1.7 Geodesics. Recall. For curve α on surface M, α can be written as components tangent and normal to M as α = α tan + α nor where

Transcription

1 1.7 Geodesics Note. A curve α(s) on surfce M cn curve in two different wys. First, α cn bend long with surfce M (the norml curvture discussed bove). Second, α cn bend within the surfce M (the geodesic curvture to be defined). Recll. For curve α on surfce M, α cn be written s components tngent nd norml to M s α α tn + α nor where α tn (ur + Γ r ij ui u j ) X r α nor (L ij u i u j ) U nd the prmeters on the right hnd side re defined in Section 5. α nor reflects the curvture of α due to the bending of M nd α tn reflects the curvture of α within M. Now α tn U {(u r + Γ r iju i u j ) X } r U 0 (recll U X 1 X 2 X X 2 ) nd α tn α α tn α + 0 α tn α + α nor α (recll α u i Xi nd X i U 0) ( α tn + α nor ) α α α 0 (recll α α (s) 1 nd d/ds). Therefore α tn is orthogonl to both U nd α. If we define w s the unit vector w U α, then α tn is multiple of w (nd w is vector tngent to M). 1

2 Definition I-7. Let α(s) be curve on M where s is rc length. The geodesic curvture of α t α(s) is the function k g k g (s) defined by α tn k g w k g ( U α ). Recll. The sclr triple product of three vectors (in R 3 ) stisfies: ( A B) C ( B C) A ( C A) B. Theorem. The geodesic curvture k g of curve α in surfce M cn be clculted s k g U α α. Proof. Since k g w α tn we hve k g w w α tn w α tn ( U α ) or k g ( α tn + α nor ) U α (since α nor is prllel to U) α U α U α α. Definition I-8. Let α α(s) be curve on surfce M. Then α is geodesic if α tn 0 ( or equivlently, if α α nor) t every point of α. 2

3 Note. A geodesic on surfce is, in sense, s stright s curve cn be on the surfce. Tht is, α hs no curvture within the surfce. For exmple, on sphere the geodesics re gret circles. Note. If α is geodesic on M then u r + Γ r ij ui u j 0 for r 1,2 nd U α α 0. (We ll use these LOTS!) Exmple (Exercise 1.7.4()). Prove tht on surfce of revolution, every meridin is geodesic. Proof. Suppose X(u,v) (f(u) cosv,f(u) sin v,g(u)). Let m(s) (f(s) cosv,f(s) sin v,g(s)) be meridin of the surfce (where we ssume the curve hs been prmeterized in terms of rclength s). Then m (s) (f (s) cosv,f (s) sinv,g (s)) m (f (s) cosv,f (s) sinv,g (s)) m m ((f (s)g (s) f (s)g (s)) sinv, ( f (s)g (s) + f (s)g (s)) cosv, 0). Now X 1 X 2 ( f(s)g (s) cosv, f(s)g (s) sinv,f (s)f(s)) 3

4 nd so U X 1 X 2 X 1 X 2 ( f(s)g (s) cosv, f(s)g (s) sinv,f (s)f(s)) (f(s)g (s)) 2 + (f (s)f(s)) 2. Therefore U m m 1 (f(s)) 2 {(g (s)) 2 + (f (s)) 2 } {(f (s)g (s) f (s)g (s))( f(s)g (s)) cosv sin v +( f (s)g (s) + f (s)g (s))( f(s)g (s)) cosv sin v, 0)} 1 (f(s)) 2 {(g (s)) 2 + (f (s)) 2 (0) 0. } Therefore m(s) is geodesic. Definition. Let X(u 1,u 2 ) be surfce nd let g ij (see pge 34) nd Γ r ij (see pge 43) be s defined in Sections 4 nd 5. The Christoffel symbols of the first kind re for i,j,k 1,2. Γ ijk Γ r ijg rk Definition. The Γ r ij of the second kind. defined in Section 1.5 re the Cristoffel symbols Note. Since Γ r ij Γr ji (see (17), pge 43) then Γ ijk Γ jik. Also, since (g ij ) 1 (g ij ), we hve Γ m ij Γ ijkg km. 4

