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In physics, Hamiltonian mechanics is a reformulation of Lagrangian mechanics that emerged in 1833. Introduced by Sir William Rowan Hamilton, Hamiltonian mechanics replaces (generalized) velocities ${\displaystyle {\dot {q}}^{i}}$ used in Lagrangian mechanics with (generalized) momenta. Both theories provide interpretations of classical mechanics and describe the same physical phenomena.

Hamiltonian mechanics has a close relationship with geometry (notably, symplectic geometry and Poisson structures) and serves as a link between classical and quantum mechanics.

Overview

Phase space coordinates (p, q) and Hamiltonian H

Let ${\displaystyle (M,{\mathcal {L}})}$ be a mechanical system with configuration space ${\displaystyle M}$ and smooth Lagrangian ${\displaystyle {\mathcal {L}}.}$ Select a standard coordinate system ${\displaystyle ({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}$ on ${\displaystyle M.}$ The quantities ${\displaystyle \textstyle p_{i}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)~{\stackrel {\text{def}}{=}}~{\partial {\mathcal {L}}}/{\partial {\dot {q}}^{i}}}$ are called momenta. (Also generalized momenta, conjugate momenta, and canonical momenta). For a time instant ${\displaystyle t,}$ the Legendre transformation of ${\displaystyle {\mathcal {L}}}$ is defined as the map ${\displaystyle ({\boldsymbol {q}},{\boldsymbol {\dot {q}}})\to \left({\boldsymbol {p}},{\boldsymbol {q}}\right)}$ which is assumed to have a smooth inverse ${\displaystyle ({\boldsymbol {p}},{\boldsymbol {q}})\to ({\boldsymbol {q}},{\boldsymbol {\dot {q}}}).}$ For a system with ${\displaystyle n}$ degrees of freedom, the Lagrangian mechanics defines the energy function $${\displaystyle E_{\mathcal {L}}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)\,{\stackrel {\text{def}}{=}}\,\sum _{i=1}^{n}{\dot {q}}^{i}{\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}^{i}}}-{\mathcal {L}}.}$$

The Legendre transform of ${\displaystyle {\mathcal {L}}}$ turns ${\displaystyle E_{\mathcal {L}}}$ into a function ${\displaystyle {\mathcal {H}}({\boldsymbol {p}},{\boldsymbol {q}},t)}$ known as the Hamiltonian. The Hamiltonian satisfies $${\displaystyle {\mathcal {H}}\left({\frac {\partial {\mathcal {L}}}{\partial {\boldsymbol {\dot {q}}}}},{\boldsymbol {q}},t\right)=E_{\mathcal {L}}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}$$ which implies that $${\displaystyle {\mathcal {H}}({\boldsymbol {p}},{\boldsymbol {q}},t)=\sum _{i=1}^{n}p_{i}{\dot {q}}^{i}-{\mathcal {L}}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t),}$$ where the velocities ${\displaystyle {\boldsymbol {\dot {q}}}=({\dot {q}}^{1},\ldots ,{\dot {q}}^{n})}$ are found from the (${\displaystyle n}$-dimensional) equation ${\displaystyle \textstyle {\boldsymbol {p}}={\partial {\mathcal {L}}}/{\partial {\boldsymbol {\dot {q}}}}}$ which, by assumption, is uniquely solvable for ⁠${\displaystyle {\boldsymbol {\dot {q}}}}$⁠. The (${\displaystyle 2n}$-dimensional) pair ${\displaystyle ({\boldsymbol {p}},{\boldsymbol {q}})}$ is called phase space coordinates. (Also canonical coordinates).

From Euler–Lagrange equation to Hamilton's equations

In phase space coordinates ⁠${\displaystyle ({\boldsymbol {p}},{\boldsymbol {q}})}$⁠, the (${\displaystyle n}$-dimensional) Euler–Lagrange equation $${\displaystyle {\frac {\partial {\mathcal {L}}}{\partial {\boldsymbol {q}}}}-{\frac {d}{dt}}{\frac {\partial {\mathcal {L}}}{\partial {\dot {\boldsymbol {q}}}}}=0}$$ becomes Hamilton's equations in ${\displaystyle 2n}$ dimensions

${\displaystyle {\frac {\mathrm {d} {\boldsymbol {q}}}{\mathrm {d} t}}={\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {p}}}},\quad {\frac {\mathrm {d} {\boldsymbol {p}}}{\mathrm {d} t}}=-{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {q}}}}.}$

The Hamiltonian ${\displaystyle {\mathcal {H}}({\boldsymbol {p}},{\boldsymbol {q}})}$ is the Legendre transform of the Lagrangian ${\displaystyle {\mathcal {L}}({\boldsymbol {q}},{\dot {\boldsymbol {q}}})}$, thus one has $${\displaystyle {\mathcal {L}}({\boldsymbol {q}},{\dot {\boldsymbol {q}}})+{\mathcal {H}}({\boldsymbol {p}},{\boldsymbol {q}})={\boldsymbol {p}}{\dot {\boldsymbol {q}}}}$$ and thus $${\displaystyle {\begin{aligned}{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {p}}}}&={\dot {\boldsymbol {q}}}\\{\frac {\partial {\mathcal {L}}}{\partial {\boldsymbol {q}}}}&=-{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {q}}}},\end{aligned}}}$$

