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 pendulum

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