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Classical Mechanics

The mathematical framework of mechanics — from Newton's laws through Lagrangian and Hamiltonian formulations to rigid body dynamics and normal modes.

Newtonian Mechanics Lagrangian Mechanics Hamiltonian Mechanics Rigid Body Dynamics Small Oscillations

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Newtonian Mechanics

Newton's laws, conservation principles, and the central force problem.

🍎 Newton's Laws & Conservation Laws

Definition — Newton's Laws

First Law: A body remains in uniform motion unless acted upon by an external force (defines inertial frames).
Second Law: \(\mathbf{F} = \frac{d\mathbf{p}}{dt} = m\mathbf{a}\) in an inertial frame.
Third Law: \(\mathbf{F}_{12} = -\mathbf{F}_{21}\) (action-reaction).

Theorem — Conservation Laws

Energy: If all forces are conservative (\(\mathbf{F} = -\nabla V\)), then \(E = T + V = \frac{1}{2}m|\dot{\mathbf{r}}|^2 + V(\mathbf{r})\) is constant.
Momentum: If \(\mathbf{F}_{\text{ext}} = \mathbf{0}\), then total momentum \(\mathbf{P} = \sum m_i\dot{\mathbf{r}}_i\) is conserved.
Angular momentum: If the total external torque \(\boldsymbol{\tau} = \mathbf{0}\), then \(\mathbf{L} = \sum \mathbf{r}_i \times \mathbf{p}_i\) is conserved.

Example

Derive the orbit equation for the Kepler problem: \(V(r) = -k/r\).

Using conservation of angular momentum \(L = mr^2\dot\theta\) and energy, the Binet substitution \(u = 1/r\) gives: \[\frac{d^2u}{d\theta^2} + u = \frac{mk}{L^2}\] Solution: \(u = \frac{mk}{L^2}(1 + e\cos(\theta-\theta_0))\), or \(r = \frac{L^2/(mk)}{1 + e\cos\theta}\). This is a conic section with eccentricity \(e = \sqrt{1 + \frac{2EL^2}{mk^2}}\).

Lagrangian Mechanics

Generalized coordinates, the Euler-Lagrange equations, and constraints.

📐 The Lagrangian Formulation

Definition — Lagrangian The Lagrangian is \(L(q_i, \dot{q}_i, t) = T - V\), where \(q_i\) are generalized coordinates, \(T\) is kinetic energy, and \(V\) is potential energy.

Theorem — Euler-Lagrange Equations The equations of motion for a system with Lagrangian \(L\) are: \[\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = 0, \qquad i = 1,\ldots,n\] These follow from Hamilton's principle: the actual path extremizes the action \(S = \int_{t_1}^{t_2}L\,dt\).

Theorem — Noether's Theorem Every continuous symmetry of the Lagrangian corresponds to a conserved quantity:

Time translation symmetry (\(\partial L/\partial t = 0\)) \(\Rightarrow\) energy conservation
Spatial translation symmetry \(\Rightarrow\) linear momentum conservation
Rotational symmetry \(\Rightarrow\) angular momentum conservation

Definition — Constraints & Lagrange Multipliers For holonomic constraints \(f_k(q_1,\ldots,q_n,t)=0\), the modified Euler-Lagrange equations are: \[\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = \sum_k \lambda_k \frac{\partial f_k}{\partial q_i}\] The multipliers \(\lambda_k\) are related to the constraint forces.

Example

Write the Lagrangian and derive the equation of motion for a simple pendulum of length \(\ell\).

Using \(\theta\) as the generalized coordinate: \(T = \frac{1}{2}m\ell^2\dot\theta^2\), \(V = -mg\ell\cos\theta\). \[L = \frac{1}{2}m\ell^2\dot\theta^2 + mg\ell\cos\theta\] Euler-Lagrange: \(\frac{d}{dt}(m\ell^2\dot\theta) + mg\ell\sin\theta = 0\), giving \(\ddot\theta + \frac{g}{\ell}\sin\theta = 0\).

Hamiltonian Mechanics

The Legendre transform, Hamilton's equations, and Poisson brackets.

