Euler-Lagrange Equation from Hamilton's Principle

1. The Principle of Least Action

In Newtonian mechanics, you solve for the motion of a particle locally. You stand at a point, calculate the forces acting on the particle, and use $\vec{F}=m\vec{a}$ to determine its acceleration, and thus where it will be an instant later.

The Principle of Least Action offers a different, "global" perspective. It imagines all possible paths a particle could take between a starting point at time $t_1$ and an ending point at time $t_2$. It then claims that the path nature actually chooses is the one that makes a specific quantity, the action, stationary (usually a minimum).

To get a feel for this new perspective and why it's so powerful, let's start with a conceptual overview.

Defining the Action and the Lagrangian

The video has given us the core idea. Now, we need to define the "action" mathematically. The action, denoted by $S$, is a functional of the system's path, $q(t)$. It is defined as the time integral of a function called the Lagrangian, $L$.

For most classical systems, the Lagrangian is defined with beautiful simplicity as the kinetic energy (T) minus the potential energy (V) of the system.

This is a crucial definition. The action is not the integral of the total energy ($T+V$). The fact that it's the difference is key and leads to all the correct physics.

The following resource explains this definition clearly and sets the stage for the derivation.

2. Derivation from Hamilton's Principle

We now have all the pieces:

1. The Physical Principle: The true path of a system between $t_1$ and $t_2$ makes the action $S$ stationary ($\delta S=0$).
2. The Mathematical Definition: $S={\int }_{t_1}^{t_2}L(q,\dot{q},t)dt$, where $L=T-V$.
3. The Mathematical Tool: From our last lesson, we know how to find the condition for a functional to be stationary—the Euler-Lagrange equation.

Let's combine them. We will now perform the same mathematical derivation as in the last lesson, but this time our functional is the action, our variable is time $t$, and our function is the path $q(t)$. The result will no longer be just a mathematical identity; it will be the equation of motion for the system.

Given your preference for rigorous derivations, following this process closely is essential. We will use the same method of path variation.

Consider the true path $q_0(t)$ and a family of varied paths:

Here, $\alpha$ is a small parameter, and $\eta (t)$ is an arbitrary function that vanishes at the endpoints, i.e., $\eta (t_1)=\eta (t_2)=0$, ensuring all paths start and end at the same points in configuration space at the same times.

The action $S$ becomes a function of $\alpha$, and the principle of stationary action requires:

Let's watch this derivation in detail.

The result of this derivation is the Euler-Lagrange Equation of Motion:

This is the central equation of Lagrangian mechanics.

For a more formal, text-based proof that you can review at your own pace, the following resource is excellent. It presents the exact same logic in a concise, mathematical format.

3. Equivalence with Newtonian Mechanics

This new formalism is elegant, but is it correct? We must verify that for a simple system, the Euler-Lagrange equation reproduces Newton's second law, $\vec{F}=m\vec{a}$.

Let's consider a single particle of mass $m$ moving in one dimension $x$ under a potential $V(x)$.

• Kinetic Energy: $T=\frac{1}{2}m{\dot{x}}^2$
• Potential Energy: $V=V(x)$
• Lagrangian: $L=T-V=\frac{1}{2}m{\dot{x}}^2-V(x)$

Now, let's compute the terms for the Euler-Lagrange equation:

1. $\frac{\partial L}{\partial x}=-\frac{\partial V}{\partial x}$. By definition, the force is the negative gradient of the potential, so $F_x=-\frac{\partial V}{\partial x}$. Thus, $\frac{\partial L}{\partial x}=F_x$.
2. $\frac{\partial L}{\partial \dot{x}}=\frac{\partial }{\partial \dot{x}}\left(\frac{1}{2}m{\dot{x}}^2\right)=m\dot{x}$. This quantity, the partial derivative of the Lagrangian with respect to a velocity, is called the generalized momentum corresponding to that coordinate. In this Cartesian case, it's just the familiar linear momentum, $p_x$.

Substituting these into the Euler-Lagrange equation $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)=\frac{\partial L}{\partial x}$ gives:

This is precisely Newton's second law. The Lagrangian formalism successfully reproduces Newtonian mechanics. The resource from Dr. Mitchell's channel also demonstrates this equivalence beautifully for the 3D case. The section in "The Lagrangian Method" PDF titled "This principle... is equivalent to F = ma" (resource LINK, section idx=1) further solidifies this point.

Conclusion

In this lesson, we have made a significant leap from a purely mathematical tool to a deep physical principle. We have shown that the entire framework of classical mechanics can be built upon a single, elegant statement: the Principle of Stationary Action.

Key Takeaways:

• The Principle of Stationary Action (Hamilton's Principle) states that the true physical path of a system is one that makes the action functional stationary.
• The Action is the time integral of the Lagrangian: $S=\int (T-V)dt$.
• The Lagrangian $L=T-V$ is a scalar quantity based on the system's energies.
• By demanding that the action be stationary ($\delta S=0$), we rigorously derive the Euler-Lagrange equation of motion: $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=\frac{\partial L}{\partial q}$.
• This formalism is entirely equivalent to Newtonian mechanics but provides a more powerful and general foundation, as it deals with scalar energies rather than vector forces.

Preview of the Next Lesson:
So far, we've only used simple Cartesian coordinates ($x,y,z$). The true power of the Lagrangian method shines when we deal with complex systems and constraints. In the next lesson, we will introduce generalized coordinates. These are coordinates that are adapted to the specific geometry of a problem (like angles for a pendulum). You will learn how to formulate the Lagrangian in any convenient coordinate system and see just how easily the Euler-Lagrange equation delivers the correct equations of motion, even for problems that would be a nightmare to solve with Newtonian forces.