Lagrangian Mechanics — Generalised Coordinates
Lagrangian Mechanics & Generalised Coordinates
Hamilton's principle (the principle of stationary action) asserts that the actual path taken by a mechanical system between two configurations at times $t_1$ and $t_2$ is the one for which the action functional $S[q]=\int_{t_1}^{t_2}L(q,\dot{q},t)\,dt$ is stationary. Here $L=T-V$ is the Lagrangian: kinetic energy minus potential energy. Applying the Euler-Lagrange equation to each generalised coordinate $q_i$ yields $$\frac{d}{dt}\frac{\partial L}{\partial\dot{q}_i}-\frac{\partial L}{\partial q_i}=Q_i^{\text{nc}},$$ where $Q_i^{\text{nc}}$ collects any non-conservative generalised forces (friction, driving forces). For conservative systems $Q_i^{\text{nc}}=0$ and the equations follow purely from $L$.
Generalised coordinates $q_1,\ldots,q_n$ are any $n$ independent parameters that uniquely specify the system's configuration. They need not be Cartesian: angles, arc lengths, normal-mode amplitudes, and relative displacements are all legitimate choices. The corresponding generalised velocities are $\dot{q}_i=dq_i/dt$. The generalised momenta conjugate to $q_i$ are $p_i=\partial L/\partial\dot{q}_i$. When $L$ does not depend on $q_i$ (a cyclic or ignorable coordinate), $p_i$ is a constant of motion.
The elegance of the Lagrangian formulation lies in its coordinate independence: the same $L=T-V$ written in any generalised coordinates automatically produces the correct equations of motion, absorbing all constraint forces and fictitious forces of curvilinear coordinates. One never needs to resolve forces along awkward directions or worry about normal forces — a dramatic practical advantage for complex systems.
Definition — Lagrangian and Action
The Lagrangian is $L(q,\dot{q},t)=T-V$, where $T$ is the kinetic energy and $V$ the potential energy. The action is $S=\int_{t_1}^{t_2}L\,dt$. The generalised momentum conjugate to $q_i$ is $p_i=\partial L/\partial\dot{q}_i$.
Theorem — Lagrange's Equations
Hamilton's principle $\delta S=0$ with fixed endpoints yields $$\frac{d}{dt}\frac{\partial L}{\partial\dot{q}_i}-\frac{\partial L}{\partial q_i}=0,\quad i=1,\ldots,n.$$ If $\partial L/\partial q_i=0$, then $q_i$ is cyclic and $p_i=\partial L/\partial\dot{q}_i=\text{const}$.
Example 1: Spherical Pendulum
A bob of mass $m$ on a massless rod of length $l$ moves on a sphere. Generalised coordinates: $(\theta,\phi)$ (polar and azimuthal angles). $$T=\frac{ml^2}{2}(\dot{\theta}^2+\sin^2\!\theta\,\dot{\phi}^2),\quad V=-mgl\cos\theta.$$ Lagrange's equations: $ml^2\ddot{\theta}=ml^2\sin\theta\cos\theta\dot{\phi}^2-mgl\sin\theta$ and $\frac{d}{dt}(ml^2\sin^2\theta\,\dot{\phi})=0$, so $p_\phi=ml^2\sin^2\theta\,\dot{\phi}=\text{const}$ (cyclic $\phi$).
Example 2: Double Pendulum
Two bobs $(m_1,l_1)$ and $(m_2,l_2)$ in a plane. Generalised coordinates $\theta_1,\theta_2$. $$L=\frac{1}{2}(m_1+m_2)l_1^2\dot{\theta}_1^2+\frac{1}{2}m_2l_2^2\dot{\theta}_2^2+m_2l_1l_2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1-\theta_2)+(m_1+m_2)gl_1\cos\theta_1+m_2gl_2\cos\theta_2.$$ The resulting equations are coupled and nonlinear, exhibiting chaotic behaviour for large amplitudes.
Example 3: Particle in a Central Force Field
In polar coordinates $(r,\phi)$: $T=\frac{m}{2}(\dot{r}^2+r^2\dot{\phi}^2)$, $V=V(r)$. $\phi$ is cyclic, so $p_\phi=mr^2\dot{\phi}=l$ (angular momentum, conserved). The $r$ equation: $$m\ddot{r}-\frac{l^2}{mr^3}=-\frac{dV}{dr}.$$ This is equivalent to 1D motion in the effective potential $V_{\text{eff}}=V(r)+l^2/(2mr^2)$.
