The Euler-Lagrange Equations and What Comes Next
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You should be comfortable with the concept of generalized coordinates and have just reviewed d'Alembert's principle, which states that the sum of the effective forces dotted with virtual displacements is zero for a system in equilibrium.
Why this matters
The Euler-Lagrange equations provide a unified, coordinate-free framework for solving mechanical problems that is often more efficient than Newton's vector laws.
While Newton's second law requires resolving forces into Cartesian components and dealing explicitly with constraint forces, the Lagrangian formulation allows us to describe a system using the minimum number of independent variables. By focusing on the scalar quantities of kinetic and potential energy, we bypass the vectorial complexity of constraints. This lecture establishes the Euler-Lagrange equations as the central result of this transition, demonstrating how they encapsulate the dynamics of any holonomic system and setting the stage for the deeper symmetries and conservation laws that define the structure of classical mechanics.
The Core Idea in Plain Language
Imagine you are trying to find the path a ball takes when thrown. In Newtonian mechanics, you must constantly calculate the forces acting on the ball at every instant, including the tension in a string if it is on a pendulum, and solve a set of coupled vector equations. This is often messy because the forces depend on the geometry of the constraints.
The Lagrangian approach changes the question. Instead of asking "what forces are pushing the ball?", we ask "what is the trade-off between its speed and its height?" We define a single quantity, the Lagrangian \( L \), which is the kinetic energy minus the potential energy. The principle behind the Euler-Lagrange equations is that nature chooses the path that makes a specific integral of this Lagrangian stationary. In practical terms, for each independent coordinate you choose to describe the system, there is one equation of motion. If you use polar coordinates for a central force problem, you get one equation for the radius and one for the angle. You do not need to calculate the constraint forces at all; they are automatically accounted for by the choice of coordinates. This shift from vector forces to scalar energies is the fundamental conceptual leap of analytical mechanics.
From d'Alembert to Euler-Lagrange
The bridge between Newton's world and Lagrange's world is d'Alembert's principle. As established in the previous lecture, for a system of \( N \) particles subject to ideal constraints, the sum of the applied forces minus the inertial forces, dotted with the virtual displacements, vanishes:
To convert this into the Euler-Lagrange form, we express the Cartesian positions \( \mathbf{r}_i \) in terms of the generalized coordinates \( q_1, \dots, q_n \). The virtual displacement \( \delta \mathbf{r}_i \) is related to the variation in generalized coordinates \( \delta q_j \) by the chain rule. Since the constraints are holonomic and ideal, the constraint forces do no work, and they disappear from the equation. The remaining terms involve the applied forces and the kinetic energy. If the applied forces are conservative, they can be derived from a potential \( V(q,t) \) that does not depend on velocity. The kinetic energy \( T \) is a quadratic function of the generalized velocities \( \dot{q}_j \). By carefully manipulating the terms involving \( \ddot{\mathbf{r}}_i \) and using the fact that the virtual displacements \( \delta q_j \) are arbitrary, we arrive at the Euler-Lagrange equations. This derivation is detailed in the cited card D-038, which shows that the condition for the system to be in dynamic equilibrium for arbitrary virtual displacements is equivalent to the following set of \( n \) second-order differential equations:
Here, \( L \) is the Lagrangian. The term \( \frac{\partial L}{\partial \dot{q}_j} \) is often called the generalized momentum conjugate to \( q_j \). The equation states that the rate of change of this generalized momentum is equal to the generalized force associated with that coordinate.
Generalized Coordinates and Momentum
One of the most powerful aspects of the Euler-Lagrange equations is the freedom to choose coordinates. In Newtonian mechanics, we are often stuck with Cartesian coordinates \( x, y, z \). In Lagrangian mechanics, we can use any set of \( n \) independent variables that fully describe the configuration of the system. For example, for a simple pendulum, we use the angle \( \theta \) rather than \( x \) and \( y \). This reduces the problem from two coupled equations to a single equation.
The concept of generalized momentum is crucial here. For a Cartesian coordinate \( x \), the generalized momentum is simply \( p_x = \frac{\partial L}{\partial \dot{x}} = m \dot{x} \), which is the standard linear momentum. However, for an angular coordinate \( \theta \), the generalized momentum is \( p_\theta = \frac{\partial L}{\partial \dot{\theta}} \). If the kinetic energy is \( T = \frac{1}{2} m l^2 \dot{\theta}^2 \), then \( p_\theta = m l^2 \dot{\theta} \), which is the angular momentum about the pivot. Thus, the Euler-Lagrange equation for an angular coordinate directly yields the conservation of angular momentum if the potential does not depend on that angle. This unification of linear and angular momentum under the single concept of generalized momentum is a significant conceptual advance.
