2.1.11 · D5Analytical Mechanics
Question bank — Hamiltonian — definition H = Σpᵢq̇ᵢ − L
True or false — justify
A valid Hamiltonian may still contain a velocity in its final form.
False. is defined as a function of ; any leftover means you skipped the inversion , so the object is ill-defined.
The Hamiltonian always equals the total energy .
False. Equality needs to be a homogeneous quadratic in (no explicit-time or linear-velocity terms) and velocity-independent; a bead on a rotating wire breaks this, giving .
If equals the energy, then is automatically conserved.
False. "" is about the form of and ; conservation is about (i.e. ). A time-dependent driving potential can make yet not conserved.
If is conserved, it must equal the total energy.
False. The rotating-wire bead has so is conserved, but . Conservation and "" are logically independent.
Hamilton's equations are second-order differential equations like Euler–Lagrange.
False. They are first-order ODEs (, ), replacing second-order equations — see Hamilton's Canonical Equations.
The conjugate momentum always equals (mass times velocity).
False. ; only for a simple term does this reduce to . For a pendulum it is (angular momentum), and with magnetic forces it even includes a vector-potential term.
The Legendre transform loses information about the original curve .
False. For a convex the transform is invertible: and encode the same physics, one parametrised by velocity-as-argument, the other by slope (momentum) — see Legendre Transform.
being conserved means every individual is conserved.
False. conservation is one global statement; a given is zero only when doesn't depend on that particular (a cyclic coordinate), which is a separate condition (see Conservation Laws & Noether's Theorem).
Spot the error
", so ."
The error is leaving in . You must write then substitute , giving — a function of .
", and since , I can just add to and stop."
You stopped one step early: the sum still holds . The defining step is elimination — invert to get and substitute everywhere.
"For the rotating bead, because so ."
Wrong: even with , , whereas . The rotational term flips sign in because the constraint is time-dependent.
"Since , we always get ."
The step uses Euler's theorem, which requires homogeneous quadratic in . If has velocity-independent pieces (moving constraints), Euler's theorem gives a different multiple and .
" contains , so depends on ."
The whole Legendre derivation shows — the and terms recombine so no survives. That vanishing is the proof .
"The Euler–Lagrange equation isn't needed to define ."
Defining needs only . But identifying (used to read off Hamilton's equations) does invoke Euler–Lagrange; without it you get but not the clean canonical equations.
Why questions
Why trade velocities for momenta at all — what do we gain?
We gain a symmetric description on Phase Space where the equations become first-order and geometric, enabling conservation laws, Poisson Brackets, and canonical transformations that Lagrangian form hides.
Why is the slope (momentum) a good variable to reparametrise by?
For a convex function, each slope hits the curve exactly once, so labelling points by their slope is a lossless, invertible relabelling — the essence of the Legendre Transform.
Why does require to be velocity-independent?
If depended on , then would include , so and the clean cancellation breaks down.
Why does an explicit time dependence in destroy conservation of ?
From the derivation ; if carries explicit , that partial is nonzero, so drifts in time and is not conserved.
Why can a bead on a rotating wire have conserved even though energy isn't?
The Lagrangian has no explicit (the rotation enters as a constant), so conserved; but the wire does work on the bead, so mechanical energy is not conserved and .
Why do we call the "conjugate" momentum rather than just "momentum"?
Because it is paired to a specific coordinate via ; for an angle it is an angular momentum, for a charged particle it includes field terms — it need not be the naive velocity.
Edge cases
For a free particle (), is still meaningful and conserved?
Yes: , and since has no explicit time or coordinate dependence, is conserved and every is conserved too (all coordinates cyclic).
What happens to the Legendre transform if is linear in a velocity (so )?
The map is not invertible (a degenerate/constrained system); cannot be solved for and the standard Hamiltonian construction fails, requiring constraint methods instead.
If a coordinate is cyclic (absent from ), what is the edge behaviour of its momentum?
Then , so is a conserved constant — a limiting "trivial" motion in that direction (link to Conservation Laws & Noether's Theorem).
At an equilibrium point where and , what do Hamilton's equations predict?
If but happens to equal , is energy conserved?
Not necessarily: is about form, but means both and hence change in time — an explicitly driven system.
What is for a system with zero kinetic energy region (particle momentarily at rest, )?
reduces to evaluated there; for a natural system with this is just (all energy potential) — a valid, non-degenerate point of the motion.
Recall check
Recall One-line self-test
Can I state, without peeking, the two independent conditions "" versus " conserved"? ::: needs homogeneous-quadratic in and velocity-independent; conserved needs . Neither implies the other.