Homework 5. Solutions 1 Calculate the Christoffel symbols of the canonical flat connection in E3 in a) cylindrical coordinates (x = r cos ϕ, y = r sin ϕ, z = h), b) spherical coordinates. r r r r r (For the case of sphere try to make calculations at least for components Γrr, Γrθ, Γrϕ, Γθθ, ..., Γϕϕ) Remark One can calculate Christoffel symbols using Levi-Civita Theorem (Homework 5). There is a third way to calculate Christoffel symbols: It is using approach of Lagrangian. This is may be the easiest and most elegant way. (see the Homework 6) In cylindrical coordinates (r, ϕ, h) we have x = r cos ϕ r = x2 + y2 y = r sin ϕ and ϕ = arctan y ( p x z = h h = z We know that in Cartesian coordinates all Christoffel symbols vanish. Hence in cylindrical coordinates (see in detail lecture notes): ∂2x ∂r ∂2y ∂r ∂2z ∂r Γr = + + = 0 , rr ∂2r ∂x ∂2r ∂y ∂2r ∂z ∂2x ∂r ∂2y ∂r ∂2z ∂r Γr =Γr = + + = sin ϕ cos ϕ + sin ϕ cos ϕ = 0 . rϕ ϕr ∂r∂ϕ ∂x ∂r∂ϕ ∂y ∂r∂ϕ ∂z − ∂2x ∂r ∂2y ∂r ∂2z ∂r x y Γr = + + = x y = r. ϕϕ ∂2ϕ ∂x ∂2ϕ ∂y ∂2ϕ ∂z − r − r − ∂2x ∂ϕ ∂2y ∂ϕ ∂2z ∂ϕ Γϕ = + + = 0 . rr ∂2r ∂x ∂2r ∂y ∂2r ∂z ∂2x ∂ϕ ∂2y ∂ϕ ∂2z ∂ϕ y x 1 Γϕ =Γϕ = + + = sin ϕ− + cos ϕ = ϕr rϕ ∂r∂ϕ ∂x ∂r∂ϕ ∂y ∂r∂ϕ ∂z − r2 r2 r ∂2x ∂ϕ ∂2y ∂ϕ ∂2z ∂ϕ x y Γϕ = + + = x− y = 0 . ϕϕ ∂2ϕ ∂x ∂2ϕ ∂y ∂2ϕ ∂z − r2 − r2 All symbols Γ·h, Γh· vanish · · r r r r r ϕ Γrh =Γhr =Γhh =Γϕh =Γhϕ =Γhr = dots = 00 2 2 2 since ∂ x = ∂ y = ∂ z = 0 ∂h∂... ∂h∂... ∂h∂... 2 2 h h ∂ z ∂h ∂h ∂h ∂ z For all symbols Γ Γ = ∂ ∂ since ∂x = ∂y = 0 and ∂y = 1. On the other hand all ∂ ∂ vanish. Hence ·· ·· all symbols Γh vanish. · · · · ·· b) spherical coordinates x = r sin cos ϕ r = x2 + y2 + z2 θ = arccos z y = r sin sin ϕ p √x2+y2+z2 ( z = r cos θ y ϕ = arctan x We already know the fast way to calculate Christoffel symbol using Lagrangian of free particle and this method work for a flat connection since flat connection is a Levi-Civita connection for Euclidean metric So perform now brute force calculations only for some components. (Then later (in homework 6) we will calculate using very quickly Lagrangian of free particle. ) 1 2 r ∂ xi Γrr = 0 since ∂2r = 0. ∂2x ∂r ∂2y ∂r ∂2z ∂r x y z Γr =Γr = + + = cos θ cos ϕ + cos θ sin ϕ sin θ = 0 , rθ θr ∂r∂θ ∂x ∂r∂θ ∂y ∂r∂θ ∂z r r − r ∂2x ∂r ∂2y ∂r ∂2z ∂r x y z Γr = + + = r sin θ cos ϕ r sin θ sin ϕ r cos θ = r θθ ∂2θ ∂x ∂2θ ∂y ∂2θ ∂z − r − r − r − ∂2x ∂r ∂2y ∂r ∂2z ∂r x y Γr =Γr = + + = sin θ sin ϕ + sin θ cos ϕ = 0 rϕ ϕr ∂r∂ϕ ∂x ∂r∂ϕ ∂y ∂r∂ϕ ∂z − r r and so on.... 2 Let be an affine connection on a 2-dimensional manifold M such that in local coordinates (u,v) it ∇ u v is given that Γuv = v, Γuv = 0. ∂ Calculate the vector field ∂ u . ∇ ∂u ∂v Using the properties of connection and definition of Christoffel symbols have ∂ ∂ ∂ ∂ u = ∂ ∂ (u) + u ∂ = ∇ ∂u ∂v ∂u ∂v ∇ ∂u ∂v ∂ ∂ ∂ ∂ ∂ ∂ ∂ + u Γu +Γv = + u v + 0 = + uv . ∂v uv ∂u uv ∂v ∂v ∂u ∂v ∂u 3 Let be an affine connection on the 2-dimensional manifold M such that in local coordinates (u,v) ∇ ∂ 2 ∂ ∂ ∂ u = (1+ u ) + u . ∇ ∂u ∂v ∂v ∂u u v Calculate the Christoffel symbols Γuv and Γuv of this connection. ∂ ∂ Using the properties of connection we have ∂ u = u ∂ + ∇ ∂u ∂v ∇ ∂u ∂v ∂ u ∂ v ∂ ∂ v ∂ u ∂ 2 ∂ ∂ ∂ ∂ (u) = u Γuv +Γuv + 1 = (1+ uΓuv) + uΓuv = 1+ u + u . ∂u ∂v ∂u ∂v · ∂v ∂v ∂u ∂v ∂u 2 v v v u Hence 1 + u =1+ uΓuv and uΓuv = u, i.e. Γuv = u and Γuv = 1. 4 a) Consider a connection such that its Christoffel symbols are symmetric in a given coordinate system: i i Γkm =Γmk. Show that they are symmetric in an arbitrary coordinate system. b∗) Show that the Christoffel symbols of connection are symmetric (in any coordinate system) if and ∇ only if XY YX [X, Y] = 0 , ∇ − ∇ − for arbitrary vector fields X, Y. c)∗ Consider for an arbitrary connection the following operation on the vector fields: S(X, Y)= XY YX [X, Y] ∇ − ∇ − and find its properties. Solution ′ ′ i i i i a) Let Γkm =Γmk. We have to prove that Γk′m′ =Γm′k′ 2 We have ′ ′ ′ i k m r i i ∂x ∂x ∂x i ∂x ∂x Γ ′ ′ = ′ ′ Γ + ′ ′ . (1) k m ∂xi ∂xk ∂xm km ∂xk ∂xm ∂xr Hence ′ ′ ′ i m k r i i ∂x ∂x ∂x i ∂x ∂x Γ ′ ′ = ′ ′ Γ + ′ ′ m k ∂xi ∂xm ∂xk mk ∂xm ∂xk ∂xr i i ∂xr ∂xr ′ ′ ′ ′ But Γkm =Γmk and ∂xm ∂xk = ∂xk ∂xm . Hence ′ ′ ′ ′ ′ i m k r i i m k r i ′ i ∂x ∂x ∂x i ∂x ∂x ∂x ∂x ∂x i ∂x ∂x i Γ ′ ′ = ′ ′ Γ + ′ ′ = ′ ′ Γ + ′ ′ =Γ ′ ′ . m k ∂xi ∂xm ∂xk mk ∂xm ∂xk ∂xr ∂xi ∂xm ∂xk km ∂xk ∂xm ∂xr k m b) The relation XY YX [X, Y] = 0 ∇ − ∇ − holds for all fields if and only if it holds for all basic fields. One can easy check it using axioms of connection X ∂ Y ∂ (see the next part). Consider = ∂xi , = ∂xj then since [∂i,∂j ] = 0 we have that k k k k XY YX [X, Y]= i∂j j ∂i =Γ ∂k Γ ∂k = (Γ Γ )∂k = 0 ∇ − ∇ − ∇ − ∇ ij − ji ij − ji k k We see that commutator for basic fields XY YX [X, Y]=0 if and only if Γ Γ = 0. ∇ − ∇ − ij − ji c) One can easy check it by straightforward calculations or using axioms for connection that S(X, Y) is a vector-valued bilinear form on vectors. In particularly S(fX,Y )= fS(X, Y) for an arbitrary (smooth) function. Show this just using axioms defining connection: S(fX,Y )= fXY Y(fX) [fX, Y]= f X Y f YX ∂YfX +[Y,fX]= ∇ − ∇ − ∇ − ∇ − f X Y f YX (∂Yf)X + ∂YfX + f [Y, X]= f( X Y YX [X, Y]) = fS(X, Y) ∇ − ∇ − ∇ − ∇ − 5 Consider the surface M in the Euclidean space En. Calculate the induced connection in the following cases a) M = S1 in E2, b) M— parabola y = x2 in E2, c) cylinder in E3. d) cone in E3. e) sphere in E3. f) saddle z = xy in E3 Solution. a) Consider polar coordinate on S1, x = R cos ϕ, y = R sin ϕ. We have to define