Math and Physics Background

Position, Velocity, and Acceleration

For a particle of constant mass \(m\), its position at time \(t\) is \(\mathbf x(t):\mathbb R\rightarrow \mathbb R^3\) (Assuming 3D space), its velocity \(\mathbf v(t) = \frac{d\mathbf x}{dt}(t)\) and its acceleration \(\mathbf a(t) = \frac{d\mathbf v}{dt}(t) = \frac{d^2\mathbf x}{dt^2}(t)\)

Newton's Law

Every object will remain at rest of in uniform motion in a straight line unless compelled to change its state by the action of an external force
The force acting on ab object is equal to the time rate-of-change of the momentum

\[f = \underset{\text{time rate of change}}{\frac{d}{dt}}\underset{\text{momentum}}{m\mathbf v}= m\frac{d\mathbf v}{dt} = m\mathbf a\]
For every action there is an equal and opposite reaction

Variational (Analytical) Mechanics

Based on fundamental energies rather than two vectorial quantities

Kinetic Energy energy due to motion
Potential Energy energy "held within" an object due to position, internal stresses, charge, etc. Potential energy has the potential to become kinetic energy

Problem Settings

Let the image of \(\mathbf f(t)\) be the path of the motion, setting some functional

\[e(\mathbf f(t), \dot{\mathbf f}(t), ...) \rightarrow \mathbb R\]

and then solve (optimize) the variational derivative to find the proper \(\mathbf f\)

Generalized Coordinates

Let the generalized coordinates be \(\mathbf q: \mathbb R \rightarrow \mathbb R^n\) be the actual variable parametrize the motion

\[\mathbf x(t) = \mathbf f(\mathbf q(t))\]

\[\frac{d\mathbf x}{dt}(t) = \underset{\text{Jacobian}}{\frac{d\mathbf f}{d\mathbf q}}\underset{\text{generalized velocity}}{\dot q(t)}\]

Lagrangian

define the Lagrangian be

\[L = T - V = \text{kinetic} - \text{potential}\]

The Principle of Least Action

Assume the end points are known, to find the path between them by finding a stationary point of the action.

Action

For constant time \(t_1, t_2\), the action is a functional equals the integral of the Lagrangian over

\[S(\mathbf q(t), \dot{\mathbf q}(t)) = \int_{t_1}^{t_2} T(\mathbf q(t), \dot{\mathbf q}(t)) - V(\mathbf q(t), \dot{\mathbf q}(t)) = dt = \int_{t_1}^{t_2}L(\mathbf q(t), \dot{\mathbf q}(t))dt\]

Stationary Point

\[S(\mathbf q + \delta\mathbf q, \dot{\mathbf q}+\delta \dot{\mathbf q}) = S(\mathbf q(t), \dot{\mathbf q}(t))\]

which means that perturbation \(\delta\) of the trajectory \(\mathbf q\) does not change the action.

The Calculus of Variations (Euler Lagrange Equations)

\[\begin{align*} S(\mathbf q + \delta\mathbf q, \dot{\mathbf q}+\delta \dot{\mathbf q}) &=\int_{t_1}^{t_2}L(\mathbf q + \delta\mathbf q, \dot{\mathbf q}+\delta \dot{\mathbf q})dt\\ &\approx \int_{t_1}^{t_2}L(\mathbf q, \dot{\mathbf q})dt + \int_{t_1}^{t_2}\frac{\partial L}{\partial\mathbf q}\delta\mathbf q + \frac{\partial L}{\partial\dot{\mathbf q}}\delta\dot{\mathbf q }dt\\ &= S(\mathbf q, \dot{\mathbf q}) + \underset{\text{first variation}}{\delta S(\mathbf q, \dot{\mathbf q})} \end{align*}\]

Therefore, principle of least action becomes

\[\delta S(\mathbf q, \dot{\mathbf q}) = 0\]

Then, use integration by parts (see APM462 notes for detailed steps)

\[\int_{t_1}^{t_2}\frac{\partial L}{\partial\mathbf q}\delta\mathbf q + \frac{\partial L}{\partial\dot{\mathbf q}}\delta\dot{\mathbf q }dt = \int_{t_1}^{t_2}(\frac{\partial L}{\partial \mathbf q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{\mathbf q}})\delta \mathbf q dt + \underset{\text{known end points so }=0}{\frac{\partial L}{\partial \dot{\mathbf q}}\delta \mathbf q \vert_{t_1}^{t_2}}\]

Therefore, we have the Euler Lagrange equation (The lemmas are in APM462 notes)

\[\frac{\partial L}{\partial \mathbf q} = \frac{d}{dt}\frac{\partial L}{\partial \dot{\mathbf q}}\]