Hamilton’s Principle
Learning Objectives:
- Understand the statement and physical meaning of Hamilton’s principle.
- Derive the Euler-Lagrange equations from the principle.
- Apply Hamilton’s principle to solve simple dynamical systems.
Key Concepts / Definitions:
- Hamilton’s Principle: The actual path taken by a system between two configurations is such that the action integral is stationary (usually a minimum).
- Action: The integral of the Lagrangian over time, $S = \int_{t_1}^{t_2} L(q, \dot{q}, t) \, dt$.
- Stationary Action: A value of the action where its first variation is zero, $\delta S = 0$.
Theory and Explanation:
Hamilton’s Principle, also known as the Principle of Stationary Action, is a cornerstone of analytical mechanics. It states that:
Out of all possible paths that a system can follow between two fixed points in configuration space and time, the actual path followed is the one that makes the action integral stationary.
This principle unifies many laws of classical mechanics and provides a natural route to derive the Euler-Lagrange equations, which are central to Lagrangian mechanics.
The action is a scalar quantity defined by:
\[S[q(t)] = \int_{t_1}^{t_2} L(q, \dot{q}, t) \, dt\]Here:
- $q(t)$ are generalized coordinates.
- $\dot{q} = \frac{dq}{dt}$ is the generalized velocity.
- $L(q, \dot{q}, t)$ is the Lagrangian of the system, typically $T - V$, where $T$ is kinetic energy and $V$ is potential energy.
If the action is stationary, then the path taken by the system satisfies:
\[\delta S = 0\]This leads to the Euler-Lagrange equation, which governs the dynamics of the system.
Mathematical Formulation:
Let $q(t)$ be a differentiable path connecting two fixed endpoints at $t = t_1$ and $t = t_2$.
The action functional is:
\[S[q(t)] = \int_{t_1}^{t_2} L(q, \dot{q}, t) \, dt\]Consider a variation $q(t) \rightarrow q(t) + \varepsilon \eta(t)$, where $\eta(t_1) = \eta(t_2) = 0$.
The variation of action is:
\[\delta S = \frac{d}{d\varepsilon} S[q + \varepsilon \eta] \bigg|_{\varepsilon=0} = \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} \eta + \frac{\partial L}{\partial \dot{q}} \dot{\eta} \right) dt\]Integrating the second term by parts and applying boundary conditions:
\[\delta S = \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} \right) \eta(t) \, dt\]For $\delta S = 0$ for arbitrary $\eta(t)$, the integrand must vanish:
\[\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0\]This is the Euler-Lagrange equation.
Solved Examples:
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Example 1:
Problem: Derive the equation of motion for a free particle using Hamilton’s principle.
Solution:
For a free particle of mass $m$, the Lagrangian is $L = \frac{1}{2} m \dot{x}^2$.Action: \(S[x(t)] = \int_{t_1}^{t_2} \frac{1}{2} m \dot{x}^2 \, dt\)
Euler-Lagrange equation: \(\frac{d}{dt} \left( m \dot{x} \right) = 0 \Rightarrow m \ddot{x} = 0\)
This implies constant velocity motion: $x(t) = vt + c$.
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Example 2:
Problem: A particle moves under a constant force $F$. Derive its equation of motion using Hamilton’s principle.
Solution:
The Lagrangian is $L = \frac{1}{2} m \dot{x}^2 + Fx$.Euler-Lagrange equation: \(\frac{d}{dt}(m \dot{x}) - F = 0 \Rightarrow m \ddot{x} = F\)
This leads to uniformly accelerated motion.
Important Points / Summary:
- Hamilton’s principle leads to the Euler-Lagrange equations.
- It is a variational reformulation of Newtonian mechanics.
- The action is stationary, not necessarily minimal.
- Applies to conservative systems and forms the foundation of field theory.
Practice Questions:
- Short Answer:
- State Hamilton’s principle in your own words.
- How is the Euler-Lagrange equation derived from the principle of stationary action?
- Numerical:
- A particle moves under a potential $V(x) = \frac{1}{2} kx^2$. Use Hamilton’s principle to find the equation of motion.
- Compute the action for a particle moving from $x=0$ to $x=a$ in time $T$ with constant velocity.
- MCQs:
- The Euler-Lagrange equation is obtained from Hamilton’s principle by:
- a) Differentiating the Lagrangian directly
- b) Rewriting Newton’s law
- c) Requiring $\delta S = 0$ for arbitrary variations
- d) Using energy conservation
Answer: c)
- In Hamilton’s principle, the variation $\eta(t)$ must:
- a) Be arbitrary
- b) Vanish at the endpoints
- c) Be a constant function
- d) Satisfy $\dot{\eta}(t_1) = 0$
Answer: b)
- The Euler-Lagrange equation is obtained from Hamilton’s principle by: