Classical MechanicsBSc · Notes

1. Three separate questions about the Hamiltonian

For any Hamiltonian problem, ask: how was HH constructed, does it equal physical mechanical energy, and is it conserved? These questions have different answers in moving coordinates and in driven or dissipative systems.

The construction is always the Legendre transformation when the regularity assumptions hold. For L=T−VL=T-V with stationary transformations and velocity-independent VV, H=T+VH=T+V expressed in q,pq,p. Conservation then follows only if the dynamics supplies it. A time-dependent potential can make this physical energy change, and drag can change it even when HH itself has no explicit time dependence.

2. Time dependence and forced Hamilton equations

For forces QiQ_i kept outside the Lagrangian, the corresponding Hamilton description is

q˙i=∂H∂pi,p˙i=−∂H∂qi+Qi(q,p,t).\boxed{\dot q_i=\frac{\partial H}{\partial p_i},\qquad \dot p_i=-\frac{\partial H}{\partial q_i}+Q_i(q,p,t).}(14.1)

In QiQ_i, replace any velocities by q˙i(q,p,t)\dot q_i(q,p,t). This is a forced Hamiltonian system. The unmodified canonical equations alone do not describe generic friction in the original physical variables. There is no universal “non-conservative Hamiltonian” obtained by adding dissipated energy to T+VT+V.

Differentiate HH along a solution:

dHdt=∂H∂t+∑i(Hqiq˙i+Hpip˙i).\frac{dH}{dt}=\frac{\partial H}{\partial t}+\sum_i\left(H_{q_i}\dot q_i+H_{p_i}\dot p_i\right).(14.2)

Insert the forced equations. The conservative products cancel, leaving

dHdt=∂H∂t+∑iQiq˙i.\boxed{\frac{dH}{dt}=\frac{\partial H}{\partial t}+\sum_iQ_i\dot q_i.}(14.3)

For an unforced system, dH/dt=∂H/∂tdH/dt=\partial H/\partial t. For stationary ordinary mechanical coordinates, ∑iQiq˙i\sum_iQ_i\dot q_i is the remaining-force power. This agrees with the Lagrangian energy balance because ∂tH∣q,p=−∂tL∣q,q˙\partial_tH|_{q,p}=-\partial_tL|_{q,\dot q} after velocity substitution. In moving coordinates, HH need not be laboratory energy and the expression must be interpreted accordingly.

If qiq_i is cyclic in HH, then p˙i=Qi\dot p_i=Q_i. With Qi=0Q_i=0, its momentum is conserved. For a regular Legendre pair, Hqi=−LqiH_{q_i}=-L_{q_i} evaluated at corresponding states; therefore a cyclic coordinate of LL is also cyclic in HH.

3. Hamiltonian and mechanical energy

For T=T2+T1+T0T=T_2+T_1+T_0 and velocity-independent VV, Chapter 10 established H=T2−T0+VH=T_2-T_0+V after expressing velocities in momenta. Thus the sufficient condition T=T2T=T_2 gives H=T+VH=T+V.

This condition is not the same as “the forces are conservative.” A moving ideal wire can do real work through its reaction. Conversely, a damped spring can still use H=p2/(2m)+kx2/2H=p^2/(2m)+kx^2/2 as its instantaneous mechanical energy, while the forced equations make that quantity decrease.

A Hamiltonian is also sensitive to the chosen Lagrangian representation. Adding a constant to the potential shifts HH by that constant without changing motion. More generally, adding a total time derivative to LL changes its canonical momentum and may change the displayed Hamiltonian. Physical predictions remain the same when the transformation is carried through consistently.

4. Applications

Example 1 · Damping in phase space

For a spring mass, use L=mx˙2/2−kx2/2L=m\dot x^2/2-kx^2/2, p=mx˙p=m\dot x, and H=p2/(2m)+kx2/2H=p^2/(2m)+kx^2/2. Drag gives Qx=−cp/mQ_x=-cp/m, so

x˙=pm,p˙=−kx−cmp.\dot x=\frac pm,\qquad \dot p=-kx-\frac cm p.(14.4)

Differentiating the first equation reproduces mx¨+cx˙+kx=0m\ddot x+c\dot x+kx=0. The energy balance is

H˙=(−cmp)pm=−cm2p2.\dot H=\left(-\frac cm p\right)\frac pm=-\frac{c}{m^2}p^2.(14.5)

For underdamping, phase trajectories spiral inward rather than following a single energy ellipse. At p=0p=0 the instantaneous loss rate is zero, even at a nonzero displacement; the acceleration then generates velocity again. The force field divergence in the (x,p)(x,p) plane is ∂xx˙+∂pp˙=−c/m\partial_x\dot x+\partial_p\dot p=-c/m, illustrating contraction of phase-space area for this dissipative model.

