Lagrangian Mechanics

sources:

• text: “Goldstein, H., Poole, C., & Safko, J. (2002). Classical Mechanics (3rd ed.). Addison Wesley.”
• text: “Landau, L. D., & Lifshitz, E. M. (1976). Mechanics (3rd ed.). Butterworth-Heinemann.”

Attempt these from memory before reading.

• Write down the Euler-Lagrange equation. Which term is the “generalised momentum”?
• What does $L = T - V$ assume about the forces and the constraints?
• State Hamilton’s principle. Is the action minimised or only stationary?
• What is a cyclic coordinate, and what is conserved when you have one?
• For a bead on a rotating hoop, when does a new stable equilibrium appear?

Newton’s second law is a vector equation. For a system of $N$ particles in three dimensions that is $3N$ coupled second-order equations, and the forces of constraint — normal forces, tension, the normal reaction of a rail — all appear in them. Most of the effort in a Newtonian solution goes into computing forces you do not care about, only to eliminate them at the end.

The Lagrangian formulation removes them at the outset by working in generalised coordinates: the smallest set of numbers that specifies the configuration once the constraints are accounted for. A bead constrained to a hoop needs one number, not three. A double pendulum needs two, not six.

The price is that the equations become second-order in the generalised coordinates with no explicit force bookkeeping. The reward is threefold:

• The constraint forces never appear.
• The choice of coordinates is free, so you pick the ones that make the problem easy.
• Symmetries become visible, and by Noether’s theorem each one hands you a conserved quantity for nothing.

The structure of the whole subject is a single scalar function — the Lagrangian — plus one equation.

The Lagrangian of a system is defined as

Where $T$ is the kinetic energy and $V$ is the potential energy.

Theorem 3.1 (Euler-Lagrange from D’Alembert). The equations of motion for a holonomic system with ideal constraints are:

Proof. Start from D’Alembert’s principle with only applied forces (ideal constraints):

Express the virtual displacement in terms of generalised coordinates:

First term (applied forces). For a conservative system, $\mathbf{F}_i^{(\mathrm{app})} = -\nabla_i V$ So:

Defining the generalised force $Q_j = \sum_i \mathbf{F}_i \cdot \frac{\partial \mathbf{r}_i}{\partial q_j}$ For conservative forces $Q_j = -\partial V/\partial q_j$.

Second term (inertia). Using $\frac{\partial \dot{\mathbf{r}}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$ (which holds when $\mathbf{r}_i = \mathbf{r}_i(q, t)$):

Now:

Using $\frac{d}{dt}\frac{\partial \mathbf{r}_i}{\partial q_j} = \frac{\partial \dot{\mathbf{r}}_i}{\partial q_j}$ and $\frac{\partial \dot{\mathbf{r}}_i}{\partial \dot{q}_j} = \frac{\partial \mathbf{r}_i}{\partial q_j}$:

Combining both terms in D’Alembert’s principle:

For conservative forces, $Q_j = -\partial V/\partial q_j$. Since $L = T - V$ and $V$ is independent of $\dot{q}_j$:

Since the $\delta q_j$ are independent (and we have $n$ degrees of freedom):

$\blacksquare$

Theorem 3.2 (Hamilton’s Principle). The actual path of a system between times $t_1$ and $t_2$ is The one that makes the action

Stationary.

Theorem 3.3 (Euler-Lagrange Equation from Hamilton’s Principle). The path $q(t)$ that makes $S$ stationary satisfies

Proof (for one degree of freedom). Consider a variation $q(t) + \epsilon \eta(t)$ where $\eta(t_1) = \eta(t_2) = 0$. The variation of the action:

Integrating the second term by parts:

The boundary term vanishes since $\eta(t_1) = \eta(t_2) = 0$. For $\delta S = 0$ for all $\eta$ By the fundamental lemma of the calculus of variations:

$\blacksquare$

Intuition. Hamilton’s principle says nature is “lazy”: out of all possible paths connecting two configurations, the actual path taken is the one that makes the action stationary. This is a profound generalisation of Fermat’s principle of least time in optics.