5 Theorem. Let X(u 1,u 2 ) be surfce nd let g ij nd Γ r ij be s defined in Sections 4 nd 5. Then nd Γ ijk X ij X k Γ ijk 1 ( gik 2 u + g jk g ) ij j u i u k Γ r ij 1 ( gik 2 gkr u j + g jk u i g ) ij u k. Proof. Since X ij Γ r ij X r + L ij U (by definition, see pge 43) then X ij X k Γ r ij X r X k + (L ij U) Xk Γ r ij g rk + 0 Γ ijk estblishing the first identity (recll g rk X r X k ). Next, g ik u j u j [ X i X k ] X ij X k + X kj X i Γ ijk + Γ kji. Permuting the indices: Now g ji u k Γ jki + Γ ikj nd g kj u i Γ kij + Γ jik. Γ ijk 1 2 (2Γ ijk) 1 2 (Γ ijk + Γ jik ) 1 2 (Γ ijk + Γ kji Γ kji + Γ kij Γ kij + Γ jik ) 1 2 {(Γ ijk + Γ kji ) + (Γ kij + Γ jik ) (Γ jki + Γ ikj )} 1 ( gik 2 u + g jk g ) ij j u i u k nd the second identity is estblished. Finlly, multiplying this identity on both sides by g kr, summing over k nd using the definition of Γ r ij we 5

6 hve Γ r ij Γ ijk g kr 1 ( gik 2 gkr u j + g jk u i g ) ij u k (recll (g ij (g ij ) 1 ), nd the third identity is estblished. Note. Since the Christoffel symbols depend only on the metric form (or First Fundmentl Form), they re prt of the intrinsic geometry of the surfce M. Definition. Let X(u 1,u 2 ) be surfce. Then the coordintes X 1 nd X 2 re orthogonl if g 12 g (This mkes sense since g ij X i X j.) Corollry. Let X(u 1,u 2 ) be surfce nd let g ij nd Γ r ij be s defined in Sections 4 nd 5. If X 1 nd X 2 re orthogonl coordintes, then Γ r ij 1 ( gir 2g rr u + g jr g ) ij j u i u r (no sums over ny of i,j,r). Proof. Since g 12 g 21 0, then g 12 g 21 0 nd g 11 1/g 11, g 22 1/g 22. The result follows from the bove theorem. Corollry. With the hypotheses of the previous corollry (with i, j, r 1,2), when j r nd when i j r Γ r ir 1 g rr 1 2g rr u i 2 u i[ln g rr] Γ r ii 1 ( g ) ii 2g rr u r. 6

7 Proof. Follows from g 12 g Note. By symmetry, Γ r ij Γr ji, nd so the previous two corollries cover ll possible cses when i,j,r {1,2} (i.e. when we del with two dimensions). In dimensions 3 nd greter (in prticulr, in the 4 dimensionl spcetime of Chpter III) we hve third cse which we stte now, nd ddress in detil lter: Theorem. In dimensions 3 nd greter, if coordintes re mutully orthogonl, then for i,j,r ll distinct, Γ r ij 0. (In the event tht one or more of i,j,r re equl, the bove corollries pply.) Note. In the cse of orthogonl coordintes, if we return to Guss nottion: g 11 E, g 12 g 21 F 0, g 22 G we hve the First Fundmentl Form (or metric form) ds 2 Edu 2 + Gdv 2 on surfce X(u,v). In this nottion, the Christoffel symbols re then Γ 1 11 E u 2E Γ 1 12 Γ1 21 E v 2E Γ 1 22 G u 2E Γ 2 22 G v 2G Γ 2 21 Γ2 12 G u 2G Γ 2 11 E v 2G. Exmple 17, pge 62. In the Eucliden plne, ds 2 du 2 + dv 2. Therefore E G 1 nd ll the Christoffel symbols re 0. Therefore 7

