Hamiltonian mechanics

The Hamiltonian mechanics , named after William Rowan Hamilton , is a branch of classical mechanics . It examines the movement in phase space . This is the set of pairs of position and momentum values ​​that can initially be freely specified for the system of particles under consideration. Then the Hamilton function determines how the locations and impulses of the particles change with time (neglecting friction ) using Hamilton's equations of motion .

The equations of motion were given in 1834 by William Rowan Hamilton .

All equations of motion that follow from an action principle can be formulated as equivalent Hamiltonian equations of motion. These have two decisive advantages:

• On the one hand one can show that the movement in phase space is equal to area (see Henri Poincaré ). From this it follows that there are eddies and stagnation points when moving in phase space , comparable to the flow of an incompressible liquid.
• On the other hand, the Hamiltonian equations of motion have a large group of transformations, the canonical transformations , which allow them to be transformed into other, sometimes solvable, Hamiltonian equations.

With the Hamiltonian equations of motion one examines integrable and chaotic motion in particular and uses it in statistical physics .

details

The Hamilton function of a system of particles is its energy as a function of phase space . It depends on the (generalized) position coordinates and on the (generalized) momentum coordinates of the particles and can also depend on the time . ${\displaystyle {\mathcal {H}}(t,q,p)}$ ${\displaystyle q=(q_{1},q_{2},\dots ,q_{n})}$ ${\displaystyle p=(p_{1},p_{2},\dots ,p_{n})}$ ${\displaystyle t}$

The number of coordinates and impulses is called the number of degrees of freedom. The phase space is -dimensional. ${\displaystyle n}$ ${\displaystyle 2n}$

The Hamilton function determines the temporal development of the particle locations and particle impulses using Hamilton's equations of motion:

${\displaystyle {\dot {q}}_{k}={\frac {d}{dt}}q_{k}={\frac {\partial {\mathcal {H}}}{\partial p_{k}}}\,,\quad {\dot {p}}_{k}={\frac {d}{dt}}p_{k}=-{\frac {\partial {\mathcal {H}}}{\partial q_{k}}}\,,\quad k=1,2,\dots ,n}$

This is a system of ordinary first order differential equations for the unknown functions of time, ${\displaystyle 2n}$ ${\displaystyle q(t),p(t)\,.}$

If the Hamilton function does not explicitly depend on, then the solution curves do not intersect and a solution curve goes through every point of the phase space. ${\displaystyle t}$

In the case of time-dependent one, the time can be understood as an additional degree of freedom with an associated momentum and the time-independent Hamilton function . Therefore, in the following we limit ourselves to time-independent Hamilton functions. However, the function is not restricted downwards and the hypersurface of constant energy is not compact, as assumed by some considerations. ${\displaystyle {\mathcal {H}}(t,q,p)}$ ${\displaystyle t=q_{0}}$ ${\displaystyle p_{0}}$ ${\displaystyle {\hat {\mathcal {H}}}(q_{0},q,p_{0},p)={\mathcal {H}}(q_{0},q,p)+p_{0}}$ ${\displaystyle {\mathcal {H}}(q_{0},q,p)+p_{0}}$ ${\displaystyle {\hat {\mathcal {H}}}=E}$

Particle in potential

For a particle of mass that moves non-relativistically in a potential , the Hamilton function is composed of kinetic and potential energy : ${\displaystyle m}$ ${\displaystyle V}$

${\displaystyle {\mathcal {H}}(\mathbf {q} ,\mathbf {p} )={\frac {\mathbf {p} ^{2}}{2\,m}}+V(\mathbf {q} )}$

The associated Hamiltonian equations of motion

${\displaystyle {\dot {q}}_{k}={\frac {p_{k}}{m}}\ ,\ {\dot {p}}_{k}=-{\frac {\partial V}{\partial q_{k}}}}$

are Newton's equations for motion in a conservative force field,

${\displaystyle m\,{\ddot {q}}_{k}=F_{k}=-{\frac {\partial V}{\partial q_{k}}}\,.}$

