N Oscillators and Normal Coordinates
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You should be comfortable with the equations of motion for a single damped, driven harmonic oscillator and understand the concept of normal modes in a two-oscillator system. Familiarity with matrix notation for quadratic forms in kinetic and potential energy is assumed.
Why this matters
Normal coordinates transform a complex, coupled system into a set of independent, simple oscillators.
In the previous lecture, we examined how energy sloshes between two coupled oscillators, creating beats. However, real physical systems—such as a guitar string, a crystal lattice, or a molecule—involve many interacting degrees of freedom. Solving the coupled differential equations for \(N\) masses directly is intractable for large \(N\). The power of normal coordinates lies in their ability to decouple the system. By finding the specific linear combinations of displacements that oscillate independently, we reduce a complex many-body problem to \(N\) independent simple harmonic oscillators. This mathematical structure is the foundation for understanding phonons in solids, vibrational modes in molecules, and ultimately the quantization of fields.
The Idea of Decoupling
Imagine a row of balls connected by springs. If you push one ball, the motion propagates down the line, but the motion is messy; every ball moves at different times and amplitudes. However, there are special patterns of motion where every ball moves in unison, just with different amplitudes. In these special patterns, the internal forces between balls do no net work to change the shape of the motion; they only provide the restoring force. These special patterns are the normal modes. The key insight is that any arbitrary motion of the system can be viewed as a superposition of these independent modes. Instead of tracking the position of every single ball, we track the "amount" of each mode present. These amounts are the normal coordinates. They are independent variables, meaning the equation for one normal coordinate does not contain any other normal coordinate. This is the essence of diagonalization: we are rotating our coordinate system to align with the natural axes of the system's energy landscape.
Quadratic Forms and Diagonalization
For small oscillations about a stable equilibrium, the kinetic energy \(T\) and potential energy \(V\) can be expressed as quadratic forms in the generalized coordinates \(\mathbf q\). Let \(\mathbf q\) be a column vector of \(N\) coordinates. The kinetic energy is given by \(T = \frac{1}{2} \dot{\mathbf q}^{\mathsf T} \mathbf M \dot{\mathbf q}\), where \(\mathbf M\) is the symmetric, positive-definite mass matrix. The potential energy is \(V = \frac{1}{2} \mathbf q^{\mathsf T} \mathbf K \mathbf q\), where \(\mathbf K\) is the symmetric, positive-semidefinite stiffness matrix. The equations of motion are derived from the Lagrangian \(L = T - V\), leading to the matrix equation \(\mathbf M \ddot{\mathbf q} + \mathbf K \mathbf q = \mathbf 0\).
To decouple this system, we seek a linear transformation \(\mathbf q = \mathbf A \boldsymbol\eta\), where \(\boldsymbol\eta\) is the vector of normal coordinates and \(\mathbf A\) is a constant transformation matrix. We choose \(\mathbf A\) such that it simultaneously diagonalizes both \(\mathbf M\) and \(\mathbf K\). Specifically, we require \(\mathbf A^{\mathsf T} \mathbf M \mathbf A = \mathbf I\) (the identity matrix) and \(\mathbf A^{\mathsf T} \mathbf K \mathbf A = \boldsymbol\Omega^2\), where \(\boldsymbol\Omega^2\) is a diagonal matrix containing the squared normal frequencies \(\omega_r^2\). This is possible because \(\mathbf M\) and \(\mathbf K\) are real symmetric matrices, and \(\mathbf M\) is positive-definite, allowing us to define an inner product in which \(\mathbf K\) is symmetric and thus diagonalizable by an orthogonal transformation.
Substituting \(\mathbf q = \mathbf A \boldsymbol\eta\) into the equations of motion and premultiplying by \(\mathbf A^{\mathsf T}\), we obtain \(\ddot{\boldsymbol\eta} + \boldsymbol\Omega^2 \boldsymbol\eta = \mathbf 0\). Since \(\boldsymbol\Omega^2\) is diagonal, this single matrix equation splits into \(N\) independent scalar equations: \(\ddot{\eta}_r + \omega_r^2 \eta_r = 0\) for \(r = 1, \dots, N\).
