Hamilton’s Principle of Least Action

Configuration space: $(q_{1}, \dots, q_{n})$ .

Extended configuration space: Is the space $(q_{1}, \dots, q_{n}, t)$ .

Holonomic systems are those that only have a holonomic constraints (only coordinates not velocity).

HPLA: For a monogenic and holonomic mechanical system the true trajectory of all possible taken between two configuration is the one that minimizes action: $I = \int_{A}^{B} (T - V) d t = \int_{A}^{B} L (q, \dot{q}, t) d t$ . I.e. $δ I = 0$ finds the stationary value.

Suppose we have a lagrangian that does not depend on $q_{j}$ explicitly, then $p_{j} = \frac{\partial L}{\partial {\dot{q}}_{j}} = C$ , the generalize conjugated momenta. Thus, $p_{j}$ is a constant of the motion.

$\frac{d}{d t} L = \sum [\frac{\partial L}{\partial q_{j}} {\dot{q}}_{j} + \frac{\partial L}{\partial {\dot{q}}_{j}} {\ddot{q}}_{j}] + \frac{\partial L}{\partial t} = \sum ({\dot{p}}_{j} {\dot{q}}_{j} + p_{j} {\ddot{q}}_{j}) + \frac{\partial L}{\partial t} = \sum \frac{d}{d t} (p_{j} {\dot{q}}_{j}) + \frac{\partial L}{\partial t} = \frac{d}{d t} \sum p_{j} {\dot{q}}_{j} + \frac{\partial L}{\partial t}$ . Then, $- \frac{\partial L}{\partial t} = \frac{d}{d t} (\sum p_{j} {\dot{q}}_{j} - L) = \frac{d H}{d t}$ . Then, when the Lagrangian has no dependence on time, energy is conserved since $\frac{d H}{d t} = 0$ .

$d H = \sum_{j} (p_{j} d {\dot{q}}_{j} + {\dot{q}}_{j} d p_{j}) - \sum (\frac{\partial L}{\partial q_{j}} d q_{j} + \frac{\partial L}{\partial {\dot{q}}_{j}} d {\dot{q}}_{j}) - \frac{\partial L}{\partial t} d t = \sum (p_{j} d {\dot{q}}_{j} + {\dot{q}}_{j} d p_{j} - {\dot{p}}_{j} d q_{j} - p_{j} d {\dot{q}}_{j}) - \frac{\partial L}{\partial t} d t = \sum ({\dot{q}}_{j} d p_{j} - {\dot{p}}_{j} d q_{j}) - \frac{\partial L}{\partial t} d t$ . Thus, $H = H (q, p, t)$ . So, ${\dot{q}}_{j} = \frac{\partial H}{\partial p_{j}}$ , ${\dot{p}}_{j} = - \frac{\partial H}{\partial q_{j}}$ , $- \frac{\partial L}{\partial t} = \frac{\partial H}{\partial t}$ . Recall, $\frac{d H}{d t} = - \frac{\partial L}{\partial t} = \frac{\partial H}{\partial t}$ . Thus, $(\frac{d}{d t} - \frac{\partial}{\partial t}) H = 0$ .

$L (q, \dot{q}) \to H = C$ .
$H (p, q) \Rightarrow H = C$ . $\frac{d}{d t} H = \sum (\frac{\partial H}{\partial q_{j}} {\dot{q}}_{j} + \frac{\partial H}{\partial p_{j}} {\dot{p}}_{j}) + \frac{\partial H}{\partial t} = \sum (\frac{\partial H}{\partial q_{j}} \frac{\partial H}{\partial p_{j}} - \frac{\partial H}{\partial p_{j}} \frac{\partial H}{\partial q_{j}}) + \frac{\partial H}{\partial t} = [H, H] + \frac{\partial H}{\partial t} = \frac{\partial H}{\partial t} = - \frac{\partial L}{\partial t}$
- ${\vec{r}}_{i} = {\vec{r}}_{i} (q)$ .
- $V = V (q)$ .