In particular, the potential energy of a one-dimensional is harmonic oscillator The Hooke's spring force in the equation of motion ${\displaystyle (n=1)}$ ${\displaystyle V(q)={\frac {1}{2}}\,m\,\omega ^{2}\,q^{2}\,.}$

${\displaystyle m\,{\ddot {q}}=-m\,\omega ^{2}\,q}$

causes the track to swing around the rest position,

${\displaystyle q(t)=A\,\cos {\bigl (}\omega \,(t-t_{0}){\bigr )}\,.}$

Here is the amplitude and a time at which this maximum deflection is passed. ${\displaystyle A}$ ${\displaystyle t_{0}}$

Free relativistic particle

For a relativistic, free particle with the energy-momentum relationship , the Hamilton function is ${\displaystyle E^{2}-\mathbf {p} ^{2}\,c^{2}=m^{2}\,c^{4}}$

${\displaystyle {\mathcal {H}}(\mathbf {q} ,\mathbf {p} )={\sqrt {m^{2}\,c^{4}+\mathbf {p} ^{2}\,c^{2}}}.}$

The Hamiltonian equations of motion say how the speed is related to the momentum and that the momentum does not change with time:

${\displaystyle {\dot {q}}_{k}={\frac {p_{k}\,c^{2}}{\sqrt {m^{2}\,c^{4}+\mathbf {p} ^{2}\,c^{2}}}}}$

${\displaystyle {\dot {p}}_{k}=0}$

If the Hamilton function does not depend on time, as in these examples, the system of particles retains its initial energy; it is then a conservation quantity.

Working principle

The Hamiltonian equations of motion follow from the Hamiltonian principle of stationary action. Of all conceivable paths in phase space,

${\displaystyle \Gamma \colon t\mapsto {\bigl (}q(t),p(t){\bigr )}\,,}$

which initially currently through the starting point ${\displaystyle {\underline {t}}}$

${\displaystyle {\bigl (}{\underline {q}},{\underline {p}}{\bigr )}={\bigl (}q({\underline {t}}),p({\underline {t}}){\bigr )}}$

and finally currently by the end point ${\displaystyle {\overline {t}}}$

${\displaystyle {\bigl (}{\overline {q}},{\overline {p}}{\bigr )}={\bigl (}q({\overline {t}}),p({\overline {t}}){\bigr )}}$

run, the path physically traversed is the one on which the effect

${\displaystyle W[\Gamma ]=\int _{\underline {t}}^{\overline {t}}\mathrm {d} t\,\left(\sum _{i=1}^{n}p_{i}(t)\,{\frac {\mathrm {d} q_{i}(t)}{\mathrm {d} t}}-{\mathcal {H}}(q(t),p(t))\right)}$

is stationary.

If you consider a single-parameter family of curves

${\displaystyle \Gamma _{\alpha }:t\mapsto {\bigl (}q(t,\alpha ),p(t,\alpha ){\bigr )}\,,}$

which initially currently through the starting point ${\displaystyle {\underline {t}}}$

${\displaystyle {\bigl (}{\underline {q}},{\underline {p}}{\bigr )}={\bigl (}q({\underline {t}},\alpha ),p({\underline {t}},\alpha ){\bigr )}}$

and finally currently by the end point ${\displaystyle {\overline {t}}}$

${\displaystyle {\bigl (}{\overline {q}},{\overline {p}}{\bigr )}={\bigl (}q({\overline {t}},\alpha ),p({\overline {t}},\alpha ){\bigr )}}$

run, the effect is for extremal, if the derivative after disappears there. ${\displaystyle W[\Gamma _{\alpha }]}$ ${\displaystyle \alpha =0}$ ${\displaystyle \alpha }$

We call this derivation a variation of the effect

${\displaystyle \delta W={\frac {\partial W[\Gamma _{\alpha }]}{\partial \alpha }}_{|_{\alpha =0}}\,.}$

Likewise is

${\displaystyle \delta q_{i}={\frac {\partial q_{i}(t,\alpha )}{\partial \alpha }}_{|_{\alpha =0}}}$

the variation of the place and

${\displaystyle \delta p_{i}={\frac {\partial p_{i}(t,\alpha )}{\partial \alpha }}_{|_{\alpha =0}}}$

the variation of momentum.