Example 4: Charged Particle in Electromagnetic Field
For a charge $q$ in fields $\mathbf{E},\mathbf{B}$ with scalar potential $\varphi$ and vector potential $\mathbf{A}$: $$L=\frac{1}{2}m\dot{\mathbf{r}}^2-q\varphi+q\dot{\mathbf{r}}\cdot\mathbf{A}.$$ The generalised momentum is $\mathbf{p}=m\dot{\mathbf{r}}+q\mathbf{A}$ (kinetic plus field momentum). Lagrange's equations reproduce the Lorentz force $m\ddot{\mathbf{r}}=q(\mathbf{E}+\dot{\mathbf{r}}\times\mathbf{B})$ — the electromagnetic field enters $L$ through the vector potential, not through a force.
Practice Problems
- Write the Lagrangian for a particle of mass $m$ constrained to move on the surface of a cone of half-angle $\alpha$ under gravity. Identify any cyclic coordinates.
- For the Lagrangian $L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)-\frac{1}{2}k(x^2+y^2)$, find all equations of motion and identify the conserved quantities.
- A bead of mass $m$ slides on a frictionless helical wire $x=a\cos\theta$, $y=a\sin\theta$, $z=b\theta$. Use $\theta$ as the single generalised coordinate and derive its equation of motion.
- Show that for the Lagrangian $L=T(\dot{q})-V(q)$ with $T$ quadratic in $\dot{q}$, the Euler-Lagrange equations are equivalent to $\dot{p}_i=\partial L/\partial q_i$.
- Two equal masses $m$ are connected by a spring (constant $k$, natural length $l_0$) and slide on a frictionless horizontal table. Use centre-of-mass and relative coordinates to decouple the Lagrangian.
Show Answer Key
1. On a cone: $z = r\cot\alpha$, so $\dot{z} = \dot{r}\cot\alpha$. $L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\phi}^2 + \dot{z}^2) - mgz = \frac{1}{2}m(\dot{r}^2(1+\cot^2\alpha) + r^2\dot{\phi}^2) - mgr\cot\alpha$. Cyclic coordinate: $\phi$ (no $\phi$ in $L$), so $p_\phi = mr^2\dot{\phi}$ is conserved (angular momentum).
2. $\frac{\partial L}{\partial \dot{x}} = m\dot{x}$, EL: $m\ddot{x} = -kx$. Similarly $m\ddot{y} = -ky$. Conserved: energy $E = T+V$ (no explicit time) and angular momentum $L_z = m(x\dot{y}-y\dot{x})$ (rotational symmetry of $V$).
3. Position: $(a\cos\theta, a\sin\theta, b\theta)$. Velocity: $(-a\sin\theta, a\cos\theta, b)\dot{\theta}$. $T = \frac{1}{2}m(a^2+b^2)\dot{\theta}^2$, $V = mgb\theta$. $L = \frac{1}{2}m(a^2+b^2)\dot{\theta}^2 - mgb\theta$. EL: $m(a^2+b^2)\ddot{\theta} = -mgb$, i.e., $\ddot{\theta} = -gb/(a^2+b^2)$.
4. $T = \frac{1}{2}\sum_{ij}M_{ij}\dot{q}_i\dot{q}_j$ (quadratic). $p_i = \frac{\partial L}{\partial \dot{q}_i} = \sum_j M_{ij}\dot{q}_j$. EL: $\dot{p}_i = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} = \frac{\partial L}{\partial q_i} = \frac{\partial T}{\partial q_i} - \frac{\partial V}{\partial q_i}$. When $M_{ij}$ is constant, $\frac{\partial T}{\partial q_i}=0$ and $\dot{p}_i = -\frac{\partial V}{\partial q_i} = \frac{\partial L}{\partial q_i}$. ✓
5. CM: $R = (x_1+x_2)/2$, relative: $r = x_1 - x_2$. Total mass $M=2m$, reduced mass $\mu=m/2$. $L = \frac{1}{2}M\dot{R}^2 + \frac{1}{2}\mu\dot{r}^2 - \frac{1}{2}k(r-l_0)^2$. CM moves freely ($\ddot{R}=0$), relative coordinate oscillates: $\mu\ddot{r} = -k(r-l_0)$ with frequency $\omega = \sqrt{k/\mu} = \sqrt{2k/m}$.