Untitled

💼 Case: consider that we know the path that minimizes the action and call it $x(t)$

We imagine deforming it slightly to a non-minimal path $x(t)+\delta x(t)$

🗒️ Note: $\delta x(t)$ is completely arbitrary except that it is assumed very small with respect to $x(t)$ and that the deformed path must obey the same boundary conditions as the original which means

$$ \delta x(t_0)=\delta x(t_1)=0 $$

For this to be the case either the integrand is positive and negative in different places and sums to zero or the integrand is zero everywhere. We want to be allowed to deform the path in any way we like so want $\delta x(t)$ to be arbitrary.

💎 Conclusion: the least action principle directly implies that $x(t)$ obeys Lagrange’s equation therefore it directly also implies Newton’s laws.

🗒️ Note: We derived the Least action for a 1D system with DoF $x$, it is straightforward to derive it for arbitrary systems. In general when $L$ depends on $N$ degrees of freedom, we label the generalized coordinates $\{ q_i \}$ with corresponding velocities $\{ \dot q_i\}$. In this case we get

$$ \frac{\text d }{\text dt}\left ( \frac{\partial L}{\partial \dot q_i} \right )=\frac{\partial L}{\partial q_i} $$

Examples

It is customary to phrase it as the minimization of

$$ \int^{x_1}_{x_0}f(\{y_i(x)\},\{y_i'(x)\},x)\,\text dx $$

This is minimized if

$$ \delta \left ( \int^{x_1}_{x_0}f(\{y_i\},\{y'_i\},x)\text dx \right )=0 $$

which is the case if

$$ \frac{\text d}{\text dx} \left ( \frac{\partial f}{\partial y'_i}\right )=\frac{\partial f}{\partial y_i} $$

💃 Example 1: what is the shortest path between two points

Untitled

🦎 Infinitesimal step

$$ \begin{aligned} \text dD&=\sqrt{\text dx^2+\text dy^2}\\ &=\sqrt{1+\left ( \frac{\text dy}{\text dx} \right )^2}\,\text dx \\ &=\sqrt{1+y'^2}\,\text dx \end{aligned} $$

👣 Total distance

$$ D=\int^{x_1}_{x_0}\sqrt{1+y'^2}\,\text dx $$