Beltrami Identity and First Integrals

The identity's descendants are everywhere a shape or a path is optimized. In general relativity, particles and light rays follow geodesics of spacetime, and the metric of a static or axisymmetric spacetime is independent of time or azimuth — cyclic coordinates whose Beltrami-type integrals give the conserved energy and angular momentum that fix planetary precession, gravitational lensing, and the timing corrections baked into every GPS receiver. Modern optimal-control theory takes the brachistochrone as its founding example: Pontryagin's maximum principle generalizes the first integral to systems with constraints, and the "quantum brachistochrone" asks for the fastest Hamiltonian steering a qubit between states, a live question in quantum computing. Minimal surfaces built from the catenary integrand shape tensile architecture — Frei Otto's cable-net roof for the 1972 Munich Olympics is a physical soap-film computation — and appear in block-copolymer gyroids and biological membranes. Seismologists and optical engineers ray-trace through the same Fermat variational principle, and energy-conserving symplectic integrators, together with the recent Hamiltonian and Lagrangian neural networks of machine learning, deliberately encode the conserved $H$ so that simulated and learned dynamics respect the very first integral Beltrami exploited.

Think of the independent variable $x$ as time. If the rule governing $y$ looks identical at every instant — no explicit $x$ anywhere in $L$ — then no moment is special, and by the deepest theorem in physics something must be conserved. That something is the "energy" $H=y'p-L$, with $p=\partial L/\partial y'$ the momentum conjugate to $y$. The Beltrami identity is just this conservation law written for a general variational problem, and it pays an immediate dividend: the Euler-Lagrange equation is second order, but a conserved quantity is one integration already performed, so the problem drops to first order and the extremal is pinned to a single level set of $H$ rather than roaming the whole $(y,y')$ plane. The first figure races the two most famous curves the identity produces against each other.

Which variable is missing tells you instantly which conservation law to reach for. If $L$ lacks explicit $x$, use Beltrami and $H$ is constant. If $L$ lacks explicit $y$ — a cyclic coordinate — use the even simpler momentum integral $\partial L/\partial y'=\text{const}$, which holds regardless of any $x$-dependence. The second figure makes the Beltrami case geometric: for an $x$-free integrand the conserved value labels a single curve in the phase plane, and the extremal is trapped on it. The catenary's integrand supplies concrete numbers.

Everything begins with a functional whose integrand is autonomous — it depends on $y$ and its slope $y'$, but not on where you are along the $x$-axis. This single structural fact is the entire hypothesis of the Beltrami identity.

where: $J[y]$ — the functional, a number assigned to each admissible curve $y$; $L(y,y')$ — the integrand, or Lagrangian, with no explicit $x$; $y'=\mathrm{d}y/\mathrm{d}x$; $a,b$ — fixed endpoints. Reading: the absence of $x$ inside $L$ is the symmetry that the identity converts into a conserved quantity. Stationarity of $J$ under fixed-endpoint variations is the Euler-Lagrange equation, the general necessary condition every extremal must satisfy. It is second order in $y$, which is exactly the cost the Beltrami identity will refund.

where: $\partial L/\partial y$ — partial derivative holding $y'$ fixed; $\partial L/\partial y'\equiv p$ — the momentum conjugate to $y$. Reading: this holds whether or not $L$ contains $x$; the Beltrami identity is what it becomes when $x$ drops out. The identity itself states that a specific combination of $L$ and its momentum is constant along every extremal. It is a first integral: the constant $C$ absorbs one of the two integrations the Euler-Lagrange equation would otherwise demand.

where: $C$ — the conserved constant, fixed by boundary or geometric data. Reading: solve this first-order equation for $y'$ and separate variables; you never touch the second-order equation directly. The same content, sign-flipped, is the Hamiltonian or energy function. Writing it this way exposes the physics: $H$ is what is conserved, and $-C$ is its value on the chosen extremal.