What Comes Next: Cyclic Coordinates
The Euler-Lagrange equations are not just a different way to write Newton's laws; they reveal the deep structure of mechanics. A key feature to look for in any Lagrangian is a "cyclic" or "ignorable" coordinate. A coordinate \( q_k \) is cyclic if it does not appear explicitly in the Lagrangian \( L \), meaning \( \frac{\partial L}{\partial q_k} = 0 \). In this case, the Euler-Lagrange equation simplifies to:
This implies that the generalized momentum \( p_k = \frac{\partial L}{\partial \dot{q}_k} \) is constant in time. This is a direct statement of a conservation law. For instance, if the potential energy of a system is independent of the position \( x \), then \( x \) is a cyclic coordinate, and the linear momentum \( p_x \) is conserved. If the potential is independent of the angle \( \phi \), then \( \phi \) is cyclic, and the angular momentum \( L_z \) is conserved. This connection between symmetry (invariance of the Lagrangian under a transformation) and conservation laws is the precursor to Noether's theorem, which will be discussed in the synthesis lecture. It shows that conservation laws are not separate postulates but are inherent consequences of the form of the Lagrangian and the symmetries of the system.
The derivation
Euler-Lagrange Equations from d'Alembert
For a system of N particles subject to ideal (workless) holonomic constraints and described by n independent generalized coordinates \( q_1,\dots,q_n \), d'Alembert's principle \( \sum_i \left(\mathbf{F}_i - m_i \ddot{\mathbf{r}}_i\right)\cdot \delta\mathbf{r}_i = 0 \) is exactly equivalent to the n Euler–Lagrange equations \[ \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}_j}\right) - \frac{\partial L}{\partial q_j} = 0, \qquad L = T - V, \] provided the applied forces derive from a potential \( V=V(q,t) \) that is independent of the generalized velocities.
A common pitfall is assuming that the Lagrangian \( L \) is unique. In fact, you can add the total time derivative of any function \( f(q,t) \) to the Lagrangian, i.e., \( L' = L + \frac{d}{dt}f(q,t) \), and the Euler-Lagrange equations remain unchanged. This is because the action integral changes only by boundary terms, which vanish for variations that are zero at the endpoints. However, this freedom does not extend to adding arbitrary functions of velocity or time that are not total derivatives. Always verify that your Lagrangian correctly captures the physics by checking the resulting equations of motion against known cases, such as free particle motion or simple harmonic oscillation. C
Worked example
Problem. A particle of mass \( m \) moves in a vertical plane under the influence of gravity \( g \). It is constrained to move on a frictionless circular hoop of radius \( R \). Let \( \theta \) be the angle measured from the lowest point of the hoop. Find the equation of motion for \( \theta \).
- Choose the generalized coordinate \( \theta \). The position of the particle can be described by \( x = R \sin \theta \) and \( y = -R \cos \theta \) (taking the center of the hoop as the origin, with \( y \) positive upwards).
- Calculate the kinetic energy \( T \). The velocity components are \( \dot{x} = R \cos \theta \, \dot{\theta} \) and \( \dot{y} = R \sin \theta \, \dot{\theta} \). Thus, \( v^2 = \dot{x}^2 + \dot{y}^2 = R^2 \dot{\theta}^2 (\cos^2 \theta + \sin^2 \theta) = R^2 \dot{\theta}^2 \). So, \( T = \frac{1}{2} m R^2 \dot{\theta}^2 \).
- Calculate the potential energy \( V \). Taking the lowest point as the reference level, the height is \( h = R(1 - \cos \theta) \). So, \( V = m g R (1 - \cos \theta) \).
- Form the Lagrangian \( L = T - V = \frac{1}{2} m R^2 \dot{\theta}^2 - m g R (1 - \cos \theta) \).
- Apply the Euler-Lagrange equation for \( \theta \). First, compute \( \frac{\partial L}{\partial \dot{\theta}} = m R^2 \dot{\theta} \). Then, \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) = m R^2 \ddot{\theta} \). Next, compute \( \frac{\partial L}{\partial \theta} = - m g R \sin \theta \).
- Substitute into the equation: \( m R^2 \ddot{\theta} - (- m g R \sin \theta) = 0 \), which simplifies to \( m R^2 \ddot{\theta} + m g R \sin \theta = 0 \).
Units check. The term \( \ddot{\theta} \) has units of \( \mathrm{rad/s^2} \) (dimensionless per time squared). The term \( \frac{g}{R} \sin \theta \) has units of \( \frac{\mathrm{m/s^2}}{\mathrm{m}} = \mathrm{1/s^2} \). Since radians are dimensionless, the units are consistent. The equation describes the angular acceleration being proportional to the sine of the angle, which is the standard equation for a simple pendulum.
Three quick tests
- If the Lagrangian \( L \) does not depend explicitly on time \( t \), what quantity is conserved?
Answer
The energy (or Hamiltonian) of the system is conserved. This follows from the fact that \( \frac{\partial L}{\partial t} = 0 \) implies that the total energy \( E = \sum \dot{q}_i p_i - L \) is constant. - For a free particle in 3D space, how many Euler-Lagrange equations do you need to write?
Answer
Three. One for each Cartesian coordinate \( x, y, z \). Each equation will be of the form \( m \ddot{x} = 0 \), \( m \ddot{y} = 0 \), \( m \ddot{z} = 0 \), reflecting the conservation of linear momentum in each direction. - Why is the Lagrangian formulation particularly useful for systems with constraints?
Answer
Because it allows us to use generalized coordinates that automatically satisfy the constraints. We do not need to calculate the constraint forces explicitly, as they do no work in ideal constraints and thus do not appear in the Lagrangian. This reduces the number of equations to solve to the number of degrees of freedom.
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.