The variation of the effect is according to the chain rule

${\displaystyle \delta W=\sum _{i=1}^{n}\int _{\underline {t}}^{\overline {t}}\mathrm {d} t{\Bigl (}\delta p_{i}(t)\,{\frac {\mathrm {d} q_{i}(t)}{\mathrm {d} t}}+p_{i}(t)\,{\frac {\mathrm {d} \delta q_{i}(t)}{\mathrm {d} t}}-\delta q_{i}(t)\,{\frac {\partial {\mathcal {H}}}{\partial q_{i}}}_{|_{(q(t),p(t))}}-\delta p_{i}(t)\,{\frac {\partial {\mathcal {H}}}{\partial p_{i}}}_{|_{(q(t),p(t))}}{\Bigr )}\,.}$

We write the second term as a complete time derivative and a term that occurs without a time derivative: ${\displaystyle \delta q_{i}}$

${\displaystyle p_{i}(t)\,{\frac {\mathrm {d} \delta q_{i}(t)}{\mathrm {d} t}}={\frac {\mathrm {d} {\bigl (}p_{i}(t)\,\delta q_{i}(t){\bigr )}}{\mathrm {d} t}}-{\frac {\mathrm {d} p_{i}(t)}{\mathrm {d} t}}\,\delta q_{i}(t)}$

The integral over the complete derivative results in the beginning and end times and disappears because then disappears, because all curves of the family go through the same starting and end points. If we finally summarize the terms with and , then the variation of the effect is ${\displaystyle {\bigl (}p_{i}(t)\,\delta q_{i}(t){\bigr )}}$ ${\displaystyle \delta q_{i}}$ ${\displaystyle \delta q_{i}}$ ${\displaystyle \delta p_{i}}$

${\displaystyle \delta W=\sum _{i=1}^{n}\int _{\underline {t}}^{\overline {t}}\mathrm {d} t{\Bigl (}-\delta q_{i}(t){\bigl (}{\frac {\mathrm {d} p_{i}(t)}{\mathrm {d} t}}+{\frac {\partial {\mathcal {H}}}{\partial q_{i}}}_{|_{(q(t),p(t))}}{\bigr )}+\delta p_{i}(t){\bigl (}{\frac {\mathrm {d} q_{i}(t)}{\mathrm {d} t}}-{\frac {\partial {\mathcal {H}}}{\partial p_{i}}}_{|_{(q(t),p(t))}}{\bigr )}{\Bigr )}\,.}$

In order for the effect to be stationary, this integral must vanish for all and all that initially and finally vanish. This is exactly the case when the factors with which they appear in the integral vanish: ${\displaystyle \delta q_{i}}$ ${\displaystyle \delta p_{i}}$

${\displaystyle 0={\frac {\mathrm {d} p_{i}(t)}{\mathrm {d} t}}+{\frac {\partial {\mathcal {H}}}{\partial q_{i}}}_{|_{(q(t),p(t))}}}$

${\displaystyle 0={\frac {\mathrm {d} q_{i}(t)}{\mathrm {d} t}}-{\frac {\partial {\mathcal {H}}}{\partial p_{i}}}_{|_{(q(t),p(t))}}}$

The effect is stationary if the Hamiltonian equations of motion apply.

Relation to the Lagrange function

The Hamilton function is the Legendre transform of the Lagrange function with respect to the velocities ${\displaystyle {\dot {q}}=({\dot {q}}_{1},{\dot {q}}_{2}\dots {\dot {q}}_{n})}$ ${\displaystyle {\mathcal {L}}(q,{\dot {q}}):}$

${\displaystyle {\mathcal {H}}(q,p)=\sum _{k=1}^{n}p_{k}\,{\dot {q}}_{k}(q,p)-{\mathcal {L}}(q,{\dot {q}}(q,p))}$

On the right-hand side, the velocities mean those functions that are obtained when defining the impulses ${\displaystyle {\dot {q}}}$ ${\displaystyle {\dot {q}}(q,p)}$

${\displaystyle p_{k}={\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{k}}}}$

dissolves according to the speeds.

