24 Jul 2025
Variational Principles, Constraints, and Least Action
Euler-Lagrange calculus, D'Alembert and modified Hamilton principles, multipliers, and constrained examples.
Calculus of variations and the Euler-Lagrange equation
Consider the functional
\[J[y]=\int_{x_1}^{x_2}F(x,y,y^{\prime})\,dx.\]Vary the curve as $y(x,\epsilon)=y(x)+\epsilon\eta(x)$ while holding its endpoints fixed:
\[\eta(x_1)=\eta(x_2)=0.\]The first variation is
\[\delta J =\int_{x_1}^{x_2}\left( \frac{\partial F}{\partial y}\eta +\frac{\partial F}{\partial y^{\prime}}\eta^{\prime} \right)dx.\]After integration by parts,
\[\delta J =\left[\frac{\partial F}{\partial y^{\prime}}\eta\right]_{x_1}^{x_2} +\int_{x_1}^{x_2}\left[ \frac{\partial F}{\partial y} -\frac{d}{dx}\left(\frac{\partial F}{\partial y^{\prime}}\right) \right]\eta\,dx.\]The boundary term vanishes because the endpoints are fixed. Since $\eta$ is otherwise arbitrary, stationarity $\delta J=0$ requires
\[\boxed{ \frac{d}{dx}\left(\frac{\partial F}{\partial y^{\prime}}\right) -\frac{\partial F}{\partial y}=0}.\]For several dependent variables, the same equation holds for each one.
Hamilton’s principle from D’Alembert’s principle
D’Alembert’s equation in generalized coordinates is
\[\sum_j\left[ Q_j-\frac{d}{dt}\frac{\partial T}{\partial\dot q_j} +\frac{\partial T}{\partial q_j} \right]\delta q_j=0.\]For conservative forces $Q_j=-\partial V/\partial q_j$ and $L=T-V$, this becomes
\[\sum_j\left[ \frac{\partial L}{\partial q_j} -\frac{d}{dt}\frac{\partial L}{\partial\dot q_j} \right]\delta q_j=0.\]Integrate from $t_1$ to $t_2$. If the comparison paths have the same configurations at both endpoints,
\[\delta q_j(t_1)=\delta q_j(t_2)=0,\]then integrating D’Alembert’s equation and reversing the time derivative by parts gives
\[\begin{aligned} 0 &=\int_{t_1}^{t_2}\sum_j\left[ \frac{\partial L}{\partial q_j} -\frac{d}{dt}\left(\frac{\partial L}{\partial\dot q_j}\right) \right]\delta q_j\,dt\\ &=\int_{t_1}^{t_2}\sum_j\left[ \frac{\partial L}{\partial q_j}\delta q_j +\frac{\partial L}{\partial\dot q_j}\delta\dot q_j \right]dt -\left[\sum_j\frac{\partial L}{\partial\dot q_j}\delta q_j\right]_{t_1}^{t_2}. \end{aligned}\]The boundary term vanishes, and the remaining integral is the first variation of the action. Hence
\[\boxed{ \delta S=0, \qquad S=\int_{t_1}^{t_2}L\,dt}.\]Thus Hamilton’s principle follows from D’Alembert’s principle for an ideal constrained conservative system.
Modified Hamilton principle
In phase space, regard $q_i(t)$ and $p_i(t)$ as independent and vary
\[S_H=\int_{t_1}^{t_2}\left(\sum_i p_i\dot q_i-H(q,p,t)\right)dt.\]Only the coordinate variations are fixed at the endpoints:
\[\delta q_i(t_1)=\delta q_i(t_2)=0;\]$\delta p_i$ need not vanish there. Direct variation and integration of $p_i\delta\dot q_i$ by parts give
\[\delta S_H =\left[\sum_i p_i\delta q_i\right]_{t_1}^{t_2} +\int_{t_1}^{t_2}\sum_i\left[ \left(\dot q_i-\frac{\partial H}{\partial p_i}\right)\delta p_i -\left(\dot p_i+\frac{\partial H}{\partial q_i}\right)\delta q_i \right]dt.\]The endpoint term is zero. Independent interior variations give both Hamilton equations.
