The Lagrangian and equations of motion

1. From forces to a scalar description

In the pendulum calculation of Unit 1, the tension disappeared when Newton's equation was projected onto the allowed tangent direction. That was useful, but we still had to differentiate Cartesian coordinates twice and project the resulting acceleration. For a system with several moving parts, this is much of the work.

Lagrangian mechanics organizes the same dynamics through kinetic and potential energy. Both are scalars: they can be calculated without resolving every force into components. The geometry has already been incorporated in the transformations $\mathbf r_a(q_1,\ldots,q_n,t)$. The method then gives one equation for each independent coordinate.

The underlying assumptions in this unit are constant particle masses, an inertial frame for calculating physical kinetic energy, and ideal holonomic constraints incorporated in independent generalized coordinates. Non-ideal forces will be retained explicitly. These qualifications matter: eliminating a coordinate through a non-integrable rolling condition is not generally a valid substitution into the formulas below.

2. Kinetic energy, potential energy, and the Lagrangian

For particles labeled by $a$, kinetic energy is

It measures energy associated with motion in the chosen inertial frame. Use the transformation equations to express every velocity in terms of $q_i,\dot q_i,t$ before constructing the Lagrangian.

A conservative applied force can be represented by a potential: $\mathbf F_a^{(c)}=-\nabla_a V$. Its work is the negative change in potential energy when the potential has no explicit time dependence. The zero of $V$ is arbitrary. For gravity near Earth's surface with upward coordinate $y$, $V=mgy$ because $-\partial V/\partial y=-mg$.

Definition. For an ordinary mechanical system with a velocity-independent potential, the Lagrangian is

$L$ has units of energy, but it is not the stored mechanical energy; that is $T+V$. The subtraction has a dynamical purpose. Differentiating $-V$ gives the applied conservative force, while derivatives of $T$ produce the inertial terms, including the geometrical terms introduced by curved coordinates. The usefulness of $L$ lies in those derivatives, not in assigning a separate substance to “kinetic energy minus potential energy.”

A coordinate $q_i$ may be a distance or an angle. Its generalized velocity is $\dot q_i=dq_i/dt$. In partial derivatives of $L$, treat $q_i$ and $\dot q_i$ as independent arguments; only after differentiating do they follow a particular time-dependent motion. For example,

The latter still depends on time through both $r(t)$ and $\dot\theta(t)$.

3. Lagrange's equations

For conservative applied forces already included in $V$,

This is a system of $n$ second-order differential equations when the velocity dependence is regular. Position and velocity initial data supply the $2n$ constants needed to select a motion. “Conservative form” here means no remaining generalized force on the right; prescribed moving constraints or explicit time dependence can still exchange energy with the system.

If forces remain outside $V$, define their generalized components by virtual work:

Then

Here and throughout Units 2–3, $Q_i$ denotes only the forces not already represented in $L$. Including gravity in $V$ and again on the right would count it twice. $Q_i\delta q_i$ is work, so $Q_i$ has dimensions of energy divided by the dimensions of $q_i$. For an angle it is a torque.

The syllabus requires the equations and their applications, not their full derivation. An optional derivation is provided separately. The proofs of conservation laws and kinetic-energy properties later in this unit remain part of the prescribed development.

4. Reading the three operations

For each coordinate, the three operations are: take a partial derivative with respect to its velocity, take the total time derivative of that result, and subtract the partial derivative with respect to the coordinate. They cannot be interchanged.

For $\partial L/\partial\dot\theta=mr^2\dot\theta$, the product rule gives

Writing only $mr^2\ddot\theta$ silently assumes $r$ is fixed. Equally, a coordinate absent from $L$ can have its velocity present: this will be the source of a conservation law, not a reason to discard its equation.

A systematic calculation begins with a coordinate convention and a sketch. Express $T$ and $V$ in those coordinates, form $L$, apply the equation to each coordinate, and inspect dimensions and limiting cases. Reactions can be recovered afterward from Newton's law if the problem asks for them.

5. Applications

Example 1 · Free motion in Cartesian coordinates

A particle of mass $m$ moves in a plane without applied force. Choose $q_1=x,q_2=y$. Then $T=m(\dot x^2+\dot y^2)/2$, $V=0$, and $L=T$.

The $x$ derivatives are $\partial L/\partial\dot x=m\dot x$ and $\partial L/\partial x=0$, so $m\ddot x=0$. The $y$ equation similarly gives $m\ddot y=0$. Integrating once gives constant velocities $v_{x0},v_{y0}$; integrating again gives

A straight line at constant speed emerges from the scalar kinetic energy. The freedom to choose $x_0,y_0$ and the two velocities reflects the four initial data for two degrees of freedom.