Besides, since ${\displaystyle {\boldsymbol {p}}=\partial {\mathcal {L}}/\partial {\dot {\boldsymbol {q}}}}$, the Euler–Lagrange equations yield ${\displaystyle {\dot {\boldsymbol {p}}}={\frac {\mathrm {d} {\boldsymbol {p}}}{\mathrm {d} t}}={\frac {\partial {\mathcal {L}}}{\partial {\boldsymbol {q}}}}=-{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {q}}}}.}$

From stationary action principle to Hamilton's equations

Let ${\displaystyle {\mathcal {P}}(a,b,{\boldsymbol {x}}_{a},{\boldsymbol {x}}_{b})}$ be the set of smooth paths ${\displaystyle {\boldsymbol {q}}:\to M}$ for which ${\displaystyle {\boldsymbol {q}}(a)={\boldsymbol {x}}_{a}}$ and ${\displaystyle {\boldsymbol {q}}(b)={\boldsymbol {x}}_{b}.}$ The action functional ${\displaystyle {\mathcal {S}}:{\mathcal {P}}(a,b,{\boldsymbol {x}}_{a},{\boldsymbol {x}}_{b})\to \mathbb {R} }$ is defined via $${\displaystyle {\mathcal {S}}=\int _{a}^{b}{\mathcal {L}}(t,{\boldsymbol {q}}(t),{\dot {\boldsymbol {q}}}(t))\,dt=\int _{a}^{b}\left(\sum _{i=1}^{n}p_{i}{\dot {q}}^{i}-{\mathcal {H}}({\boldsymbol {p}},{\boldsymbol {q}},t)\right)\,dt,}$$ where ⁠${\displaystyle {\boldsymbol {q}}={\boldsymbol {q}}(t)}$⁠, and ${\displaystyle {\boldsymbol {p}}=\partial {\mathcal {L}}/\partial {\boldsymbol {\dot {q}}}}$ (see above). A path ${\displaystyle {\boldsymbol {q}}\in {\mathcal {P}}(a,b,{\boldsymbol {x}}_{a},{\boldsymbol {x}}_{b})}$ is a stationary point of ${\displaystyle {\mathcal {S}}}$ (and hence is an equation of motion) if and only if the path ${\displaystyle ({\boldsymbol {p}}(t),{\boldsymbol {q}}(t))}$ in phase space coordinates obeys the Hamilton's equations.

Basic physical interpretation

A simple interpretation of Hamiltonian mechanics comes from its application on a one-dimensional system consisting of one nonrelativistic particle of mass m. The value ${\displaystyle H(p,q)}$ of the Hamiltonian is the total energy of the system, in this case the sum of kinetic and potential energy, traditionally denoted T and V, respectively. Here p is the momentum mv and q is the space coordinate. Then $${\displaystyle {\mathcal {H}}=T+V,\qquad T={\frac {p^{2}}{2m}},\qquad V=V(q)}$$ T is a function of p alone, while V is a function of q alone (i.e., T and V are scleronomic).

In this example, the time derivative of q is the velocity, and so the first Hamilton equation means that the particle's velocity equals the derivative of its kinetic energy with respect to its momentum. The time derivative of the momentum p equals the Newtonian force, and so the second Hamilton equation means that the force equals the negative gradient of potential energy.

Example

A spherical pendulum consists of a mass m moving without friction on the surface of a sphere. The only forces acting on the mass are the reaction from the sphere and gravity. Spherical coordinates are used to describe the position of the mass in terms of (r, θ, φ), where r is fixed, r = ℓ.

The Lagrangian for this system is $${\displaystyle L={\frac {1}{2}}m\ell ^{2}\left({\dot {\theta }}^{2}+\sin ^{2}\theta \ {\dot {\varphi }}^{2}\right)+mg\ell \cos \theta .}$$

Thus the Hamiltonian is $${\displaystyle H=P_{\theta }{\dot {\theta }}+P_{\varphi }{\dot {\varphi }}-L}$$ where $${\displaystyle P_{\theta }={\frac {\partial L}{\partial {\dot {\theta }}}}=m\ell ^{2}{\dot {\theta }}}$$ and $${\displaystyle P_{\varphi }={\frac {\partial L}{\partial {\dot {\varphi }}}}=m\ell ^{2}\sin ^{2}\!\theta \,{\dot {\varphi }}.}$$ In terms of coordinates and momenta, the Hamiltonian reads $${\displaystyle H=\underbrace {\left} _{T}+\underbrace {{\Big }} _{V}={\frac {P_{\theta }^{2}}{2m\ell ^{2}}}+{\frac {P_{\varphi }^{2}}{2m\ell ^{2}\sin ^{2}\theta }}-mg\ell \cos \theta .}$$ Hamilton's equations give the time evolution of coordinates and conjugate momenta in four first-order differential equations, $${\displaystyle {\begin{aligned}{\dot {\theta }}&={P_{\theta } \over m\ell ^{2}}\\{\dot {\varphi }}&={P_{\varphi } \over m\ell ^{2}\sin ^{2}\theta }\\{\dot {P_{\theta }}}&={P_{\varphi }^{2} \over m\ell ^{2}\sin ^{3}\theta }\cos \theta -mg\ell \sin \theta \\{\dot {P_{\varphi }}}&=0.\end{aligned}}}$$ Momentum ⁠${\displaystyle P_{\varphi }}$⁠, which corresponds to the vertical component of angular momentum ⁠${\displaystyle L_{z}=\ell \sin \theta \times m\ell \sin \theta \,{\dot {\varphi }}}$⁠, is a constant of motion. That is a consequence of the rotational symmetry of the system around the vertical axis. Being absent from the Hamiltonian, azimuth ${\displaystyle \varphi }$ is a cyclic coordinate, which implies conservation of its conjugate momentum.