\begin{aligned} \vec{H} & = \sum p_{j} {\dot{q}}_{j} - L (q, \dot{q}, t) \\ = \sum p_{j} {\dot{q}}_{j} - \sum \frac{1}{2} m_{i} {\dot{\vec{r}}}_{i}^{2} + V (q) \\ = \sum p_{j} {\dot{q}}_{j} - \sum \frac{1}{2} m_{i} (\sum_{j} \frac{\partial {\vec{r}}_{i}}{\partial q_{j}} {\dot{q}}_{j}) (\sum_{k} \frac{\partial {\vec{r}}_{i}}{\partial q_{k}} {\dot{q}}_{k}) + V (q) \\ = \sum p_{j} {\dot{q}}_{j} - \sum \frac{1}{2} m_{i} (\sum_{j k} \frac{\partial {\vec{r}}_{i}}{\partial q_{j}} \cdot \frac{\partial {\vec{r}}_{i}}{\partial q_{k}} {\dot{q}}_{j} {\dot{q}}_{k}) + V (q) \\ = \sum p_{j} {\dot{q}}_{j} - \sum_{j k} (\sum_{i} \frac{1}{2} m_{i} (\frac{\partial {\vec{r}}_{i}}{\partial q_{j}} \cdot \frac{\partial {\vec{r}}_{i}}{\partial q_{k}})) {\dot{q}}_{j} {\dot{q}}_{k} + V (q) \\ = \sum \frac{\partial L}{\partial {\dot{q}}_{j}} {\dot{q}}_{j} - \sum_{j k} (\sum_{i} \frac{1}{2} m_{i} (\frac{\partial {\vec{r}}_{i}}{\partial q_{j}} \cdot \frac{\partial {\vec{r}}_{i}}{\partial q_{k}})) {\dot{q}}_{j} {\dot{q}}_{k} + V (q) \\ = \sum \frac{\partial T}{\partial {\dot{q}}_{j}} {\dot{q}}_{j} - \sum_{j k} (\sum_{i} \frac{1}{2} m_{i} (\frac{\partial {\vec{r}}_{i}}{\partial q_{j}} \cdot \frac{\partial {\vec{r}}_{i}}{\partial q_{k}})) {\dot{q}}_{j} {\dot{q}}_{k} + V (q) \\ = 2 T - \sum_{j k} (\sum_{i} \frac{1}{2} m_{i} (\frac{\partial {\vec{r}}_{i}}{\partial q_{j}} \cdot \frac{\partial {\vec{r}}_{i}}{\partial q_{k}})) {\dot{q}}_{j} {\dot{q}}_{k} + V (q) \\ = T + V . \end{aligned}

Thus, the Hamiltonian is only the system’s energy when:

${\vec{r}}_{i} = {\vec{r}}_{i} (q)$ .
$V = V (q)$ .

Since $T$ is a second order homogenous function: $\frac{\partial T}{\partial {\dot{q}}_{j}} {\dot{q}}_{j} = 2 T$ .

Euler’s Theorem of Homogenous Functions

Let $f (x_{1}, \dots, x_{m})$ be a homogeneous function such that $f (t x_{1}, \cdot, t x_{m}) = t^{n} f (x_{1}, \dots, x_{m})$ . Then, $\sum_{i}^{m} x_{i} \frac{\partial f}{\partial x_{i}} n f (x_{1}, \dots, x_{m})$ .

Uniqueness of Lagrangian

$L = T - V$ . Recall: $δ I = δ \int_{a}^{b} L (q, \dot{q}, t) d t$ . Consider: $δ \int_{a}^{b} [L (q, \dot{q}, t) + \frac{d F (q, t)}{d t}] d t$ with $L_{1} (q, \dot{q}, t) = [L (q, \dot{q}, t) + \frac{d F (q, t)}{d t}]$ .

Then, $δ I_{1} = δ \int_{a}^{b} L_{1} d t = δ I + δ \int_{a}^{b} \frac{d F (q, t)}{d t} d t = δ I$ . This gives a canonical transformation $L_{1} = L + \frac{d}{d t} F (q, t)$ .

Classical Noether Theorem

Consider an infinitesimal transformation, Noether’s Transformation, in the $(q, t)$ space by $\overset{―}{t} = t + ε τ (q, t)$ and ${\overset{―}{q}}_{i} = q_{i} + ε η_{i} (q, t)$ .

