Lagrangian formalism and generalised coordinates

In one sentence

Write $L = T - V$ in whatever coordinates suit the constraints, apply the Euler–Lagrange equations, and the constraint forces never appear — which is the entire reason the formalism exists.

Why the exam asks it

Because setting up $L$ correctly is most of the work and the rest is mechanical, and because two things get misremembered: the sign in $T-V$, and the fact that a generalised momentum need not be mass times velocity.

The idea

For a system with generalised coordinates $q_1,\dots,q_n$ — any parameters that fix the configuration — the Lagrangian is

$$L(q,\dot q, t) = T - V,$$

and Hamilton's principle says the motion makes the action $S=\int L\,dt$ stationary. That is the calculus of variations from the previous chapter with $t$ as the independent variable, so the Euler–Lagrange equations follow immediately:

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right) - \frac{\partial L}{\partial q_i} = 0, \qquad i=1,\dots,n .$$

$n$ second-order equations, so $2n$ constants, fixed by initial positions and velocities.

Why generalised coordinates

A holonomic constraint is one expressible as $f(q,t)=0$. Each such constraint removes one degree of freedom, and choosing coordinates that already respect it means the constraint forces — tension, normal reaction — never enter the equations. A pendulum in Cartesian coordinates needs the rod's tension; in the angle $\theta$ it does not appear at all.

That elimination is the whole advantage, and it is why the method is worth learning beyond Newton's laws.

Non-holonomic constraints — inequalities, or velocity constraints not integrable to $f(q,t)=0$, such as a rolling ball — are not handled this way and need multipliers or other methods. Options sometimes present one as if it were holonomic.

The pendulum, done properly

One coordinate $\theta$, with $T = \tfrac12 m\ell^2\dot\theta^2$ and $V = -mg\ell\cos\theta$:

$$L = \tfrac12 m\ell^2\dot\theta^2 + mg\ell\cos\theta .$$

Then $\partial L/\partial\dot\theta = m\ell^2\dot\theta$ and $\partial L/\partial\theta = -mg\ell\sin\theta$, so

$$\ddot\theta = -\frac{g}{\ell}\sin\theta .$$

For small $\theta$, $\sin\theta\approx\theta$ gives simple harmonic motion with $\omega=\sqrt{g/\ell}$ and period $2\pi\sqrt{\ell/g}$ — independent of amplitude only in that approximation. The exact period grows with amplitude and diverges as the swing approaches the vertical.

Cyclic coordinates and conserved momenta

The generalised momentum conjugate to $q_i$ is

$$p_i = \frac{\partial L}{\partial \dot q_i}.$$

If $L$ does not contain $q_i$ — a cyclic or ignorable coordinate — then Euler–Lagrange gives $\dot p_i = 0$ and $p_i$ is conserved.

For a central force, $L = \tfrac12 m(\dot r^2 + r^2\dot\varphi^2) - V(r)$ has no $\varphi$, so

$$p_\varphi = mr^2\dot\varphi$$

is constant — angular momentum, and Kepler's second law is exactly this statement. Note that $p_\varphi$ is not $m\dot\varphi$: the conjugate momentum to an angle is an angular momentum, with the wrong dimensions to be a linear one.

This is the cleanest instance of Noether's theorem: a continuous symmetry of $L$ gives a conserved quantity. Translation invariance gives linear momentum, rotation invariance gives angular momentum, and invariance under time translation gives energy.

Where intuition breaks

"$L = T + V$." $T - V$. The Hamiltonian is the one that is often $T+V$, and swapping them is the commonest error in the topic.

"The generalised momentum is mass times velocity." $p_i = \partial L/\partial\dot q_i$, which for an angle is an angular momentum, and in a magnetic field is $m\mathbf{v} + q\mathbf{A}$ rather than $m\mathbf{v}$.

"Every constraint reduces the degrees of freedom by one." Every holonomic one does. Non-holonomic constraints restrict the motion without reducing the configuration space.

"Constraint forces must be computed." Choosing coordinates that respect the constraints eliminates them entirely. If you need them, use multipliers deliberately.

"The pendulum period is independent of amplitude." Only in the small-angle approximation. The exact period involves an elliptic integral and grows without bound as the amplitude approaches $\pi$.

"Hamilton's principle says the action is minimised." Stationary. As in the calculus of variations chapter, that is not the same thing, and for long enough times the action is a saddle rather than a minimum.

"Adding a total time derivative to $L$ changes the motion." It does not — $L$ and $L + \frac{dF(q,t)}{dt}$ give the same equations. $L$ is not unique, which is occasionally the point of a question.

The exam's angle

1. Count degrees of freedom first: coordinates minus holonomic constraints. That fixes how many equations you need.
2. Choose coordinates that satisfy the constraints automatically. The whole method rests on it.
3. Write $T$ and $V$ separately, then $L = T-V$. Most errors happen here rather than in the differentiation.
4. Scan for cyclic coordinates before doing any work — each one is a conserved momentum and a first integral for free.
5. Check dimensions of each conjugate momentum, which catches the angular-versus-linear confusion immediately.

The night before

• $L = T - V$; Hamilton's principle makes $\int L\,dt$ stationary.
• $\frac{d}{dt}\frac{\partial L}{\partial\dot q_i} = \frac{\partial L}{\partial q_i}$; $n$ second-order equations.
• Holonomic constraints $f(q,t)=0$ each remove one degree of freedom, and good coordinates make constraint forces vanish. Non-holonomic ones do not work this way.
• Pendulum: $L = \tfrac12 m\ell^2\dot\theta^2 + mg\ell\cos\theta$, $\ddot\theta = -\frac g\ell\sin\theta$, small-angle period $2\pi\sqrt{\ell/g}$.
• $p_i = \partial L/\partial\dot q_i$; cyclic coordinate $\Rightarrow$ conserved momentum.
• Central force: $\varphi$ cyclic, $p_\varphi = mr^2\dot\varphi$ conserved — Kepler's second law.
• Noether: continuous symmetry $\Rightarrow$ conservation law.
• $L$ is not unique: adding $dF/dt$ changes nothing.