1 Lagrangian Mechanics
Lagrangian mechanics is a mathematical framework for studying mechanical systems similar to [[newtonian-mechanics]] and Hamiltonian Mechanics. The foundational idea underpinning Lagrangian mechanics is the simplification of problems to a scalar representation of a physical system’s position in phase space.
## Motivation To motivate this idea recall from [[newtonian-mechanics]] that we can describe a system of \(N\) point masses \(m_{1}, \dots, m_{N}\) with 3-dimensional Cartesian coordinates \(\vec{r}_{i}:=(x_{i},y_{i},z_{i})\) for \(i=1, \dots, N\) requiring \(3N\) coordinates. The velocity of each particle is given by the time derivative \(\vec{v}_{i}=\dot{\vec{r}}_{i}=d\vec{r}_{i} /dt\). From Newton’s second law we have \[
F=m \vec{a}=m\ddot{ \vec{r}},
\] thus for an \(N\)-particle system in 3-dimensions we have \(3N\) second order Ordinary Differential Equations in the positions of the particles to solve for. Lagrangian mechanics avoids forces and instead considers energies in the system with the Lagrangian summarizing the dynamics of the entire system. Since, in the absence of an electromagnetic field, the non-relativistic Lagrangian equation is given by the difference of kinetic energy \(K\) and potential energy \(U\) \[
\mathcal{L}=K-U.
\] We note that kinetic energy is given by \[
K=\frac{1}{2}\sum_{k=1}^{N}m_{k}v^2_{k}
\] where \(v_{k}^2=\vec{v}_{k}\cdot \vec{v}_{k}\) and is therefore a function of only velocities and not positions. Similarly, potential energy \(U\) is only a function of position for conservative forces (e.g. Newton’s gravity).
1.0.1 Intuitive Example - Defining the Lagrangian
To gain intuition for the system let us consider the introductory example of a pendulum on a string with no resistance. Figure: Diagram of a swinging pendulum with no friction.
The kinetic and potential energy of the system are \[ \begin{align} K & =\frac{1}{2}mv^2=\frac{1}{2}m\dot{s}^2=\frac{1}{2}ml^2\dot{\theta}^2 \\ U & = mgy= mgl\cos\theta, \end{align} \] where we have used \(\theta=\frac{s}{l}\) to rewrite our derivatives in terms of angle \(\theta\), thus the Lagrangian \(\mathcal{L}\) is \[ \mathcal{L}=\frac{1}{2}ml^2\dot{\theta}^2+mgl\cos \theta. \] Lagrangian Euler-Lagrange Equation
We are thus able to write the Euler-Lagrange equation which is given by \[ \frac{d}{dt} \frac{\partial\mathcal{L}}{\partial \dot{\theta}}= \frac{{\partial L}}{\partial \theta}. \] We can compute \[ \begin{align} \frac{{\partial \mathcal{L}}}{\partial \theta} & =-mgl\sin \theta =\mathcal{F}\\ \frac{\partial\mathcal{L}}{\partial \dot{\theta}} & =ml^2\dot{\theta}=\mathcal{P}\implies \frac{d}{dt} \frac{\partial\mathcal{L}}{\partial \dot{\theta}} =ml^2\ddot{\theta}, \end{align} \] where \(\mathcal{P}\) is generalized momentum at the point and \(\mathcal{F}\) the generalized force, thus we see that the Euler-Lagrange equation gives the familiar Newtonian force is equal to change in momentum.