Clairut’s Relation: Geodesics on Surfaces of Revolution

One of the recurring themes in mathematics is how symmetry simplifies problems. In differential geometry, surfaces of revolution – shapes like spheres, cylinders, cones, or donuts, formed by spinning a curve around an axis – possess a fundamental rotational symmetry. It turns out this symmetry provides a powerful shortcut for understanding the “straightest paths,” or geodesics, on these surfaces. This shortcut is encapsulated in a neat result known as Clairaut’s relation. It’s a classic piece of mathematics that elegantly connects the geometry of the path to a conserved quantity, which, as we’ll see, is essentially angular momentum, echoing Noether’s theorem. Let’s unpack this.

Setting the Stage: Surfaces of Revolution

First, let’s quickly recall how we describe a surface of revolution mathematically. We start with a profile curve in a half-plane, say the $rz$ -plane with $r \ge 0$ . Let’s parameterize this curve by $u$ , so its coordinates are $(r(u), z(u))$ . We then rotate this curve around the $z$ -axis. A point on the resulting surface can be described by the parameter $u$ (telling us which point on the profile curve we started with) and the angle of rotation $v$ (telling us how far we’ve rotated it). The Cartesian coordinates $(x, y, z)$ are given by:

$\displaystyle \mathbf{X}(u,v) = (r(u)\cos v, r(u)\sin v, z(u))$

Here, $r(u)$ is the distance from the axis of rotation for a given $u$ .

To talk about geodesics (shortest paths), we need to measure distances on the surface. This is done using the metric, or the first fundamental form, which tells us the infinitesimal squared distance $ds^2$ for a small change $du$ and $dv$ . By calculating the tangent vectors $\mathbf{X}_u = \partial \mathbf{X} / \partial u$ and $\mathbf{X}_v = \partial \mathbf{X} / \partial v$ and their dot products, we find the metric coefficients:

$\displaystyle E(u) = \mathbf{X}_u \cdot \mathbf{X}_u = (r'(u))^2 + (z'(u))^2$

$\displaystyle F(u,v) = \mathbf{X}_u \cdot \mathbf{X}_v = 0$

$\displaystyle G(u) = \mathbf{X}_v \cdot \mathbf{X}_v = (r(u))^2$

So, the metric is: $\displaystyle ds^2 = E(u) du^2 + G(u) dv^2$ The crucial point here, stemming directly from the rotational symmetry, is that $E$ and $G$ depend only on $u$ and not on the rotation angle $v$ . The local geometry is the same all the way around a circle of latitude (a “parallel” curve where $u$ is constant). This independence from $v$ is the key simplification. Also, $F=0$ tells us the coordinate curves ( $u=$ constant and $v=$ constant, i.e., parallels and meridians) are orthogonal, which is convenient.

What’s a Geodesic, Anyway?

Intuitively, a geodesic is the path you’d follow on the surface if you tried to walk “straight ahead” without turning left or right relative to the surface. Think of stretching a rubber band between two points on a balloon. Mathematically, it’s a curve $\gamma(s)$ (parameterized by arc length $s$ ) whose acceleration vector $\gamma''(s)$ has no component tangent to the surface; any acceleration is purely normal. This is captured by the geodesic equation $\nabla_{\gamma'} \gamma' = 0$ , where $\nabla$ is the covariant derivative on the surface. In local coordinates $(u, v)$ , this becomes a system of second-order ODEs involving Christoffel symbols $\Gamma^k_{ij}$ , which encode the surface’s curvature:

$\displaystyle \frac{d^2 u}{ds^2} + \Gamma^1_{uu} \left(\frac{du}{ds}\right)^2 + 2\Gamma^1_{uv} \frac{du}{ds}\frac{dv}{ds} + \Gamma^1_{vv} \left(\frac{dv}{ds}\right)^2 = 0$

$\displaystyle \frac{d^2 v}{ds^2} + \Gamma^2_{uu} \left(\frac{du}{ds}\right)^2 + 2\Gamma^2_{uv} \frac{du}{ds}\frac{dv}{ds} + \Gamma^2_{vv} \left(\frac{dv}{ds}\right)^2 = 0$

Solving these directly can be cumbersome.

The Variational Approach and Noether’s Insight

An alternative, often more insightful, way to find geodesics is through the calculus of variations. Geodesics are paths that locally minimize (or more generally, make stationary) the path length functional $L[\gamma] = \int ds$ . It’s often easier to work with the energy functional $E[\gamma] = \frac{1}{2} \int (ds/dt)^2 dt$ , which gives the same paths (parameterized by constant speed). For our surface of revolution, the integrand (the “Lagrangian”) is:

$\displaystyle L(u, \dot{u}, v, \dot{v}) = \frac{1}{2} \left( E(u) \dot{u}^2 + G(u) \dot{v}^2 \right)$

where dots denote differentiation with respect to the parameter $t$ .

The paths that make the action $\int L dt$ stationary satisfy the Euler-Lagrange equations. For our coordinates $u$ and $v$ , these are:

$\displaystyle \frac{\partial L}{\partial u} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{u}} \right) = 0$

$\displaystyle \frac{\partial L}{\partial v} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{v}} \right) = 0$

Now, look at the Lagrangian $L$ . Because $E$ and $G$ depend only on $u$ , the Lagrangian $L$ does not depend explicitly on $v$ . The coordinate $v$ is “cyclic” or “ignorable”. This is the mathematical manifestation of the rotational symmetry.