The Eigenvalue Problem
The normal frequencies \(\omega_r\) and the mode shapes (columns of \(\mathbf A\)) are determined by the generalized eigenvalue problem. Assuming a solution of the form \(\mathbf q(t) = \mathbf a e^{i\omega t}\), we substitute this into the equation of motion \(\mathbf M \ddot{\mathbf q} + \mathbf K \mathbf q = \mathbf 0\) to get \((\mathbf K - \omega^2 \mathbf M) \mathbf a = \mathbf 0\). For a non-trivial solution \(\mathbf a \neq \mathbf 0\) to exist, the determinant of the coefficient matrix must vanish:
This is a polynomial equation of degree \(N\) in \(\omega^2\). The roots \(\omega_1^2, \dots, \omega_N^2\) are the squared normal frequencies. For each root \(\omega_r^2\), we solve for the corresponding eigenvector \(\mathbf a_r\). These eigenvectors define the relative displacements of the masses in the \(r\)-th normal mode. A crucial property of this system is that eigenvectors corresponding to distinct eigenvalues are orthogonal with respect to the mass matrix \(\mathbf M\). That is, \(\mathbf a_r^{\mathsf T} \mathbf M \mathbf a_s = 0\) for \(r \neq s\). We can normalize them such that \(\mathbf a_r^{\mathsf T} \mathbf M \mathbf a_s = \delta_{rs}\), where \(\delta_{rs}\) is the Kronecker delta. This orthogonality ensures that the normal coordinates are independent and that energy does not transfer between modes in the absence of damping or driving.
General Solution and Superposition
Once the normal coordinates are identified, the general solution for the displacement of the system is a superposition of the normal modes. The displacement vector \(\mathbf q(t)\) is given by:
where each normal coordinate \(\eta_r(t)\) evolves independently as a simple harmonic oscillator: \(\eta_r(t) = C_r \cos(\omega_r t + \phi_r)\). The constants \(C_r\) and \(\phi_r\) are determined by the initial conditions of the system. The vector \(\mathbf a_r\) is the \(r\)-th eigenvector, normalized such that \(\mathbf a_r^{\mathsf T} \mathbf M \mathbf a_r = 1\). This decomposition is unique. It allows us to analyze complex motions by breaking them down into their fundamental building blocks. For example, if the system is initially displaced in a way that matches a specific eigenvector, only that mode will be excited. If the initial displacement is arbitrary, multiple modes will be excited, and the resulting motion will be a complex combination of sinusoidal oscillations at different frequencies.
The derivation
Normal Coordinates for N Coupled Oscillators
For a conservative system of \(N\) degrees of freedom executing small oscillations about a stable equilibrium, with quadratic kinetic energy \(T=\tfrac12\,\dot{\mathbf q}^{\mathsf T}\mathbf M\,\dot{\mathbf q}\) and potential \(V=\tfrac12\,\mathbf q^{\mathsf T}\mathbf K\,\mathbf q\) (\(\mathbf M\) symmetric positive-definite, \(\mathbf K\) symmetric positive-semidefinite), there exists a single real linear transformation \(\mathbf q=\mathbf A\,\boldsymbol\eta\) that simultaneously diagonalizes \(\mathbf M\) and \(\mathbf K\), giving \(\mathbf A^{\mathsf T}\mathbf M\mathbf A=\mathbf I\) and \(\mathbf A^{\mathsf T}\mathbf K\mathbf A=\boldsymbol\Omega^2=\mathrm{diag}(\omega_1^2,\dots,\omega_N^2)\). In the normal coordinates \(\eta_r\) the Lagrangian separates into \(N\) uncoupled harmonic oscillators \(\ddot\eta_r+\omega_r^2\eta_r=0\), whose frequencies \(\omega_r^2\) are the roots of \(\det(\mathbf K-\omega^2\mathbf M)=0\).
A common misconception is that the normal coordinates are simply the physical displacements of the masses. They are not. They are abstract linear combinations of the physical coordinates. The physical displacement of mass \(i\) is \(\sum_{r=1}^N a_{ir} \eta_r(t)\). Confusing the mode shape \(\mathbf a_r\) with the coordinate \(\eta_r\) leads to errors in calculating energy and interpreting the physical motion. Remember, \(\eta_r\) is the "amplitude" of the \(r\)-th mode, not the position of a specific mass. C
Worked example
Problem. Consider a system of two identical masses \(m = 1.0\ \mathrm{kg}\) connected by three identical springs of stiffness \(k = 100\ \mathrm{N/m}\). The outer springs are attached to fixed walls, and the inner spring connects the two masses. Find the normal frequencies and the mode shapes.