8 geodesic α stisfies u r + Γ r iju i u j 0 for r 1,2, or u r 0 for r 1,2. Tht is, u 1 u 0 nd u 2 v 0. Therefore u(s) s + b nd v(s) cs + d for some, b, c, d. Therefore, geodesics in the Eucliden plne re stright lines. Note. We will show in Theorem I-9 tht the shortest pth on surfce joining two points is geodesic. This theorem, combined with the previous exmple PROVES tht the shortest distnce between two points in plne is stright line. Oddly enough, you ve probbly never seen this PROVEN before! Exmple 18, pge 62. Consider sphere of rdius r with geogrphic coordintes (like ltitude nd longitude) u nd v. Then the sphere is given by X(u,v) (r cos ucosv,r sin ucosv,r sin v) (see Exmple 7, pge 23). The metric form is (see pge 33) ds 2 r 2 cos 2 vdu 2 + r 2 dv 2 (since there is no dudv term, F g 12 g 21 0 nd these coordintes re orthogonl). Therefore E r 2 cos 2 v nd G r 2 ( constnt). Then E u G u G v 0 nd the nonzero Christoffel symbols re Γ 1 12 Γ1 21 E v 2E 2r2 cosv sin v 2r 2 cos 2 v tn v Γ 2 11 E v 2G 2r2 cosv sin v cosv sinv. 2r 2 It is shown in Exercise (t the end of this section) tht this implies geodesics re gret circles. 8

9 Note. In Exmple 19 pge 62, it is shown tht the Eucliden plne when equipped with polr coordintes (which re orthogonl coordintes) yields geodesics which re lines (s expected). Note. In generl, to determine the geodesics for surfce, requires tht one solve differentil equtions. This cn be difficult (sometimes impossible to do in terms of elementry functions). In Chpter III we will compute some geodesics in 4-dimensionl spcetime (in fct, plnets nd light follow geodesics if 4-D spcetime). Theorem I-9. Let α(s), s [,b] be curve on the surfce M : X(u 1,u 2 ), where s is rclength. If α is the shortest possible curve on M connecting its two end points, then α is geodesic. Ide of Proof. We will vry α(s) by slight mount ε. Then compring the rclength of α from α() to α(b) with the rclength of the slightly vried curve from α() to α(b) nd ssuming α to yield the miniml rclength, we will show tht α stisfies eqution (32) (pge 59) nd is therefore geodesic. Proof. Let α(s) X(u 1 (s),u 2 (s)). Consider the fmily of curves α ε (s) of the form U i (s,ε) u i (s) + εv i (s) for i 1,2, s [,b] where v i re smooth functions with v i () v i (b) 0 for i 1,2 (so (U 1,U 2 ) still joins α() nd α(b)), (U 1,U 2 ) M, but otherwise v i re rbitrry. 9

10 Let L(ε) denote the length of α ε : where λ(s,ε) L(ε) b λ(s,ε) ds {g ij (U 1,U 2 ) Ui s U j } 1/2 s (the squre root of the metric form of M long α ε ). Now L hs minimum t ε 0 so d dε [L(ε)] d dε [ b ] b λ(s,ε) ds (since λ nd λ/ ε re continuous) stisfies Now λ ε ε L (0) b [ ( g ij(u 1,U 2 ) Ui s [λ(s, 0)]ds 0. ε U j ) 1/2 ] s ε [λ(s,ε)]ds 1 { 2 (λ(s,ε)) 1 ε [g ij(u 1,U 2 )] Ui U j s s +g ij (U 1,U 2 ) [ ] [ ]} U i U j ε s s + g ij(u 1,U 2 ) Ui U j s ε s {( 1 2λ(s, ε) U 1[g ij(u 1,U 2 )] U1 ε + ) U 2[g ij(u 1,U 2 )] U2 ε Ui U j [ ]} s s + 2g ij(u 1,U 2 ) Ui U j s ε s {( ) 1 2λ(s, ε) U k[g ij(u 1,U 2 )] Uk U i U j ε s s +2g ij (U 1,U 2 ) Ui 2 U j } s ε s {( ) 1 gij U i U j 2λ(s, ε) U kvk s s + 2g U i 2 U j } ij s ε s 10