Deriving Hamilton's equations

Hamilton's equations can be derived by a calculation with the Lagrangian ⁠${\displaystyle {\mathcal {L}}}$⁠, generalized positions q, and generalized velocities ⋅q, where ⁠${\displaystyle i=1,\ldots ,n}$⁠. Here we work off-shell, meaning ⁠${\displaystyle q^{i}}$⁠, ⁠${\displaystyle {\dot {q}}^{i}}$⁠, ⁠${\displaystyle t}$⁠ are independent coordinates in phase space, not constrained to follow any equations of motion (in particular, ${\displaystyle {\dot {q}}^{i}}$ is not a derivative of ⁠${\displaystyle q^{i}}$⁠). The total differential of the Lagrangian is: $${\displaystyle \mathrm {d} {\mathcal {L}}=\sum _{i}\left({\frac {\partial {\mathcal {L}}}{\partial q^{i}}}\mathrm {d} q^{i}+{\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}^{i}}}\,\mathrm {d} {\dot {q}}^{i}\right)+{\frac {\partial {\mathcal {L}}}{\partial t}}\,\mathrm {d} t\ .}$$ The generalized momentum coordinates were defined as ⁠${\displaystyle p_{i}=\partial {\mathcal {L}}/\partial {\dot {q}}^{i}}$⁠, so we may rewrite the equation as: $${\displaystyle {\begin{aligned}\mathrm {d} {\mathcal {L}}=&\sum _{i}\left({\frac {\partial {\mathcal {L}}}{\partial q^{i}}}\,\mathrm {d} q^{i}+p_{i}\mathrm {d} {\dot {q}}^{i}\right)+{\frac {\partial {\mathcal {L}}}{\partial t}}\mathrm {d} t\\=&\sum _{i}\left({\frac {\partial {\mathcal {L}}}{\partial q^{i}}}\,\mathrm {d} q^{i}+\mathrm {d} (p_{i}{\dot {q}}^{i})-{\dot {q}}^{i}\,\mathrm {d} p_{i}\right)+{\frac {\partial {\mathcal {L}}}{\partial t}}\,\mathrm {d} t\,.\end{aligned}}}$$

After rearranging, one obtains: $${\displaystyle \mathrm {d} \!\left(\sum _{i}p_{i}{\dot {q}}^{i}-{\mathcal {L}}\right)=\sum _{i}\left(-{\frac {\partial {\mathcal {L}}}{\partial q^{i}}}\,\mathrm {d} q^{i}+{\dot {q}}^{i}\mathrm {d} p_{i}\right)-{\frac {\partial {\mathcal {L}}}{\partial t}}\,\mathrm {d} t\ .}$$

The term in parentheses on the left-hand side is just the Hamiltonian ${\textstyle {\mathcal {H}}=\sum p_{i}{\dot {q}}^{i}-{\mathcal {L}}}$ defined previously, therefore: $${\displaystyle \mathrm {d} {\mathcal {H}}=\sum _{i}\left(-{\frac {\partial {\mathcal {L}}}{\partial q^{i}}}\,\mathrm {d} q^{i}+{\dot {q}}^{i}\,\mathrm {d} p_{i}\right)-{\frac {\partial {\mathcal {L}}}{\partial t}}\,\mathrm {d} t\ .}$$

One may also calculate the total differential of the Hamiltonian ${\displaystyle {\mathcal {H}}}$ with respect to coordinates ⁠${\displaystyle q^{i}}$⁠, ⁠${\displaystyle p_{i}}$⁠, ⁠${\displaystyle t}$⁠ instead of ⁠${\displaystyle q^{i}}$⁠, ⁠${\displaystyle {\dot {q}}^{i}}$⁠, ⁠${\displaystyle t}$⁠, yielding: $${\displaystyle \mathrm {d} {\mathcal {H}}=\sum _{i}\left({\frac {\partial {\mathcal {H}}}{\partial q^{i}}}\mathrm {d} q^{i}+{\frac {\partial {\mathcal {H}}}{\partial p_{i}}}\mathrm {d} p_{i}\right)+{\frac {\partial {\mathcal {H}}}{\partial t}}\,\mathrm {d} t\ .}$$

One may now equate these two expressions for ⁠${\displaystyle d{\mathcal {H}}}$⁠, one in terms of ⁠${\displaystyle {\mathcal {L}}}$⁠, the other in terms of ⁠${\displaystyle {\mathcal {H}}}$⁠: $${\displaystyle \sum _{i}\left(-{\frac {\partial {\mathcal {L}}}{\partial q^{i}}}\mathrm {d} q^{i}+{\dot {q}}^{i}\mathrm {d} p_{i}\right)-{\frac {\partial {\mathcal {L}}}{\partial t}}\,\mathrm {d} t\ =\ \sum _{i}\left({\frac {\partial {\mathcal {H}}}{\partial q^{i}}}\mathrm {d} q^{i}+{\frac {\partial {\mathcal {H}}}{\partial p_{i}}}\mathrm {d} p_{i}\right)+{\frac {\partial {\mathcal {H}}}{\partial t}}\,\mathrm {d} t\ .}$$