If one can invert the definition of the impulses and solve for the velocities, then the Hamiltonian equations of motion apply if and only if the Euler-Lagrange equations of the effect

${\displaystyle W[\Gamma ]=\int _{\underline {t}}^{\overline {t}}\mathrm {d} t\,{\mathcal {L}}(q(t),{\dot {q}}(t))}$

are fulfilled. Because the partial derivative of after the impulses results according to the chain rule and the definition of the impulses ${\displaystyle {\mathcal {H}}}$

${\displaystyle {\frac {\partial {\mathcal {H}}}{\partial p_{i}}}={\dot {q}}_{i}+\sum _{j}p_{j}{\frac {\partial {\dot {q}}_{j}}{\partial p_{i}}}-\sum _{j}{\frac {\partial {\dot {q}}_{j}}{\partial p_{i}}}\underbrace {\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{j}}} _{p_{j}}={\dot {q}}_{i}}$

The derivation according to the location coordinates also results

${\displaystyle {\frac {\partial {\mathcal {H}}}{\partial q_{i}}}=\sum _{j}p_{j}{\frac {\partial {\dot {q}}_{j}}{\partial q_{i}}}-{\frac {\partial {\mathcal {L}}}{\partial q_{i}}}-\sum _{j}{\frac {\partial {\dot {q}}_{j}}{\partial q_{i}}}\underbrace {\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{j}}} _{p_{j}}=-{\frac {\partial {\mathcal {L}}}{\partial q_{i}}}}$

The Euler-Lagrange equation says

${\displaystyle {\frac {\partial {\mathcal {L}}}{\partial q_{i}}}={\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{i}}}={\dot {p}}_{i}\,.}$

So the Hamiltonian equations of motion hold when the Euler-Lagrange equation holds. Conversely, the Euler-Lagrange equation applies if the Hamiltonian equations of motion apply.

For example, the free relativistic particle depends on the Lagrangian

${\displaystyle {\mathcal {L}}=-m\,c^{2}{\sqrt {1-{\dot {\mathbf {q} }}^{2}/c^{2}}}}$

the momentum according to

${\displaystyle \mathbf {p} ={\frac {m{\dot {\mathbf {q} }}}{\sqrt {1-{\dot {\mathbf {q} }}^{2}/c^{2}}}}}$

on the speed. Conversely, the speed is therefore the function

${\displaystyle {\dot {\mathbf {q} }}={\frac {\mathbf {p} \,c^{2}}{\sqrt {m^{2}\,c^{4}+\mathbf {p} ^{2}\,c^{2}}}}}$

of the momentum. The Hamilton function of the free, relativistic particle already given results in the above equation for inserted. ${\displaystyle {\mathcal {H}}}$

If the Lagrangian does not depend explicitly on the time, then Noether's theorem says that the energy

${\displaystyle E(q,{\dot {q}})=\sum _{k}{\dot {q}}_{k}{\frac {\partial {\mathcal {L}}}{\partial {\dot {q}}_{k}}}-{\mathcal {L}}}$

retains its initial value on the physical paths. The comparison with the Legendre transformation shows that the Hamilton function is about this energy, for which the velocities are to be understood as a function of the momenta:

${\displaystyle {\mathcal {H}}(q,p)=E(q,{\dot {q}}(q,p))}$

Poisson bracket

The value of a phase space function changes over time on orbits because it explicitly depends on and because the point on the path changes: ${\displaystyle \Phi (t,q,p)}$ ${\displaystyle (q(t),p(t))}$ ${\displaystyle t}$

${\displaystyle {\frac {\mathrm {d} \Phi (t,q(t),p(t))}{\mathrm {d} t}}={\frac {\partial \Phi }{\partial t}}+\sum _{i}{\Bigl (}{\frac {\partial \Phi }{\partial q_{i}}}{\frac {\mathrm {d} q_{i}}{\mathrm {d} t}}+{\frac {\partial \Phi }{\partial p_{i}}}{\frac {\mathrm {d} p_{i}}{\mathrm {d} t}}{\Bigr )}}$

