# Hamilton’s Principle of Least Action

Configuration space: $$(q_1,\cdots,q_n)$$.

Extended configuration space: Is the space $$(q_1,\cdots,q_n,t)$$.

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: $$I = \int_A^B (T-V)dt = \int_A^B\mathcal{L}(q,\dot{q},t)dt$$. I.e. $$\delta I = 0$$ finds the stationary value.

Suppose we have a lagrangian that does not depend on $$q_j$$ explicitly, then $$p_j = \frac{\partial \mathcal{L}}{\partial\dot{q}_j} = \mathbb{C}$$, the generalize conjugated momenta. Thus, $$p_j$$ is a constant of the motion.

$$\frac{d}{dt}\mathcal{L} = \sum\left[\frac{\partial\mathcal{L}}{\partial q_j}\dot{q}_j + \frac{\partial\mathcal{L}}{\partial \dot{q}_j}\ddot{q}_j\right] + \frac{\partial\mathcal{L}}{\partial t} = \sum\left(\dot{p}_j\dot{q}_j + p_j\ddot{q}_j\right) + \frac{\partial\mathcal{L}}{\partial t} = \sum\frac{d}{dt}\left(p_j\dot{q}_j\right) + \frac{\partial\mathcal{L}}{\partial t} = \frac{d}{dt}\sum p_j\dot{q}_j + \frac{\partial\mathcal{L}}{\partial t}$$. Then, $$-\frac{\partial\mathcal{L}}{\partial t} = \frac{d}{dt}\left(\sum p_j\dot{q}_j - \mathcal{L}\right) = \frac{d\mathcal{H}}{dt}$$. Then, when the Lagrangian has no dependence on time, energy is conserved since $$\frac{d\mathcal{H}}{dt} = 0$$.

$$d\mathcal{H} = \sum_j (p_jd\dot{q}_j + \dot{q}_jdp_j) - \sum \left(\frac{\partial\mathcal{L}}{\partial q_j}dq_j + \frac{\partial\mathcal{L}}{\partial\dot{q}_j}d\dot{q}_j\right) - \frac{\partial\mathcal{L}}{\partial t}dt = \sum (p_j d\dot{q}_j + \dot{q}_jdp_j - \dot{p}_jdq_j - p_jd\dot{q}_j) - \frac{\partial\mathcal{L}}{\partial t}dt = \sum (\dot{q}_jdp_j - \dot{p}_jdq_j) - \frac{\partial\mathcal{L}}{\partial t}dt$$. Thus, $$\mathcal{H} = \mathcal{H}(q,p,t)$$. So, $$\dot{q}_j = \frac{\partial\mathcal{H}}{\partial p_j}$$, $$\dot{p}_j = -\frac{\partial\mathcal{H}}{\partial q_j}$$, $$-\frac{\partial\mathcal{L}}{\partial t} = \frac{\partial\mathcal{H}}{\partial t}$$. Recall, $$\frac{d\mathcal{H}}{dt} = -\frac{\partial\mathcal{L}}{\partial t} = \frac{\partial\mathcal{H}}{\partial t}$$. Thus, $$\left(\frac{d}{dt}-\frac{\partial}{\partial t}\right)\mathcal{H} = 0$$.