Problem. Derive the equation of motion for a simple pendulum of length $l$ and mass $m$.

Solution. Take $\theta$ as the generalised coordinate. The position of the bob is $(l\sin\theta, -l\cos\theta)$.

Euler-Lagrange equation:

For small angles ($\sin\theta \approx \theta$): $\ddot{\theta} + \frac{g}{l}\theta = 0$ Giving simple Harmonic motion with $\omega = \sqrt{g/l}$. $\blacksquare$

Problem. Derive the equations of motion for a double pendulum: mass $m_1$ on rod $l_1$ Mass $m_2$ on rod $l_2$ attached to $m_1$.

Solution. Generalised coordinates: angles $\theta_1, \theta_2$ from the vertical. Position of $m_1$:

Position of $m_2$:

Velocities:

Kinetic energy:

Potential energy:

The Euler-Lagrange equations for $\theta_1$ and $\theta_2$ yield two coupled second-order ODEs. For equal masses and lengths ($m_1 = m_2 = m$, $l_1 = l_2 = l$):

$\blacksquare$

Problem. Two masses $m_1$ and $m_2$ ($m_1 > m_2$) are connected by a massless inextensible string over a frictionless pulley. Find the acceleration using the Lagrangian.

Problem. A bead of mass $m$ slides without friction on a circular hoop of radius $R$. The hoop rotates about a vertical diameter with constant angular velocity $\omega$. Find the equilibrium positions and their stability.

Definition. A coordinate $q_j$ is cyclic (or ignorable) if it does not appear explicitly in the Lagrangian: $\partial L / \partial q_j = 0$.

Theorem 3.4. If $q_j$ is cyclic, the conjugate generalised momentum $p_j = \partial L / \partial \dot{q}_j$ is a constant of motion.

Proof. The Euler-Lagrange equation for a cyclic coordinate is:

$\blacksquare$

Intuition. Cyclic coordinates correspond to symmetries of the system. Each symmetry gives a conserved quantity --- this is the essence of Noether’s theorem (Section 5).

When holonomic constraints cannot be eliminated by coordinate choice, introduce Lagrange multipliers $\lambda_a$:

The multipliers $\lambda_a$ are proportional to the constraint forces.

Definition. The energy function (also called the Jacobi integral) is:

Theorem 3.5. If $L$ does not depend explicitly on time, then $h$ is conserved. Furthermore, if the transformation $\mathbf{r}_i = \mathbf{r}_i(q)$ does not depend explicitly on time and $V$ is velocity-independent, then $h = T + V$ (the total energy).

Proof. Taking the total time derivative:

The $\ddot{q}_j$ terms cancel. Using the Euler-Lagrange equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j} = \frac{\partial L}{\partial q_j}$:

If $\partial L/\partial t = 0$ Then $dh/dt = 0$.

For the second part, when $\mathbf{r}_i = \mathbf{r}_i(q)$ (scleronomic) and $V = V(q)$:

Is a homogeneous quadratic form in $\dot{q}_j$. By Euler’s theorem for homogeneous functions:

Since $\partial L/\partial \dot{q}_j = \partial T/\partial \dot{q}_j$ (as $V$ is velocity-independent):

$\blacksquare$

The clearest single picture of what the formalism buys is the rotating hoop of worked example 3.7, whose equilibria bifurcate as the hoop speeds up.

   stable equilibria of the bead
   versus rotation rate omega

   theta_eq
    pi |                 *  *  *
       |              *
    pi/2             *
      |            *
      0 *-------*            <- bottom stays stable below
      +-------------------------------- omega
                 g/sqrt(R)      omega^2 R > g

Below $\omega^2 R = g$ the only equilibrium is the bottom of the hoop. Above it, the bottom becomes unstable and two new stable equilibria appear at $\cos\theta = g/(R\omega^2)$ — a pitchfork bifurcation, and the same mechanism that keeps a rider upright on a rotating fairground cylinder.

