Let $\mathcal{S}$ be a surface of revolution $$x^2+y^2=R^2(z).$$ By Pythagoras, the length element is $$\sqrt{dx^2+dy^2+dz^2}$$ and converting to cylindrical coordinates it is $$\sqrt{r^2d\theta^2+dz^2+dr^2}$$ which is equivalent to $$dz\sqrt{r^2\left(\frac{d\theta}{dz}\right)^2+\left(\frac{dr}{dz}\right)^2+1}=\mathcal{L}dz.$$ Applying the Euler-Lagrange equation in $\theta$ gives $$ \frac{\partial\mathcal{L}}{\partial \theta}=0=\frac{d}{d\theta}\left(\frac{\partial\mathcal{L}}{\partial \theta'}\right)$$ which means $$ \frac{\partial\mathcal{L}}{\partial \theta'}=\frac{\theta'}{\sqrt{\left(r\theta'\right)^2+r'^2+1}}=constant=\sqrt{C}, $$ where prime denotes derivative with respect to $z$. I am asked to show that the conserved quantity, called Clairault's integral, is given by $$r\sin\theta.$$ I don't know what Clairault's integral is. Clearly, after some algebra, the separable differential equation $$\frac{d\theta}{dz}=\sqrt{\frac{C\left(r'^2+1\right)}{1-Cr^2}} $$ is involved, which can be solved by direct integration, but I'm not sure how to proceed. What is Clairault's integral and how do we simplify the conserved quantity? For reference, this problem is taken from Chapter 5 of Mark Levi's book Classical Mechanics with Calculus of Variations and Optimal Control.


Sketched proof:

  1. We use cylindrical coordinates $(r, \varphi, z)$ with standard metric $$g~=~\mathrm{d}r^2 + r^2\mathrm{d}\varphi^2 + \mathrm{d}z^2. \tag{1}$$ (Note that we have renamed OP's $\theta$ variable as $\varphi$, because Ref. 1 uses the notation $\theta$ for another purpose, cf. eq. (7) below.)

  2. The induced metric on the surface of revolution $r=R(z)$ is $$h~=~ R(z)^2\mathrm{d}\varphi^2 + (1+R^{\prime}(z)^2)\mathrm{d}z^2.\tag{2}$$

  3. The arc-length of a (not necessarily geodesic) curve $\gamma$ is given by the integral $$ S~=~\int_{\lambda_i}^{\lambda_f}\! d\lambda~L, \qquad L~:=~ \sqrt{L_0} \tag{3},$$ $$L_0~:=~ R(z)^2\dot{\varphi}^2+ (1+R^{\prime}(z)^2)\dot{z}^2,\tag{4}$$ where $\lambda$ is the parametrization parameter, and dot denotes differentiation wrt. $\lambda$.

  4. In classical mechanics, the quantities $S$ and $L$ play the role of action and Lagrangian, respectively. One may show that the Euler-Lagrange (EL) equations for the square root Lagrangian $L$ and the non-square-root Lagrangian $L_0$ are both the geodesic equations, although the latter is affinely parametrized, cf. my Phys.SE answer here.

  5. Note that the Lagrangian $$L~=~\frac{ds}{d\lambda}\tag{5}$$ has an interpretation as a derivative of arc-length wrt. the parametrization $\lambda$.

  6. Clairault's integral is apparently the canonical/conjugate momentum $$p_{\varphi} ~:=~\frac{\partial L}{\partial \dot{\varphi}}~\stackrel{(3)+(4)}{=}~\frac{R(z)^2\dot{\varphi}}{L}~\stackrel{(5)}{=}~R(z)^2\frac{d\varphi}{ds}~\stackrel{(7)}{=}~R(z)\sin\theta . \tag{6}$$ In the last step of eq. (6), we used the fact that $$\sin\theta ~=~R(z)\frac{d\varphi}{ds}, \tag{7}$$ where $\theta$ denotes the angle between the geodesic and the meridian.

  7. The variable $\varphi$ is cyclic. Therefore $p_{\varphi}$ is constant along a geodesic. Physically $p_{\varphi}$ has an interpretation as angular momentum along the $z$-axis of a non-relativistic point particle with mass $m=2$.


  1. Mark Levi, Classical Mechanics with Calculus of Variations and Optimal Control: An Intuitive Introduction, Problem 5.9, p. 208.

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