Variational Mechanics and Symmetries

An action principle reorganizes Newtonian dynamics by selecting the same trajectories as $m\mathbf a=\mathbf F$ whenever the forces admit a regular Lagrangian. It does not change the predictions in that domain. Its power is structural: constraints can be built into coordinates, equivalent coordinate descriptions are handled uniformly, and continuous symmetries expose conserved quantities.

Required background. Calculus and Linear Algebra provides multivariable differentiation, linear maps, inner products, and basis changes. Helpful background. Ordinary Differential Equations provides initial-value problems, linearization, and asymptotic solution methods.

Stationary action and the Euler–Lagrange equations

For generalized coordinates $q^i$ and a Lagrangian $L(q,\dot q,t)$, define

$$S[q]=\int_{t_1}^{t_2}L(q,\dot q,t)\,dt.$$

Vary the path as $q^i\mapsto q^i+\epsilon\eta^i$, holding its endpoints fixed. To first order,

$$\delta S=\int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q^i}\eta^i +\frac{\partial L}{\partial\dot q^i}\dot\eta^i \right)dt.$$

Integration by parts moves the derivative off $\eta^i$. The endpoint term vanishes because $\eta^i(t_1)=\eta^i(t_2)=0$, leaving

$$\delta S=\int_{t_1}^{t_2} \left[ \frac{\partial L}{\partial q^i} -\frac{d}{dt}\left(\frac{\partial L}{\partial\dot q^i}\right) \right]\eta^i\,dt.$$

Arbitrary interior variations therefore give

$$\frac{d}{dt}\frac{\partial L}{\partial\dot q^i} -\frac{\partial L}{\partial q^i}=0.$$

For Cartesian particles with $L=\sum_a m_a\dot{\mathbf x}_a^2/2-V$, this becomes $m_a\ddot{\mathbf x}_a=-\nabla_aV$: exactly Newton’s equation for conservative forces. This endpoint-fixed derivation and its assumptions are given in Tong 2005, §2.1, pp. 10–12, PDF. The action is stationary, not necessarily minimal.

The same result follows from virtual work: for conservative generalized forces $Q_i=-\partial_iV$, d’Alembert’s condition $(d/dt)(\partial T/\partial\dot q^i)-\partial T/\partial q^i=Q_i$ is precisely the Euler–Lagrange equation for $L=T-V$. This independently checks equivalence with Newtonian mechanics.

Stationarity is a first-variation statement. Whether a stationary path minimizes or saddles the action also depends on the interval and endpoint conditions. A dynamically stable harmonic oscillator supplies a useful counterexample to “least action.” For $L=m(\dot q^2-\omega^2q^2)/2$, vary the equilibrium path $q=0$ on $0\le t\le T$ by $q=\epsilon a\sin(\pi t/T)$, where $a$ is a fixed length. The endpoints remain fixed, but direct integration gives

$$S[\epsilon a\sin(\pi t/T)]-S[0] =\frac{\epsilon^2ma^2T}{4} \left(\frac{\pi^2}{T^2}-\omega^2\right).$$

For $T\gt\pi/\omega$ this variation lowers the action; sufficiently rapid sine variations raise it. The equilibrium is therefore a saddle of this fixed-endpoint action even though small dynamical perturbations remain bounded. Infer the equations from $\delta S=0$ and test dynamical stability separately.

Symmetry and conserved quantities

Suppose $\delta q^i=\epsilon\xi^i(q,t)$ changes the Lagrangian by a total derivative, $\delta L=\epsilon\,dF/dt$. The first-variation identity along an Euler–Lagrange trajectory is

$$\frac{\delta L}{\epsilon} =\frac{d}{dt}\left(p_i\xi^i\right).$$

Comparing the two total derivatives gives

$$Q=p_i\xi^i-F, \qquad p_i=\frac{\partial L}{\partial\dot q^i}, \qquad \frac{dQ}{dt}=0.$$

Thus the general boundary term $F$ follows directly from the on-shell variation. Exact invariance is the specialization $F=0$; Tong 2005, §2.4, pp. 23–26, PDF derives that case for spatial translations and rotations. A coordinate absent from $L$ is cyclic, so its conjugate momentum is constant.

Time translation also shifts the independent variable, so it is not one of the fixed-time variations just used. Its conserved quantity follows directly instead: along an Euler–Lagrange trajectory,

$$\frac{d}{dt}\left(p_i\dot q^i-L\right)=-\frac{\partial L}{\partial t}.$$

Therefore a time-independent Lagrangian conserves the energy $E=p_i\dot q^i-L$, as derived separately in Tong 2005, §2.4, p. 26, PDF.

If the velocity Hessian $W_{ij}=\partial^2L/(\partial\dot q^i\partial\dot q^j)$ is invertible, the Legendre transform

$$H(q,p,t)=p_i\dot q^i-L$$

gives Hamilton’s equations $\dot q^i=\partial H/\partial p_i$ and $\dot p_i=-\partial H/\partial q^i$; see Tong 2005, §4.1, pp. 80–88, PDF. This is another description, not another physical law. A singular Hessian signals constraints and forbids naïve inversion.

The distinction between physical motion and its coordinate representation is especially useful here. A regular change of generalized coordinates alters the component form of the Euler–Lagrange equations but not their solution curves on configuration space. By contrast, adding a total derivative changes the canonical momenta even though it leaves those curves fixed. Canonical quantities must therefore be interpreted together with the chosen Lagrangian, while the configuration-space trajectory is common to the equivalent descriptions. The Hamiltonian also need not equal physical energy when $L$ depends explicitly on time; time-translation symmetry, rather than the name “Hamiltonian,” is what supplies a conserved energy.

Checks and limits

• $L$ has energy dimension and $S$ energy times time; both Euler–Lagrange terms have generalized-force dimension.
• The sign $L=T-V$ gives $m\ddot{\mathbf x}=-\nabla V$.
• Adding $dK(q,t)/dt$ changes only the fixed-endpoint boundary term and leaves the trajectories unchanged, although canonical momenta shift.
• For time-independent $V$ and quadratic $T$, the Hamiltonian is $H=T+V$.

Dry friction, impacts, and general velocity-dependent forces need additional structure; not every force is generated by an ordinary $L(q,\dot q,t)$. Degenerate Lagrangians require constrained mechanics.

Exercise: a cyclic angle

A mass $m$ moves in a plane under $V(r)$. Use the Lagrangian to obtain angular-momentum conservation and the radial equation.

Solution

With

$$L=\frac m2(\dot r^2+r^2\dot\phi^2)-V(r),$$

$\phi$ is cyclic, so $\ell=mr^2\dot\phi$ is constant. The radial equation is

$$m\ddot r=mr\dot\phi^2-V'(r) =\frac{\ell^2}{mr^3}-V'(r).$$

Equivalently, $m\ddot r=-V_{\rm eff}'(r)$ with $V_{\rm eff}=V+\ell^2/(2mr^2)$. The centrifugal term has force dimension and disappears smoothly as $\ell\to0$.

The Newtonian Gravitational Field and Potential now supplies the central conservative interaction used throughout gravitation.

References

• Tong, David. Classical Dynamics. University of Cambridge lecture notes, 2004/2005. §§2.1 and 2.4, pp. 10–12 and 23–26; §4.1, pp. 80–88, PDF.