Hamilton's principle

See also more rigorous derivation Euler–Lagrange equation

Requiring that the true trajectory q(t) be a stationary point of the action functional ${\displaystyle {\mathcal {S}}}$ is equivalent to a set of differential equations for q(t) (the Euler–Lagrange equations), which may be derived as follows.

Let q(t) represent the true evolution of the system between two specified states q₁ = q(t₁) and q₂ = q(t₂) at two specified times t₁ and t₂, and let ε(t) be a small perturbation that is zero at the endpoints of the trajectory

${\displaystyle {\boldsymbol {\varepsilon }}(t_{1})={\boldsymbol {\varepsilon }}(t_{2})\ {\stackrel {\mathrm {def} }{=}}\ 0}$

To first order in the perturbation ε(t), the change in the action functional ${\displaystyle \delta {\mathcal {S}}}$ would be

${\displaystyle \delta {\mathcal {S}}=\int _{t_{1}}^{t_{2}}\;\left[L(\mathbf {q} +{\boldsymbol {\varepsilon }},{\dot {\mathbf {q} }}+{\dot {\boldsymbol {\varepsilon }}})-L(\mathbf {q} ,{\dot {\mathbf {q} }})\right]dt=\int _{t_{1}}^{t_{2}}\;\left({\boldsymbol {\varepsilon }}\cdot {\frac {\partial L}{\partial \mathbf {q} }}+{\dot {\boldsymbol {\varepsilon }}}\cdot {\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\right)\,dt}$

where we have expanded the Lagrangian L to first order in the perturbation ε(t).

Applying integration by parts to the last term results in

${\displaystyle \delta {\mathcal {S}}=\left[{\boldsymbol {\varepsilon }}\cdot {\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\right]_{t_{1}}^{t_{2}}+\int _{t_{1}}^{t_{2}}\;\left({\boldsymbol {\varepsilon }}\cdot {\frac {\partial L}{\partial \mathbf {q} }}-{\boldsymbol {\varepsilon }}\cdot {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\right)\,dt}$

The boundary conditions ${\displaystyle {\boldsymbol {\varepsilon }}(t_{1})={\boldsymbol {\varepsilon }}(t_{2})\ {\stackrel {\mathrm {def} }{=}}\ 0}$ causes the first term to vanish

${\displaystyle \delta {\mathcal {S}}=\int _{t_{1}}^{t_{2}}\;{\boldsymbol {\varepsilon }}\cdot \left({\frac {\partial L}{\partial \mathbf {q} }}-{\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\mathbf {q} }}}}\right)\,dt}$

Hamilton's principle requires that this first-order change ${\displaystyle \delta {\mathcal {S}}}$ is zero for all possible perturbations ε(t), i.e., the true path is a stationary point of the action functional ${\displaystyle {\mathcal {S}}}$ (either a minimum, maximum or saddle point). This requirement can be satisfied if and only if

These equations are called the Euler–Lagrange equations for the variational problem.

The conjugate momentum p_k for a generalized coordinate q_k is defined by the equation

${\displaystyle p_{k}\ {\overset {\mathrm {def} }{=}}\ {\frac {\partial L}{\partial {\dot {q}}_{k}}}}$.

An important special case of the Euler–Lagrange equation occurs when L does not contain a generalized coordinate q_k explicitly,

${\displaystyle {\frac {\partial L}{\partial q_{k}}}=0\quad \Rightarrow \quad {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {q}}_{k}}}=0\quad \Rightarrow \quad {\frac {dp_{k}}{dt}}=0\,,}$

that is, the conjugate momentum is a constant of the motion.

In such cases, the coordinate q_k is called a cyclic coordinate. For example, if we use polar coordinates t, r, θ to describe the planar motion of a particle, and if L does not depend on θ, the conjugate momentum is the conserved angular momentum.

Trivial examples help to appreciate the use of the action principle via the Euler–Lagrange equations. A free particle (mass m and velocity v) in Euclidean space moves in a straight line. Using the Euler–Lagrange equations, this can be shown in polar coordinates as follows. In the absence of a potential, the Lagrangian is simply equal to the kinetic energy

${\displaystyle L={\frac {1}{2}}mv^{2}={\frac {1}{2}}m\left({\dot {x}}^{2}+{\dot {y}}^{2}\right)}$

in orthonormal (x,y) coordinates, where the dot represents differentiation with respect to the curve parameter (usually the time, t). Therefore, upon application of the Euler–Lagrange equations,

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {x}}}}\right)-{\frac {\partial L}{\partial x}}=0\qquad \Rightarrow \qquad m{\ddot {x}}=0}$

And likewise for y. Thus the Euler–Lagrange formulation can be used to derive Newton's laws.

In polar coordinates (r, φ) the kinetic energy and hence the Lagrangian becomes

${\displaystyle L={\frac {1}{2}}m\left({\dot {r}}^{2}+r^{2}{\dot {\varphi }}^{2}\right).}$

The radial r and φ components of the Euler–Lagrange equations become, respectively

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {r}}}}\right)-{\frac {\partial L}{\partial r}}=0\qquad \Rightarrow \qquad {\ddot {r}}-r{\dot {\varphi }}^{2}=0}$

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {\varphi }}}}\right)-{\frac {\partial L}{\partial \varphi }}=0\qquad \Rightarrow \qquad {\ddot {\varphi }}+{\frac {2}{r}}{\dot {r}}{\dot {\varphi }}=0.}$

The solution of these two equations is given by

${\displaystyle r={\sqrt {(at+b)^{2}+c^{2}}}}$

${\displaystyle \varphi =\tan ^{-1}\left({\frac {at+b}{c}}\right)+d}$

for a set of constants a, b, c, d determined by initial conditions. Thus, indeed, the solution is a straight line given in polar coordinates: a is the velocity, c is the distance of the closest approach to the origin, and d is the angle of motion.

The action principle can be extended to obtain the equations of motion for fields, such as the electromagnetic field or gravity.

The Einstein equation utilizes the Einstein–Hilbert action as constrained by a variational principle.

The path of a body in a gravitational field (i.e. free fall in space time, a so-called geodesic) can be found using the action principle.

In quantum mechanics, the system does not follow a single path whose action is stationary, but the behavior of the system depends on all imaginable paths and the value of their action. The action corresponding to the various paths is used to calculate the path integral, that gives the probability amplitudes of the various outcomes.

Although equivalent in classical mechanics with Newton's laws, the action principle is better suited for generalizations and plays an important role in modern physics. Indeed, this principle is one of the great generalizations in physical science. In particular, it is fully appreciated and best understood within quantum mechanics. Richard Feynman's path integral formulation of quantum mechanics is based on a stationary-action principle, using path integrals. Maxwell's equations can be derived as conditions of stationary action.