DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

66 Chapter 2. Surfaces: Local Theoryb ′ (t) =− cot u 0 a(t), b(0) =1.We solve this system by differentiating the second equation again and substituting the first:Thus, the solution isb ′′ (t) =− cot u 0 a ′ (t) =− cos 2 u 0 b(t), b(0) = 1.a(t) =sin u 0 sin ( (cos u 0 )t ) , b(t) =cos ( (cos u 0 )t ) .Note that ‖X(α(t))‖ 2 = Ea(t) 2 +2Fa(t)b(t)+Gb(t) 2 = sin 2 u 0 for all t. That is, the original vectorX 0 rotates as we parallel translate it around the latitude circle, and its length is preserved. Aswe see in Figure 4.2, the vector rotates clockwise as we proceed around the latitude circle (in theP2π cos u 0u 0X 0Figure 4.2upper hemisphere). But this makes sense: If we just take the covariant derivative of the tangentvector to the circle, it points upwards (cf. Figure 3.1), so the vector field must rotate clockwise tocounteract that effect in order to remain parallel. ▽Example 4 (Foucault pendulum). Foucault observed in 1851 that the swing plane of a pendulumlocated on the latitude circle u = u 0 precesses with a period of T =24/ cos u 0 hours. Wecan use the result of Example 3 to explain this. We imagine the earth as fixed and “transport” theswinging pendulum once around the circle in 24 hours. If we make the pendulum very long andthe swing rather short, the motion will be “essentially” tangential to the surface of the earth. If wemove slowly around the circle, the forces will be “essentially” normal to the sphere: In particular,letting R denote the radius of the earth, the tangential component of the centripetal accelerationis (cf. Figure 3.1)( ) 2π 2(R sin u 0 ) cos u 0 ≤ 2π2 R24 24 2 ≈ 0.0558 ft/sec 2 ≈ 0.17%g.Thus, the “swing vector field” is, for all practical purposes, parallel along the curve. Therefore, it2πturns through an angle of 2π cos u 0 in one trip around the circle, so it takes(2π cos u 0 )/24 = 24cos u 0hours to return to its original swing plane. ▽