Classical Action
=================================

:Date: 11 Aug 2019



What is the problem ?
-------------------------

- Equivalent theory of Newton's Law, but more fundamental.

Derivation
-------------

**Step 1** Simplest scenario in nature: A particle travels from spacetime point :math:`(t_a,x_a)` to :math:`(t_b,x_b)`, then the path :math:`x` is a function of time :math:`t`, i.e. :math:`x(t)`

Define **Action**, which is a Functional of :math:`x`

.. math::
	\begin{align*}
	& \text{Action }S[x]=\int_{t_a}^{t_b} dt \; L(x,\dot{x},t)\\
	& where \\
	& \text{Lagrangian }L = {1\over 2} m\dot{x}^2-V(x,t) 
	\end{align*}

**Step 2** Equivalent to Newton's Law, the **Principle of Least Action** states that: The actual path that the particle would take is the one leads to **minimum Action**: :math:`\delta S=0`

Therefore, to find :math:`x(t)`, we write down

.. math::
	\begin{align*}
	& \delta S  \equiv S[x+\delta x]-S[x]
	          = ... = \left. \delta x {\partial L \over \partial \dot{x}} 
	           \right|_{t_a}^{t_b} - \int_{t_a}^{t_b} dt \; \delta x \left( {d\over dt}{\partial L \over \partial \dot{x}}-{\partial L \over \partial x} \right) \\
	& \delta S=0  \Rightarrow \boxed{ 
		{d\over dt}{\partial L \over \partial \dot{x}}-{\partial L \over \partial x} = 0  \;\;\;\; \text{Euler-Lagrange Equation}
	}\\
	& \text{From here we can solve }x(t)\text{ trivially}
	\end{align*}

Example 1: Newton's Law
---------------------------------

.. math::
	\begin{align*}
	& L = {1\over 2} m\dot{x}^2-V(x,t) 
	 \Rightarrow {d\over dt}{m \dot{x}}+{\partial V \over \partial x} = 0\\
	& \text{Since } F=- {\partial V \over \partial x}
	 \Rightarrow {m \ddot{x}} = F\\
	\end{align*}