The Curvature and Geodesics of the Torus http://www.rdrop.com/~half/math/torus/index.xhtml Mark L. Irons ([email protected]) The torus is a standard example in introductory discussions of the curvature of surfaces. However, calculation of some measures of its curvature are hard to find in the literature. This paper offers full calculation of the torus’s shape operator, Riemann tensor, and related tensorial objects. In addition, we examine the torus’s geodesics by comparing a solution of the geodesic equation with results obtained from the Clairaut parameter- ization. Families of geodesics are classified. Open questions are considered. The connection form and parallel transport on the torus are investigated in an appendix. 1. The Line Element and Metric Our model of a torus has major radius c and minor radius a. We only consider the ring torus, for which c>a. We use a u,v coördinate system for which planes of constant u pass through the torus’s axis. = ( + ) x c a cos v cos u We parameterize the surface x by x(u, v) = y = (c + a cos v) sin u . z = a sin v Mark L. Irons 17 November 2005 1 THE CURVATURE AND GEODESICS OF THE TORUS We begin by calculating the coefficients E, F, and G of the first fundamental form. xu = (−(c + a cos v) sin u,(c + a cos v) cos u,0) xv = (−a cos u sin v, − a sin u sin v,a cos v) E = x u $ xu = (−(c + a cos v) sin u)2 + ((c + a cos v) cos u)2 + 0 = (c + a cos v)2 F = xu $ xv = (−(c + a cos v) sin u)(−a sin v cos u) + ((c + a cos v) cos u)(−a sin v sin u) + (0)(a cos v) = a sin v cos u sin u(c + a cos v) − a sin vcos u sin u(c + a cos v) = 0 G = xv $ x v = (−a sin v cos u)2 + (−a sin vsin u)2 + (a cos v)2 = a2 sin2vcos2u + a2 sin2v sin2u + a2 cos2v = a2 sin2v + a2 cos2v = a2 This gives us the line element ds2 = (c + a cos v)2du2 + a2dv2 and metric: 2 1 ( + ) 2 0 = c a cos v 0 ij = (c+acos v) gij 2 , g 1 . 0 a 0 a2 For later computations we’ll need the partial derivatives of the metric: 00 −2a sin v(c + a cos v) 0 g = , and g = ij,u 00 ij,v 00 2. The Shape Operator The normal to the surface is N = (cos u cos v,sinucos v,sinv). Taking the partial derivatives of N with respect to u and v gives the shape operator in those directions: − S(x u ) = Nu = (− sin u cos v,cosucos v,0) − S(xv ) = Nv = (− cos u sin v, − sin u sin v,cosv) Mark L. Irons 17 November 2005 2 THE CURVATURE AND GEODESICS OF THE TORUS Comparing these to xu and xv, the partial derivatives of the parameterization x, we find that they are multiples: =− cos v S(xu ) c + a cos v xu =−1 S(xv ) axv The Gaussian curvature K is the determinant of S, and the mean curvature H is the trace of S. − cos v = c+a cos v 0 = cos v K 1 a(c+a cos v) 0 − a 1 cos v 1 1 a cos v c+a cos v c+2a cos v = (− + − ) = − − = H 2 c a cos v a 2 a(c+a cos v) a(c+a cos v) 2a(c+a cos v) 3. The Curvature Tensor The Christoffel symbols of the second kind £u = 1 [guu(g + g − g ) + guv(g + g − g )] uu 2 uu,u uu,u uu,u vu,u vu,u uu,v = 1 [guu(0 + 0 − 0) + 0(g + g − g )] 2 vu,u vu,u uu,v = 0 £u = 1 [guu(g + g − g ) + g uv(g + g − g )] uv 2 uv,u uu,v uv,u vv,u vu,v uv,v = 1 [guu(0 + g − 0) + 0(g + g − g )] 2 uu,v vv,u vu,v uv,v = 1 guug 2 uu,v − + = 1 2a sin v(c a cos v) 2 (c + a cos v)2 =− asin v (c + a cos v) £u = 1 [guu(g + g − g ) + guv(g + g − g )] vu 2 uu,v uv,u vu,u vu,v vv,u vu,v = 1 [guu(g + 0 − 0) + 0(g + g − g )] = 1 guug = £u 2 uu,v vu,v vv,u vu,v 2 uu,v uv =− asin v (c + a cos v) £u = 1 [guu(g + g − g ) + guv(g + g − g )] vv 2 uv,v uv,v vv,u vv,v vv,v vv,v = 1 [guu(0 + 0 − 0) + 0(g + g − g )] 2 vv,v vv,v vv,v = 0 Mark L. Irons 17 November 2005 3 THE CURVATURE AND GEODESICS OF THE TORUS £v = 1 [0(g + g − g ) + gvv(0 + 0 − g )] uu 2 uu,u uu,u uu,u uu,v =−1gvvg 2 uu,v =−1 1 (−2asin v(c + a cos v)) 2 a2 = 1 + a sin v(c a cos v) £v = 1 [gvu(g + g − g ) + g vv(g + g − g )] uv 2 uv,u uu,v uv,u vv,u vu,v uv,v = 1 [0(g + g − g ) + gvv(0 + 0 − 0)] 2 uv,u uu,v uv,u = 0 £v = 1 [gvu(g + g − g ) + gvv(g + g − g )] vu 2 uu,v uv,u vu,u vu,v vv,u vu,v = 1 [0(g + g − g ) + gvv(0 + 0 − 0)] 2 uu,v uv,u vu,u = 0 £v = 1 [gvu(g + g − g ) + gvv(g + g − g )] vv 2 uv,v uv,v vv,u vv,v vv,v vv,v = 1 [0(g + g − g ) + gvv(0 + 0 − 0)] 2 uv,v uv,v vv,u = 0 Partial derivatives of the nonzero Christoffel symbols: u u −2 −1 £uv,v = £vu,v =− (asin v)(−1)(c + a cos v) (−a sin v) + (c + a cos v) (a cos v) − − =−(asin v)2(c + a cos v) 2 − (c + a cos v) 1(a cos v) 2 =− (asin v) − a cos v (c + a cos v)2 (c + a cos v) £v = 1 − + + = 1 + 2 − 2 uu,v a [sin v( a sinv) (c a cos v) cos v] a [c cos v a cos v a sin v] The Riemann tensor i =− i Throughout this section we use the identity R jkl Rjlk. u u u u u u v u u u v R uuu = £uu,u − £ uu,u − £ uu£uu − £vu£uu + £uu£uu + £ vu£uu u = 0 − 0 − 0 − £vu % 0 + 0 + 0 = 0 u u u u u u u v u u u v R uuv =−Ruvu = £uv,u − £ uu,v − £uv£uu − £vv£uu + £uu£uv + £vu£uv = 0 − 0 − 0 − 0 + 0 + 0 = 0 u u u u u u v u u u v R uvv = £uv,v − £uv,v − £uv£uv − £vv£uv + £uv£uv + £vv£uv u u u 2 u 2 = £uv,v − £uv,v − (£uv ) − 0 + (£uv ) + 0 = 0 Mark L. Irons 17 November 2005 4 THE CURVATURE AND GEODESICS OF THE TORUS u u u u u u v u u u v R vuu = £vu,u − £vu,u − £uu£vu − £vu£vu + £uu£vu + £vu£vu = 0 − 0 − 0 − 0 + 0 + 0 = 0 u u u u u u u v u u u v R vuv =−Rvvu = £vv,u − £vu,v − £uv£vu − £ vv£vu + £uu £vv + £vu£vv u u 2 = 0 − £vu,v − (£vu ) − 0 + 0 − 0 2 2 =− − asin v − a cos v − a sin v c + a cos v c + a cos v c + a cos v 2 2 = a sin v + a cos v − a sin v c + a cos v c + a cos v c + a cos v = a cos v c + a cos v u u u u u u v u u u v R vvv = £vv,v − £vv,v − £uv£vv − £vv£vv + £uv£vv + £vv£vv = 0 − 0 − 0 − 0 + 0 + 0 = 0 v v v v u v v v u v v R uuu = £uu,u − £uu,u − £uu£uu − £vu£uu + £uu£uu + £vu£uu = 0 − 0 − 0 − 0 + 0 + 0 = 0 v v v v v u v v v u v v R uuv =−Ruvu = £uv,u − £ uu,v − £uv£uu − £vv£uu + £uu£uv + £vu£uv v v u = 0 − £uu,v − 0 − 0 + £uu£uv + 0 =−1 − + + + 1 + − a sin v a(sin v( a sin v) cos v(c a cos v)) a sin v(c a cos v) c + a cos v =−1 − 2 + + + 2 a[ asin v cos v(c a cos v) a sin v] =−1 + a cos v(c a cos v) v v v v u v v v u v v R uvv = £uv,v − £uv,v − £uv£uv − £vv£uv + £uv£uv + £vv£uv = 0 − 0 − 0 − 0 + 0 + 0 = 0 v v v v u v v v u v v R vuu = £vu,u − £vu,u − £uu£vu − £vu£vu + £uu£vu + £vu£vu v u v u = 0 − 0 − £uu£vu − 0 + £uu£vu + 0 = 0 v v v v v u v v v u v v R vuv =−Rvvu = £vv,u − £vu,v − £uv£vu − £ vv£vu + £uu £vv + £vu£vv = 0 − 0 − 0 − 0 + 0 + 0 = 0 v v v v u v v v u v v R vvv = £vv,v − £vv,v − £uv£vv − £vv£vv + £uv£vv + £vv£vv = 0 − 0 − 0 − 0 + 0 + 0 = 0 The Ricci tensor = m Rij R imj = m = 1 + Ruu R umu a cos v(c a cos v) = m = a cos v Rvv R vmv c + a cos v 1 a cosv(c + a cos v) 0 Rij = a cos v 0 (c+a cos v) Mark L.