User:EditingPencil/sandbox/action

Let $$C= \{ q(t)\,|\,t_1\leq t \leq t_2 \} $$ be a trajectory for a particle. Then, the action is given by:

$$\Phi [C] = \text{Action of trajectory C} = \int_{t_1}^{t_2} L(q(t),\dot q(t),t)\, dt$$

Let a variation of the trajectory be given as: $$C' = \{ q'(t)=q(t)+\delta q(t)\, |\,t_1+\delta t_1\leq t \leq t_2 +\delta t_2\} $$, then the change in action is given by:

$$\begin{align} \delta\Phi[C]&=\Phi[C^{\prime}]-\Phi[C]\\ &=\int_{t_{1}+\delta t_1}^{t_{2}+\delta t_2}d t L(q(t)+\delta q(t),\dot{q}(t)+\delta\dot{q}(t))-\int_{t_{1}}^{t_{2}}d t L(q(t),\dot{q}(t),t)\\ &= \int_{t_{1}}^{t_{2}} [ L(q(t)+\delta q(t),\dot{q}(t)+\delta \dot q(t),t)- L(q(t),\dot{q}(t),t) ] \, dt +\,\int_{t_{2}}^{t_{2}}L(q_{s}^{\prime}\,\dot{q}_{s}^{\prime}\,t)\,d t\,-\,\int_{t_{1}}^{t_{1}}L(q_{s}^{\prime}\,\dot{q}_{s}^{\prime},t)\,d t\\ &=\int_{t_{1}}^{t_{2}} \left(\frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot q} \delta \dot q\right) \, dt +\,\int_{t_{2}}^{t_{2}+\delta t_2}L(q^{\prime},\dot{q}^{\prime},t)\,d t\,-\,\int_{t_{1}}^{t_{1}+\delta t_1}L(q^{\prime},\dot{q}^{\prime},t) \,d t\\ &=\int_{t_{1}}^{t_{2}}d t\left[\frac{\partial L}{\partial q}\,\delta q(t)+\frac{\partial L}{\partial\dot{q}}\ \frac{d}{d t}\,\delta q(t)\right]+\bigg[L\Delta t\bigg]_{t_{1}}^{t_{2}}\\ &=\int_{t_{1}}^{t_{2}}d t\left[\frac{\partial L}{\partial q}-\frac{d}{d t}\left(\frac{\partial L}{\partial\dot{q}}\right)\right]\delta q(t)+\frac{\partial L}{\partial\dot{q}}\delta q(t)\biggr|_{t_{1}}^{t_{2}}+L\Delta t\biggr|_{t_{1}}^{t_{2}} \end{align}  $$

Defining the total variation of path as $$\Delta q = q'(t')-q(t)= \delta q(t) + \dot q (t) \delta t $$ and momentum as $$\frac{\partial L}{\partial \dot q}=p $$, we get:

$$\begin{align} \Delta\Phi&=\int_{t_{1}}^{t_{2}} d t \left[\frac{\partial L}{\partial q}-\frac{d}{d t}\,\frac{\partial L}{\partial\dot{q}}\right]\delta q(t)+\bigg[p\Delta q-(p\dot q - L)\Delta t\bigg]_{t_{1}}^{t_{2}}\\ &=\int_{t_{1}}^{t_{2}} d t \left[\frac{\partial L}{\partial q}-\frac{d}{d t}\,\frac{\partial L}{\partial\dot{q}}\right]\delta q(t)+\bigg[p\Delta q-H \Delta t\bigg]_{t_{1}}^{t_{2}}\\ \end{align}    $$

Hamilton's action principle
Considering a special class of variation of path that leaves the end-points and terminal times unchanged, ie. $$\Delta t_i=\Delta q(t_i)=0. $$ For such actions, the change in action functional is given by:

$$\begin{align} \Delta_\sigma\Phi&=\int_{t_{1}}^{t_{2}} d t \left[\frac{\partial L}{\partial q}-\frac{d}{d t}\,\frac{\partial L}{\partial\dot{q}}\right]\delta q(t) \end{align}    $$

From Lagrange's equation of motion, it follows that the infinitesimal change in action functional vanishes if the given trajectory is a solution for trajectory of the particle.

Weiss action principle
Using Lagrange's equations of motion, we have the following value for the change in action functional:

$$\begin{align} \Delta\Phi&=\int_{t_{1}}^{t_{2}} d t \left[\frac{\partial L}{\partial q}-\frac{d}{d t}\,\frac{\partial L}{\partial\dot{q}}\right]\delta q(t)+\bigg[p\Delta q-H\Delta t\bigg]_{t_{1}}^{t_{2}}\\ &=\bigg[p\Delta q-H \Delta t\bigg]_{t_{1}}^{t_{2}}\\ \end{align}    $$

Hence, Hamilton's action principle can be extended to Weiss action principle as the dynamical trajectory in configuration space is that which only provides end-point contributions to $$\Delta \Phi $$.