Lagrangian and Hamiltonian Mechanics

Canonical transformations

A transformation (q,p)\to(Q,P) is canonical when it preserves the form of Hamilton's equations, which is equivalent to requiring that the new variables satisfy the fundamental Poisson brackets \{Q,P\}=1 and \{Q,Q\}=\{P,P\}=0. For linear forms, \{Q,P\}=\sum_i\left(\frac{\partial Q}{\partial q_i}\frac{\partial P}{\partial p_i}-\frac{\partial Q}{\partial p_i}\frac{\partial P}{\partial q_i}\right), so a transformation of the form Q=\vec a\cdot\vec p, P=\vec b\cdot\vec q has bracket -\vec a\cdot\vec b and is canonical only if \vec a\cdot\vec b=-1. Canonical transformations are also generated by four types of generating function: F_1(q,Q), F_2(q,P), F_3(p,Q) and F_4(p,P). For type 1 the relations are p=\partial F_1/\partial q and P=-\partial F_1/\partial Q, so the minus sign on the new momentum must be kept. For example, if F_1=\lambda qQ with a constant \lambda, then p=\lambda Q and P=-\lambda q, so p/P=-Q/q. For type 4, Q=\partial F_4/\partial P and q=-\partial F_4/\partial p; a check of the Poisson bracket with these relations confirms the transformation. A type-3 function F_3(p,Q) does not exist when p and Q are functionally dependent, for example when the transformation makes the product pQ a constant, so the choice of generating-function type must be checked first.

Coupled oscillator normal modes

A set of masses joined by springs obeys linear coupled equations m\ddot x_i=-\sum_j K_{ij}x_j, where K is a symmetric stiffness matrix. Trying x_j=A_je^{i\omega t} turns the problem into the eigenvalue equation \det(K-\omega^2 m I)=0. The eigenvalues \omega^2 give the normal mode frequencies and the eigenvectors give the pattern of motion in each mode. In a chain of springs, the diagonal entry of K is the sum of the spring constants attached to that mass, and the off-diagonal entry is minus the constant of the spring joining the two masses. Cross terms in the potential can be removed by an orthogonal rotation of coordinates, such as p=(x+y)/\sqrt{2} and q=(x-y)/\sqrt{2}, because such a rotation leaves the kinetic term and the sum x^2+y^2 unchanged. After the rotation the Hamiltonian separates into independent one-dimensional oscillators, and the ground-state energy is the sum of their zero-point energies \frac{1}{2}\hbar\omega. For a two-mass system the two eigenvalues \lambda_\pm of K/m satisfy \lambda_++\lambda_-=\mathrm{tr}K/m and \lambda_+\lambda_-=\det K/m^2, so the frequencies can be checked quickly from the trace and determinant before solving the full characteristic equation.

Gauge total derivative in Lagrangian

Adding the total time derivative \frac{dg(q_i,t)}{dt} to a Lagrangian does not change the Euler–Lagrange equations, because such a term contributes only boundary terms to the action variation with fixed endpoints. The added term is not free of velocities, however: \frac{dg}{dt}=\sum_j\frac{\partial g}{\partial q_j}\dot q_j+\frac{\partial g}{\partial t}, so the canonical momenta shift by p_i\to p_i+\partial g/\partial q_i. A momentum that is conserved before the change therefore need not stay conserved after it. The new Hamiltonian differs from the old one only by -\partial g/\partial t, because the velocity-dependent parts cancel in H=\sum p_i\dot q_i-L. As a concrete case, a term \alpha\,df(q)/dt changes the canonical momentum by \alpha f'(q) but leaves the equation of motion unchanged.

Hamilton's principal function

Hamilton's principal function is the action S=\int L\,dt evaluated along the actual path between fixed endpoints. Its dimension is energy multiplied by time, and it satisfies \partial S/\partial q=p and \partial S/\partial t=-H when expressed as a function of endpoint coordinates and times. To evaluate it directly, one integrates T-V along the trajectory, and it is often convenient to compute the kinetic and potential integrals separately. For a body dropped from rest with the potential zero at the ground, S=\int_0^T(\frac{1}{2}m\dot z^2-mgz)\,dt, and eliminating the fall time T with gT^2=2h gives S=-\frac{m}{3}\sqrt{2g}\,h^{3/2}, a negative value whenever the potential term dominates. For a projectile returning to the same horizontal level, the parabolic height y(t) and the time of flight give kinetic and potential integrals that are each proportional to mv_0^3/g, so the action is a pure numerical multiple of mv_0^3/g.

Hamilton's equations from Lagrangian

Starting from L(q,\dot q,t), the momenta are defined by p_i=\partial L/\partial\dot q_i. These must be inverted to express velocities in terms of (q,p,t), and the Hamiltonian H=\sum_i p_i\dot q_i-L is then written in those variables. Hamilton's equations \dot q_i=\partial H/\partial p_i and \dot p_i=-\partial H/\partial q_i follow directly from this form. When the Lagrangian contains a cross product of velocities, the momenta mix the velocities: for L=m\dot x\dot y-V, one gets p_x=m\dot y and p_y=m\dot x, so \dot x=p_y/m and \dot y=p_x/m and the Hamiltonian contains the product p_xp_y. Conversely, from a given H(q,p) one can derive the second-order equation of motion by differentiating \dot q=\partial H/\partial p and substituting \dot p=-\partial H/\partial q; comparing this with a stated equation of motion fixes unknown functions in the potential, such as a q-dependent coefficient.

Small oscillations of rolling hoop

The number of degrees of freedom of a constrained system is the number of independent generalized coordinates left after all holonomic constraints are imposed. A rigid body rolling without slipping on a straight line gives constraints that tie the centre position to the rotation angle and fix the centre height, so a planar rolling hoop has only one independent coordinate. A small particle constrained to move on the inner surface of such a hoop removes one of its two planar coordinates, and its remaining angle is an independent coordinate. Adding the two bodies and subtracting all constraints gives the total number of degrees of freedom, which equals the number of normal modes of small oscillation about equilibrium.

Small oscillations of sliding pendulum

A pendulum whose pivot slides horizontally on a smooth rail has two generalized coordinates, the pivot position x and the string angle \theta. With the bob at X=x+l\sin\theta and Y=l\cos\theta, the kinetic energy contains the cross term ml\cos\theta\,\dot x\dot\theta, so the momenta mix the velocities: p_x=(M+m)\dot x+ml\cos\theta\,\dot\theta and p_\theta=ml^2\dot\theta+ml\cos\theta\,\dot x. A coordinate that does not appear in the Lagrangian is cyclic, and its conjugate momentum is conserved. Here the Lagrangian does not contain x, so p_x is conserved, while p_\theta is not conserved because the gravitational potential depends on \theta. For small oscillations, the potential is expanded to second order in \theta and the coupled linear equations are solved with the normal-mode method.