Θ13 = R1301 θ˜ θ˜ + R1302 θ˜ θ˜ + R1303 θ˜ θ˜ + R1312 θ˜ θ˜ +R θ˜∧1 θ˜3 + R θ˜∧2 θ˜3, ∧ ∧ 1313 ∧ 1323 ∧ by comparing the above result with (182), we have the following components of Riemann tensor = = R1312 H,ζζ¯ , and R1313 H,ζ¯ζ¯ .
δǫ The components of the Ricci tensor can be computed via Rαβ = γ Rǫαδβ. We use the symmetry of Riemann tensor = where R3121 H,ζζ¯. The complex conjugate of this component is = = R¯3121 R2131 H,ζζ¯ . There is only one nonzero component of Ricci tensor in pseudo-orthonormal basis = 23 + 32 = R11 γ R3121 γ R2131 2H,ζζ¯ . The Ricci tensor is given by α β Ric = Rαβθ˜ θ˜ , ⊗ α α µ we recall from (56) where θ˜ = ω µdx , to rewrite the tensor Ricci in coordinate basis as
α µ β ν Ric = Rαβ(ω µdx ) (ω νdx ), ⊗ we have only R11 , 0, then it results in 1 1 µ ν Ric = R ω µω ν dx dx , 11 ⊗ 1 and with ω µ = kµ from (159) the Ricci tensor in cordinate basis is given by − µ ν Ric = 2H, ¯ kµkν dx dx , ζζ ⊗ where we can identify the components of Ricci tensor in coordinate basis as
Rµν = 2H,ζζ¯ kµkν. (183) 37 If we use (160) we have that the matrix of Ricci tensor in coordinate basis is 1 1 0 0 1− 1 00 (Rµν) = H, ¯ − . ζζ 0 0 00 0 0 00 µν µ µν because g kµkν = kµk = 0, we have that the scalar curvature R = g Rµν is zero. Then, according to Einstein’ General Relativity, the curvature os spacetime is related to the distribution of matter, where specifically the components of the Ricci tensor are directly related to the local energy-momentum tensor Tµν by Einstein’s field equations. We have from (183) that the energy-momentum tensor Tµν of Brinkmann spacetime is given by
Tµν = ρ kµkν. (184) The above tensor is the energy-momentum tensor of null dust, where ρ denotes the energy density. The energy- momentum tensor (184) has a similar structure as that of dust (pressure-free perfect fluid) this explains the notation null dust. Null dust or pure radiation or incoherent radiation, represents a simple matter source in General Relativity and describes the flux of massless particles, for example photons, with a fixed propagation direction given by the null vector field kµ.
9. Conclusion The purpose of this manuscript was to provide a self-contained review of manipulate some objects such as con- nection coefficients and curvature tensors of a spacetime ( , g) in non-coordinate basis with concepts of vierbein field. In a traditional manner the authors that introduce the GeneralM Relativity [1, 2, 3, 4, 5, 6], have taught to calculate curvature tensors directly from metric tensor gµν in coordinate basis. However for an advanced way, some times it is necessary to calculate the connection coefficients and curvature tensors with aid of Cartan’s structure equations which µ deal with vierbein fields eα , the ’square root’ of the metric field tensor. So, we have seen that the non-coordinate bases, orthonormal and pseudo-orthonormal (complex null tetrad), are useful for pedagogical study of conection co- efficients and curvature in language of differential forms and also they are useful to manipulating and pratice the Cartan’s structure equations. This manuscript has the introduction of complex null tetrad, the pseudo-orthonormal basis, that is useful to the formalism of Newman-Penrose applied to General Relativity. The Newman-Penrose for- malism adopts a tetrad in which the two axes u (advanced) and v (retarded) along the direction of propagation are chosen to be lightlike, while the two complex axes ζ and ζ¯ are transverse to direction of propagation. It provides a particular powerful way to deal with fields that propagates at the speed of light as we have seen with the example of Brinkmann spacetime [12, 13, 14, 17].
References
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