PHYS 652: Astrophysics 1 Appendix to Lecture 2 An Alternative Lagrangian In class we used an alternative Lagrangian L = g γδ ẋ γ ẋ δ, instead of the traditional L = g γδ ẋ γ ẋ δ. Here is the justification why either works correctly, i.e., why the expression given in eq. (223) is a Lagrangian that generates the geodesic equation. We prove that by applying the Lagrange s equations x α d dλ ẋ α = 0. to the expression in eq. (223), and recovering the geodesic equation. d dλ x α = g γδ,α ẋ γ ẋ δ, = g αδ ẋ δ + g γα ẋ γ ẋ ) α ( ẋ α = g αδ,γ ẋ δ ẋ γ + g αδ ẍ δ + g γα,δ ẋ γ ẋ δ + g γα ẍ γ = (g αδ,γ + g γα,δ ) ẋ δ ẋ γ + 2g αδ ẍ δ, because we are at liberty to rename dummy variables (ones which are summed over), and to exchange indices of the metric tensor, since it is symmetric. The Lagrange equation therefore reads: x α d dλ ẋ α = (g αδ,γ + g γα,δ ) ẋ δ ẋ γ + 2g αδ ẍ δ g γδ,α ẋ γ ẋ δ = = (g αδ,γ + g γα,δ g γδ,α )ẋ δ ẋ γ + 2g αδ ẍ δ = 0. Now multiply both sides by 1 2 gνα to isolate the second derivative term: ẍ ν + 1 2 gνα (g αδ,γ + g γα,δ g γδ,α ) ẋ δ ẋ γ = 0, where we have used g να g αδ = δ ν δ (Kronecker delta). But, by definition so, we finally have Γ ν βγ = 1 2 gνα (g αδ,γ + g γα,δ g γδ,α ), ẍ ν = Γ ν βγẋδ ẋ γ, which is the geodesic equation we derived in class (eq. (31)). This proves that the Lagrangian in eq. (223) also generates the geodesic equation (the factor of 1/2, or any other positive constant, does not affect the Lagrange s equations). 1
PHYS 652: Astrophysics 2 Example of Metric Conversion Let us see how convert from one space metric to another, i.e., use eq. (4). For example, given the space metric in Cartesian coordinates (x 1,x 2,x 3 ) = (x,y,z) δ = 1 0 0 0 1 0 0 0 1 let us find the space metric in spherical coordinates (x 1,x 2,x 3 ) = (r,θ,φ). Cartesian coordinates are given in terms of spherical as:. x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, or x 1 = x 1 sin x 2 cos x 3, x 2 = x 1 sin x 2 sin x 3, x 3 = x 1 cos x 2. Then, x 1 x 1 = sinx 2 cos x 3, x 1 x 2 = x 1 cos x 2 cos x 3, x 1 x 3 = x 1 sin x 2 sin x 3, x 2 x 1 = sin x 2 sin x 3, x 3 x 1 = cos x 2, x 2 x 2 = x 1 cos x 2 sin x 3, x 3 x 2 = x 1 sin x 2, x 2 x 3 = x 1 sin x 2 cos x 3, x 3 = 0. x 3 2