The physically traversed paths satisfy the Hamiltonian equations of motion:

${\displaystyle {\frac {\mathrm {d} \Phi (t,q(t),p(t))}{\mathrm {d} t}}={\frac {\partial \Phi }{\partial t}}+\sum _{i}{\Bigl (}{\frac {\partial \Phi }{\partial q_{i}}}{\frac {\partial {\mathcal {H}}}{\partial p_{i}}}-{\frac {\partial \Phi }{\partial p_{i}}}{\frac {\partial {\mathcal {H}}}{\partial q_{i}}}{\Bigr )}}$

With the Poisson bracket for two phase space functions introduced by Siméon Denis Poisson and ${\displaystyle \Phi }$ ${\displaystyle \Psi }$

${\displaystyle {\bigl \{}\Phi ,\Psi {\bigr \}}=\sum _{i}{\Bigl (}{\frac {\partial \Phi }{\partial q_{i}}}{\frac {\partial \Psi }{\partial p_{i}}}-{\frac {\partial \Phi }{\partial p_{i}}}{\frac {\partial \Psi }{\partial q_{i}}}{\Bigr )}}$

so it applies

${\displaystyle {\frac {\mathrm {d} \Phi (t,q(t),p(t))}{\mathrm {d} t}}={\frac {\partial \Phi }{\partial t}}+{\bigl \{}\Phi ,{\mathcal {H}}{\bigr \}}\,.}$

Written with Poisson brackets , the formula picture of the Hamiltonian equations of motion resembles the Heisenberg equations of motion of quantum mechanics .

The phase space coordinates have Poisson brackets as coordinate functions

${\displaystyle \{q_{i},q_{j}\}=0=\{p_{i},p_{j}\}\,,\,\{q_{i},p_{j}\}=\delta _{ij}\,.}$

In quantum mechanics, after canonical quantization, the canonical commutation relations correspond to them.

The Poisson bracket is antisymmetric, linear and satisfies the product rule and the Jacobi identity. For all numbers and and all phase space functions holds ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle \Psi ,\Phi ,\Lambda }$

• ${\displaystyle \{\Psi ,\Phi \}=-\{\Phi ,\Psi \}\,,}$
• ${\displaystyle \{\Psi ,a\,\Phi +b\,\Lambda \}=a\,\{\Psi ,\Phi \}+b\,\{\Psi ,\Lambda \}\,,}$
• ${\displaystyle \{\Psi ,\Phi \,\,\Lambda \}=\{\Psi ,\Phi \}\,\Lambda +\Phi \,\{\Psi ,\Lambda \}\,,}$
• ${\displaystyle \{\Psi ,\{\Phi ,\Lambda \}\}+\{\Phi ,\{\Lambda ,\Psi \}\}+\{\Lambda ,\{\Psi ,\Phi \}\}=0\,.}$

The differentiable phase space functions form a Lie algebra with Poisson brackets as the Lie product.

Hamiltonian River

The vector field belongs to every (time-independent) phase space function ${\displaystyle \Phi }$ ${\displaystyle v_{\Phi }\,,}$

${\displaystyle v_{\Phi }(\Psi )=\{\Psi ,\Phi \}\,,}$

which derives phase space functions along the curves that solve the Hamilton equations with . ${\displaystyle \Psi }$ ${\displaystyle {\mathcal {H}}=\Phi }$

The figure of the initial values of the solution curves on the on Hamiltonian associated river. ${\displaystyle \Phi _{t}}$ ${\displaystyle (q(0),p(0))}$ ${\displaystyle (q(t),p(t))}$ ${\displaystyle \Phi }$

Symplectic structure

The phase space with its Poisson bracket is a symplectic manifold with the symplectic form

