The Euler-Lagrange Equation
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You should be comfortable with partial differentiation and the concept of a functional from the previous lecture on the calculus of variations.
Why this matters
The Euler-Lagrange equation is the bridge between the abstract action principle and the concrete equations of motion.
While Newton’s laws require vector addition of forces, the Lagrangian approach reduces dynamics to a scalar optimization problem. Once you master the Euler-Lagrange equation, you can derive the motion of complex systems—such as double pendulums or charged particles in electromagnetic fields—without ever explicitly solving for constraint forces. This shift from vector mechanics to scalar variational mechanics is the foundational step toward modern theoretical physics.
The Core Idea
Imagine a particle moving through space. Instead of asking "what force pushes it?", we ask "which path makes the action stationary?" The action is a number calculated by integrating the Lagrangian (kinetic energy minus potential energy) over time. The Euler-Lagrange equation is simply the mathematical condition that says: "If you nudge the path slightly, the action does not change to first order." It is the condition for a stationary point (usually, but not always, a minimum). In plain language, nature chooses the path that makes the action stationary. This equation tells us how the velocity and position must relate to each other at every instant to satisfy this condition.
Generalized Coordinates
Newtonian mechanics uses Cartesian coordinates \((x, y, z)\), which are often awkward for constrained systems. Lagrangian mechanics uses generalized coordinates \(q_i\), which can be angles, distances, or any independent parameters that fully describe the system's configuration. For a system with \(N\) degrees of freedom, we define \(N\) generalized coordinates \(q_1, q_2, \dots, q_N\). Their time derivatives are the generalized velocities \(\dot{q}_i\). The Lagrangian \(L\) is a function of these coordinates, their velocities, and time: \(L(q_i, \dot{q}_i, t)\). The beauty of this formalism is that the coordinates can be curvilinear or even abstract, as long as they are independent.
Because the coordinates are independent, we can vary each \(q_i\) separately. This independence is crucial; it gives exactly one equation of motion per degree of freedom, although those equations are in general coupled to one another.
The Euler-Lagrange Equation
From the calculus of variations, we derived that for the action \(S = \int_{t_1}^{t_2} L \, dt\) to be stationary, the integrand must satisfy a specific differential equation for each coordinate. This is the Euler-Lagrange equation. It states that the time derivative of the partial derivative of the Lagrangian with respect to velocity must equal the partial derivative of the Lagrangian with respect to position.
Notice the structure: the left-hand side involves a total time derivative of a quantity depending on velocity, while the right-hand side (moved to the left) involves a partial derivative with respect to position. This equation is a second-order ordinary differential equation for \(q_i(t)\). It is the direct analogue of Newton’s second law \(F = ma\), but expressed in terms of energy rather than force.
Connection to Newton's Law
To see how this recovers Newtonian mechanics, consider a single particle in one dimension with potential energy \(V(x)\). The Lagrangian is \(L = T - V = \frac{1}{2}m\dot{x}^2 - V(x)\). We apply the Euler-Lagrange equation with \(q = x\). First, compute the partial derivative with respect to velocity: \(\frac{\partial L}{\partial \dot{x}} = m\dot{x}\). This quantity is the momentum \(p\). Next, take the total time derivative: \(\frac{d}{dt}(m\dot{x}) = m\ddot{x}\). Now compute the partial derivative with respect to position: \(\frac{\partial L}{\partial x} = -\frac{dV}{dx}\). Substituting these into the Euler-Lagrange equation gives \(m\ddot{x} - (-\frac{dV}{dx}) = 0\), which simplifies to \(m\ddot{x} = -\frac{dV}{dx}\). Since the force is \(F = -\frac{dV}{dx}\), we recover \(F = ma\). This demonstrates that the Euler-Lagrange equation is not a new law of nature, but a more general formulation of the same dynamics.
The derivation
The Euler-Lagrange Equation
Derives that a stationary action integral requires each generalized coordinate to satisfy d/dt(dL/dq-dot) - dL/dq = 0.
A common pitfall is confusing the total derivative \(\frac{d}{dt}\) with the partial derivative \(\frac{\partial}{\partial t}\). In the Euler-Lagrange equation, the derivative acting on \(\frac{\partial L}{\partial \dot{q}_i}\) is a total time derivative, meaning you must account for the fact that \(q_i\) and \(\dot{q}_i\) both change with time. If the Lagrangian has explicit time dependence, \(\frac{\partial L}{\partial t}\) is not zero, but this does not appear directly in the Euler-Lagrange equation itself; it only affects the conservation of the Hamiltonian. Always distinguish between explicit and implicit time dependence when differentiating. C
Worked example
Problem. A particle of mass \(m = 2.0\ \mathrm{kg}\) moves in one dimension under the influence of a harmonic potential \(V(x) = \frac{1}{2}kx^2\) with spring constant \(k = 50\ \mathrm{N/m}\). The particle is released from rest at \(x(0) = 0.1\ \mathrm{m}\). Use the Euler-Lagrange equation to find the equation of motion and the angular frequency of oscillation.
- Write the Lagrangian: \(L = T - V = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\).
- Compute \(\frac{\partial L}{\partial \dot{x}} = m\dot{x}\).
- Compute \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) = m\ddot{x}\).
- Compute \(\frac{\partial L}{\partial x} = -kx\).
- Apply the Euler-Lagrange equation: \(m\ddot{x} - (-kx) = 0 \implies m\ddot{x} + kx = 0\).
- Rearrange to standard form: \(\ddot{x} + \frac{k}{m}x = 0\). The angular frequency is \(\omega = \sqrt{\frac{k}{m}}\).
- Substitute values: \(\omega = \sqrt{\frac{50}{2.0}} = \sqrt{25} = 5.0\ \mathrm{rad/s}\).
Units check. \(k\) has units \(\mathrm{N/m} = \mathrm{kg/s^2}\). \(m\) has units \(\mathrm{kg}\). Thus \(\frac{k}{m}\) has units \(\frac{\mathrm{kg/s^2}}{\mathrm{kg}} = \mathrm{s^{-2}}\). The square root gives \(\mathrm{s^{-1}}\), which is radians per second. The units are consistent.
Three quick tests
- What is the physical interpretation of the term \(\frac{\partial L}{\partial \dot{q}_i}\)?
Answer
It is the generalized momentum conjugate to the coordinate \(q_i\). - If the Lagrangian does not depend explicitly on a coordinate \(q_j\), what can you say about \(\frac{\partial L}{\partial \dot{q}_j}\)?
Answer
It is conserved (constant in time) because \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right) = \frac{\partial L}{\partial q_j} = 0\). - Why do we use generalized coordinates instead of Cartesian coordinates in Lagrangian mechanics?
Answer
Generalized coordinates automatically incorporate constraints, reducing the number of variables needed to describe the system and eliminating the need to solve for constraint forces explicitly.
Problem set for PU-201 →
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.