We used this in Eq without explaining it. Where does it come from? We know that the derivative of a scalar is a covariant vector, df

Transcription

1 Lecture 19: Covariant derivative of contravariant vector The covariant derivative of a (contravariant) vector is V µ ; ν = ν Vµ + µ ν V. (19.1) We used this in Eq without exlaining it. Where does it come from? We know that the derivative of a scalar is a covariant vector, df ϕ, µ = µ ϕ. Now, suose the scalar is the contraction of 2 vectors: ϕ = A µ B µ (19.2) then by definition (and the roduct rule) ν ϕ = A µ, ν B µ + A µ B µ, ν A µ ; ν Bµ + A µ B µ ; ν = A µ ; ν + A B µ + A µ B µ, ν (19.3) From Eq we have A µ ; ν B µ + A µ B µ ; ν = A µ ; ν Bµ + A B µ + A µ B µ, ν or A µ B µ ; ν = A µ Bµ, ν + A B µ = A µ B µ, ν + µ ν B (19.4) and since A µ is arbitrary, we may say B µ ; ν = Bµ, ν + µ ν B. (19.5) Covariant derivative of tensor By similar maniulations, we can identify the covariant derivative of a contravarient second-rank tensor----we write ϕ = A µ B ν T µν (19.6) and use the roduct rule again to write κ ϕ = A µ, κ B ν T µν + A µ B ν, κ T µν + A µ B ν T µν, κ to find A µ ; κ B ν T µν + A µ B ν ; κ T µν + A µ B ν T µν ; κ (19.7) 85

2 Geodesic coordinates T µν ; κ = T µν, κ + µ and so forth. T ν + ν T µ (19.8) Geodesic coordinates Suose we locally change coordinates to a system x µ = b µ x, with (linear) transformation coefficients b µ chosen such that the new metric at that oint is g µν df = g αβ b µ α bν β = ηµν. (19.9) Equations 19.9 constitute 1 inhomogeneous equations for 16 unknowns b µ, whose determinant, det g αβ 1 = g, is non-zero. Therefore they can always be solved, leaving 6 free arameters. These are in fact the 6 arameters of the Lorentz transformation (3 boost, 3 rotation) which, as we already know, leave the Minkoski metric unchanged. Moreover, we can secify the coordinates further so that in the new system, all first derivatives of the new metric, g µν, κ vanish at the oint a. The coordinates that do this are x µ = b µ x Γµ κ x a x κ a κ ! Λµ κλ x a x κ a κ x λ a λ + where the coefficients Γ µ κ and Λµ κλ are manifestly symmetric in their lower indices, hence reresent 2 and 8 indeendent arameters, resectively. Problem: An object with 3 indices that run from to 3 obviously has 64 comonents. Show that if the object is fully symmetric in the 3 indices, then there are but 2 indeendent comonents. Hence show that Λ µ κλ has 8 indeendent comonents. Problem: Find the relation between the coefficients Γ µ κ and the Christoffel symbols µ κ (a) defined at the oint a in terms of the (derivatives of the) old metric g µν. Since there are 2 first derivatives of the metric tensor, we can obviously choose the coefficients Γ µ κ to set the derivatives of the new metric tensor equal to zero at one oint. Since then the 86

3 Lecture 19: Christoffel symbol vanishes, the motion of a test body in these new coordinates is unaccelerated, i.e. freely falling. What about the second derivatives of the metric tensor? There are 1 distinct comonents, g µν,, (two distinct airs of symmetric indices, giving 1 1 = 1), but only 8 indeendent arameters Λ µ κλ, hence there will be 2 quantities involving second derivatives of the metric tensor, that cannot be made to vanish at a oint by a coordinate transformation. In a frame where the first derivatives of the metric tensor can be chosen to vanish at a oint, the Christoffel symbols also vanish at that oint, hence d 2 x µ d 2 =. (19.1 ) τ The new coordinates at that oint are freely falling, hence the name geodesic. We have soken before of arallel transort, and concluded that when a vector A µ is transorted an amount δx κ arallel to itself, the change in A µ, arising from the change in the coordinates, is δa µ = κµ A δx κ. (19.11 ) We can think of the covariant derivative as the difference between the ordinary derivative and the change that would occur if the vector were merely arallel-transorted; hence the change in a contravariant vector under arallel transort is δa µ = µ A δx κ. (19.12 ) Finally, we note that the 4-velocity U µ df dx µ = is always arallel-transorted; moreover, the contraction of the velocity and A µ is a scalar that is invariant under arallel dτ transort δ U µ A µ = U µ κ A µ A µ µ κ U δx κ =. (19.13) To ut it another way, the angle between a vector that is arallel-transorted and the velocity is always constant. 87

4 Let us arallel-transort a vector around an infinitesimal closed curve arameterized by x µ () : if ~ Vµ () V µ x κ (), we find ~ Vµ () V~ µ () = d dv~ µ ( ) d = d κµ ( ) V ~ ( ) dxκ d. (19.14 ) Let x µ ( = ) = a µ ; we can then exand in Taylor s series about =: V ~µ ( ) V µ (a) + (a) V (a) x κ ( ) a κ (19.15 ) and (x( )) (a) + x λ ( ) a λ x λ (a). (19.16) Therefore δx λ ( ) = x λ ( ) a λ, we find V λ λ ( µ ) V µ ( a) d ( a) + ( x ( ) a ) λ ( a) i µν µν κ κ α V( a) + ( x ( ) a ) κ ( a) Vα( a) Thus, exanding in owers of V ~µ () V µ (a) (a) V (a) + (a) α λ (a) V α (a) + V (a) λ (a) d dxν d + (19.18 ) (19.17) d dxν d δxλ ( ) + O (δx ν δx λ ) We dro the O (δx ν δx λ ) term in Eq because we shall ultimately take the neighborhood of a λ to be arbitrarily small. Now, since the arameterization describes a closed curve, we have ν dx ν d dx d = Thus we have to evaluate ( ) (19.19 ) 88

5 Lecture 19: ( ) ( ) ( ) ( ) ( ) ν λ λ ν λ dx x a dx x (19.2) We see immediately that, because d(x ν x λ ) = x λ dx ν + x ν dx λ is a erfect derivative, dx x = dx x ν λ λ ν Thus we have an exression of the form (19.21) δ 1 V = µ Rµνλ ( a) V ( a) dx x 2 ν λ (19.22) where equation defines the Riemann curvature tensor: R µνλ = λ ν λµ + α λα α λµ να (19.23) Clearly, R µνλ = if the sace is flat, i.e. if the first and second derivatives of the metric tensor vanish, since then the Christoffel symbols and their first derivatives vanish. Does this mean that in a freely falling system the curvature tensor is zero? No, because while the Christoffel symbols vanish, their (ordinary) derivatives will not. Thus we can, in rincile, distinguish between a flat sace and a freely falling system in a curved sace, by the non-vanishing of the curvature in the latter case. We note that R µνλ is a tensor by construction, since everything else in Eq is a tensor. We also note that it is antisymmetric in λν. If one lowers the to index to roduce the 4th rank covariant tensor R κµ νλ = g R µνλ, we find the latter satisfies four identities: R µν R κ µν R µν R R µν R µν R µν + R κµ ν + R κν µ. By virtue of these it is ossible to show that only 2 of the comonents of the tensor R µνλ are indeendent (see Ohanian and Ruffini,. 334ff), hence they may be identified with the 2 non-trivial comonents of the second derivative of the metric tensor. In fact, u to a constant multilier R µνλ is unique. 89

6 9