Lagrange’s method of undetermined multipliers
For holonomic constraints
\[f_\alpha(q,t)=0, \qquad \alpha=1,\ldots,k,\]introduce multipliers $\lambda_\alpha(t)$ and the augmented Lagrangian
\[L_{\mathrm a}=L+\sum_\alpha\lambda_\alpha f_\alpha.\]Variation with respect to $q_i$ and $\lambda_\alpha$ yields
\[\boxed{ \frac{d}{dt}\frac{\partial L}{\partial\dot q_i} -\frac{\partial L}{\partial q_i} =\sum_\alpha\lambda_\alpha \frac{\partial f_\alpha}{\partial q_i}}, \qquad \boxed{f_\alpha=0}.\]The multiplier terms are the generalized constraint forces. The sign of a multiplier depends on whether $+\lambda_\alpha f_\alpha$ or $-\lambda_\alpha f_\alpha$ is chosen; physical forces do not depend on that convention.
Simple pendulum by a multiplier
Use Cartesian coordinates with $y$ positive upward:
\[L=\frac m2(\dot x^2+\dot y^2)-mgy, \qquad f=x^2+y^2-l^2=0.\]The multiplier equations are
\[m\ddot x=2\lambda x, \qquad m\ddot y=-mg+2\lambda y.\]Write $x=l\sin\theta$ and $y=-l\cos\theta$, with $\theta=0$ at the lowest point. Eliminating $\lambda$ gives
\[\boxed{\ddot\theta+\frac gl\sin\theta=0}.\]Because the constraint force is $2\lambda(x,y)$ and the string tension points inward,
\[\boxed{T=-2\lambda l=m\left(l\dot\theta^2+g\cos\theta\right)}.\]Rolling hoop on an inclined plane
Let $s$ be distance measured down a plane of angle $\alpha$, and choose the sense of $\theta$ so rolling without slipping is
\[f=s-R\theta=0.\]For a body of mass $M$, radius $R$, and centre-of-mass moment of inertia $I$,
\[L=\frac12M\dot s^2+\frac12I\dot\theta^2+Mgs\sin\alpha.\]The multiplier equations are
\[M\ddot s-Mg\sin\alpha=\lambda, \qquad I\ddot\theta=-\lambda R.\]Using $\ddot s=R\ddot\theta$ gives
\[\boxed{ \ddot s=\frac{g\sin\alpha}{1+I/(MR^2)}}.\]For a thin hoop, $I=MR^2$, so
\[\boxed{\ddot s=\frac12g\sin\alpha},\]with units $\mathrm{m\,s^{-2}}$. The multiplier is negative for the chosen $s$ direction, corresponding to static friction acting up the plane.
Principle of least action
Hamilton’s action $S=\int L\,dt$ is stationary for paths with fixed times and fixed configuration endpoints. At fixed energy, Maupertuis’ form of the principle is
\[\boxed{ \delta W=0, \qquad W=\int_{q_1}^{q_2}\sum_i p_i\,dq_i}.\]The name “least action” is traditional. The required value is stationary and need not always be a minimum.
Solved Problems
1. Friction required for a hoop to roll without slipping
A thin hoop rolls down an incline of angle $\alpha$ without slipping. Find its acceleration, the static-friction force, and the minimum coefficient of static friction. Evaluate the acceleration and coefficient for $\alpha=30^\circ$.
Let $s$ increase down the plane. Static friction $f$ acts up the plane, so translation gives
\[Mg\sin\alpha-f=M\ddot s.\]The hoop has $I=MR^2$. Taking angular acceleration in the rolling sense as positive,
\[fR=I\ddot\theta, \qquad \ddot s=R\ddot\theta.\]Therefore
\[fR=MR^2\frac{\ddot s}{R}, \qquad f=M\ddot s.\]Substitution into the translational equation yields
\[2M\ddot s=Mg\sin\alpha,\]so
\[\boxed{\ddot s=\frac12g\sin\alpha}, \qquad \boxed{f=\frac12Mg\sin\alpha}.\]The normal reaction is $N=Mg\cos\alpha$. The no-slip condition $f\leq\mu_sN$ requires
\[\boxed{\mu_s\geq\frac12\tan\alpha}.\]At $\alpha=30^\circ$,
\[\ddot s=\frac12(9.81)\sin30^\circ =2.4525\ \mathrm{m\,s^{-2}},\] \[\mu_{s,\min}=\frac12\tan30^\circ=0.2887.\]Static friction supplies angular acceleration but does no work at the instantaneous point of contact. The result applies only while the inequality above is satisfied.