🔄 Hamilton's Equations

Definition — Hamiltonian via Legendre Transform Define the conjugate momentum \(p_i = \frac{\partial L}{\partial \dot{q}_i}\). The Hamiltonian is: \[H(q_i, p_i, t) = \sum_i p_i\dot{q}_i - L\] For natural systems (time-independent constraints, \(V\) independent of \(\dot{q}\)), \(H = T + V\) = total energy.

Theorem — Hamilton's Canonical Equations \[\dot{q}_i = \frac{\partial H}{\partial p_i}, \qquad \dot{p}_i = -\frac{\partial H}{\partial q_i}\] These \(2n\) first-order ODEs are equivalent to the \(n\) second-order Euler-Lagrange equations.

Definition — Poisson Bracket For phase space functions \(f, g\): \[\{f, g\} = \sum_i\left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\right)\] Time evolution: \(\dot{f} = \{f, H\} + \frac{\partial f}{\partial t}\). A quantity is conserved iff \(\{f, H\} = 0\) and \(\partial f/\partial t = 0\).

Theorem — Fundamental Poisson Brackets \[\{q_i, q_j\} = 0, \qquad \{p_i, p_j\} = 0, \qquad \{q_i, p_j\} = \delta_{ij}\] These are the classical analogs of the canonical commutation relations in quantum mechanics.

Example

For a particle in 1D with \(H = \frac{p^2}{2m} + \frac{1}{2}kq^2\), find \(\{q, H\}\) and \(\{p, H\}\).

\(\{q, H\} = \frac{\partial q}{\partial q}\frac{\partial H}{\partial p} - \frac{\partial q}{\partial p}\frac{\partial H}{\partial q} = 1 \cdot \frac{p}{m} - 0 = \frac{p}{m} = \dot{q}\). \(\{p, H\} = \frac{\partial p}{\partial q}\frac{\partial H}{\partial p} - \frac{\partial p}{\partial p}\frac{\partial H}{\partial q} = 0 - 1\cdot kq = -kq = \dot{p}\). Confirming Hamilton's equations: \(\dot{q} = p/m\) and \(\dot{p} = -kq\) (harmonic oscillator).

Rigid Body Dynamics

Euler angles, the inertia tensor, and Euler's equations for rigid body rotation.

🌀 Inertia Tensor & Euler's Equations

Definition — Moment of Inertia Tensor For a rigid body, the inertia tensor about the center of mass is: \[I_{ij} = \int \rho(\mathbf{r})\left(|\mathbf{r}|^2\delta_{ij} - r_i r_j\right)dV\] The angular momentum is \(\mathbf{L} = \mathbf{I}\boldsymbol{\omega}\) and kinetic energy is \(T_{\text{rot}} = \frac{1}{2}\boldsymbol{\omega}^T\mathbf{I}\boldsymbol{\omega}\).

The principal axes are the eigenvectors of \(\mathbf{I}\); the corresponding eigenvalues \(I_1, I_2, I_3\) are the principal moments of inertia. In the principal axis frame, \(\mathbf{I}\) is diagonal.

Theorem — Euler's Equations of Rotation In the body-fixed principal axis frame: \[\begin{aligned} I_1\dot\omega_1 - (I_2 - I_3)\omega_2\omega_3 &= \tau_1 \\ I_2\dot\omega_2 - (I_3 - I_1)\omega_3\omega_1 &= \tau_2 \\ I_3\dot\omega_3 - (I_1 - I_2)\omega_1\omega_2 &= \tau_3 \end{aligned}\] These are nonlinear coupled ODEs for the angular velocity components in the body frame.

Definition — Euler Angles The orientation of a rigid body is parameterized by three Euler angles \((\phi, \theta, \psi)\):

\(\phi\): precession about the space \(z\)-axis
\(\theta\): nutation (tilt of the body axis)
\(\psi\): spin about the body symmetry axis

The rotation matrix is \(R = R_z(\phi)R_x(\theta)R_z(\psi)\) (in the ZXZ convention).

Example

A torque-free symmetric top has \(I_1 = I_2 \neq I_3\). Describe the motion.

