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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 $\mathbb{R}^2$. Let $f,g\in C^\infty(\mathbb{R})$, and observe that $\varphi:\mathbb{R}^2\rightarrow \mathbb{R}^3$ given by $\varphi(u,v)=(f(v)\cos u,f(v)\sin u,g(v))$ is an immersion assuming $f'(v)^2+g'(v)^2\not= 0$ and $f(v)\not= 0$. 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 $$ (g_{ij})=\left( \begin{array}{cc} f^2 & 0 \\ 0 & f'^2+g'^2 \end{array} \right), $$ and the local equations of a geodesic $\gamma$ are $$ \left\{ \begin{array}{l} \frac{d^2 u}{dt^2} + \frac{2ff'}{f^2}\frac{du}{dt}\frac{dv}{dt}=0 \\ \frac{d^2 v}{dt^2}-\frac{ff'}{f'^2+g'^2}\left( \frac{du}{dt} \right)^2 + \frac{f'f'' + g'g''}{f'^2+g'^2} \left( \frac{dv}{dt} \right)^2 = 0. \end{array} \right. $$

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.

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  • $\begingroup$ I realize that I probably should reparametrize away the $f'^2+g'^2$ term, but I'd prefer not to -- since this stuff confuses me so easily, I want to stick as closely to the problem itself as possible. $\endgroup$ Mar 17 '11 at 9:10
  • $\begingroup$ I believe your expression of $\frac{d}{dt} |\gamma'(t)|^2$ is incorrect, since $v$ is a function of time and $f$ is a function of $v$ hence when you write $f'$, you should have written $(f(v))'=f'(v)v'$ $\endgroup$
    – gpr1
    Jan 23 '19 at 17:06
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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".

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  • $\begingroup$ Thank you, this is very instructive. It sometimes seems like notation is the hardest part about (basic) Riemannian geometry! $\endgroup$ Mar 17 '11 at 17:02
  • $\begingroup$ I just stumbled upon this problem in do Carmo's book, and I believe after the substitution you proposed we still have a term $2u'u''f(v)^2+4v'(u')^2f(v)f'(v)$ which would be cancelled if we considered the first equation. (but then the second equation and the proposed interpretation are not equivalent!) $\endgroup$
    – gpr1
    Jan 23 '19 at 17:08

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