Legendre transforms and canonical momenta

Hold position and time fixed and differentiate the Lagrangian with respect to each velocity. Those slopes define canonical momenta. For an ordinary Cartesian kinetic energy, the result is mass times velocity. A term linear in velocity shifts the slope without changing its second derivative. Thus canonical momentum can contain a position-dependent contribution even when the mechanical velocity and kinetic energy are unchanged.

The Legendre construction subtracts the original function from its tangent pairing. First form $p_i\dot q_i-L$, then replace every velocity using the inverted momentum equations. Leaving a velocity in the final expression means the exchange has not been completed. Positions, momenta, and time are the independent arguments of the Hamiltonian; an expression that secretly still depends on velocity cannot be differentiated as though those arguments were independent.

The relevant Jacobian is the matrix of second velocity derivatives of the Lagrangian. A nonzero determinant guarantees a local velocity-to-momentum inverse by the inverse function theorem. Positive definiteness is sufficient but stronger than necessary: a regular classical Lagrangian can have an indefinite Hessian. A local inverse also does not establish that the same inverse is single-valued across the entire velocity domain. Branches and coordinate charts still matter.

Consider a Lagrangian independent of one velocity. Its associated momentum vanishes regardless of that velocity, so no inverse can recover the missing information. That relation is a primary constraint, not permission to divide by zero or discard the coordinate immediately. Singular systems require the constrained Hamiltonian procedure developed later. Notice also that a regular transform need not preserve convexity properties familiar from optimization; here the immediate issue is equivalent local equations of motion.

For $L=m\dot q^2/2+a q\dot q-kq^2/2$, the momentum is $p=m\dot q+a q$. Eliminating velocity gives a shifted square in the Hamiltonian. The linear term is a total time derivative when a is constant, so it changes the momentum convention while leaving the physical acceleration unchanged. Comparing $p$ directly with $m\dot q$ would therefore compare different quantities. This modest example prepares the distinction between canonical and kinetic momenta in electromagnetism.

For a time-independent regular Lagrangian, the associated energy function is conserved along solutions. In the present example it equals ordinary kinetic plus potential energy after the momentum shift is undone. Time-dependent coordinate choices or explicit time-dependent interactions require more care: the Hamiltonian still generates the chosen canonical evolution, but its numerical value need not equal a conserved laboratory energy. Always state the frame and explicit time dependence before calling a Hamiltonian simply the energy.