This is where Noether’s theorem comes into play. It tells us that for every continuous symmetry of the Lagrangian, there’s a corresponding conserved quantity. Since $L$ is independent of $v$ (symmetric under rotations, i.e., changes in $v$ ), the quantity conjugate to $v$ , which is $p_v = \partial L / \partial \dot{v}$ , must be conserved. Let’s compute it:

$\displaystyle p_v = \frac{\partial}{\partial \dot{v}} \left[ \frac{1}{2} \left( E(u) \dot{u}^2 + G(u) \dot{v}^2 \right) \right] = G(u) \dot{v}$

Substituting $G(u) = r(u)^2$ , we get the conserved quantity:

$\displaystyle p_v = r(u)^2 \dot{v} = \text{constant}$

This quantity, $r^2 \dot{v}$ , is precisely the angular momentum per unit mass of a particle moving along the geodesic, measured relative to the $z$ -axis ( $r$ is the distance to the axis, and $r\dot{v}$ is the azimuthal component of velocity). So, the rotational symmetry directly implies the conservation of angular momentum along geodesics. The second Euler-Lagrange equation confirms this: since $\partial L / \partial v = 0$ , it simplifies to $d/dt (\partial L / \partial \dot{v}) = 0$ , meaning $p_v$ is constant.

Clairaut’s Relation Emerges

Now we connect this conserved quantity to the path’s geometry. Let $\psi$ be the angle between the geodesic’s tangent vector and the tangent to the parallel circle (a curve of constant $u$ ) at the same point. If we parameterize the geodesic by arc length $s$ , so its speed $||\gamma'(s)|| = \sqrt{E(u')^2 + G(v')^2} = 1$ (where primes now mean $d/ds$ ), a geometric calculation shows that the component of the unit tangent vector along the parallel direction is given by: $\displaystyle \cos \psi = \frac{G v'}{\sqrt{G}} = \sqrt{G} v' = r(u) v'$ (Similarly, the component along the meridian direction is $\sin \psi = \sqrt{E} u'$ .)

We found from Noether’s theorem (or the Euler-Lagrange equation) that $r^2 v'$ is constant along the geodesic. Let’s call this constant $C$ . $\displaystyle r(u(s))^2 v'(s) = C$ Now substitute $v' = \cos \psi / r(u)$ :

$\displaystyle r(u)^2 \left( \frac{\cos \psi}{r(u)} \right) = C$ $\displaystyle r(u) \cos \psi = C$

This is Clairaut’s Relation. It states that for any geodesic on a surface of revolution, the product of the radius $r$ of the parallel through a point on the geodesic and the cosine of the angle $\psi$ the geodesic makes with that parallel is constant along the entire geodesic.

The constant $C$ is determined by the geodesic’s starting point and direction. It has a clear physical meaning: $C = r^2 v'$ is the conserved angular momentum per unit mass. The relation tells us how the path must bend as it moves closer to or farther from the axis of rotation. If $r$ decreases, $|\cos \psi|$ must increase, so $\psi$ approaches $0$ or $\pi$ (the path becomes more parallel to the equator). If $r$ increases, $|\cos \psi|$ must decrease, so $\psi$ approaches $\pi/2$ (the path becomes more aligned with the meridians). The minimum radius $r_{min}$ the geodesic can reach (unless $C=0$ ) occurs when $|\cos \psi|=1$ , giving $r_{min} = |C|$ . If $C=0$ , then $\cos \psi = 0$ (unless $r=0$ ), meaning $\psi = \pi/2$ , and the geodesic must be a meridian.

Examples in Practice

Clairaut’s relation is very useful for analyzing geodesics:

Cylinder: $r(u) = R$ (constant). Clairaut gives $R \cos \psi = C$ . Thus, $\cos \psi$ is constant. Geodesics make a constant angle with the parallels – they are helices (including circles where $\psi=0$ , $C=R$ , and vertical lines where $\psi=\pi/2$ , $C=0$ ).
Sphere: $r(u) = R \sin u$ (where $u$ is the polar angle). Clairaut gives $R \sin u \cos \psi = C$ . We know geodesics are great circles. At the equator ( $u=\pi/2$ ), $r=R$ , so $C = R \cos \psi_{eq}$ . A great circle reaches a maximum latitude $u_{min}$ where it runs parallel to the equator ( $\psi=0$ ), so $R \sin u_{min} = |C|$ . Thus $\sin u_{min} = |\cos \psi_{eq}|$ , consistent with great circle geometry. Meridians correspond to $C=0$ .
Cone: $r(u) = u \sin \alpha$ (where $u$ is distance from apex, $\alpha$ is half-angle). Clairaut gives $u \sin \alpha \cos \psi = C$ . Geodesics on a cone become straight lines when unrolled. The relation shows that if $C \ne 0$ , the geodesic never reaches the apex ( $u=0$ ) and has a minimum distance $u_{min} = |C|/\sin \alpha$ . If $C=0$ , the geodesic is a generator line (meridian).

Conclusion

Clairaut’s relation $r \cos \psi = C$ is a prime example of how symmetry leads to conservation laws, which in turn simplify the description of motion (or in this case, paths). It elegantly translates the conservation of angular momentum, guaranteed by rotational symmetry via Noether’s theorem, into a simple geometric constraint on geodesics. It provides a first integral for the geodesic equations, making the analysis of these “straightest paths” on surfaces like cylinders, spheres, and cones much more tractable than solving the full second-order ODEs directly. It’s a beautiful piece of classical geometry with deep connections to physics.