Since these calculations are off-shell, one can equate the respective coefficients of ⁠${\displaystyle \mathrm {d} q^{i}}$⁠, ⁠${\displaystyle \mathrm {d} p_{i}}$⁠, ⁠${\displaystyle \mathrm {d} t}$⁠ on the two sides: $${\displaystyle {\frac {\partial {\mathcal {H}}}{\partial q^{i}}}=-{\frac {\partial {\mathcal {L}}}{\partial q^{i}}}\quad ,\quad {\frac {\partial {\mathcal {H}}}{\partial p_{i}}}={\dot {q}}^{i}\quad ,\quad {\frac {\partial {\mathcal {H}}}{\partial t}}=-{\partial {\mathcal {L}} \over \partial t}\ .}$$

On-shell, one substitutes parametric functions ${\displaystyle q^{i}=q^{i}(t)}$ which define a trajectory in phase space with velocities ⁠${\displaystyle {\dot {q}}^{i}={\tfrac {d}{dt}}q^{i}(t)}$⁠, obeying Lagrange's equations: $${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}^{i}}}-{\frac {\partial {\mathcal {L}}}{\partial q^{i}}}=0\ .}$$

Rearranging and writing in terms of the on-shell ${\displaystyle p_{i}=p_{i}(t)}$ gives: $${\displaystyle {\frac {\partial {\mathcal {L}}}{\partial q^{i}}}={\dot {p}}_{i}\ .}$$

Thus Lagrange's equations are equivalent to Hamilton's equations: $${\displaystyle {\frac {\partial {\mathcal {H}}}{\partial q^{i}}}=-{\dot {p}}_{i}\quad ,\quad {\frac {\partial {\mathcal {H}}}{\partial p_{i}}}={\dot {q}}^{i}\quad ,\quad {\frac {\partial {\mathcal {H}}}{\partial t}}=-{\frac {\partial {\mathcal {L}}}{\partial t}}\,.}$$

In the case of time-independent ${\displaystyle {\mathcal {H}}}$ and ⁠${\displaystyle {\mathcal {L}}}$⁠, i.e. ⁠${\displaystyle \partial {\mathcal {H}}/\partial t=-\partial {\mathcal {L}}/\partial t=0}$⁠, Hamilton's equations consist of 2n first-order differential equations, while Lagrange's equations consist of n second-order equations. Hamilton's equations usually do not reduce the difficulty of finding explicit solutions, but important theoretical results can be derived from them, because coordinates and momenta are independent variables with nearly symmetric roles.

Hamilton's equations have another advantage over Lagrange's equations: if a system has a symmetry, so that some coordinate ${\displaystyle q_{i}}$ does not occur in the Hamiltonian (i.e. a cyclic coordinate), the corresponding momentum coordinate ${\displaystyle p_{i}}$ is conserved along each trajectory, and that coordinate can be reduced to a constant in the other equations of the set. This effectively reduces the problem from n coordinates to (n − 1) coordinates: this is the basis of symplectic reduction in geometry. In the Lagrangian framework, the conservation of momentum also follows immediately, however all the generalized velocities ${\displaystyle {\dot {q}}_{i}}$ still occur in the Lagrangian, and a system of equations in n coordinates still has to be solved.

The Lagrangian and Hamiltonian approaches provide the groundwork for deeper results in classical mechanics, and suggest analogous formulations in quantum mechanics: the path integral formulation and the Schrödinger equation.

Properties of the Hamiltonian

• The value of the Hamiltonian ${\displaystyle {\mathcal {H}}}$ is the total energy of the system if and only if the energy function ${\displaystyle E_{\mathcal {L}}}$ has the same property. (See definition of ⁠${\displaystyle {\mathcal {H}}}$⁠).
• ${\displaystyle {\frac {d{\mathcal {H}}}{dt}}={\frac {\partial {\mathcal {H}}}{\partial t}}}$ when ⁠${\displaystyle \mathbf {p} (t)}$⁠, ⁠${\displaystyle \mathbf {q} (t)}$⁠ form a solution of Hamilton's equations. Indeed, ${\textstyle {\frac {d{\mathcal {H}}}{dt}}={\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {p}}}}\cdot {\dot {\boldsymbol {p}}}+{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {q}}}}\cdot {\dot {\boldsymbol {q}}}+{\frac {\partial {\mathcal {H}}}{\partial t}},}$ and everything but the final term cancels out.
• ${\displaystyle {\mathcal {H}}}$ does not change under point transformations, i.e. smooth changes ${\displaystyle {\boldsymbol {q}}\leftrightarrow {\boldsymbol {q'}}}$ of space coordinates. (Follows from the invariance of the energy function ${\displaystyle E_{\mathcal {L}}}$ under point transformations. The invariance of ${\displaystyle E_{\mathcal {L}}}$ can be established directly).
• ${\displaystyle {\frac {\partial {\mathcal {H}}}{\partial t}}=-{\frac {\partial {\mathcal {L}}}{\partial t}}.}$ (See § Deriving Hamilton's equations).
• ⁠${\displaystyle -{\frac {\partial {\mathcal {H}}}{\partial q^{i}}}={\dot {p}}_{i}={\frac {\partial {\mathcal {L}}}{\partial q^{i}}}}$⁠. (Compare Hamilton's and Euler-Lagrange equations or see § Deriving Hamilton's equations).
• ${\displaystyle {\frac {\partial {\mathcal {H}}}{\partial q^{i}}}=0}$ if and only if ⁠${\displaystyle {\frac {\partial {\mathcal {L}}}{\partial q^{i}}}=0}$⁠.A coordinate for which the last equation holds is called cyclic (or ignorable). Every cyclic coordinate ${\displaystyle q^{i}}$ reduces the number of degrees of freedom by ⁠${\displaystyle 1}$⁠, causes the corresponding momentum ${\displaystyle p_{i}}$ to be conserved, and makes Hamilton's equations easier to solve.