- Define the coordinates \(x_1\) and \(x_2\) as the displacements of the left and right masses from equilibrium. The kinetic energy is \(T = \frac{1}{2} m \dot{x}_1^2 + \frac{1}{2} m \dot{x}_2^2\). The potential energy is \(V = \frac{1}{2} k x_1^2 + \frac{1}{2} k (x_2 - x_1)^2 + \frac{1}{2} k x_2^2\). Simplifying \(V\), we get \(V = \frac{1}{2} k (2x_1^2 - 2x_1 x_2 + 2x_2^2)\). Thus, \(\mathbf M = m \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) and \(\mathbf K = k \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}\).
- Solve the eigenvalue problem \(\det(\mathbf K - \omega^2 \mathbf M) = 0\). This gives \(\det \begin{pmatrix} 2k - m\omega^2 & -k \\ -k & 2k - m\omega^2 \end{pmatrix} = 0\). Let \(\lambda = \omega^2\). The determinant is \((2k - m\lambda)^2 - k^2 = 0\). So, \(2k - m\lambda = \pm k\).
- Case 1: \(2k - m\lambda = k \implies m\lambda = k \implies \omega_1^2 = k/m\). With \(k=100\) and \(m=1\), \(\omega_1 = \sqrt{100} = 10\ \mathrm{rad/s}\). The eigenvector \(\mathbf a_1\) satisfies \((\mathbf K - 100 \mathbf M)\mathbf a_1 = 0\). \(\begin{pmatrix} 100 & -100 \\ -100 & 100 \end{pmatrix} \begin{pmatrix} a_{11} \\ a_{12} \end{pmatrix} = \mathbf 0 \implies a_{11} = a_{12}\). Normalizing with \(\mathbf a_1^{\mathsf T} \mathbf M \mathbf a_1 = 1\), we get \(m(a_{11}^2 + a_{12}^2) = 1 \implies 2 a_{11}^2 = 1 \implies a_{11} = 1/\sqrt{2}\). So \(\mathbf a_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}\).
- Case 2: \(2k - m\lambda = -k \implies m\lambda = 3k \implies \omega_2^2 = 3k/m\). With \(k=100\) and \(m=1\), \(\omega_2 = \sqrt{300} = 10\sqrt{3} \approx 17.32\ \mathrm{rad/s}\). The eigenvector \(\mathbf a_2\) satisfies \((\mathbf K - 300 \mathbf M)\mathbf a_2 = 0\). \(\begin{pmatrix} -100 & -100 \\ -100 & -100 \end{pmatrix} \begin{pmatrix} a_{21} \\ a_{22} \end{pmatrix} = \mathbf 0 \implies a_{21} = -a_{22}\). Normalizing, \(m(a_{21}^2 + a_{22}^2) = 1 \implies 2 a_{21}^2 = 1 \implies a_{21} = 1/\sqrt{2}\). So \(\mathbf a_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).
Units check. \(\omega\) has units of \(\mathrm{rad/s}\). \(\sqrt{k/m} = \sqrt{\mathrm{N/m} / \mathrm{kg}} = \sqrt{(\mathrm{kg \cdot m/s^2}) / (\mathrm{m \cdot kg})} = \sqrt{1/s^2} = 1/s\). Correct.
Three quick tests
- What is the physical interpretation of the normal coordinates \(\eta_r\)?
Answer
They are the amplitudes of the independent normal modes, representing the "amount" of each mode present in the system's motion. - Why must the mass matrix \(\mathbf M\) be positive-definite?
Answer
To ensure that the kinetic energy is always positive for any non-zero velocity, which is a physical requirement for a stable system with real masses. - How are the eigenvectors \(\mathbf a_r\) normalized?
Answer
Such that \(\mathbf a_r^{\mathsf T} \mathbf M \mathbf a_s = \delta_{rs}\), ensuring orthogonality and unit mass in the normal coordinate system.
Problem set for PU-103 →
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.