From Discrete Chains to the Continuum Limit
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You should be comfortable with the concept of normal modes and the diagonalization of coupled oscillators as established in the previous lecture. Familiarity with basic calculus, specifically limits and partial derivatives, is required to follow the transition from discrete sums to integrals.
Why this matters
The continuum limit reveals that the discrete mathematics of coupled oscillators is a finite approximation of the continuous field theory that governs waves in strings and membranes.
While the discrete chain model is computationally tractable and physically intuitive for small systems, it fails to capture the smooth propagation of energy in macroscopic objects. By taking the limit where the number of masses approaches infinity and their spacing approaches zero, we recover the wave equation. This transition is not merely a mathematical curiosity; it is the foundational step that connects classical mechanics to field theory, demonstrating that the Lagrangian formalism scales seamlessly from particles to fields.
The Intuition of the Limit
Imagine a heavy rope strung with beads. If you have only ten beads, the motion is jerky; energy hops from one bead to the next in distinct steps. If you have a million beads, the motion appears perfectly smooth. The "continuum limit" is the mathematical process of removing the beads and replacing them with a continuous string. In this limit, the discrete coordinates \( q_i \) become a continuous function \( y(x, t) \). The discrete mass \( m \) becomes a linear mass density \( \mu \), and the discrete spring constant \( k \) becomes a tension \( T \). The key insight is that the equations of motion do not change their fundamental nature; they simply change their notation from difference equations to differential equations. The physics remains the same: local interactions propagate disturbances through the system.
From Discrete Coordinates to a Field
Consider a linear chain of \( N \) identical masses \( m \), each connected to its neighbours by springs of constant \( k \). The displacement of the \( i \)-th mass from its equilibrium position is \( q_i(t) \). The spacing between masses is \( a \). To take the continuum limit, we introduce a continuous spatial coordinate \( x = i a \). The discrete displacement \( q_i \) is replaced by the field variable \( y(x, t) \), such that \( y(i a, t) = q_i(t) \).
The kinetic energy of the discrete system is \( T = \frac{1}{2} \sum_{i=1}^{N} m \dot{q}_i^2 \). In the continuum limit, the sum becomes an integral. Since \( dx = a \), we have \( \sum_i \dots \approx \frac{1}{a} \int \dots dx \). The mass \( m \) is related to the linear density \( \mu \) by \( m = \mu a \). Substituting these, the kinetic energy density becomes \( \frac{1}{2} \mu \dot{y}^2 \), and the total kinetic energy is:
Similarly, the potential energy stored in the springs depends on the relative displacement \( q_{i+1} - q_i \). Using a Taylor expansion, \( q_{i+1} - q_i \approx a \frac{\partial y}{\partial x} \). The potential energy of one spring is \( \frac{1}{2} k (q_{i+1} - q_i)^2 \approx \frac{1}{2} k a^2 \left( \frac{\partial y}{\partial x} \right)^2 \). Summing over all springs and converting to an integral, we identify the tension \( T_{\text{tension}} = k a \). The potential energy becomes:
The Lagrangian is the difference \( L = T - V \), now expressed as an integral of a Lagrangian density \( \mathcal{L} \).
The Euler-Lagrange Equation for Fields
In particle mechanics, the Euler-Lagrange equation is \( \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} = 0 \). For a field \( y(x,t) \), the Lagrangian is an integral over space. The principle of least action requires that the variation \( \delta S = 0 \) for arbitrary variations \( \delta y(x,t) \) that vanish at the boundaries.
Applying the calculus of variations to the functional \( S = \int dt \int dx \, \mathcal{L}(y, \dot{y}, y') \), we obtain the field Euler-Lagrange equation. The time derivative term involves \( \frac{\partial \mathcal{L}}{\partial \dot{y}} \), and the spatial derivative term involves integration by parts, which shifts the derivative onto the variation. The resulting equation is:
For our string, \( \mathcal{L} = \frac{1}{2} \mu \dot{y}^2 - \frac{1}{2} T_{\text{tension}} (y')^2 \). Note that \( \mathcal{L} \) does not depend explicitly on \( y \), so \( \frac{\partial \mathcal{L}}{\partial y} = 0 \). Calculating the other terms:
Substituting these into the field Euler-Lagrange equation:
Assuming \( \mu \) and \( T_{\text{tension}} \) are constant, this simplifies to the standard wave equation:
The wave speed is \( v = \sqrt{T_{\text{tension}}/\mu} \). This matches the result from the discrete chain in the long-wavelength limit, confirming the consistency of the continuum approximation.
