Christoffel Symbols. 1 In General Topologies. Joshua Albert. September 28, W. First we say W : λ n = x µ (λ) so that the world

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1 Christoffel Symbols Joshua Albert September 28, 22 In General Topoloies We have a metric tensor nm defined by, Note by some handy theorem that for almost any continuous function F (L), equation 2 still holds. Now we work out an explict form of equation 2. d s 2 ab dx a dx b () which tells us how the distance is measured between two points in a manifold M. Note ab is a function of only x a and x b. Say we wish to investiate what an observer will experience as she moves on a world line W in M, then we will take intuition from classical mechanics and extremize the action of the Laranian L alon W. First we say W : λ n x µ (λ) so that the world line is parametrized. Here λ plays a role similar to time in classical mechanics except our time in incorporated into the 4-position x c. Now we define the Laranian as, ds 2 L We need to extremize the followin functional for some affine parameter λ, W L From the familiar Euler-Larane equations we et the result, d L L (2) ẋ c x c ds 2 L ab ẋ a ẋ b Where we note that the metric tensor is independent of λ. Thus, L ẋ ẋ a c ab ẋ ẋ b ẋ b + c ab ẋ ẋ a c 2 ab ẋ a ẋ c ẋ b 2 ab δ a c ẋ b 2 cb ẋ b Where we used the symmetry of nm. Note that indices are arbitrary and that we can swap them at will for each other if it simplifies the procedure. Now we apply the λ derivative. d L dẋ b 2 ẋ c cb cb x d 2 cb ẍ b cb d ẋ d ẋ b dx d ẋ b

2 Now for the riht hand side, L x c ab c ẋ a ẋ b tensorial object then it must invariantly transform via equation 3. And so we bein by expandin the metric tensor and then applyin the partial derivative, like i j k i j A i i A j j i j. Then from equation 2 we have, 2 cb ẍ b cb d ẋ d ẋ b ab c ẋ a ẋ b 2 e c cb ẍ b + e c (2 cb a ab c )ẋ a ẋ b 2δ e b ẍ b + e c ( cb a + c a b ab c )ẋ a ẋ b ẍ e + 2 e c ( b c a + c a b ab c )ẋ a ẋ b a b 2 e c ( Ab b b c + Aa a A c c c a Aa a A b b ab ) x a x b x c We need to deal with these partial derivatives, (5) ẍ e + Γ e ab ẋ a ẋ b Where Γ e ab e c ( 2 b c a + c a b ab c ) are called the Christoffel Symbols of the second kind. 2 Γ e ab is not a tensor A tensor is simply a eometrical object which represents some locally isomorphic operation on a manifold M. This implies that the null space is preserved in all representations of the same tensor. Let s try to take Γ e ab into a different representation via, A i i A j j i j x i x j x i x j i j 2 x i x i + x i x i x j x j i j + x i 2 x j x i x j i j x j x k i j x k x j 2 x i 2 x i i j + A i i A j j A k k i j k Where we used symmetry of i and j to combine terms. Therefore, equation 5 becomes, where, χ e a b A e e Aa a A b b χ e ab (3) A t s x t x s (4) is the jacobian of the new representation, x t, with respect to the old representation, x s. If Γ e ab is indeed a Christoffel Symbols of the first kind are almost never seen or used. a b 2 e c 2 x b (2 x b x a b c + A b b b c a x c x b c a + A c c c a b 2 x a 2 x a x c ab A a a ab c ) Carryin the factors in and seein the Kronecker deltas that appear we et, 2

3 a b 2 (2 2 x b δ e x b x a b + Ab b b c a e c A a x c x b a δ e a + Ac c c a b e c 2 x a 2 A b x a x c b A c b δe a Aa a ab c e c ) Or, 2 (2 2 x e + A b x b x a b b c a e c A e x c x b a + c a b e c 2 2 x e x a x b A a a A b b A c c ab c e c ) x a x b A c c A e c + 2 (Ab b A a a b c a A e e e c + A a a A b b c a b A e e e c A a a A b b ab c A e e e c ) and, t t (9) x ρ cosθ sinφ () y ρ sinθ sinφ () z ρ cosφ (2) t t (3) ρ x 2 + y 2 + z 2 (4) θ tan y x (5) a b x a x b A c c A e c + A a a A b b A e e Γe a b ) (6) φ cos z ρ (6) 3 Christoffel Symbols of Flat Space- Time in Spherical Coordinates Say we have a Minkowski space-time with euclidean coordinates x µ (t,x, y, z ), which has metric, ab (7) ds 2 ab dx a dx b dt 2 dx 2 dy 2 dz 2 (8) Let s now look in spherical coordinates x σ (t,ρ,θ,φ). We have the relations, The jacobian A n m x µ of which is, x σ t x µ A n m ρ x µ (7) θ x µ φ x µ cosθ sinφ sinθ sinφ cosφ ρ sinθ sinφ ρ cosθ sinφ ρ cosθ cosφ ρ sinθ cosφ ρ sinφ (8) Clearly then, in the spherical coordinate system the unit vectors would be just the rows of our above jacobian. That is 3