Hamiltonian as the total system energy

In its application to a given system, the Hamiltonian is often taken to be $${\displaystyle {\mathcal {H}}=T+V}$$

where ${\displaystyle T}$ is the kinetic energy and ${\displaystyle V}$ is the potential energy. Using this relation can be simpler than first calculating the Lagrangian, and then deriving the Hamiltonian from the Lagrangian. However, the relation is not true for all systems.

The relation holds true for nonrelativistic systems when all of the following conditions are satisfied $${\displaystyle {\frac {\partial V({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}{\partial {\dot {q}}_{i}}}=0\;,\quad \forall i}$$ $${\displaystyle {\frac {\partial T({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}{\partial t}}=0}$$ $${\displaystyle T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})=\sum _{i=1}^{n}\sum _{j=1}^{n}{\biggl (}c_{ij}({\boldsymbol {q}}){\dot {q}}_{i}{\dot {q}}_{j}{\biggr )}}$$

where ${\displaystyle t}$ is time, ${\displaystyle n}$ is the number of degrees of freedom of the system, and each ${\displaystyle c_{ij}({\boldsymbol {q}})}$ is an arbitrary scalar function of ${\displaystyle {\boldsymbol {q}}}$.

In words, this means that the relation ${\displaystyle {\mathcal {H}}=T+V}$ holds true if ${\displaystyle T}$ does not contain time as an explicit variable (it is scleronomic), ${\displaystyle V}$ does not contain generalised velocity as an explicit variable, and each term of ${\displaystyle T}$ is quadratic in generalised velocity.

Proof

Preliminary to this proof, it is important to address an ambiguity in the related mathematical notation. While a change of variables can be used to equate ${\displaystyle {\mathcal {L}}({\boldsymbol {p}},{\boldsymbol {q}},t)={\mathcal {L}}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}$, it is important to note that ${\displaystyle {\frac {\partial {\mathcal {L}}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}{\partial {\dot {q}}_{i}}}\neq {\frac {\partial {\mathcal {L}}({\boldsymbol {p}},{\boldsymbol {q}},t)}{\partial {\dot {q}}_{i}}}}$. In this case, the right hand side always evaluates to 0. To perform a change of variables inside of a partial derivative, the multivariable chain rule should be used. Hence, to avoid ambiguity, the function arguments of any term inside of a partial derivative should be stated.

Additionally, this proof uses the notation ${\displaystyle f(a,b,c)=f(a,b)}$ to imply that ${\displaystyle {\frac {\partial f(a,b,c)}{\partial c}}=0}$.

Starting from definitions of the Hamiltonian, generalized momenta, and Lagrangian for an ${\displaystyle n}$ degrees of freedom system $${\displaystyle {\mathcal {H}}=\sum _{i=1}^{n}{\biggl (}p_{i}{\dot {q}}_{i}{\biggr )}-{\mathcal {L}}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}$$ $${\displaystyle p_{i}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)={\frac {\partial {\mathcal {L}}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}{\partial {\dot {q}}_{i}}}}$$ $${\displaystyle {\mathcal {L}}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)=T({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)-V({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}$$

Substituting the generalized momenta into the Hamiltonian gives $${\displaystyle {\mathcal {H}}=\sum _{i=1}^{n}\left({\frac {\partial {\mathcal {L}}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}{\partial {\dot {q}}_{i}}}{\dot {q}}_{i}\right)-{\mathcal {L}}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}$$

Substituting the Lagrangian into the result gives $${\displaystyle {\begin{aligned}{\mathcal {H}}&=\sum _{i=1}^{n}\left({\frac {\partial \left(T({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)-V({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)\right)}{\partial {\dot {q}}_{i}}}{\dot {q}}_{i}\right)-\left(T({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)-V({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)\right)\\&=\sum _{i=1}^{n}\left({\frac {\partial T({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}{\partial {\dot {q}}_{i}}}{\dot {q}}_{i}-{\frac {\partial V({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}{\partial {\dot {q}}_{i}}}{\dot {q}}_{i}\right)-T({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)+V({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)\end{aligned}}}$$

Now assume that $${\displaystyle {\frac {\partial V({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}{\partial {\dot {q}}_{i}}}=0\;,\quad \forall i}$$

and also assume that $${\displaystyle {\frac {\partial T({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}{\partial t}}=0}$$

Applying these assumptions results in $${\displaystyle {\begin{aligned}{\mathcal {H}}&=\sum _{i=1}^{n}\left({\frac {\partial T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}{\partial {\dot {q}}_{i}}}{\dot {q}}_{i}-{\frac {\partial V({\boldsymbol {q}},t)}{\partial {\dot {q}}_{i}}}{\dot {q}}_{i}\right)-T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})+V({\boldsymbol {q}},t)\\&=\sum _{i=1}^{n}\left({\frac {\partial T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}{\partial {\dot {q}}_{i}}}{\dot {q}}_{i}\right)-T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})+V({\boldsymbol {q}},t)\end{aligned}}}$$

Next assume that T is of the form $${\displaystyle T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})=\sum _{i=1}^{n}\sum _{j=1}^{n}{\biggl (}c_{ij}({\boldsymbol {q}}){\dot {q}}_{i}{\dot {q}}_{j}{\biggr )}}$$

where each ${\displaystyle c_{ij}({\boldsymbol {q}})}$ is an arbitrary scalar function of ${\displaystyle {\boldsymbol {q}}}$.