From Euler's equations with \(\tau = 0\) and \(I_1 = I_2\): \(\dot\omega_3 = 0\) (constant spin), and \(\dot\omega_1 = \Omega\omega_2\), \(\dot\omega_2 = -\Omega\omega_1\) where \(\Omega = (I_3-I_1)\omega_3/I_1\). Solution: \(\omega_1 + i\omega_2 = Ae^{-i\Omega t}\). The angular velocity precesses around the body symmetry axis with frequency \(|\Omega|\). In space, the body axis precesses around \(\mathbf{L}\).

Small Oscillations & Normal Modes

Linearization near equilibrium, the eigenvalue problem for normal modes, and coupled oscillators.

📊 Normal Mode Analysis

Near a stable equilibrium \(\mathbf{q}_0\), expand \(T \approx \frac{1}{2}\dot{\mathbf{q}}^T \mathbf{M} \dot{\mathbf{q}}\) and \(V \approx \frac{1}{2}\mathbf{q}^T \mathbf{K} \mathbf{q}\), where \(\mathbf{M}\) (mass matrix) and \(\mathbf{K}\) (stiffness matrix) are symmetric positive definite.

Theorem — Normal Modes The normal mode frequencies \(\omega_k\) are determined by the generalized eigenvalue problem: \[\det(\mathbf{K} - \omega^2\mathbf{M}) = 0\] Each eigenvalue \(\omega_k^2\) gives a normal mode frequency, and the corresponding eigenvector \(\mathbf{a}_k\) gives the mode shape. The general solution is: \[\mathbf{q}(t) = \sum_k c_k\mathbf{a}_k\cos(\omega_k t + \phi_k)\]

Definition — Normal Coordinates The normal coordinates \(\eta_k\) decouple the system: \(\ddot\eta_k + \omega_k^2\eta_k = 0\). They are related to the original coordinates by \(\mathbf{q} = \mathbf{P}\boldsymbol{\eta}\) where \(\mathbf{P}\) is the modal matrix whose columns are the eigenvectors \(\mathbf{a}_k\).

Example

Find the normal modes of two identical masses \(m\) connected by springs of constant \(k\) in the configuration wall—spring—mass—spring—mass—spring—wall.

Equations of motion: \(m\ddot{x}_1 = -2kx_1 + kx_2\), \(m\ddot{x}_2 = kx_1 - 2kx_2\). Mass matrix: \(\mathbf{M} = mI\). Stiffness matrix: \(\mathbf{K} = k\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\). Eigenvalue problem: \(\det\begin{pmatrix}2k-m\omega^2 & -k \\ -k & 2k-m\omega^2\end{pmatrix}=0\). \((2k-m\omega^2)^2 = k^2\), giving \(\omega_1^2 = k/m\) (symmetric mode: \(\mathbf{a}_1 = (1,1)^T\)) and \(\omega_2^2 = 3k/m\) (antisymmetric mode: \(\mathbf{a}_2 = (1,-1)^T\)).

Key Takeaways

Conservation laws follow from symmetries: Noether's theorem unifies energy, momentum, and angular momentum conservation.
The Lagrangian formulation handles constraints naturally via generalized coordinates.
Hamilton's equations transform mechanics into a first-order system on phase space; Poisson brackets encode the algebraic structure.
Euler's equations for rigid bodies are nonlinear — even torque-free symmetric tops exhibit precession.
Normal modes decouple coupled oscillators into independent harmonic motions via the generalized eigenvalue problem.

Practice Problems

Problem 1

A particle moves under the central force \(F(r) = -kr\). Find the orbit and show it is an ellipse centered at the origin.

Show Solution ▼

The potential is \(V = kr^2/2\) (isotropic harmonic oscillator). Using \(u = 1/r\) and the orbit equation, or solving in Cartesian coordinates: \(x = A\cos(\omega t + \alpha)\), \(y = B\cos(\omega t + \beta)\) with \(\omega = \sqrt{k/m}\). This traces an ellipse centered at the origin (not the focus, unlike Kepler orbits).

Problem 2

Write the Lagrangian for a bead of mass \(m\) sliding on a frictionless wire in the shape of a parabola \(z = \alpha r^2\) under gravity, using \(r\) as the generalized coordinate.