where: $H$ — the variational Hamiltonian, the energy belonging to $L$; $p=\partial L/\partial y'$ — the conjugate momentum. Reading: $H$ is built from $L$ by a Legendre-type combination; its constancy is the whole point. When $L$ does depend explicitly on $x$, the identity is replaced by an evolution law for $H$ that isolates exactly the offending term. This equation is the honest general statement of which the Beltrami identity is the $\partial L/\partial x=0$ special case.

where: $\partial L/\partial x$ — the explicit $x$-dependence of the integrand; it vanishes in Beltrami's case. Reading: $H$ drifts at exactly the rate the integrand changes with $x$, so a hidden clock in $L$ destroys the conservation law. Finally, the sibling first integral for the other missing variable. If $y$ itself is absent — a cyclic or ignorable coordinate — the Euler-Lagrange equation immediately delivers a conserved momentum, simpler than Beltrami and immune to $x$-dependence.

where: $\partial L/\partial y'$ — the conserved momentum. Reading: match the missing variable to its integral — absent $x$ gives Beltrami's $H$, absent $y$ gives the momentum.

The identity's status is entirely dictated by which variables the integrand does and does not contain.

1 · The brachistochrone descent time (fully worked)

A bead slides without friction from A = (0, 0) to a point B under gravity $g=9.80\ \mathrm{m/s^2}$, starting from rest. Let B lie at the lowest point of the cycloidal arch, with a vertical drop of 1.00 m. Find the cycloid parameter, the descent time along the brachistochrone, and compare it with a straight ramp between the same two points.

Step 1. The brachistochrone is the cycloid $x=R(\theta-\sin\theta)$, $y=R(1-\cos\theta)$ (with $y$ measured downward), obtained from $L=\sqrt{(1+y'^2)/(2gy)}$ by one use of Beltrami. Its lowest point is at $\theta=\pi$, where $y=2R$. Setting the drop $2R=1.00\ \mathrm{m}$ gives R = 0.50 m, and the horizontal span is $R\pi=\mathbf{1.571\ m}$.

Step 2. Along the cycloid the speed is $v=\sqrt{2gy}$ (energy conservation) and the arc element is $\mathrm{d}s=2R\sin(\theta/2)\,\mathrm{d}\theta$, so $\mathrm{d}t=\mathrm{d}s/v=\sqrt{R/g}\,\mathrm{d}\theta$. Integrating to $\theta=\pi$ gives the remarkably clean $t_{\text{cyc}}=\pi\sqrt{R/g}=\pi\sqrt{0.50/9.80}=\pi(0.2259)=\mathbf{0.710\ s}$. (This same integral, giving a time independent of the starting height on the arch, is the tautochrone property Huygens used in his pendulum clock.)

Step 3. Compare a straight ramp of length $\ell=\sqrt{1.571^2+1.00^2}=1.862\ \mathrm{m}$. The along-ramp acceleration is $a=g\,(h/\ell)=9.80(1.00/1.862)=5.263\ \mathrm{m/s^2}$, and from rest $\ell=\tfrac12 a t^2$ gives $t_{\text{str}}=\sqrt{2\ell/a}=\sqrt{2(1.862)/5.263}=\mathbf{0.841\ s}$.

Step 4. The cycloid beats the straight line by $0.841/0.710=1.185$, about 16 % faster, even though it is the longer path — the extra initial steepness buys speed early, and Beltrami is what turned the messy time integral into the cycloid in the first place. Brachistochrone 0.710 s vs straight ramp 0.841 s.

2 · The catenary sag and a first-integral check

A uniform chain hangs as $y=a\cosh(x/a)$ with $a=2.00\ \mathrm{m}$, suspended from equal-height supports at $x=\pm 3.00\ \mathrm{m}$. Find the sag and the chain's length, and verify numerically that the Beltrami constant is genuinely constant.

Step 1. The sag is the rise from the lowest point to a support: $\Delta y=a[\cosh(3/a)-1]=2.00[\cosh(1.5)-1]=2.00(2.35241-1)=\mathbf{2.70\ m}$.