Connection to Normal Modes
In the discrete system, we found that the normal modes are standing waves with specific frequencies \( \omega_r \). In the continuum limit, the number of modes becomes infinite, and the discrete set of frequencies becomes a continuous spectrum. The normal coordinate \( \xi_r \) in the discrete case corresponds to the amplitude of a specific Fourier mode in the continuum.
The decoupling result from D-132, where \( \xi_r(t) = C_r \cos(\omega_r t + \phi_r) \), generalizes to a superposition of continuous modes. The total displacement \( y(x,t) \) can be written as an integral over wave numbers \( k \):
Here, the dispersion relation is \( \omega(k) = v k \), which is linear for a string. This linear dispersion is a direct consequence of the quadratic form of the Lagrangian density. If the potential energy had higher-order terms in \( y' \), the dispersion would be non-linear, leading to phenomena like solitons. The continuum limit thus provides a bridge between the discrete algebraic eigenvalue problem and the continuous spectral analysis of differential operators.
The derivation
Decoupling into Normal Coordinates
For a conservative system of n coupled oscillators about a stable equilibrium, with symmetric positive-definite mass matrix M and symmetric positive-semidefinite stiffness matrix K, there exists a single linear change of coordinates q = A ξ that simultaneously diagonalizes both matrices — ATM A = I and ATK A = Ω = diag(ω12, …, ωn2) — reducing the coupled equations M q̈ + K q = 0 to n independent scalar harmonic oscillators ξ̈r + ωr2 ξr = 0. The ξr are the normal coordinates.
A common misconception is that the continuum limit always yields a linear wave equation. This is only true if the potential energy is quadratic in the strain. If the springs have non-linear force-displacement relationships, the continuum limit will include higher-order spatial derivatives (e.g., \( \frac{\partial^4 y}{\partial x^4} \)), leading to non-linear wave equations. The validity of the simple wave equation relies on the assumption of small displacements and Hookean springs. C
Worked example
Problem. A string of length \( L = 2.0 \, \mathrm{m} \) has a linear mass density \( \mu = 0.05 \, \mathrm{kg/m} \) and is under a tension \( T_{\text{tension}} = 100 \, \mathrm{N} \). Calculate the speed of a transverse wave propagating along the string and the frequency of the fundamental mode (first harmonic).
- Determine the wave speed \( v \) using the relation derived from the Euler-Lagrange equation: \( v = \sqrt{\frac{T_{\text{tension}}}{\mu}} \).
- Calculate \( v = \sqrt{\frac{100 \, \mathrm{N}}{0.05 \, \mathrm{kg/m}}} \).
- For a string fixed at both ends, the fundamental mode has a wavelength \( \lambda_1 = 2L \).
- Use the wave relation \( v = f_1 \lambda_1 \) to find the fundamental frequency \( f_1 = \frac{v}{2L} \).
Units check. Tension is \( \mathrm{N} = \mathrm{kg \cdot m/s^2} \). Dividing by \( \mu \) (\( \mathrm{kg/m} \)) gives \( \mathrm{m^2/s^2} \). The square root is \( \mathrm{m/s} \), which is correct for speed. Frequency is \( \mathrm{m/s} \) divided by \( \mathrm{m} \), giving \( \mathrm{s^{-1}} \) or \( \mathrm{Hz} \).
Three quick tests
- What replaces the discrete mass \( m \) in the continuum limit?
Answer
The linear mass density \( \mu \), defined as mass per unit length. - Why does the potential energy term in the Lagrangian density involve \( (\frac{\partial y}{\partial x})^2 \)?
Answer
It represents the strain in the string, which is proportional to the slope of the displacement curve. - How does the dispersion relation \( \omega(k) \) for a string differ from that of a rigid rod?
Answer
For a string, \( \omega(k) \) is linear (\( \omega \propto k \)), whereas for a rigid rod, it is typically quadratic or higher order due to bending stiffness.
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.