Lagrangian Mechanics

A point with minimal action is $S$ defined through the condition $$\begin{equation} \left. \frac{d}{ds}S(q_s)\right|_{s=0} = 0 \end{equation}$$ where $$q_s = q + s\delta q$$ for all $\delta q \colon [t_0,t_1]\rightarrow\mathbb{R}^n$ with $$\delta q(t_0)=\delta q(t_1)=0.$$ We want to derive $$\begin{equation} F=ma \quad \Leftrightarrow \quad \left.\frac{d}{ds} S(q_s)\right|_{s=0}=0 \; \textrm{for all} \; \delta q\colon [t_0,t_1]\rightarrow\mathbb{R}^n \; \textrm{with}\; \delta q(t_0)=\delta q(t_1)=0. \end{equation}$$ To do this, we now use the definition of the action and the chain rule: $$\begin{split} \left.\frac{d}{ds}S(q_s)\right|_{s=0} &= \frac{d}{ds}\left.\int_{t_0}^{t_1} \frac{1}{2}m\dot{q}_s(t)\cdot\dot{q}_s(t)-V(q_s(t))\,dt\right|_{s=0} \\ \\ &= \left.\int_{t_0}^{t_1}\frac{d}{ds}\left[ \frac{1}{2}m\dot{q}_s(t) \cdot \dot{q}_s(t) - V(q_s(t))\right]dt\right|_{s=0} \\ \\ &= \left.\int_{t_0}^{t_1}\left[ m\dot{q}_s \cdot \frac{d}{ds}\dot{q}_s(t)- \nabla V(q_s(t)) \cdot \frac{d}{ds}q_s(t)\right]dt\right|_{s=0} \\ \end{split}$$ Next we note that $$\frac{d}{ds} q_s (t) = \delta q(t)$$ so $$\frac{d}{ds} \dot{q}_s (t) = \frac{d}{ds} \frac{d}{dt} {q}_s (t) = \frac{d}{dt} \frac{d}{ds} {q}_s (t) = \frac{d}{dt} \delta{q}(t) .$$ Therefore we have $$\begin{split} \left.\frac{d}{ds}S(q_s)\right|_{s=0} &= \int_{t_0}^{t_1}\left[ m\dot{q} \cdot \frac{d}{dt} \delta q(t) - \nabla V(q(t)) \cdot \delta q(t)\right]dt . \end{split}$$ Finally, we can integrate by parts and note that the boundary terms vanish because $\delta q=0$ at $t_1$ and $t_0$: $$\begin{split} \frac{d}{ds}S(q_s)\rvert_{s=0} &= \left.\int_{t_0}^{t_1}\left[ -m\ddot{q}(t)-\nabla V(q(t))\right] \cdot \delta q(t) dt\right. . \\ \end{split}$$ Looking at this, we can see that the variation of the action vanishes for all possible variations $\delta q$ if and only if the term in brackets is exactly zero: $$-m\ddot{q}(t)-\nabla V(q(t)) = 0.$$ Therefore, the correct path $q$ with a minimal amount of action $S$ is the one that follows from the equation $$F=ma.$$

The Lagrangian in classical mechanics is given by the kinetic energy $\cal T$ minus the potential energy $\cal U$:

$$\cal L=T-U .$$

Therefore, a harmonic oscillator in the Lagrangian framework is characterized by the action

$$S = \int dt \left( \frac{m}{2} \left(\frac{dx}{dt}\right) - \frac{k}{2}x^2 \right) ,$$ where $L= \frac{m}{2} \left(\frac{dx}{dt}\right) - \frac{k}{2}x^2$ is the Lagrangian.

Starting from this action, we can derive the equation of motion by using the Euler-Lagrange equation.

$$\frac{\partial {\cal L}}{\partial x}-\frac{d}{dt}(\frac{\partial {\cal L}}{\partial dot{x}})= 0$$

$$\Rightarrow m \frac{d^2}{dt^2} x -kx= 0$$

The Lagrangian for the pendulum can be written $${\cal L=T-U}=\frac{mr^2\dot{\theta}^2}{2}+mgr\cos\theta$$ The equation of motion is $$\frac{\partial {\cal L}}{\partial \theta}-\frac{d}{dt}(\frac{\partial {\cal L}}{\partial dot{\theta}})= -mgr\sin\theta-mr^2\frac{d\dot\theta}{dt}=0$$ giving: $$\ddot{\theta}+\frac{g}{r}\sin\theta=0$$

Small Amplitude Approximation

If $\theta <<1$ then $\sin\theta\approx\theta$ and $$\frac{d^2\theta}{dt^2}+\frac{g}{r}\theta$$ This solution describes simple harmonic motion $$\theta=A\cos(\omega t-\phi)$$ where $$\omega=\sqrt{\frac{g}{r}}$$ and the constants $A$ and $\phi$ are determined from the initial condition E.g. if $\theta=0,\;t=0$ then $\phi=\pm \frac{\pi}{2}$. If $d\theta/dt=0,\;t=0$ then $\phi=0,$ or $\pi$ (depending on whether the pendulum is moving to the left or right initially) The period is $$\tau=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{r}{g}}$$

Finite Amplitude

If the amplitude is not small we have to solve the nonlinear equation $$\frac{d^2\theta}{dt^2}+\frac{g}{r}\sin \theta=0$$ (nonlinear because sin $\theta$ is a nonlinear function of the dependent variable $\theta$.)

A frictionless pulley with two masses, $m_1$ and $m_2$, hanging from it.

We have $$\begin{align*} K &= \frac{1}{2}(m_1+m_2)(\frac{d}{dt}(\ell-x))^2 = \frac{1}{2}(m_1+m_2)\dot{x}^2 \\ V &= -m_1 g x - m_2 g(\ell-x) \\ \text{so } L &= K-V = \frac{1}{2}(m_1+m_2)\dot{x}^2 + m_1gx+m_2g(\ell-x) \end{align*}$$ The configuration space is $Q=(0,\ell)$, and $x\in(0,\ell)$. Moreover $TQ=(0,\ell)\times\mathbb{R}\ni(x,\dot{x})$. As usual $L:TQ\rightarrow\mathbb{R}$. Note that solutions of the Euler-Lagrange equations will only be defined for \emph{some} time $t\in\mathbb{R}$, as eventually the solutions reaches the ``edge'' of $Q$.

The momentum is: $$p = \frac{\partial L}{\partial\dot{x}} = (m_1+m_2)\dot{x}$$ and the force is: $$F = \frac{\partial L}{\partial x} = (m_1-m_2)g$$ The Euler-Lagrange equations say $$\begin{align*} \dot{p} &= F \\ (m_1+m_2)\ddot{x} &= (m_1-m_2)g \\ \ddot{x} &= \frac{m_1-m_2}{m_1+m_2}g \end{align*}$$ So this is like a falling object in a downwards gravitational acceleration $a=\left(\frac{m_1-m_2}{m_1+m_2}\right)g$.

We integrate the expression for $\ddot{x}$ twice to obtain the general solution to the motion $x(t)$. Note that $\ddot{x}=0$ when $m_1=m_2$, and $\ddot{x}=g$ if $m_2=0$.