the connection on S1 E2 ∂ ϕ induced by the canonical flat connection on . It suffices to define ∂ =Γϕϕ∂ϕ. ∇ ∂ϕ ∂ϕ Recall the general rule. Let r(uα): xi = xi(uα) is embedded surface in Euclidean space En. The basic ∂ ∂r(u) vectors = . To take the induced covariant derivative XY for two tangent vectors X, Y we take ∂uα ∂uα ∇ a usual derivative of vector Y along vector X (the derivative with respect to canonical flat connection: in Cartesian coordiantes is just usual derivatives of components) then we take the tangent component of the answer, since in general derivative of vector Y along vector X is not tangent to surface: 2 ∂ γ ∂ (canonical) ∂ ∂ r(u) ∂ =Γαβ = ∂ = ∇ ∂uα ∂uβ ∂uγ ∇ α ∂uβ ∂uα∂uβ tangent tangent 3 ∂ ( ∂ β ) is just usual derivative in Euclidean space since for canonical connection all Christoffel ∇canonical α ∂u symbols vanish.) In the case of 1-dimensional manifold, curve it is just tangential acceleration!: 2r ∂ u ∂ (canonical) ∂ d (u) a ∂ =Γuu = ∂ = = tangent ∇ ∂u ∂u ∂u ∇ u ∂u du2 tangent tangent For the circle S1, (x = R cos ϕ, y = R sin ϕ), in E2. We have ∂ ∂x ∂ ∂y ∂ ∂ ∂ rϕ = = + = R sin ϕ + R cos ϕ , ∂ϕ ∂ϕ ∂x ∂ϕ ∂y − ∂x ∂y ∂ ϕ (canonic.) ∂ ∂ =Γ ∂ϕ = ∂ϕ = rϕ = ∂ϕ ϕϕ ∂ϕ ∇ ∂ϕ ∇ tangent ∂ϕ tangent ∂ ∂ ∂ ∂ ∂ ∂ ( R sin ϕ) + (R cos ϕ) = R cos ϕ R sin ϕ = 0, ∂ϕ − ∂x ∂ϕ ∂y − ∂x − ∂y tangent tangent ∂ ∂ since the vector R cos ϕ R sin ϕ is orthogonal to the tangent vector rϕ. In other words it means that − ∂x − ∂y acceleration is centripetal: tangential acceleration equals to zero. ϕ We see that in coordinate ϕ, Γϕϕ = 0. Additional work: Perform calculation of Christoffel symbol in stereographic coordinate t: 2tR2 R(t2 R2) x = ,y = − . R2 + t2 t2 + R2 In this case 2 ∂ ∂x ∂ ∂y ∂ 2R 2 2 ∂ ∂ rt = = + = (R t ) + 2tR , ∂t ∂t ∂x ∂t ∂y (R2 + t2)2 − ∂x ∂x ∂ t (canonic.) ∂ ∂ =Γ ∂t = ∂t = rt =(rtt) = ∂t tt ∂t tangent ∇ ∂t ∇ tangent ∂t tangent 4t 2R2 ∂ ∂ rt + 2t + 2R −t2 + R2 (R2 + t2)2 − ∂x ∂y tangent In this case rtt is not orthogonal to velocity: to calculate (rtt)tangent we need to extract its orthogonal component: (rtt) = rtt rtt, nt n tangent −h i We have r 1 2 2 nt = = 2tR∂x +(t R )∂y , r R2 + t2 − | | 3 4R where rt, n = 0. Hence rtt, nt = 2− 2 2 and h i h i (t +R ) (rtt) = rtt rtt, nt n = tangent −h i 2 3 4t 2R ∂ ∂ 4R 1 2 2 2t rt + 2t + 2R + 2tR∂x +(t R )∂y = − rt −t2 + R2 (R2 + t2)2 − ∂x ∂y (t2 + R2)2 · R2 + t2 − t2 + R2 We come to the answer: 2t t 2t ∂ ∂t = − ∂t, i.e.Γ = − ∇ t t2 + R2 tt t2 +R2 4 Of course we could calculate the Christoffel symbol in stereographic coordinates just using the fact that we ϕ already know the Christoffel symbol in polar coordinates: Γϕϕ = 0, hence dt dϕ dϕ d2ϕ dt d2ϕ dt Γt = Γϕ + = tt dϕ dx dx ϕϕ dt2 dϕ dt2 dϕ It is easy to see that t = R tan π + ϕ , i.e.