I’m having problems with exercise 1 of chapter 3 of do Carmo’s “Riemannian Geometry”. Here is the background:

Let (u,v) be the coordinates on R2. Let f,g∈C∞(R), and observe that φ:R2→R3 given by φ(u,v)=(f(v)cosu,f(v)sinu,g(v)) is an immersion assuming f′(v)2+g′(v)2≠0 and f(v)≠0. The image is the surface of revolution generated by the curve (f(v),g(v)) being rotated about the z-axis. The induced metric is

(gij)=(f200f′2+g′2),

and the local equations of a geodesic γ are

{d2udt2+2ff′f2dudtdvdt=0d2vdt2−ff′f′2+g′2(dudt)2+f′f″Then, do Carmo says: Obtain the following geometric meaning of the equations above: the second equation is, except for meridians (u=u_0) and parallels (v=v_0), equivalent to the fact that the “energy” |\gamma'(t)|^2 of a geodesic is constant along \gamma; the first equation signifies that if \beta(t) is the oriented angle, \beta(t)<\pi, of \gamma with a parallel P intersecting \gamma at \gamma(t), then r\cos \beta is constant, where r is the radius of P.

This last paragraph is what's confusing me. First of all, I've seen the energy of a path as \int_a^b |\gamma'(t)|^2 dt, so maybe that's why he put "energy" in quotes. But also, geodesics have constant speed! So this should be constant along all geodesics too (regardless of whether they're meridians or parallels). But I figured that maybe I should just blindly plug and chug since that seems to work scarily often in Riemannian geometry, so I found |\gamma'(t)|^2, took its t-derivative, and substituted in the second equation in the system for geodesics. Here it is:

|\gamma'(t)|^2 = \left\langle \frac{d \gamma}{dt} , \frac{d \gamma}{dt} \right\rangle = u'^2 g_{11} + 2u'v'g_{12} + v'^2 g_{22} = u'^2f^2+v'^2(f'^2+g'^2)

so

\frac{d}{dt} |\gamma'(t)|^2 = 2u'u''f^2+2u'^2ff' + 2v'v''(f'^2+g'^2)+v'^2(2f'f''+2g'g'')

= 2u'u''f^2+2u'^2ff' + 2v'(ff'u'^2-(f'f''+g'g'')v'^2)) + 2v'v''(f'^2+g'^2)+2v'^2(f'f''+g'g'')

= 2u'u''f^2+2u'^2ff'(1+v')+2v'v''(f'^2+g'^2)+2v'^2(f'f''+g'g'')(1-v')and this looks hopeless.

**Answer**

What you did wrong: the function f should be treated as f = f(\gamma(t)) = f(u(t),v(t)). So \frac{d}{dt}f = \partial_u f \frac{d}{dt}u + \partial_vf\frac{d}{dt}v using the chain rule. By the parametrization, you \partial_uf = 0, while \partial_v f is what you wrote f' originally. (Whereas in the above you implicitly wrote \frac{d}{dt}f^2 = 2f f', which is wrong.)

This is an instance where you let notation get in your way. If would be clearer if you reserve the \prime for f', the derivative of the function f relative to the parameter v and g' for the derivative of the function f relative to parameter v, and use explicitly either \frac{d}{dt} or \cdot to denote time derivatives along the geodesic (as the geodesic equations that you originally copied down does).

For deriving the "conservation of energy", it may help if instead of the second equation in the two geodesic equations (also, I think there may be a sign error in it, but I am not 100% certain), you look at that equation multiplied by (f'^2 + g'^2)\frac{dv}{dt}, that is

(f'^2 + g'^2) \dot{v} \ddot{v} + ff'\dot{v}(\dot{u})^2 + (f'f'' + g'g'')(\dot{v})^3 = 0

Incidentally, the second thing about angle against the parallels is called "Clairaut's relation".

**Attribution***Source : Link , Question Author : Aaron Mazel-Gee , Answer Author : Willie Wong*