1. $$\mathcal{L}(q,\dot{q})\rightarrow \mathcal{H} = \mathbb{C}$$.
2. $$\mathcal{H}(p,q)\Rightarrow \mathcal{H} = \mathbb{C}$$. $$\frac{d}{dt}\mathcal{H} = \sum\left(\frac{\partial\mathcal{H}}{\partial q_j}\dot{q}_j + \frac{\partial\mathcal{H}}{\partial p_j}\dot{p}_j\right) + \frac{\partial\mathcal{H}}{\partial t} = \sum\left(\frac{\partial\mathcal{H}}{\partial q_j}\frac{\partial\mathcal{H}}{\partial p_j} - \frac{\partial\mathcal{H}}{\partial p_j}\frac{\partial\mathcal{H}}{\partial q_j}\right) + \frac{\partial\mathcal{H}}{\partial t} = \left[\mathcal{H},\mathcal{H}\right] + \frac{\partial\mathcal{H}}{\partial t} = \frac{\partial\mathcal{H}}{\partial t} = -\frac{\partial\mathcal{L}}{\partial t}$$
• $$\vec{r}_i = \vec{r}_i(q)$$.
• $$V = V(q)$$.
\begin{align*} \vec{H} &= \sum p_j\dot{q}_j - \mathcal{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 \left(\sum_j\frac{\partial\vec{r}_i}{\partial q_j}\dot{q}_j\right)\left(\sum_k\frac{\partial\vec{r}_i}{\partial q_k}\dot{q}_k\right) + V(q) \\ &= \sum p_j\dot{q}_j - \sum \frac{1}{2}m_i \left(\sum_{jk}\frac{\partial\vec{r}_i}{\partial q_j}\cdot\frac{\partial\vec{r}_i}{\partial q_k}\dot{q}_j\dot{q}_k\right) + V(q) \\ &= \sum p_j\dot{q}_j - \sum_{jk}\left(\sum_i \frac{1}{2}m_i \left(\frac{\partial\vec{r}_i}{\partial q_j}\cdot\frac{\partial\vec{r}_i}{\partial q_k}\right)\right)\dot{q}_j\dot{q}_k + V(q) \\ &= \sum \frac{\partial\mathcal{L}}{\partial\dot{q}_j}\dot{q}_j - \sum_{jk}\left(\sum_i \frac{1}{2}m_i \left(\frac{\partial\vec{r}_i}{\partial q_j}\cdot\frac{\partial\vec{r}_i}{\partial q_k}\right)\right)\dot{q}_j\dot{q}_k + V(q) \\ &= \sum \frac{\partial T}{\partial\dot{q}_j}\dot{q}_j - \sum_{jk}\left(\sum_i \frac{1}{2}m_i \left(\frac{\partial\vec{r}_i}{\partial q_j}\cdot\frac{\partial\vec{r}_i}{\partial q_k}\right)\right)\dot{q}_j\dot{q}_k + V(q) \\ &= 2T - \sum_{jk}\left(\sum_i \frac{1}{2}m_i \left(\frac{\partial\vec{r}_i}{\partial q_j}\cdot\frac{\partial\vec{r}_i}{\partial q_k}\right)\right)\dot{q}_j\dot{q}_k + V(q) \\ &= T + V. \end{align*}

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

1. $$\vec{r}_i = \vec{r}_i(q)$$.
2. $$V = V(q)$$.

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

## Euler’s Theorem of Homogenous Functions

Let $$f(x_1,\cdots,x_m)$$ be a homogeneous function such that $$f(tx_1, \cdot, tx_m) = t^nf(x_1,\cdots,x_m)$$. Then, $$\sum_i^m x_i\frac{\partial f}{\partial x_i}nf(x_1,\cdots,x_m)$$.

## Uniqueness of Lagrangian

$$\mathcal{L} = T - V$$. Recall: $$\delta I = \delta\int_a^b\mathcal{L}(q,\dot{q},t)dt$$. Consider: $$\delta \int_a^b\left[\mathcal{L}(q,\dot{q},t) + \frac{d F(q,t)}{d t}\right]dt$$ with $$\mathcal{L}_1(q,\dot{q},t) = \left[\mathcal{L}(q,\dot{q},t) + \frac{d F(q,t)}{d t}\right]$$.

Then, $$\delta I_1 = \delta\int_a^b L_1 dt = \delta I + \delta\int_a^b\frac{dF(q,t)}{dt}dt = \delta I$$. This gives a canonical transformation $$\mathcal{L}_1 = \mathcal{L} + \frac{d}{dt}F(q,t)$$.

## Classical Noether Theorem

Consider an infinitesimal transformation, Noether’s Transformation, in the $$(q,t)$$ space by $$\overline{t} = t + \varepsilon \tau(q,t)$$ and $$\overline{q}_i = q_i + \varepsilon\eta_i(q,t)$$.

$$\frac{d\overline{q}_i}{d \overline{t}} = \frac{\dot{q}_i + \varepsilon\dot{\eta}_i}{1 + \varepsilon\dot{\tau}} = \frac{(\dot{q}_i + \varepsilon\dot{\eta}_i)(1-\varepsilon\dot{\tau})}{1 - \varepsilon^2\dot{\tau}^2} \approx \dot{q}_i + \varepsilon(\dot{\eta}_i - \dot{q}_i\dot{\tau})$$