Example 2 · Motion in a uniform gravitational field

Use upward $y$ and horizontal $x$. With $V=mgy$,

For $x$, $m\ddot x=0$. For $y$, $\partial L/\partial y=-mg$ and $d(\partial L/\partial\dot y)/dt=m\ddot y$, so $m\ddot y+mg=0$. Therefore

If $v_{y0}=9.8\,\mathrm{m\,s^{-1}}$, the highest point occurs when $\dot y=v_{y0}-gt=0$: $t=1\,\mathrm s$. The rise is $9.8-4.9=4.9\,\mathrm m$. Mass cancels because inertial and gravitational masses have been identified in the model.

Example 3 · The planar pendulum

A bob of mass $m$ is attached to a fixed pivot by a massless rod of length $a$. Let $\theta$ be measured from the downward vertical. The transformations are $x=a\sin\theta$, $y=-a\cos\theta$. Squaring $\dot x=a\cos\theta\dot\theta$ and $\dot y=a\sin\theta\dot\theta$ and adding gives $v^2=a^2\dot\theta^2$.

Choose the lowest point as zero potential:

Now $\partial L/\partial\dot\theta=ma^2\dot\theta$ and $\partial L/\partial\theta=-mga\sin\theta$. Thus

For positive small $\theta$, acceleration is negative: gravity restores the bob toward the lowest point. Replacing $\sin\theta$ by $\theta$ requires small angles in radians. It gives $\omega=\sqrt{g/a}$ and period $2\pi\sqrt{a/g}$, not the exact period at arbitrary amplitude. The rod reaction never entered $L$ because it performs no virtual work.

Example 4 · The Atwood machine

Two masses $m_1,m_2$ share a taut, massless string over an ideal fixed pulley. Choose $q$ downward for $m_1$; then the downward coordinate of $m_2$ is $b-q$. Their speeds have equal magnitude $|\dot q|$.

Dropping the constant,

Lagrange's equation gives $(m_1+m_2)\ddot q-(m_1-m_2)g=0$. Hence

For $m_1=4\,\mathrm{kg}$ and $m_2=1\,\mathrm{kg}$, $\ddot q=0.6g=5.88\,\mathrm{m\,s^{-2}}$. The heavier mass accelerates downward. If the pulley has appreciable rotational inertia or the string stretches, this kinetic energy is incomplete; changing the physical model requires changing $T$ and possibly the number of coordinates.

6. Newtonian and Lagrangian descriptions

The method is most economical when geometry makes the reactions awkward but the energies simple. For a free particle, Newton's equation may be shorter. A formalism is a way of organizing mechanics, not a different force law.

7. Results and physical interpretation

The minus sign in $L=T-V$ gives the correct conservative force sign. Independent coordinates reduce the number of equations; they do not remove the effects of geometry. Before treating a result as physical, check coordinate conventions, excluded forces, and whether the constraints are still satisfied. For a string pendulum, for example, a computed negative tension signals failure of the taut-string model.

Exercises

Hints and solutions are collected separately.

1. 7.1 Why is the Lagrangian not the total mechanical energy?

2. 7.2 If $q$ is an angle, what are the dimensions of $Q_q$ and $p_q$?

3. 7.3 For $L=\tfrac12m(q)\dot q^2-V(q)$, find the equation of motion.

4. 7.4 For upward vertical coordinate $y$, construct $L$ for a freely falling mass and find its acceleration.

5. 7.5 A pendulum of length $0.98\,\mathrm m$ makes small oscillations. Find its period for $g=9.8\,\mathrm{m\,s^{-2}}$.

6. 7.6 Find the acceleration of an ideal Atwood machine with masses $5$ and $3\,\mathrm{kg}$.

7. 7.7 Construct the polar Lagrangian for a free particle and obtain both equations of motion.

8. 7.8 Why does adding a constant to $V$ leave the motion unchanged?

9. 7.9 A mass slides without friction on $y=cx^2$ under gravity. Construct $L$ using $x$ and find its equation.

10. 7.10 Explain, using a pendulum, what information is eliminated and what is retained by using generalized coordinates.

11. 7.11 If gravity is included in $V$, should $Q_y=-mg$ also be used?

12. 7.12 For a string pendulum, why can a correct angular equation still cease to describe the motion?