Differentiating this with respect to ${\displaystyle {\dot {q}}_{l}}$, ${\displaystyle l\in }$, gives $${\displaystyle {\begin{aligned}{\frac {\partial T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}{\partial {\dot {q}}_{l}}}&=\sum _{i=1}^{n}\sum _{j=1}^{n}{\biggl (}{\frac {\partial \left}{\partial {\dot {q}}_{l}}}{\biggr )}\\&=\sum _{i=1}^{n}\sum _{j=1}^{n}{\biggl (}c_{ij}({\boldsymbol {q}}){\frac {\partial \left}{\partial {\dot {q}}_{l}}}{\biggr )}\end{aligned}}}$$

Splitting the summation, evaluating the partial derivative, and rejoining the summation gives $${\displaystyle {\begin{aligned}{\frac {\partial T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}{\partial {\dot {q}}_{l}}}&=\sum _{i\neq l}^{n}\sum _{j\neq l}^{n}{\biggl (}c_{ij}({\boldsymbol {q}}){\frac {\partial \left}{\partial {\dot {q}}_{l}}}{\biggr )}+\sum _{i\neq l}^{n}{\biggl (}c_{il}({\boldsymbol {q}}){\frac {\partial \left}{\partial {\dot {q}}_{l}}}{\biggr )}+\sum _{j\neq l}^{n}{\biggl (}c_{lj}({\boldsymbol {q}}){\frac {\partial \left}{\partial {\dot {q}}_{l}}}{\biggr )}+c_{ll}({\boldsymbol {q}}){\frac {\partial \left}{\partial {\dot {q}}_{l}}}\\&=\sum _{i\neq l}^{n}\sum _{j\neq l}^{n}{\biggl (}0{\biggr )}+\sum _{i\neq l}^{n}{\biggl (}c_{il}({\boldsymbol {q}}){\dot {q}}_{i}{\biggr )}+\sum _{j\neq l}^{n}{\biggl (}c_{lj}({\boldsymbol {q}}){\dot {q}}_{j}{\biggr )}+2c_{ll}({\boldsymbol {q}}){\dot {q}}_{l}\\&=\sum _{i=1}^{n}{\biggl (}c_{il}({\boldsymbol {q}}){\dot {q}}_{i}{\biggr )}+\sum _{j=1}^{n}{\biggl (}c_{lj}({\boldsymbol {q}}){\dot {q}}_{j}{\biggr )}\end{aligned}}}$$

Summing (this multiplied by ${\displaystyle {\dot {q}}_{l}}$) over ${\displaystyle l}$ results in $${\displaystyle {\begin{aligned}\sum _{l=1}^{n}\left({\frac {\partial T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}{\partial {\dot {q}}_{l}}}{\dot {q}}_{l}\right)&=\sum _{l=1}^{n}\left(\left(\sum _{i=1}^{n}{\biggl (}c_{il}({\boldsymbol {q}}){\dot {q}}_{i}{\biggr )}+\sum _{j=1}^{n}{\biggl (}c_{lj}({\boldsymbol {q}}){\dot {q}}_{j}{\biggr )}\right){\dot {q}}_{l}\right)\\&=\sum _{l=1}^{n}\sum _{i=1}^{n}{\biggl (}c_{il}({\boldsymbol {q}}){\dot {q}}_{i}{\dot {q}}_{l}{\biggr )}+\sum _{l=1}^{n}\sum _{j=1}^{n}{\biggl (}c_{lj}({\boldsymbol {q}}){\dot {q}}_{j}{\dot {q}}_{l}{\biggr )}\\&=\sum _{i=1}^{n}\sum _{l=1}^{n}{\biggl (}c_{il}({\boldsymbol {q}}){\dot {q}}_{i}{\dot {q}}_{l}{\biggr )}+\sum _{l=1}^{n}\sum _{j=1}^{n}{\biggl (}c_{lj}({\boldsymbol {q}}){\dot {q}}_{l}{\dot {q}}_{j}{\biggr )}\\&=T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})+T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})\\&=2T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})\end{aligned}}}$$

This simplification is a result of Euler's homogeneous function theorem.

Hence, the Hamiltonian becomes $${\displaystyle {\begin{aligned}{\mathcal {H}}&=\sum _{i=1}^{n}\left({\frac {\partial T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}{\partial {\dot {q}}_{i}}}{\dot {q}}_{i}\right)-T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})+V({\boldsymbol {q}},t)\\&=2T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})-T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})+V({\boldsymbol {q}},t)\\&=T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})+V({\boldsymbol {q}},t)\end{aligned}}}$$

Application to systems of point masses

For a system of point masses, the requirement for ${\displaystyle T}$ to be quadratic in generalised velocity is always satisfied for the case where ${\displaystyle T({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)=T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}$, which is a requirement for ${\displaystyle {\mathcal {H}}=T+V}$ anyway.

