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CM.4 · Constraints, variational equations and cyclic coordinates

GRE · GRE Subject Test · GRE 物理 · 知识点 26

训练
26.1

约束、变分方程与循环坐标

A pendulum tension changes throughout a swing, but its angular equation can be derived without solving for tension.

Prerequisites: 1, 47.

  • Derive Euler–Lagrange motion after imposing a holonomic constraint 完整约束
  • Use a cyclic coordinate 循环坐标 to identify conserved canonical momentum
  • Linearise a stable equilibrium and check the resulting frequency
词汇 训练
English 中文 拼音
holonomic constraint/ˌhɒləˈnɒmɪk kənˈstreɪnt/ 完整约束 wán zhěng yuē shù
cyclic coordinate/ˈsaɪklɪk kəʊˈɔːdɪnət/ 循环坐标 xún huán zuò biāo
26.2

Choose independent coordinates

A constraint removes an independent coordinate before you form the kinetic energy. For a fixed-length pendulum choose θ from the downward vertical: x=l sinθ and y=−l cosθ. Differentiating both coordinates gives v²=l²θdot², not l² sin²θ θdot². Thus T=ml²θdot²/2 and V=mgl(1−cosθ), with zero potential at the bottom. The fixed length is a holonomic constraint: it is a relation among coordinates and possibly time. A rolling velocity constraint needs its own analysis; do not automatically treat every constraint as a coordinate substitution.

26.3

Derive the equation

For ideal constraints and the usual conservative system, the stationary-action equation is d/dt(∂L/∂qdot)−∂L/∂q=0 with L=T−V. The variations vanish at the two time endpoints. A pendulum gives ∂L/∂θdot=ml²θdot and ∂L/∂θ=−mgl sinθ, so ml²θddot+mgl sinθ=0. The momentum derivative acts on every time-dependent factor: for a varying radius, d(mr²θdot)/dt contains 2mr rdot θdot. The variational method does not mean mechanical energy is conserved in a time-dependent system.

26.4

Find cyclic momentum

For planar central motion, L=m(rdot²+r²θdot²)/2−V(r). The angle θ is cyclic because L has no explicit θ dependence, even though L depends on θdot. Its canonical momentum pθ=mr²θdot is constant. The radial equation is m rddot=mrθdot²−dV/dr. The first term is part of the coordinate acceleration, not a new outward real force in an inertial frame. Conservation of pθ makes angular speed increase as r decreases; constant angular speed is a different, externally driven situation.

26.5

Linearise with conditions

Near a stable equilibrium q0, expand V to second order and use a constant local inertia M: V≈V(q0)+V″(q0)(q−q0)²/2. Then the small-displacement frequency is sqrt(V″/M). For a pendulum sinθ≈θ in radians, so ω=sqrt(g/l); the approximation requires small amplitude. A negative V″ gives instability, not an oscillation with an imaginary measurable frequency. Check the full nonlinear equation first, then state the approximation and compare units: V″/M must have units of inverse time squared.

26.6

Worked method

For a fixed pendulum length l, use theta from downward vertical: x = l sin theta, y = -l cos theta. Both Cartesian velocities give $T=ml^2\dot\theta^2/2$ and $V=mgl(1-\cos\theta)$. The Euler-Lagrange equation 欧拉—拉格朗日方程 is

$$\frac d{dt}(ml^2\dot\theta)+mgl\sin\theta=0.$$
Divide by $ml^2$ before linearising:
$$\ddot\theta+(g/l)\sin\theta=0,\qquad\omega=\sqrt{g/l}\quad(|\theta|\ll1).$$
The approximation uses radians. Endpoint variations vanish in the stationary-action derivation.

Constraints, variational equations and cyclic coordinates: GRE original diagram
Constraints, variational equations and cyclic coordinates: original GRE teaching diagram.
词汇 训练
English 中文 拼音
Euler-Lagrange equation 欧拉—拉格朗日方程 ōu lā — lā gé lǎng rì fāng chéng
26.7

Check conditions and vocabulary

A cyclic coordinate is absent from L itself; its velocity need not be absent. Do not discard the time derivative of a changing r² factor.

holonomic constraint: A constraint expressible as a relation among coordinates and possibly time.

cyclic coordinate: A coordinate absent explicitly from the Lagrangian, with conserved canonical momentum under the Euler–Lagrange equation.

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