$\frac{d {\overset{―}{q}}_{i}}{d \overset{―}{t}} = \frac{{\dot{q}}_{i} + ε {\dot{η}}_{i}}{1 + ε \dot{τ}} = \frac{({\dot{q}}_{i} + ε {\dot{η}}_{i}) (1 - ε \dot{τ})}{1 - ε^{2} {\dot{τ}}^{2}} \approx {\dot{q}}_{i} + ε ({\dot{η}}_{i} - {\dot{q}}_{i} \dot{τ})$

Consider: $\int_{{\overset{―}{t}}_{1}}^{{\overset{―}{t}}_{2}} L (\overset{―}{q}, \dot{\overset{―}{q}}, \dot{\overset{―}{t}}) d \overset{―}{t} = \int_{t_{1}}^{t_{2}} L (q, \dot{q}, t) d t + ε \int_{t_{1}}^{t_{2}} \frac{d f (q, t)}{d t} d t + O (ε^{2})$ . For some function $f (q, t)$ . Then,

\begin{aligned} L (\overset{―}{q} (t), \dot{\overset{―}{q}} (t), \overset{―}{t} (t)) \frac{d \overset{―}{t}}{d t} & = L (q, \dot{q}, t) + ε \frac{d f}{d t} \\ {[L (q, \dot{q}, t) + \sum \frac{\partial L}{\partial q_{i}} ε η_{i} + \frac{\partial L}{\partial {\dot{q}}_{i}} ε ({\dot{η}}_{i} - {\dot{q}}_{i} \dot{τ})] + \frac{\partial L}{\partial t} ε τ} (1 + ε \dot{τ}) & = L (q, \dot{q}, t) + ε \frac{d f}{d t} \\ ε \sum [\frac{\partial L}{\partial q_{i}} η_{i} + \frac{\partial L}{\partial {\dot{q}}_{i}} ({\dot{η}}_{i} - {\dot{q}}_{i} \dot{τ})] + ε \frac{\partial L}{\partial t} τ + ε L \dot{τ} & = ε \frac{d f}{d t} \\ \sum [\frac{\partial L}{\partial q_{i}} η_{i} + \frac{\partial L}{\partial {\dot{q}}_{i}} ({\dot{η}}_{i} - {\dot{q}}_{i} \dot{τ})] + \frac{\partial L}{\partial t} τ + L \dot{τ} & = \frac{d f}{d t} \\ \sum (η_{i} - {\dot{q}}_{i} τ) [\frac{\partial L}{\partial q_{i}} - \frac{d}{d t} \frac{\partial L}{\partial {\dot{q}}_{i}}] + \frac{d}{d t} [L τ + \sum \frac{\partial L}{\partial {\dot{q}}_{i}} (η_{i} - {\dot{q}}_{i} τ)] & = \frac{d f}{d t} \\ \sum \frac{d}{d t} [L τ + \sum \frac{\partial L}{\partial {\dot{q}}_{i}} (η_{i} - {\dot{q}}_{i} τ)] & = \frac{d f}{d t} \\ \frac{d}{d t} [\sum (L τ + \sum \frac{\partial L}{\partial {\dot{q}}_{i}} (η_{i} - {\dot{q}}_{i} τ)) - f] & = 0 \\ \frac{d}{d t} [F] & = 0 \\ F (q, \dot{q}, t) & = C \end{aligned}

When Noether’s tranformation is known, you can integrate to get the gauge term: $f$ .

Example #1

Consider the planar central force problem. $L = \frac{1}{2} m ({\dot{x}}^{2} + {\dot{y}}^{2}) - V (\sqrt{x^{2} + y^{2}})$ . Also written as: $L = \frac{1}{2} m ({\dot{r}}^{2} + r^{2} {\dot{θ}}^{2}) - V (r)$ . So, $\overset{―}{t} = t + η τ (q, t)$ and ${\overset{―}{q}}_{i} = q_{i} + ε η_{i} (q, t)$ . If we rotate the system, then in this transformation $τ = 0$ , $r^{'} = r \Rightarrow η_{r} = 0$ , $θ^{'} = θ + ε \to η_{θ} = 1$ . Plutting this in, we find, $\frac{d f}{d t} = 0$ . Further, $f (t) - m r^{2} \dot{θ} η_{θ} = f (t) - m r^{2} \dot{θ}$ . Since, $\frac{d}{d t} (f (t) - m r^{2} \dot{θ}) = 0$ , $f (t) = m r^{2} \dot{θ} + C$ . Further, $F = C - m r^{2} \dot{θ}$ . Thus, $m r^{2} \dot{θ}$ is a constant of the motion.

Energy dissipated by resistor: $\frac{1}{2} I^{2} R = \frac{1}{2} R {\dot{Q}}^{2}$ . $F = \frac{1}{2} m {\dot{q}}_{i}$ . $\frac{d}{d t} \frac{\partial T}{\partial {\dot{q}}_{j}} - \frac{\partial T}{\partial q_{j}} = Q_{j}^{(a)}$ .