Show Solution ▼

\(\dot{z} = 2\alpha r\dot{r}\). \(T = \frac{1}{2}m(\dot{r}^2 + r^2\dot\theta^2 + \dot{z}^2) = \frac{1}{2}m(\dot{r}^2(1+4\alpha^2 r^2) + r^2\dot\theta^2)\). If constrained to a vertical plane (\(\dot\theta=0\)): \(L = \frac{1}{2}m(1+4\alpha^2 r^2)\dot{r}^2 - mg\alpha r^2\). Euler-Lagrange gives \(m(1+4\alpha^2 r^2)\ddot{r} + 4m\alpha^2 r\dot{r}^2 + 2mg\alpha r = 0\).

Problem 3

For the 1D Hamiltonian \(H = \frac{p^2}{2m} + \lambda q^4\), show that \(\{L_z, H\} \neq 0\) in general (i.e., angular momentum is not conserved). Compute \(\{q^2, H\}\).

Show Solution ▼

\(\{q^2, H\} = \frac{\partial(q^2)}{\partial q}\frac{\partial H}{\partial p} - \frac{\partial(q^2)}{\partial p}\frac{\partial H}{\partial q} = 2q\cdot\frac{p}{m} - 0 = \frac{2qp}{m}\). Since this is nonzero, \(q^2\) is not conserved. The anharmonic potential \(\lambda q^4\) breaks the symmetry that would be needed for additional conservation laws.

Problem 4

A uniform disk of mass \(M\) and radius \(R\) rolls without slipping down an incline of angle \(\alpha\). Find the acceleration using the Lagrangian method.

Show Solution ▼

Let \(x\) be the distance along the incline. Rolling constraint: \(\dot{x} = R\dot\theta\). \(T = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}I\dot\theta^2 = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}\cdot\frac{1}{2}MR^2\cdot\frac{\dot{x}^2}{R^2} = \frac{3}{4}M\dot{x}^2\). \(V = -Mgx\sin\alpha\). Euler-Lagrange: \(\frac{3}{2}M\ddot{x} = Mg\sin\alpha\), so \(\ddot{x} = \frac{2}{3}g\sin\alpha\) (slower than sliding due to rotational inertia).

Problem 5

Find the normal mode frequencies for a double pendulum (two identical masses \(m\) on strings of length \(\ell\)) in the small-angle approximation.

Show Solution ▼

Small-angle Lagrangian: \(T = \frac{1}{2}m\ell^2(2\dot\theta_1^2 + \dot\theta_2^2 + 2\dot\theta_1\dot\theta_2)\), \(V = mg\ell(3 - 2\cos\theta_1 - \cos\theta_2) \approx mg\ell(\theta_1^2 + \frac{1}{2}\theta_2^2)\). Mass matrix: \(\mathbf{M}=m\ell^2\begin{pmatrix}2&1\\1&1\end{pmatrix}\), stiffness: \(\mathbf{K}=mg\ell\begin{pmatrix}2&0\\0&1\end{pmatrix}\). Solving \(\det(\mathbf{K}-\omega^2\mathbf{M})=0\): \(\omega^2 = \frac{g}{\ell}(2 \pm \sqrt{2})\).

Self-Assessment Quiz

1. Noether's theorem states that time-translation symmetry implies conservation of:

A Linear momentum

B Angular momentum

C Energy

D Charge

2. The Euler-Lagrange equation for a free particle (\(V = 0\)) in Cartesian coordinates gives:

A Circular motion

B Uniform rectilinear motion

C Exponential growth

D Simple harmonic motion

3. The fundamental Poisson bracket \(\{q_i, p_j\}\) equals:

A 0

B \(\delta_{ij}\)

C \(-\delta_{ij}\)

D \(q_i p_j\)

4. For a torque-free symmetric top with \(I_1 = I_2 \neq I_3\), which component of angular velocity is constant in the body frame?

A \(\omega_1\)

B \(\omega_2\)

C \(\omega_3\) (along the symmetry axis)

D \(|\boldsymbol{\omega}|\)

5. In normal mode analysis, the number of normal modes for a system with \(n\) degrees of freedom is:

A \(n-1\)

B \(n\)

C \(2n\)

D Depends on the potential