I thought this video was really fun and wrote up some of the derivations that the video went over quickly.

Introduction and Basic Principles

Maupertuis’ principle of least action states that the action, defined as:

\[ S_0 = \sum mvs \]

where $m$ is the mass, $v$ is the velocity, and $s$ is the distance, reaches a minimum along the actual path of motion.

Euler later generalized this to a continuous form:

\[ S = \int mv ds\]

This formulation holds only for systems with constant energy.

Consider a true path $x(t)$ and a nearby path $x(t) + \eta(t)$, where $\eta(t)$ is a small perturbation. Similar to how, at the minimum, of a parabola if you move a small distance the value of the function doesn’t change much, you can show that the actions of these two paths are the same up to first order in $\eta$.

As with any minimum, the first-order variation in the action vanishes:

\[ \delta S = 0 \]

where $\delta S$ represents the first variation of the action functional (some theorem in calculus of variations).

Full Derivation of Lagrangian Equation

Starting from

\[ \delta S = 0\] with \[ S = \int mv ds\] , we get

\[ \delta \int mv ds = 0.\]

Using $v = \frac{ds}{dt}$,

\[ \delta \int m v^2 dt = 0.\]

Since kinetic energy is $T = \frac{1}{2} mv^2$,

\[ \delta \int 2T dt = 0.\]

We also know that the total energy $E = T + V$ which means

\[ \delta \int (T - V + E) dt = 0.\]

which splits into

\[ \delta \int (T - V) dt + \delta \int E dt = 0.\]

Since energy is constant,

\[ \delta \int (T - V) dt + \delta(Et)= 0.\]

Applying the variation product rule:

\[ \delta \int (T - V) dt + E \delta t + t \delta E = 0.\]

Euler showed that:

  1. Energy is constant along each path ($t \delta E = 0$)
  2. We only consider paths with equal duration ($E \delta t = 0$)

Therefore:

\[ \delta \int (T - V) dt = 0.\]

This is Hamilton’s principle, where $L = T - V$ is the Lagrangian:

\[ \delta \int L dt = 0.\]

Deriving $F=ma$

Consider a path $y(t)$ with perturbation $\eta(t)$, giving $q(t) = y(t) + \eta(t)$. The action variation is:

\[ \delta S = S[q(t)] - S[y(t)] \]

Computing the action for both paths:

\[\begin{align*} S[q(t)] = \int_{t_1}^{t_2} \left( \frac{1}{2} m \left( \frac{dy}{dt} \right)^2 - V(y) \right) \, dt + \int_{t_1}^{t_2} \left( m \frac{dy}{dt} \frac{d\eta}{dt} - \eta V'(y) \right) \, dt \end{align*}\]

\[ S[y(t)] = \int_{t_1}^{t_2} \left( \frac{1}{2} m \left( \frac{dy}{dt} \right)^2 - V(y) \right) \, dt \]

After cancellation, we get

\[ \delta S = \int_{t_1}^{t_2} \left( m \frac{dy}{dt} \frac{d\eta}{dt} - \eta V’(y) \right) \, dt \] which doing some integration by parts (also using something called variation must vanish at endpoints) gives us

\[\delta S = \int_{t_1}^{t_2} \left( -m \frac{d^2y}{dt^2} - \frac{dV}{dy} \right) \eta dt = 0\]

Since $\eta$ is arbitrary, we must have:

\[m \frac{d^2y}{dt^2} = -\frac{dV}{dy}\]

which, voila, is just $F = ma$ which is Newton’s second law.

Show how Euler-Lagrange’s Equation implies Fermat’s Principle of Least Time

todo

Write out the Lagrangian for a double pendulum

todo