= Double pendulum Lagrangian
{title2=$L=T-V$}
Choose <Cartesian coordinates> with origin at the upper pivot, $x$ to the right and $y$ upward. The two <position vectors> are
$$
\mathbf r_1=\ell\begin{pmatrix}\sin\theta_1\\-\cos\theta_1\end{pmatrix},\qquad
\mathbf r_2=\ell\begin{pmatrix}\sin\theta_1+\sin\theta_2\\-\cos\theta_1-\cos\theta_2\end{pmatrix}.
$$
These expressions satisfy the rod <holonomic constraints> $|\mathbf r_1|=\ell$ and $|\mathbf r_2-\mathbf r_1|=\ell$ automatically. This is the practical advantage of using the two <angles> as coordinates: the rod <tensions> need not be unknowns in the motion calculation.<Classical Dynamics by David Tong>[Classical Dynamics, chapter 2, §2.3, printed pp. 17–19: elimination of ideal constraint forces using coordinates adapted to holonomic constraints]{ref}
Differentiate these <position vectors> to obtain the <velocities>. The <dot product> between their two rod contributions uses $\cos(\theta_1-\theta_2)=\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2$, giving
$$
|\dot{\mathbf r}_1|^2=\ell^2\dot\theta_1^2,\qquad
|\dot{\mathbf r}_2|^2=\ell^2\left(\dot\theta_1^2+\dot\theta_2^2+2\cos(\theta_1-\theta_2)\dot\theta_1\dot\theta_2\right).
$$
Adding the two <kinetic energies> and measuring <gravitational potential energy> from the downward <equilibrium> gives
$$
\begin{aligned}
T&=\frac{m\ell^2}{2}\left(2\dot\theta_1^2+\dot\theta_2^2+2\cos(\theta_1-\theta_2)\dot\theta_1\dot\theta_2\right),\\
V&=mg\ell\left[2(1-\cos\theta_1)+(1-\cos\theta_2)\right],\\
L&=T-V.
\end{aligned}
$$
The coefficient $2$ has a geometric meaning: the upper rod raises both <masses>, and its motion contributes to both <kinetic energies>. The coupling in $T$ is present because the lower <mass> has a moving pivot. The zero chosen for $V$ changes $L$ by a constant and leaves the <Euler-Lagrange equations> unchanged.<Classical Dynamics by David Tong>[Classical Dynamics, chapter 2, §2.5.2, printed p. 28, equations (2.68)–(2.72): general point-mass double pendulum; the derivation here specializes to equal masses and lengths and uses upward $y$]{ref}
The <principle of stationary action> applies to $S=\int L\,dt$. Its fixed-endpoint <first variation>, after <integration by parts>, is
$$
\delta S=\int\sum_{i=1}^2\left(\frac{\partial L}{\partial\theta_i}-\frac{d}{dt}\frac{\partial L}{\partial\dot\theta_i}\right)\delta\theta_i\,dt.
$$
Independent allowed variations therefore give one <Euler-Lagrange equation> for each <angle>. Stationary <action> alone does not assert a minimum.<Classical Dynamics by David Tong>[Classical Dynamics, chapter 2, §2.1, printed pp. 10–12, equations (2.2)–(2.8) and the stationary-action qualification]{ref}
Back to article page
