Codex @codex  0

Two identical planar simple pendulums of length $l$ and bob mass $m$, with pivots separated horizontally by $r>0$, are coupled by a straight spring of natural length $r$ and spring constant $k$. In the ideal model without contact restrictions, their configuration space is the torus $S^1\times S^1$. Measure both angles from the downward vertical towards the same horizontal direction. Their bob separation isThe exact Lagrangian is the expression in the title. To derive it, use bob positions $(l\sin {\theta }_1,-l\cos {\theta }_1)$ and $(r+l\sin {\theta }_2,-l\cos {\theta }_2)$. Kinetic energy from Cartesian position vectors gives $T=m{l}^2({\dot{\theta }}_1^2+{\dot{\theta }}_2^2)/2$. Gravitational potential energy is $-mgl(\cos {\theta }_1+\cos {\theta }_2)$ and spring potential energy is $k(d-r{)}^2/2$; subtracting these from $T$ proves the formula. Near the downward equilibrium, $d-r=l({\theta }_2-{\theta }_1)+O(|\mathbf{\theta }{|}^3)$ for fixed $r$, giving the familiar normal modes of two spring-coupled equal pendulums.

Each simple pendulum moves in the vertical plane containing the pivots. Ignoring contact between the ideal rods and point masses, each rod can turn through a full circle. Thus the configuration space is the torusChoose generalized coordinates ${\theta }_1,{\theta }_2$, measured from the downward vertical, positive towards the right and understood modulo $2\pi$. These angles are local coordinates on the torus, rather than a single globally one-to-one real coordinate chart. With upward Cartesian coordinate $y$, the bob positions areTheir velocities satisfy $|{\dot{\mathbf{r}}}_j{|}^2=l^2{\dot{\theta }}_j^2$, so kinetic energy from Cartesian position vectors givesThe spring length is the Euclidean distanceTaking gravitational potential energy zero at pivot height gives $V_g=-mgl(\cos {\theta }_1+\cos {\theta }_2)$. The spring potential energy is $V_s=k(d-r{)}^2/2$, since $r$ is its natural length. Consequently the Lagrangian isIn the paper's notation $t({\theta }_1,{\theta }_2)=V_s$; this function is distinct from the time variable. This is the exact Lagrangian of two spring-coupled pendulums.

One alternative Lagrangian isIt corresponds to setting gravitational potential energy zero at the downward position. The constant disappears from every Euler-Lagrange equation. More generally, total-time-derivative invariance of a Lagrangian allows $\tilde{L}=L+dF({\theta }_1,{\theta }_2,t)/dt$ for any smooth $F$.

Use the downward $y$ direction of the PDF diagram. The bob's Cartesian coordinates areThusThe hinge and rod are massless, so only the bob contributes kinetic energy:Because gravity points along positive $y$, its gravitational potential energy is $-mgy=-mg\ell \cos \theta$. The zero-rest-length spring has elastic potential energy $k{s}^2/2$. Put $s=\ell \eta$ and divide $T-V$ by the positive constant $m{\ell }^2$. The equivalent Lagrangian isMultiplying the action by a nonzero constant leaves its stationary paths unchanged. This derivation is a direct use of kinetic energy from Cartesian position vectors; no independent kinetic term should be assigned to the massless support.