4 ˆt dx µ d t ˆρ dx µ dρ cosθ sinφ sinθ sinφ cosφ (9) (2) m n A a m Ab n ab (24) 2 cosθ sinφ sinθ sinφ cosφ ρ sinθ sinφ ρ cosθ sinφ ρ cosθ cosφ ρ sinθ cosφ ρ sinφ (25) (26) ˆθ dx µ dθ ρ sinθ sinφ ρ cosθ sinφ (2) Calculation of the square of the jacobian yields the columns, ˆφ dx µ dφ ρ cosθ cosφ ρ sinθ cosφ ρ sinφ (22) (23) A a m Ab Now, usin our rules for the lowerin and raisin of tensor objects, m n x a x b x m x n ab we can convert the flat space-time metric in euclidean coordinates to the spherical coordinate system. Note here that usin the covariant metric, that is with indices on the bottom, then we must have the euclidean coordinates in terms of the spherical coordinates. Applyin m n A a m Ab n ab we et, A a m Ab () (cosθ sin 2 φ ρ sin 2 θ sin 2 φ +ρ cosθ cos 2 φ) ( ρ sinθ cosθ sin 2 φ ρ 2 sinθ cosθ sinφ) (ρ cos 2 θ sinφ cosφ ρ 2 sin 2 θ sinφ cosφ ρ 2 cosθ sinφ cosφ) 4

5 will be zero because the Minkowski metric is diaonal. () (cosθ ρ sin 2 θ ) sin 2 φ + ρ cosθ cos 2 φ) ( ρ sinθ cosθ (sin 2 A a Ab ab φ + ρ sinφ) (((cosθ ρ)ρ cosθ ρ 2 sin 2 θ ) sinφ cosφ) A a Ab ab (((cosθ ρ sin 2 θ ) sin 2 φ + ρ cosθ cos 2 φ) 2 A a m Ab 2 + (ρ sinθ cosθ (sin 2 φ + ρ sinφ)) 2 () + (((cosθ ρ)ρ cosθ ρ 2 sin 2 θ ) sinφ cosφ) 2 ) (sinθ cosθ sin 2 φ + ρ sinθ cosθ sin 2 φ +ρ sinθ cosθ ) ( ρ sin 2 θ sin 2 φ + ρ 2 cos 2 θ sin 2 φ) (ρ sinθ cosθ sinφ cosφ + ρ 2 sinθ cosθ sinφ cosφ ρ 2 sinθ sinφ cosφ) 22 A a 2 Ab 2 ab ((sinθ cosθ (sin 2 φ + ρ sin 2 φ + ρ)) 2 + (ρ( sin 2 θ + ρ cos 2 θ ) sin 2 φ) 2 () (sinθ cosθ (sin 2 φ + ρ sin 2 + (ρ(sinθ cosθ + ρ sinθ cosθ ρ sinθ ) sinφ cosφ) 2 ) φ + ρ)) (ρ( sin 2 θ + ρ cos 2 θ ) sin 2 φ) ρ 2 sin 2 φ (ρ(sinθ cosθ + ρ sinθ cosθ ρ sinθ ) sinφ cosφ) 33 A a 3 Ab 3 ab A a m Ab 3 () (cosθ sinφ cosφ ρ sinφ cosφ) ( ρ sinθ sinφ cosφ) (ρ cosθ cos 2 φ + ρ 2 sin 2 φ) (((cosθ ρ) sinφ cosφ) 2 + ( ρ sinθ sinφ cosφ) 2 + (ρ cosθ cos 2 φ + ρ 2 sin 2 φ) 2 ) ρ 2 () ((cosθ ρ) sinφ cosφ) ( ρ sinθ sinφ cosφ) (ρ cosθ cos 2 φ + ρ 2 sin 2 φ) Now we can calculate the full metric. All off diaonals 5