Consider the kinetic energy for a system of N point masses. If it is assumed that ${\displaystyle T({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)=T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}$, then it can be shown that ${\displaystyle {\dot {\mathbf {r} }}_{k}({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)={\dot {\mathbf {r} }}_{k}({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}$ (See Scleronomous § Application). Therefore, the kinetic energy is $${\displaystyle T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})={\frac {1}{2}}\sum _{k=1}^{N}{\biggl (}m_{k}{\dot {\mathbf {r} }}_{k}({\boldsymbol {q}},{\boldsymbol {\dot {q}}})\cdot {\dot {\mathbf {r} }}_{k}({\boldsymbol {q}},{\boldsymbol {\dot {q}}}){\biggr )}}$$

The chain rule for many variables can be used to expand the velocity $${\displaystyle {\begin{aligned}{\dot {\mathbf {r} }}_{k}({\boldsymbol {q}},{\boldsymbol {\dot {q}}})&={\frac {d\mathbf {r} _{k}({\boldsymbol {q}})}{dt}}\\&=\sum _{i=1}^{n}\left({\frac {\partial \mathbf {r} _{k}({\boldsymbol {q}})}{\partial q_{i}}}{\dot {q}}_{i}\right)\end{aligned}}}$$

Resulting in $${\displaystyle {\begin{aligned}T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})&={\frac {1}{2}}\sum _{k=1}^{N}\left(m_{k}\left(\sum _{i=1}^{n}\left({\frac {\partial \mathbf {r} _{k}({\boldsymbol {q}})}{\partial q_{i}}}{\dot {q}}_{i}\right)\cdot \sum _{j=1}^{n}\left({\frac {\partial \mathbf {r} _{k}({\boldsymbol {q}})}{\partial q_{j}}}{\dot {q}}_{j}\right)\right)\right)\\&=\sum _{k=1}^{N}\sum _{i=1}^{n}\sum _{j=1}^{n}\left({\frac {1}{2}}m_{k}{\frac {\partial \mathbf {r} _{k}({\boldsymbol {q}})}{\partial q_{i}}}\cdot {\frac {\partial \mathbf {r} _{k}({\boldsymbol {q}})}{\partial q_{j}}}{\dot {q}}_{i}{\dot {q}}_{j}\right)\\&=\sum _{i=1}^{n}\sum _{j=1}^{n}\left(\sum _{k=1}^{N}\left({\frac {1}{2}}m_{k}{\frac {\partial \mathbf {r} _{k}({\boldsymbol {q}})}{\partial q_{i}}}\cdot {\frac {\partial \mathbf {r} _{k}({\boldsymbol {q}})}{\partial q_{j}}}\right){\dot {q}}_{i}{\dot {q}}_{j}\right)\\&=\sum _{i=1}^{n}\sum _{j=1}^{n}{\biggl (}c_{ij}({\boldsymbol {q}}){\dot {q}}_{i}{\dot {q}}_{j}{\biggr )}\end{aligned}}}$$

This is of the required form.

Conservation of energy

If the conditions for ${\displaystyle {\mathcal {H}}=T+V}$ are satisfied, then conservation of the Hamiltonian implies conservation of energy. This requires the additional condition that ${\displaystyle V}$ does not contain time as an explicit variable.

$${\displaystyle {\frac {\partial V({\boldsymbol {q}},{\boldsymbol {\dot {q}}},t)}{\partial t}}=0}$$

With respect to the extended Euler-Lagrange formulation (See Lagrangian mechanics § Extensions to include non-conservative forces), the Rayleigh dissipation function represents energy dissipation by nature. Therefore, energy is not conserved when ${\displaystyle R\neq 0}$. This is similar to the velocity dependent potential.

In summary, the requirements for ${\displaystyle {\mathcal {H}}=T+V={\text{constant of time}}}$ to be satisfied for a nonrelativistic system are

1. ${\displaystyle V=V({\boldsymbol {q}})}$
2. ${\displaystyle T=T({\boldsymbol {q}},{\boldsymbol {\dot {q}}})}$
3. ${\displaystyle T}$ is a homogeneous quadratic function in ${\displaystyle {\boldsymbol {\dot {q}}}}$

Hamiltonian of a charged particle in an electromagnetic field

A sufficient illustration of Hamiltonian mechanics is given by the Hamiltonian of a charged particle in an electromagnetic field. In Cartesian coordinates the Lagrangian of a non-relativistic classical particle in an electromagnetic field is (in SI Units): $${\displaystyle {\mathcal {L}}=\sum _{i}{\tfrac {1}{2}}m{\dot {x}}_{i}^{2}+\sum _{i}q{\dot {x}}_{i}A_{i}-q\varphi ,}$$ where q is the electric charge of the particle, φ is the electric scalar potential, and the A_i are the components of the magnetic vector potential that may all explicitly depend on ${\displaystyle x_{i}}$ and ⁠${\displaystyle t}$⁠.

This Lagrangian, combined with Euler–Lagrange equation, produces the Lorentz force law $${\displaystyle m{\ddot {\mathbf {x} }}=q\mathbf {E} +q{\dot {\mathbf {x} }}\times \mathbf {B} \,,}$$ and is called minimal coupling.

The canonical momenta are given by: $${\displaystyle p_{i}={\frac {\partial {\mathcal {L}}}{\partial {\dot {x}}_{i}}}=m{\dot {x}}_{i}+qA_{i}.}$$

The Hamiltonian, as the Legendre transformation of the Lagrangian, is therefore: $${\displaystyle {\mathcal {H}}=\sum _{i}{\dot {x}}_{i}p_{i}-{\mathcal {L}}=\sum _{i}{\frac {\left(p_{i}-qA_{i}\right)^{2}}{2m}}+q\varphi .}$$

This equation is used frequently in quantum mechanics.

