Question

Only need prob 4
3. Geodesics on the sphere. Consider the 2-sphere with coordinates (0,0) and metric dsa = aʼ(do? + sin? 0d$2) Note that a is the constant radius of the sphere. (a) Show that lines of constant longitude (ø = po = constant) are geodesics. (b) Show that the only line of constant latitude (0 = 0o = constant) that is a geodesic is the equator (@0 = 7/2).

Answer 1

Answer ③. we're ds=ar (dot sinho do) - goo= ah , goo=asino goo - 960=0 from these we can calculate the Christoffel symbols buriy mi žgim (ad; @ma + Oxe Gm; – Omn Gix) by which we get. only non zero Christoffel symbors - coto O .sinoCOSO (As parameter) The geodesic eqn is given by e + Fiesta e centro so, we get the geodesic equ as O corean.) -(1) - sinacos 2 Co.ew.) - (2) de + 200+0 do dos ao longitude (0=0o = cont) are given by (a) The lines of const

So, the o. em. W gives 0 LH-S-0-0-0 = Rinos <the 8 ear (2) ginen R +)-S Oto LH-S-O + O So, au lines of const congitude are geodesilos - (B) o for lines of const- latitude, (o=do=const). : 0. гар у теа9. LH-S= 0 - sino coso = 0 = sin 20 = 0 » O=o or 8 = 5/2 (equator) 0-com (2) reas UR.HAS Lil--S = O geodesic hence, line provides of const. latitude ois a 0=o or oth (equator)

® The geodesic ear. to the vectro field ū becomes dva Ⓡear : 40 d x 0 - dve - sino coso vesco ..) e ean: dureret ho v edrolo 20 dvø + core vozo -.(a) Diff. wat a, ) mest be oms. sinocosolution= 0 → od one sino cos 6 (-coto vo) -o (oring wy) sinocosol 2 davo diz + costo voza + 4 Loc..(5) Kocosto

Similarly diff. Wort 2 of ean. (4) become drvo + kry co ..(6) da2 tkrv 0 general sol. nQ(a) = A sin(kx) + B cos(kx) vº (4) , C n (k) + Dos (K+) h tu as to have a for simplicity we take =(4,0) :. initial conditions are vQ(120) - => Bas VW (120) = 0 => D=0 = voca) = Asin(ka) + cos((a)? --(7) VⓇ(X) = Csin(kx) J but we also need to satisfy em B) e saj Pluggin ew. I) in 3 to (3) & a given 150, CE - Tesco = - the (K=(ra) Sino cos vin) = (cos (rose), -stro sin Cucosoj) |