Consider: $$\int_{\overline{t}_1}^{\overline{t}_2}\mathcal{L}(\overline{q},\dot{\overline{q}},\dot{\overline{t}})d\overline{t} = \int_{t_1}^{t_2}\mathcal{L}(q,\dot{q},t)dt + \varepsilon\int_{t_1}^{t_2}\frac{df(q,t)}{dt}dt + \mathcal{O}(\varepsilon^2)$$. For some function $$f(q,t)$$. Then,

\begin{align*} \mathcal{L}(\overline{q}(t),\dot{\overline{q}}(t), \overline{t}(t))\frac{d\overline{t}}{dt} &= \mathcal{L}(q,\dot{q},t) + \varepsilon\frac{df}{dt} \\ \left\{\left[\mathcal{L}(q,\dot{q},t) + \sum\frac{\partial\mathcal{L}}{\partial q_i}\varepsilon\eta_i + \frac{\partial\mathcal{L}}{\partial \dot{q}_i}\varepsilon(\dot{\eta}_i - \dot{q}_i\dot{\tau})\right] + \frac{\partial\mathcal{L}}{\partial t}\varepsilon\tau\right\}(1+\varepsilon\dot{\tau}) &= \mathcal{L}(q,\dot{q},t) + \varepsilon\frac{df}{dt} \\ \varepsilon\sum\left[\frac{\partial\mathcal{L}}{\partial q_i}\eta_i + \frac{\partial\mathcal{L}}{\partial \dot{q}_i}(\dot{\eta}_i - \dot{q}_i\dot{\tau})\right] + \varepsilon\frac{\partial\mathcal{L}}{\partial t}\tau + \varepsilon\mathcal{L}\dot{\tau} &=\varepsilon\frac{df}{dt} \\ \sum\left[\frac{\partial\mathcal{L}}{\partial q_i}\eta_i + \frac{\partial\mathcal{L}}{\partial \dot{q}_i}(\dot{\eta}_i - \dot{q}_i\dot{\tau})\right] + \frac{\partial\mathcal{L}}{\partial t}\tau + \mathcal{L}\dot{\tau} &=\frac{df}{dt} \\ \sum(\eta_i-\dot{q}_i\tau)\left[\frac{\partial\mathcal{L}}{\partial q_i} - \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{q}_i}\right] + \frac{d}{dt}\left[\mathcal{L}\tau + \sum\frac{\partial\mathcal{L}}{\partial\dot{q}_i}(\eta_i-\dot{q}_i\tau)\right] &= \frac{df}{dt} \\ \sum\frac{d}{dt}\left[\mathcal{L}\tau + \sum\frac{\partial\mathcal{L}}{\partial\dot{q}_i}(\eta_i-\dot{q}_i\tau)\right] &= \frac{df}{dt} \\ \frac{d}{dt}\left[\sum\left(\mathcal{L}\tau + \sum\frac{\partial\mathcal{L}}{\partial\dot{q}_i}(\eta_i-\dot{q}_i\tau)\right) - f\right] &= 0 \\ \frac{d}{dt}\left[\mathcal{F}\right] &= 0 \\ \mathcal{F}(q,\dot{q},t) &= \mathbb{C} \end{align*}

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

## Example #1

Consider the planar central force problem. $$\mathcal{L} = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - V(\sqrt{x^2 + y^2})$$. Also written as: $$\mathcal{L} = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) - V(r)$$. So, $$\overline{t} = t + \eta\tau(q,t)$$ and $$\overline{q}_i = q_i + \varepsilon\eta_i(q,t)$$. If we rotate the system, then in this transformation $$\tau=0$$, $$r'=r\Rightarrow\eta_r = 0$$, $$\theta' = \theta + \varepsilon\rightarrow\eta_\theta = 1$$. Plutting this in, we find, $$\frac{df}{dt} = 0$$. Further, $$f(t) - mr^2\dot{\theta}\eta_\theta = f(t) - mr^2\dot{\theta}$$. Since, $$\frac{d}{dt}\left(f(t)-mr^2\dot{\theta}\right) = 0$$, $$f(t) = mr^2\dot{\theta} + \mathbb{C}$$. Further, $$\mathcal{F} = \mathbb{C} - mr^2\dot{\theta}$$. Thus, $$mr^2\dot{\theta}$$ is a constant of the motion.

Energy dissipated by resistor: $$\frac{1}{2}I^2R = \frac{1}{2}R\dot{Q}^2$$. $$\mathcal{F} = \frac{1}{2}m\dot{q}_i$$. $$\frac{d}{dt}\frac{\partial T}{\partial\dot{q}_j} - \frac{\partial T}{\partial q_j} = Q_j^{(a)}$$.

Created: 2024-05-30 Thu 21:18

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