${\displaystyle \omega =\sum _{i}\mathrm {d} q_{i}\,\mathrm {d} p_{i}\,.}$

Applied to the vector fields belonging to and , this two-form results in the Poisson bracket of the two functions: ${\displaystyle \Phi }$ ${\displaystyle \Psi }$

${\displaystyle \omega \,(v_{\Phi },v_{\Psi })=\{\Phi ,\Psi \}}$

The symplectic form is invariant under any Hamiltonian flow. This means the following: If initially a two-dimensional surface is given in phase space, then it is mapped onto the surface over time by the Hamiltonian flow of a phase space function . The size of the initial area measured with the symplectic shape corresponds to the size at each later time. Hamiltonian river is equal area: ${\displaystyle F}$ ${\displaystyle \Phi }$ ${\displaystyle \Phi _{t}(F)}$

${\displaystyle \int _{F}\omega =\int _{\Phi _{t}(F)}\omega }$

Since the surface element is invariant, the volume element is also invariant under a Hamiltonian flow. This finding is Liouville's theorem. The volume of an area of the phase space does not change with Hamiltonian time evolution: ${\displaystyle \omega }$ ${\displaystyle \omega ^{n}=n!\,\mathrm {d} ^{n}q\,\mathrm {d} ^{n}p}$ ${\displaystyle B}$

${\displaystyle \int _{B}\omega ^{n}=\int _{\Phi _{t}(B)}\omega ^{n}}$

In particular, the area within which the system is initially due to the measurement errors remains the same size. However, one cannot conclude from this that initial ignorance does not increase. In the case of chaotic movements, initial values, which initially differed only in small measurement errors, can be distributed over a large area with many small holes like whipped cream. Even whipping cream does not increase its microscopically determined volume.

Canonical transformation

The Hamilton equations are simplified if the Hamilton function does not depend on a variable, for example . Then there is a symmetry: the Hamilton function is invariant under the shift of Conversely, if there is a symmetry (in the vicinity of a point that is not a fixed point), the position and momentum variables can be chosen in such a way that the Hamilton function of a Variable does not depend. Then it's easy ${\displaystyle q_{1}\,,}$ ${\displaystyle q_{1}\,.}$ ${\displaystyle q_{1}}$ ${\displaystyle p_{1}(t)=p_{1}(0)\,.}$

Integrable movement

The equations of motion are integrable if the Hamilton function only depends on the momenta. Then the impulses are constant and the derivatives of the Hamilton function with respect to the impulses are the temporally constant velocities with which the coordinates increase linearly, ${\displaystyle q}$

${\displaystyle {\dot {q}}_{k}={\frac {\partial {\mathcal {H}}}{\partial p_{k}}}=\omega _{k}(p)\,,\ {\dot {p}}_{k}=-{\frac {\partial {\mathcal {H}}}{\partial q_{k}}}=0\,,\ p_{k}(t)=p_{k}(0)\,,\ q_{k}(t)=\omega _{k}(p)\,t+q_{k}(0)\,.}$

If the phase space area of ​​constant energy is also compact, then the coordinates are the angles on a torus which , when enlarged, name the same point, ${\displaystyle {\mathcal {H}}(q,p)=E}$ ${\displaystyle q_{k}}$ ${\displaystyle 2\pi }$

${\displaystyle q_{k}\sim q_{k}+2\pi \,.}$

The phase space of such an integrable system consists of -dimensional tori , around which the solution curves of the Hamiltonian equations wind. ${\displaystyle n}$

Connection with quantum mechanics

As used in the mechanics, the Hamiltonian the time evolution is determined, the determined Hamiltonian the time evolution in quantum mechanics. It is obtained for many quantum mechanical systems from the Hamilton function of the corresponding classical system by canonical quantization , by reading the algebraic expression for as a function of operators and that satisfy the canonical commutation relations. ${\displaystyle {\mathcal {H}}(q,p)}$ ${\displaystyle q}$ ${\displaystyle p}$

swell

• VI Arnol'd : Mathematical Methods of Classical Mechanics. Springer-Verlag, 1989, ISBN 0-387-96890-3 .

See also

• Field theory (physics)
• Hamilton-Jacobi formalism