PHYS 652: Astrophysics 3 From eq. (4), we have ds 2 = δ dx i dx j = δ x i x k x j x l dx k dx l [ = (dx 1 ) 2 x i x j ] δ x 1 x 1 + dx 1 dx [δ 2 x i x j ] x 1 x 2 + dx 1 dx [δ 3 x i x j ] x 1 x 3 ] + (dx 2 ) [δ 2 x i x j ] x 2 x 2 + dx 2 dx [δ 3 x i x j ] x 2 x 3 ] + dx 3 dx [δ 2 x i x j ] x 3 x 2 + (dx 3 ) [δ 2 x i x j ] x 3 x 3 = (dx 1 ) 2 [ sin 2 x 2 cos 2 x 3 + sin 2 x 2 sin 2 x 3 + cos 2 x 2] [ + dx 2 dx 1 x i x j δ x 2 x 1 [ + dx 3 dx 1 x i x j δ x 3 x 1 + dx 1 dx 2 [ x 1 sin x 2 cos x 2 cos 2 x 3 + x 1 sin x 2 cos x 2 sin 2 x 3 x 1 cos x 2 sin x 3] + dx 1 dx 3 [ x 1 sin 2 x 2 sin x 3 cos x 3 + x 1 sin 2 x 2 sin x 3 cos x 3] + dx 2 dx 1 [ x 1 sin x 2 cos x 2 cos 2 x 3 + x 1 sin x 2 cos x 2 sin 2 x 3 x 1 cos x 2 sin x 3] + (dx 2 ) 2 [ (x 1 ) 2 cos 2 x 2 cos 2 x 3 + (x 1 ) 2 cos 2 x 2 sin 2 x 3 + (x 1 ) 2 sin x 3] + dx 2 dx 3 [ (x 1 ) 2 cos x 2 sinx 2 cos x 3 sinx 3 + (x 1 ) 2 cos x 2 sin x 2 cos x 3 sin x 3] + dx 3 dx 1 [ x 1 sin 2 x 2 sin x 3 cos x 3 + x 1 sin 2 x 2 sin x 3 cos x 3] + dx 3 dx 2 [ (x 1 ) 2 cos x 2 sinx 2 cos x 3 sinx 3 + (x 1 ) 2 cos x 2 sin x 2 cos x 3 sin x 3] + (dx 3 ) 2 [ (x 1 ) 2 sin 2 x 2 sin 2 x 3 + (x 1 ) 2 sin 2 x 2 cos 2 x 3] = (dx 1 ) 2 + (x 1 ) 2 (dx 2 ) 2 + (x 1 ) 2 sin 2 x 2 (dx 3 ) 2 = dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 = p 11 (dr) 2 + p 22 (dθ) 2 + p 33 (dφ) 2 = p 11 (dx 1 ) 2 + p 22 (dx 2 ) 2 + p 33 (dx 3 ) 2 = p dx i dx j. Reading off diagonal components of the metric, we have p 11 = 1, p 22 = r 2, p 33 = r 2 sin 2 θ, so, the space metric for spherical coordinates is 1 0 0 1 0 0 p = 0 r 2 0 or p = 1 0 0 0 r 2 sin 2 r 0 2 1 θ 0 0 r 2 sin 2 θ. Deriving the geodesic equation in spherical coordinates. Let us now compute the geodesic in 3D flat space, expressed in spherical coordinates. This should be an analog to geodesics in flat space in Cartesian coordinates: ẍ α = 0. This can be done in at least two ways. Method 1: Brute force computing Christoffel symbols and substituting them into the geodesic equation. From the eq. (14), Christoffel symbols for the spherical space are given by Γ k = 1 2 pkl (p il,j + p lj,i p,l ). 3
PHYS 652: Astrophysics 4 Since p 11 = 1 all of its derivatives vanish. Also, because of symmetry (look at the definition given in eq. (14) and recall that the metric tensor is symmetric). Therefore, we have Γ k 1j = Γ k j1 = 1 2 pkl (p 1l,j + p lj,1 p 1j,l ) = 1 2 pkl p lj,1 = 1 2 Utilizing the following properties of the metric p : (p k2 p 2j,1 + p k3 p 3j,1 ). p is diagonal p = 0, p,k = 0, when i j, p 11 = const. p 11,i = 0, p p (φ) = p (x 3 ) p,3 = 0, p 22 = p 22 (r) = p 22 (x 1 ) p 22,2 = 0, p 22,3 = 0, p 33 = p 33 (r,θ) = p 33 (x 1,x 2 ) p 33,3 = 0. after some bookkeeping, we obtain Γ 1 1j = Γ 1 j1 = 0, Γ 1 22 = 1 2 p1l (p 2l,2 + p l2,2 p 22,l ) = 1 2 p11 p 22,1 = r, Γ 1 23 = Γ 1 32 = 1 2 p1l (p 2l,3 + p l3,2 p 23,l ) = 1 2 p1l p l3,2 = 0, Γ 1 33 = 1 2 p1l (p 3l,3 + p l3,3 p 33,l ) = 1 2 p11 p 33,1 = r sin 2 θ, Γ 2 1j = 1 2 p22 p 2j,1, Γ 2 11 = 0, Γ 2 12 = Γ 2 21 = 1 2 p22 p 22,1 = 1 1 2 r 22r = 1 r, Γ 2 13 = Γ 2 31 = 0, Γ 2 22 = 0, Γ 2 23 = Γ 2 32 = 1 2 p2l (p 2l,3 + p l3,2 p 23,l ) = 1 2 p22 (p 22,3 + p 23,2 p 23,2 ) = 0, Γ 2 33 = 1 2 p2l (p 3l,3 + p l3,3 p 33,l ) = 1 2 p22 p 33,2 = 1 1 2 r 2(2r2 sin θ cos θ) = sinθ cos θ, Γ 3 = 1 2 p3l (p il,j + p lj,i p,l ) = 1 2 p33 (p i3,j + p 3j,i p,3 ) = 1 2 p33 (p i3,j + p 3j,i ), Γ 3 11 = 0, Γ 3 12 = Γ 3 21 = 0, Γ 3 13 = Γ 3 31 = 1 2 p33 (p 13,3 + p 33,1 ) = 1 2 p33 p 33,1 = 1 1 2 r 2 sin 2 θ (2r sin2 θ) = 1 r, Γ 3 22 = 0, Γ 3 23 = Γ 3 32 = 1 2 p33 (p 23,3 + p 33,2 ) = 1 2 p33 p 33,2 = 1 1 2 r 2 sin 2 θ (2r2 sin θ cos θ) = cot θ, Γ 3 33 = 1 2 p33 (p 33,3 + p 33,3 ) = 0. 4
PHYS 652: Astrophysics 5 Geodesic equation in spherical coordinates then becomes (recall x 1 = r, x 2 = θ, x 2 = φ): ẍ 1 = r = Γ 1 γδẋ γ ẋ δ = Γ 1 22(ẋ 2 ) 2 Γ 1 33(ẋ 3 ) 2 = r θ 2 + r sin 2 θ φ 2, ẍ 2 = θ = Γ 2 γδẋ γ ẋ δ = 2Γ 2 12ẋ 1 ẋ 2 Γ 2 33(ẋ 3 ) 2 = 2 1 rṙ θ + sin θ cos θ φ 2, ẍ 3 = φ = Γ 3 γδẋ γ ẋ δ = 2Γ 3 13ẋ 1 ẋ 3 2Γ 3 23ẋ 2 ẋ 3 = 2 1 rṙ φ 2cot θ θ φ, Method 2: Using a Lagrangian L = g γδ ẋ γ ẋ δ. The alternative Lagrangian mentioned earlier becomes L = p ẋ i ẋ j = ṙ 2 + r 2 θ2 + r 2 sin 2 θ φ 2, so applying the Lagrange equations yields, for each coordinate r, θ, φ: x l d dλ ẋ l = 0, r d dλ ṙ = 2r θ + 2r sin 2 θ φ 2 2 r = 0, = r = r θ 2 + r sin 2 θ φ 2, θ d dλ θ = 2r 2 sin θ cos θ φ 2 4rṙ θ 2r 2 θ = 0, = 1 θ = 2 rṙ θ + sinθ cos θ φ 2, φ d dλ φ = 4rṙ sin2 θ φ 4r 2 sin θ cos θ θ φ 2r 2 sin 2 θ φ = 0, = φ = 2 1 rṙ φ 2cot θ θ φ. This set of equations represents motion in flat space, as described by spherical coordinates, and therefore should describe straight lines. This is fairly easy to see for purely radial motion in the x y plane, θ = π/2 and φ = const., so the RHS of all three geodesic equations above vanish, and we recover a straight (radial) line r = 0. In a more general case, it is less trivial to show that the equations above represent straight lines. As mentioned in class, using this alternative Lagrangian allows one to readily read off Christoffel symbols. From the equation above, they are readily identified as Γ 1 22 = r, Γ 1 33 = r sin 2 θ, Γ 2 12 = Γ 2 21 = 1 r, Γ 2 33 = sin θ cos θ, Γ 3 13 = Γ 3 31 = 1 r, Γ 3 23 = Γ 3 32 = cot θ, just as we computed by brute force. The factor 2 in front of Christoffel symbols Γ i jk which have unequal lower indices (j k) reflects the fact that because of symmetry both Γ i jk and Γi kj are counted. 5