Under gauge transformation: $${\displaystyle \mathbf {A} \rightarrow \mathbf {A} +\nabla f\,,\quad \varphi \rightarrow \varphi -{\dot {f}}\,,}$$ where f(r, t) is any scalar function of space and time. The aforementioned Lagrangian, the canonical momenta, and the Hamiltonian transform like: $${\displaystyle L\rightarrow L'=L+q{\frac {df}{dt}}\,,\quad \mathbf {p} \rightarrow \mathbf {p'} =\mathbf {p} +q\nabla f\,,\quad H\rightarrow H'=H-q{\frac {\partial f}{\partial t}}\,,}$$ which still produces the same Hamilton's equation: $${\displaystyle {\begin{aligned}\left.{\frac {\partial H'}{\partial {x_{i}}}}\right|_{p'_{i}}&=\left.{\frac {\partial }{\partial {x_{i}}}}\right|_{p'_{i}}({\dot {x}}_{i}p'_{i}-L')=-\left.{\frac {\partial L'}{\partial {x_{i}}}}\right|_{p'_{i}}\\&=-\left.{\frac {\partial L}{\partial {x_{i}}}}\right|_{p'_{i}}-q\left.{\frac {\partial }{\partial {x_{i}}}}\right|_{p'_{i}}{\frac {df}{dt}}\\&=-{\frac {d}{dt}}\left(\left.{\frac {\partial L}{\partial {{\dot {x}}_{i}}}}\right|_{p'_{i}}+q\left.{\frac {\partial f}{\partial {x_{i}}}}\right|_{p'_{i}}\right)\\&=-{\dot {p}}'_{i}\end{aligned}}}$$

In quantum mechanics, the wave function will also undergo a local U(1) group transformation during the Gauge Transformation, which implies that all physical results must be invariant under local U(1) transformations.

Relativistic charged particle in an electromagnetic field

The relativistic Lagrangian for a particle (rest mass ${\displaystyle m}$ and charge ⁠${\displaystyle q}$⁠) is given by:

$${\displaystyle {\mathcal {L}}(t)=-mc^{2}{\sqrt {1-{\frac {{{\dot {\mathbf {x} }}(t)}^{2}}{c^{2}}}}}+q{\dot {\mathbf {x} }}(t)\cdot \mathbf {A} \left(\mathbf {x} (t),t\right)-q\varphi \left(\mathbf {x} (t),t\right)}$$

Thus the particle's canonical momentum is $${\displaystyle \mathbf {p} (t)={\frac {\partial {\mathcal {L}}}{\partial {\dot {\mathbf {x} }}}}={\frac {m{\dot {\mathbf {x} }}}{\sqrt {1-{\frac {{\dot {\mathbf {x} }}^{2}}{c^{2}}}}}}+q\mathbf {A} }$$ that is, the sum of the kinetic momentum and the potential momentum.

Solving for the velocity, we get $${\displaystyle {\dot {\mathbf {x} }}(t)={\frac {\mathbf {p} -q\mathbf {A} }{\sqrt {m^{2}+{\frac {1}{c^{2}}}{\left(\mathbf {p} -q\mathbf {A} \right)}^{2}}}}}$$

So the Hamiltonian is $${\displaystyle {\mathcal {H}}(t)={\dot {\mathbf {x} }}\cdot \mathbf {p} -{\mathcal {L}}=c{\sqrt {m^{2}c^{2}+{\left(\mathbf {p} -q\mathbf {A} \right)}^{2}}}+q\varphi }$$

This results in the force equation (equivalent to the Euler–Lagrange equation) $${\displaystyle {\dot {\mathbf {p} }}=-{\frac {\partial {\mathcal {H}}}{\partial \mathbf {x} }}=q{\dot {\mathbf {x} }}\cdot ({\boldsymbol {\nabla }}\mathbf {A} )-q{\boldsymbol {\nabla }}\varphi =q{\boldsymbol {\nabla }}({\dot {\mathbf {x} }}\cdot \mathbf {A} )-q{\boldsymbol {\nabla }}\varphi }$$ from which one can derive $${\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {m{\dot {\mathbf {x} }}}{\sqrt {1-{\frac {{\dot {\mathbf {x} }}^{2}}{c^{2}}}}}}\right)&={\frac {\mathrm {d} }{\mathrm {d} t}}(\mathbf {p} -q\mathbf {A} )={\dot {\mathbf {p} }}-q{\frac {\partial \mathbf {A} }{\partial t}}-q({\dot {\mathbf {x} }}\cdot \nabla )\mathbf {A} \\&=q{\boldsymbol {\nabla }}({\dot {\mathbf {x} }}\cdot \mathbf {A} )-q{\boldsymbol {\nabla }}\varphi -q{\frac {\partial \mathbf {A} }{\partial t}}-q({\dot {\mathbf {x} }}\cdot \nabla )\mathbf {A} \\&=q\mathbf {E} +q{\dot {\mathbf {x} }}\times \mathbf {B} \end{aligned}}}$$

The above derivation makes use of the vector calculus identity: $${\displaystyle {\tfrac {1}{2}}\nabla \left(\mathbf {A} \cdot \mathbf {A} \right)=\mathbf {A} \cdot \mathbf {J} _{\mathbf {A} }=\mathbf {A} \cdot (\nabla \mathbf {A} )=(\mathbf {A} \cdot \nabla )\mathbf {A} +\mathbf {A} \times (\nabla \times \mathbf {A} ).}$$

An equivalent expression for the Hamiltonian as function of the relativistic (kinetic) momentum, ⁠${\displaystyle \mathbf {P} =\gamma m{\dot {\mathbf {x} }}(t)=\mathbf {p} -q\mathbf {A} }$⁠, is $${\displaystyle {\mathcal {H}}(t)={\dot {\mathbf {x} }}(t)\cdot \mathbf {P} (t)+{\frac {mc^{2}}{\gamma }}+q\varphi (\mathbf {x} (t),t)=\gamma mc^{2}+q\varphi (\mathbf {x} (t),t)=E+V}$$