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 \((t_a,x_a)\) to \((t_b,x_b)\), then the path \(x\) is a function of time \(t\), i.e. \(x(t)\)
Define Action, which is a Functional of \(x\)
\[\begin{split}\begin{align*}
& \text{Action }S[x]=\int_{t_a}^{t_b} dt \; L(x,\dot{x},t)\\
& where \\
& \text{Lagrangian }L = {1\over 2} m\dot{x}^2-V(x,t)
\end{align*}\end{split}\]
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: \(\delta S=0\)
Therefore, to find \(x(t)\), we write down
\[\begin{split}\begin{align*}
& \delta S \equiv S[x+\delta x]-S[x]
= ... = \left. \delta x {\partial L \over \partial \dot{x}}
\right|_{t_a}^{t_b} - \int_{t_a}^{t_b} dt \; \delta x \left( {d\over dt}{\partial L \over \partial \dot{x}}-{\partial L \over \partial x} \right) \\
& \delta S=0 \Rightarrow \boxed{
{d\over dt}{\partial L \over \partial \dot{x}}-{\partial L \over \partial x} = 0 \;\;\;\; \text{Euler-Lagrange Equation}
}\\
& \text{From here we can solve }x(t)\text{ trivially}
\end{align*}\end{split}\]
Example 1: Newton’s Law
\[\begin{split}\begin{align*}
& L = {1\over 2} m\dot{x}^2-V(x,t)
\Rightarrow {d\over dt}{m \dot{x}}+{\partial V \over \partial x} = 0\\
& \text{Since } F=- {\partial V \over \partial x}
\Rightarrow {m \ddot{x}} = F\\
\end{align*}\end{split}\]