PHYS 652: Astrophysics 6 It is not advisable to compute the geodesic equation from the traditional Lagrangian L = gγδ ẋ γ ẋ δ, as it will quickly lead to some extremely cumbersome algebra. The three Lagrange s equation should eventually reduce to the geodesic equations we derived above (because the two are equivalent in terms of producing the same result) but it quickly becomes obvious which approach is preferable. Applying the Geodesic Equation Let us compute the geodesic equation on the surface of the 3D sphere. The radius is then constant r = R, the coordinates are (x 1,x 2 ) = (θ,φ), and the metric is ( ) R 2 0 p = 0 R 2 sin 2. θ The Lagrangian again is L = p ẋ i ẋ j = R 2 θ2 + R 2 sin 2 θ φ 2, where i,j = 1,2. Applying the Lagrange equations yields, for each coordinate θ and φ: x l d dλ ẋ l = 0, θ d dλ θ = 2R 2 sin θ cos θ φ 2 2R 2 θ = 0, = θ = sin θ cos θ φ2, φ d dλ φ = 4R2 sin θ cos θ θ φ 2r 2 sin 2 θ φ = 0, = φ = 2cot θ θ φ. The second equation reduces to φ 2cot θ θ φ = 0 φsin 2 θ 2sin θ cos θ θ φ = 0 d ( φsin θ) 2 = 0, dt where the the conserved term in parentheses is the angular momentum. We know that the geodesics on the surface of the sphere must be a part of a great circle the circle which contains the two points and whose radius is the radius of the sphere (its center also coincides with the center of the sphere). We can check the two special cases, and make sure they are correct: 1. Equator: for the two points along the equator the shortest distance will be also along the equator. We need to show that such a curve φ = c 1 λ + φ 0, and θ = π/2 satisfies the geodesic equation. Plug in the geodesic equation and obtain So, the equator is a geodesic. φ = c 1 λ + φ 0, φ = c1, φ = 0, θ = π 2, θ = 0, θ = 0, θ = sin θ cos θ φ 2 = sin π 2 cos π 2 c 1 2 = 0, φ = 2cot θ θ φ = 2cot π 2 0c 1 = 0. 6
PHYS 652: Astrophysics 7 2. Meridian: for the two points along the same meridian (arc of the great circle connecting the two poles) the shortest distance should also be along the meridian. We need to show that such a curve φ = φ 0, and θ = c 2 λ + θ 0 satisfies the geodesic equation. Plug in the geodesic equation and obtain So, the meridian is a geodesic. φ = φ 0, φ = 0, φ = 0, θ = c 2 λ + θ 0, θ = c2, θ = 0, θ = sin θ cos θ φ 2 = sin (c 2 λ + θ 0 )cos (c 2 λ + θ 0 ) 0 2 = 0, φ = 2cot θ θ φ = 2cot (c 2 λ + θ 0 )c 2 0 = 0. 7