Multiple representations of the same fact, for the same system:

• Words. Spin the hoop fast enough and the bead prefers a position off the bottom.
• Equation. $\ddot\theta = \sin\theta(\omega^2\cos\theta - g/R)$.
• Equilibria. $\sin\theta = 0$ always; $\cos\theta = g/(R\omega^2)$ when $\omega^2 R > g$.
• Stability. The second derivative of the effective potential changes sign at $\omega^2 R = g$, which is why the bottom turns over.

The Lagrangian formulation is like giving nature a GPS route planner. Instead of tracking every force individually, you describe the landscape of kinetic and potential energy, and nature automatically picks the path that makes the action stationary. Think of a ball rolling down a hill: you do not need to compute each force component if you know the shape of the terrain. The Lagrangian L = T - V encodes this terrain, and the Euler-Lagrange equations are nature’s optimization algorithm. Cyclic coordinates are like flat stretches of the landscape where nothing changes in a particular direction, so the corresponding momentum stays constant. The whole framework is coordinate-independent, meaning you can describe the same physical system using any convenient set of variables without changing the physics.

The electromagnetic Lagrangian is the one to remember. It shows that $T - V$ is not the definition of a Lagrangian but a special case, and that the general object is whatever scalar makes the action stationary.

Mix these with material from the generalised coordinates page, Noether’s theorem, and the ODE module — selecting the method is the part blocked practice never trains.

1. (Constraints.) A particle slides on a sphere of radius $R$ under gravity. Write the Lagrangian, show the constraint is holonomic, and eliminate it. At what angle does the particle leave the sphere?
2. (Noether.) Show that translational symmetry of $L$ implies conservation of total momentum, and rotational symmetry implies conservation of angular momentum, using $\frac{d}{dt}\left(\sum_j \dot q_j \frac{\partial L}{\partial \dot q_j}
   • L\right)$ for the generator of the symmetry.
3. (ODEs.) Linearise the double pendulum equations about the equilibrium $\theta_1 = \theta_2 = 0$, find the normal mode frequencies, and show they reduce to $\sqrt{g/l}$ and $\sqrt{2g/l}$ when the two bobs are identical.
4. (Calculus of variations.) Show that the shortest path between two points on a cylinder is a helix, by writing the arc length as an action and applying the Euler-Lagrange equation.
5. (Energy.) For the rotating hoop, compute $h$, show it is conserved, and show explicitly that $h \neq T + V$.

• Generalised Coordinates and Constraints: Generalised coordinates and constraints provide the foundation for formulating the Lagrangian.

• Hamiltonian Mechanics: The Hamiltonian formalism extends Lagrangian mechanics to phase space and provides energy conservation.

• Noether’s Theorem: Noether’s theorem connects symmetries of the Lagrangian to conserved quantities.

• Calculus

• Linear Algebra

Mistake 1: Including constraint forces in the Lagrangian The Lagrangian formulation eliminates constraint forces (like normal forces or tension in a string) by choosing appropriate generalised coordinates. A common error is adding the constraint force back into the Euler-Lagrange equations. Constraint forces do no work in holonomic systems and should not appear in $L = T - V$.

Mistake 2: Assuming $L = T - V$ always gives the correct equations The Lagrangian $L = T - V$ is valid when the potential $V$ is velocity-independent and the coordinate transformation is scleronomic (time-independent). For rheonomic constraints or velocity-dependent potentials, the correct Lagrangian may differ. Always verify the conditions under which $L = T - V$ applies.

Mistake 3: Forgetting that the action is stationary, not necessarily minimised Hamilton’s principle states the action is stationary ($\delta S = 0$), not necessarily minimised. The actual path can be a saddle point or even a local maximum of the action. The stationarity condition is what yields the Euler-Lagrange equations, and the second variation determines whether the path is a minimum.

• Classical Field Theory
• Generalised Coordinates and Constraints
• Hamiltonian Mechanics
• Hamiltonian Mechanics: Advanced Topics