2. Tension in a pendulum released from rest
A simple pendulum of mass $m$, length $l$, and release angle $\theta_0$ starts from rest. Use the constraint-force result and energy conservation to obtain the tension at any later angle $\theta$. Then find the bottom tension for $m=0.20\ \mathrm{kg}$ and $\theta_0=60^\circ$.
With the lowest point as zero potential, energy conservation gives
\[\frac12ml^2\dot\theta^2+mgl(1-\cos\theta) =mgl(1-\cos\theta_0).\]Hence
\[l\dot\theta^2=2g(\cos\theta-\cos\theta_0).\]The inward radial equation, equivalently obtained from the multiplier, is
\[T-mg\cos\theta=ml\dot\theta^2.\]Eliminating $\dot\theta$ gives
\[\boxed{T(\theta)=mg(3\cos\theta-2\cos\theta_0)}.\]At the bottom, $\theta=0$, so
\[T_{\mathrm b}=mg(3-2\cos\theta_0) =(0.20)(9.81)(3-1) =\boxed{3.924\ \mathrm N}.\]At release, the same expression gives $T_0=mg\cos\theta_0$, as required because the initial radial speed is zero. The string can enforce the assumed constraint only where $T\geq0$.
Descriptive Questions
- Derive the Euler-Lagrange equation for a functional $J[y]=\int F(x,y,y^{\prime})\,dx$, including the endpoint conditions on admissible variations.
- Compare Hamilton’s configuration-space principle with the modified Hamilton principle in phase space, stating which endpoint variations vanish in each.
- Explain how Lagrange multipliers represent generalized constraint forces and why the sign of a multiplier depends on the chosen constraint convention.
- Distinguish Hamilton’s fixed-time stationary action from Maupertuis’ fixed-energy abbreviated action.
Numerical Problems
- For $J[y]=\tfrac12\int_0^1\big[(y^{\prime})^2+y^2\big]\,dx$ with $y(0)=0$ and $y(1)=1$, the extremal obeys $y^{\prime\prime}-y=0$. Find $y(0.5)$. Answer: $y(x)=\sinh x/\sinh(1)$, so $y(0.5)=0.4434$.
- A free particle of mass $3.0\ \mathrm{kg}$ follows the stationary path from $x=0$ at $t=0$ to $x=6.0\ \mathrm m$ at $t=2.0\ \mathrm s$. Find its speed and action $S=\int L\,dt$. Answer: $3.000\ \mathrm{m\,s^{-1}}$, $27.00\ \mathrm{J\,s}$.
- The phase-space action has $H=p^2/(2m)+m\omega^2q^2/2$. For $m=2.0\ \mathrm{kg}$, $\omega=3.0\ \mathrm{s^{-1}}$, $q(0)=0$, and $p(0)=m\omega A$ with $A=0.40\ \mathrm m$, find $t$, $q$, and $p$ after one-quarter period. Answer: $0.5236\ \mathrm s$, $0.4000\ \mathrm m$, $0$.
- Linearize the multiplier equation for a simple pendulum of length $0.75\ \mathrm m$ and find its small-oscillation period for $g=9.81\ \mathrm{m\,s^{-2}}$. Answer: $1.737\ \mathrm s$.
Maxima verification: rolling and pendulum residuals, with all numerical values.
References
- Calculus of variations — Wikipedia
- H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Chapters 2 and 8, Pearson (2002).
- L. D. Landau and E. M. Lifshitz, Mechanics, 3rd ed., Sections 2 and 44, Butterworth-Heinemann (1976).
- I. M. Gelfand and S. V. Fomin, Calculus of Variations, Chapters 1–2, Dover (2000).
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