Example 2 · A changing spring stiffness

Suppose an external mechanism prescribes k=k(t)>0k=k(t)>0. Choose xx from a fixed equilibrium position. Then L=mx˙2/2−k(t)x2/2L=m\dot x^2/2-k(t)x^2/2 and

H=p22m+12k(t)x2.H=\frac{p^2}{2m}+\frac12k(t)x^2.(14.6)

There is no remaining QxQ_x in this model. Hamilton's equations give x˙=p/m\dot x=p/m, p˙=−k(t)x\dot p=-k(t)x, and

H˙=12k˙(t)x2.\dot H=\frac12\dot k(t)x^2.(14.7)

The force is the gradient of an instantaneous position potential, but the potential depends explicitly on time. Increasing stiffness while the spring is stretched raises its energy. It is therefore useful to distinguish a force derivable from V(x,t)V(x,t) from a time-independent conservative system with conserved mechanical energy.

Example 3 · The rotating wire revisited

For a bead at signed position qq on a horizontal wire rotating with fixed Ω\Omega, the Lagrangian is L=mq˙2/2+mΩ2q2/2L=m\dot q^2/2+m\Omega^2q^2/2. The momentum is p=mq˙p=m\dot q, so

H=p22m−12mΩ2q2.H=\frac{p^2}{2m}-\frac12m\Omega^2q^2.(14.8)

Hamilton's equations give q˙=p/m\dot q=p/m, p˙=mΩ2q\dot p=m\Omega^2q, and hence q¨=Ω2q\ddot q=\Omega^2q. This HH is conserved. The laboratory kinetic energy is Elab=p2/(2m)+mΩ2q2/2E_{\rm lab}=p^2/(2m)+m\Omega^2q^2/2; differentiating gives

E˙lab=pm(mΩ2q)+mΩ2qpm=2Ω2qp.\dot E_{\rm lab}=\frac pm(m\Omega^2q)+m\Omega^2q\frac pm=2\Omega^2qp.(14.9)

The motor can supply or remove power through the rotating constraint. This is a direct counterexample to blindly identifying an autonomous reduced Hamiltonian with laboratory energy.

Example 4 · An equivalent Lagrangian with shifted momentum

Start with L=mx˙2/2−V(x)L=m\dot x^2/2-V(x) and add df/dtdf/dt with f=axtf=ax t, where aa is a constant force. Since df/dt=atx˙+axdf/dt=at\dot x+ax,

L′=12mx˙2−V(x)+atx˙+ax.L'=\frac12m\dot x^2-V(x)+at\dot x+ax.(14.10)

The equation remains Newton's law: d(mx˙+at)/dt−[−V′(x)+a]=mx¨+V′(x)=0d(m\dot x+at)/dt-[-V'(x)+a]=m\ddot x+V'(x)=0. The new momentum is P=mx˙+atP=m\dot x+at, so x˙=(P−at)/m\dot x=(P-at)/m. The Legendre transformation gives

H′=(P−at)22m+V(x)−ax.H'=\frac{(P-at)^2}{2m}+V(x)-ax.(14.11)

Hamilton's equations yield x˙=(P−at)/m\dot x=(P-at)/m and P˙=−V′(x)+a\dot P=-V'(x)+a. Differentiating the momentum definition gives mx¨=P˙−a=−V′(x)m\ddot x=\dot P-a=-V'(x), as required. This optional representation example shows that canonical quantities depend on how the same dynamics is written; the mechanical energy remains mx˙2/2+V(x)m\dot x^2/2+V(x).

5. What belongs in an answer about non-conservative systems?

State the conservative Lagrangian, identify the remaining generalized forces, construct HH using its momenta, and include those forces in the momentum equations. If the question asks about energy, evaluate H˙\dot H and identify whether HH is physical mechanical energy. Do not replace the force model by an accumulated “energy loss” inside an ordinary phase-space function.

Special reformulations can embed dissipative dynamics in time-dependent or enlarged Hamiltonian systems, but their momenta and energy interpretation differ. Such constructions are beyond what is needed to handle the non-conservative systems in this syllabus.

Exercises

Hints and solutions are collected separately.

  1. 14.1 State sufficient conditions for H=T+VH=T+V.

  2. 14.2 Write the forced Hamilton equations and derive their energy balance.

  3. 14.3 For a damped free particle, express the drag force in canonical variables.

  4. 14.4 A damped oscillator has m=2m=2, c=3c=3, p=4p=4 in SI units. Find its instantaneous energy loss rate.

  5. 14.5 For H=p2/(2m)+k(t)x2/2H=p^2/(2m)+k(t)x^2/2, find H˙\dot H without assuming energy conservation.

  6. 14.6 Give an autonomous Hamiltonian that differs from laboratory energy.

  7. 14.7 If HH has no qiq_i dependence but Qi=F0Q_i=F_0 is constant, what happens to pip_i?

  8. 14.8 Calculate the phase-plane divergence for a damped oscillator.

  9. 14.9 Why is adding accumulated dissipated work to HH not a general canonical construction?

  10. 14.10 For L↦L+df(q,t)/dtL\mapsto L+df(q,t)/dt, find the momentum change.

  11. 14.11 For f=axtf=axt, verify that the transformed Hamiltonian reproduces Newton’s equation.

  12. 14.12 Can HH equal mechanical energy but fail to be conserved?

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