Step 2. The length is $s=\int_{-3}^{3}\sqrt{1+y'^2}\,\mathrm{d}x=\big[a\sinh(x/a)\big]_{-3}^{3}=2a\sinh(1.5)=2(2.00)(2.12928)=\mathbf{8.52\ m}$.

Step 3. Check the first integral $y/\sqrt{1+y'^2}=a$. At the support $x=3$: $y'=\sinh(1.5)=2.129$, so $\sqrt{1+y'^2}=\cosh(1.5)=2.352$, and $y/\sqrt{1+y'^2}=(2.00\times 2.352)/2.352=\mathbf{2.00\ m}=a$. The same value emerges at every $x$ because $\cosh^2-\sinh^2=1$ — the Beltrami constant really is a constant of the motion. Sag 2.70 m, length 8.52 m, Beltrami constant a = 2.00 m.

3 · A great-circle geodesic by the cyclic-coordinate integral

On a sphere (a surface of revolution) the arc-length integrand is independent of the azimuthal angle $\phi$, making $\phi$ cyclic. Show that this yields Clairaut's relation and find the highest latitude reached by a great circle that crosses the equator heading 45° from the meridian.

Step 1. Because $\phi$ is cyclic, equation (6) gives the momentum integral $\partial L/\partial\phi'=\text{const}$, which for the sphere reduces to Clairaut's relation $r\sin\psi=\text{const}$, where $r$ is the distance from the polar axis and $\psi$ is the angle the geodesic makes with the local meridian.

Step 2. At the equator $r=R_s$ (sphere radius) and $\psi=45^\circ$, so the constant is $R_s\sin 45^\circ$. At the geodesic's northern vertex it runs due east, $\psi=90^\circ$, where $r=r_{\text{vertex}}$ and $r_{\text{vertex}}\cdot 1=R_s\sin 45^\circ$.

Step 3. Writing $r=R_s\cos\lambda$ with $\lambda$ the latitude gives $\cos\lambda_{\max}=\sin 45^\circ$, so $\lambda_{\max}=90^\circ-45^\circ=\mathbf{45^\circ}$. A great circle crossing the equator at azimuth $\alpha$ tops out at latitude $90^\circ-\alpha$ — the rule every long-haul flight planner uses. λ_max = 45° for a 45° equator crossing.

The Beltrami identity carries an Italian geometer's name for a result whose seed was planted a century before him and whose full meaning was harvested two decades after his death — a small parable of how physics attributes its ideas.

Texts. Stone & Goldbart, Mathematics for Physics, §1.2 · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 22 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 22 · Gelfand & Fomin, Calculus of Variations, §4 · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 2 & 8 · Nakahara, Geometry, Topology and Physics, Ch. 1 (for the Noether reading). Key numerical anchors: the identity follows from computing $\frac{\mathrm{d}}{\mathrm{d}x}(L-y'\,\partial L/\partial y')$ and applying Euler-Lagrange; the catenary $y=a\cosh((x-x_0)/a)$ satisfies $y/\sqrt{1+y'^2}=a$, with sag $=2.70$ m and length $=8.52$ m for $a=2.00$ m, $x=\pm3$ m; the cycloid descent time is $t=\pi\sqrt{R/g}=0.710$ s and the straight-ramp time is $0.841$ s for $R=0.50$ m, $g=9.80\ \mathrm{m/s^2}$; for quadratic kinetic energy Euler's homogeneous-function theorem gives $H=T+V$; and Clairaut's relation yields $\lambda_{\max}=90^\circ-\alpha$.

Prerequisites. Euler-Lagrange Equation · First Variation and Stationarity · Functionals and Function Spaces · Fundamental Lemma of the Calculus of Variations. Leads to. The Brachistochrone Problem · Hamilton's Principle of Stationary Action · Lagrangian Mechanics · Geodesics as Variational Problems.

Ask the tutor: Hand me any variational integrand and I will spot the missing variable, write down the right first integral, and walk it all the way to the extremal curve.