PX285 - F3b - pendulum

considering a simple pendulum of length, $l$ , and $θ$ is the angle between the pendulum and the equilibrium position at a given time
writing down the lagrangian, $L$ , and using it to calculate the hamiltonian, $H$ and identifying the momentum:

\begin{aligned} T & = \frac{1}{2} I {\dot{θ}}^{2} \\ \frac{1}{2} m v^{2} & = \frac{1}{2} m l^{2} {\dot{θ}}^{2} \\ ⟹ v & = l \dot{θ} \\ V & = - m g l \cos θ \\ ∴ L & = \frac{1}{2} m l^{2} {\dot{θ}}^{2} + m g l \cos θ \\ (1) & p & = \frac{\partial L}{\partial \dot{θ}} = m l^{2} \dot{θ} \end{aligned}

note: here, $p \neq m v$ because this is the angular momentum
the definition of $H$ from first principles:

H = - L + p \dot{θ} = - L + m l^{2} {\dot{θ}}^{2}

substituting for $L :$

H = \frac{1}{2} m l^{2} {\dot{θ}}^{2} - m g l \cos θ

substituting using $(1) :$

H (θ, p) = \frac{1}{2} m l^{2} {(\frac{p}{m l^{2}})}^{2} - m g l \cos θ = \frac{p^{2}}{2 m l^{2}} - m g l \cos θ

using hamilton's equations:

\begin{aligned} (2) & \dot{θ} & = \frac{p}{m l^{2}} \\ (3) & \dot{p} & = - m g l \sin θ \end{aligned}

if $θ << 1 :$

\dot{p} \approx - m g l θ

on phase plane:

\begin{aligned} d θ & = \frac{p}{m l^{2}} d t \\ d p & = - m g l \sin θ d t \end{aligned}

deriving the equations of motion:

\ddot{θ} = \frac{\dot{p}}{m l^{2}} = \frac{- m g l}{m l^{2}} \sin θ = - ω^{2} \sin θ

where, $ω = \sqrt{\frac{g}{l}}$

for $θ << 1 :$

\ddot{θ} \approx - ω^{2} θ

this is SHM with solutions:

θ = A \cos (ω t + ϕ) = A \cos ψ

⟹ p = m l^{2} \dot{θ} = - A m l^{2} ω \sin (ω t + ϕ)

defining $\tilde{p}$ as:

\tilde{p} = \frac{p}{m l^{2} ω} = - A \sin ψ

a circular trajectory on the $\tilde{p} - θ$ plane with clockwise rotation
an elliptical trajectory on the $p - θ$ plane
the slope:

\frac{d p}{d θ} = \frac{d p}{d t} \frac{d t}{d θ} = \frac{\dot{p}}{\dot{θ}} = - \frac{m g l \sin θ}{\frac{p}{m l^{2}}} = \frac{m^{2} l^{3} g \sin θ}{p}

the slope is negative
if pendulum oscillates back and forth: $E < E^{*} = 2 m g l$ , $- θ_{m a x} < θ < θ_{m a x}$ , $θ_{m a x} < π$
if $θ = \pm θ_{m a x} : p = 0$
if $E > E^{*} = 2 m g l :$ no restriction on $θ$ , $θ = \pm π : p > 0 :$

\frac{d p}{d t} \propto \frac{\sin θ}{p} \to 0

if $E = E^{*}$ , the pendulum can come to rest at $θ = \pm π$
let, $θ = π - ϵ :$

\dot{θ} = - \dot{ϵ} = \frac{p}{m l^{2}}

in the late stages, $ϵ << 1 :$

\dot{p} = - m g l ϵ

\ddot{ϵ} = - \frac{1}{m l^{2}} \dot{p} = \frac{m g l}{m l^{2}} ϵ = ω^{2} ϵ

note: this is not SHM as there is a $^{'} +^{'}$ , not $^{'} -^{'}$
the solution:

ϵ = A e^{ω t} + B e^{- ω t}

the first term is diverging from upright, while the second is converging to it
here, $ϵ$ is getting smaller:

ϵ = B e^{- ω t}

seeking a expression for $\frac{d p}{d θ} :$

\begin{aligned} \frac{d p}{d t} & = - m g l ϵ = - m g l B e^{- ω t} \\ \frac{d θ}{d t} & = - \dot{ϵ} = B ω e^{- ω t} \\ ∴ \frac{d p}{d θ} & = - \frac{m g l}{ω} \end{aligned}

the gradient is a non-zero negative constant

disconnected trajectories

not all states of a given energy are necessarily accessible, this is called breaking of ergodicity
ergodic systems explore all sates consistent with energy conservation
considering a rollercoaster in a 'mexican hat' potential, where, $E^{*} = m g h^{*}$
for $E > E^{*}$
this is ergodic as it can explore all states consistent with energy, $E$