11 since Uk ε vk. With ε 0, Uj ε vj nd λ(s, 0) 1 (s is rclength on α α 0 ) we hve λ ε (s, 0) 1 ( ) gij U i U j + 2g 2 U kvk ik U i v k nd since ε 0 implies U i u i, then λ ε (s, 0) 1 2 ( gij u k vk u i u j + 2g ik u i v k nd so L (0) 1 b ( ) gij 2 u k ui u j v k + 2g ik u i v k ds 0. Now by Integrtion by Prts b 2g ik u i v k ds Therefore Let u 2g ik u i nd dv v k ds. Then du s [2g iku i ]ds nd v { } b 2g ik u i v k s [2g iku i ]v k ds 0 L (0) b ) v k (s)ds v k. s [2g iku i ]v k ds since v k () v k (b) 0. b ( gij b ) ds u k ui u j v k s [2g iku i ]v k ( gij u k ui u j ) s [2g iku i ] v k ds Since the integrl must be zero for ll rbitrry v k, then the remining prt of the integrnd must be zero: 1 g ij 2 u k ui u j s [g iku i ] 0 11

12 for k 1,2. Now when ε 0, U i u i nd Therefore s [g iku i ] s [g ik(u 1,u 2 )u i ] for k 1,2 implies du 1 u 1 ds + g ik du 2 ) u i + g u 2 ik (u 1,u 2 ) dui ds ds u 1 u1 + g ) ik u 2 u2 u i + g ik (u 1,u 2 )u i ( gik ( gik ( gik u j uj ) u i + g mk u m g ik u j uj u i + g mk u m. 1 g ij 2 u k ui u j s [g iku i ] 0 1 g ij 2 u k ui u j g ik u j ui u j g mk u m 0 for k 1,2, or using the nottion of eqution (35) (pge 60) { } 1 2 (Γ ikj + Γ jki ) (Γ kji + Γ ijk ) u i u j g mk u m 0 for k 1,2. Since Γ ikj u i u j Γ jki u i u j (interchnging dummy vribles i nd j) nd Γ kji Γ jki (symmetry in the first nd second coordintes) then {1 } 2 (Γ ikj + Γ jki ) Γ kji u i u j nd the bove eqution ( ) becomes ( ) { } 1 2 (Γ jki + Γ jki ) Γ jki u i u j 0 Γ ijk u i u j + g mk u m 0 for k 1,2. Multiplying by g kr nd summing over k: Γ ijk g kr u i u j + g kr g mk u m 0 12

13 for r 1,2 or Γ r iju i u j + u r 0 for r 1,2. This is eqution (32) nd therefore α is geodesic of M. Note. Agin, Theorem I-9 long with Exmple 17 shows tht the shortest distnce between two points in the Eucliden plne is stright line. Theorem I-9 long with Exmple 18 show tht the shortest distnce between two points on sphere is prt of gret circle (explining pprently unusul routes on interntionl irline flights). Note. The converse of Theorem I-9 is not true. Tht is, there my be geodesic joining points which does not minimize distnce. (Recll tht we set L (0) 0, but did not check L (0) - we my hve mximum of L!) For exmple, we cn trvel the six miles from Johnson City to Jonesboro (long very smll piece of geodesic), or we cn trvel in the opposite direction long very lrge piece of geodesic ( 24,000 miles) nd trvel round the world to get to Jonesboro (NOT minimum distnce). Note. Not ll surfces my llow one to crete geodesic joining rbitrry points. For exmple, the Eucliden plne minus the origin does not dmit geodesic from (1,1) to ( 1, 1). Note. In the next theorem, we prove tht for ny point on surfce, there is unique (directed) geodesic through tht point in ny direction. 13

14 Theorem I-10. Given point P on surfce M nd unit tngent vector v t P, there exists unique geodesic α such tht α(0) P nd α (0) v. Proof. Let P X(u 1 0,u2 0 ) nd v vi Xi (u 1 0,u2 0 ). We need two functions u r (t), r 1,2 where { u r + Γ r ij ui u j 0 for r 1,2 u r (0) u r 0, ur (0) v r for r 1,2. This is system of two ordinry differentil equtions in two unknown functions, ech with two initil conditions. Such system of IVPs hs unique solution (check out the chpter of n ODEs book entitled Existence nd Uniqueness Theorems ) u r (t) for r 1,2. We now only need to estblish tht t represents rclength. With s equl to rclength, ( ) 2 ds E dt ( ) 2 ( du du + 2F dt dt ) dv + G dt ( ) 2 dv g iju i u j f(t) dt is the metric form nd if we show this quntity is 1, then t s nd t equls rclength (we need α(0) v to eliminte the negtive sign; this is insured by the initil conditions). Well, f(0) g ij (u 1 0,u2 0 )ui (0)u j (0) X i (u 1 0,u2 0 ) X j (u 1 0,u2 0 )ui (0)u j (0) X i (u 1 0,u2 0 ) X ) ) j (u 1 0,u2 0 )vi v j (v i Xi (u 1 0,u2 0 ) (v j Xj (u 1 0,u2 0 ) v v v 2 1. Next, f (t) g ij u k uk u i u j + g ij u i u j + g ij u i u j. 14

15 Since g ij u k Γ ikj + Γ jki (eqution (35b), pge 60) Γ r ik g rj + Γ r jk g ri (eqution (33), pge 59) g jr Γ r ik + g irγ r jk (symmetry of g ij) then f (t) (g jr Γ r ik + g irγ r jk )ui u j u k + g rj u r u j + g ir u i u r g ir u i (u r + Γ r jku j u k ) + g rj u j (u r + Γ r iku i u k ) 0 (from the first condition of the ODE). ( ) 2 ds Therefore f(t) is constnt nd f(t) 1. Hence f(t) 1 dt nd t s (tht is, t is rclength). Therefore α(s) X(u 1 (s),u 2 (s)) is the desired geodesic. Exmple (Exercise ()). If M hs metric form ds 2 Edu 2 + Gdv 2 with E u G u 0, then geodesic on M stisfies du dv h G E E h 2 for some constnt h (see Exercise ). Use this bove eqution to show tht geodesic on the geogrphic sphere stisfies X(u,v) (R cosucos v,rsin ucos v,rsin v) du dv where h is constnt. h sec 2 v R2 h 2 sec 2 v 15 h sec 2 v R2 h 2 h 2 tn 2 v

16 Solution. First, X 1 ( R sin ucosv,r cosucos v, 0) X 2 ( R cosusin v, R sinusin v,rcos v) E g 11 X 1 X 1 R 2 cos 2 v G g 22 X 2 X 2 R 2 sin 2 v + R 2 cos 2 v R 2. Then du dv h R 2 R2 cos 2 v R 2 cos 2 v h 2 hr R cosv since v ( π/2,π/2) R 2 cos 2 v h2 h sec v cosv R 2 h 2 sec 2 v h sec 2 R2 h 2 (1 + tn 2 v) h sec 2 v R2 h 2 h 2 tn 2 v. Exmple (Exercise (b)). Substitute w h tn v nd integrte the bove eqution to obtin cos(u u 0 ) + γ tnv 0 where u 0 nd γ re constnts. Solution. With w h tnv, dw h sec 2 v dv nd so h 2 sec 2 v u R2 h 2 h 2 tn 2 v dv ( ) 1 R2 h 2 w 2dw w cos 1 + u 0. R2 h 2 Therefore cos(u u 0 ) w R2 h 2 h tnv R2 h 2. 16

17 With γ h/ R 2 h 2 we hve cos(u u 0 ) + γ tn v 0. Exmple (Exercise (c)). Show tht the eqution given in (b) when written in Crtesin coordintes is liner eqution of the form αx+βy+γz 0 nd so represents the intersection of the sphere with plne pssing through the origin (nd therefore the geodesic is gret circle). Solution. Multiplying by R cosv on ech side of the eqution gives R cosv cos(u u 0 )+γr sin v 0 or R cosv(cosucos u 0 +sin usin u 0 )+ γr sinv 0 or (cos u 0 )Rcos ucosv + (sin u 0 )Rsin ucos v + γr sin v 0. In cylindricl coordintes, ρ, θ, ϕ, we hve the reltionships x r cosθ sin ϕ, y ρ sin θ sin ϕ, nd z ρ cos ϕ. Here, our R corresponds to cylindricl coordintes ρ nd our u corresponds to cylindricl coordintes θ. However, our v does not correspond to ϕ of cylindricl coordintes but insted corresponds to π/2 ϕ (see my Clculus 3 notes Our coordintes relte to rectngulr coordintes s x R cos u sin(π/2 v) R cosucos v, y R sin usin(π/2 v) R sin ucos v, nd z R cos(π/2 v) R sin v. So the bove eqution is of the form αx + βy + γz 0 where α cosu 0, β sin u 0, nd γ Lst updted: June 19, 2016