When can the Lagrangian be squared without changing the stationary path?

Question

Assume that the path $y(x)$ makes the functional $$ S[y] = \int _a ^b L(y, y', x) dx$$ stationary. Under what conditions does $$ \int _a ^b L(y, y', x)^2 dx$$ have the same stationary path? And what other functions can be applied to $L$ without changing the stationary path?

The question is inspired by ordinary calculus: The $x$ that makes $f(x)$ extremal also makes $f(x)^2$ extremal, and often it is convenient to minimize the latter.

Oh, and this very technique is used in this script on page 13 and 14. To give a brief summary, the goal there is to find the geodesic curve that minimizes

$$ \int (g_{i j} \frac{dx^{i}}{dt}\frac{dx^{j}}{dt})^{1/2} dt $$

where $x^i$ are curvilinear coordinates $g_{i j}$ is the metric. And then the author says that it is easy to realize that we can also simply use

$$ \int g_{i j} \frac{dx^{i}}{dt}\frac{dx^{j}}{dt} dt$$

Answer 1

1. OP is asking:

   When does the Lagrangian $L(q,\dot{q},t)$ lead to the same solutions for the Euler-Lagrange (EL) eqs. as the squared Lagrangian $L^2$?

   Well, that's a good question. There is a natural generalization:

   When does the Lagrangian $L(q,\dot{q},t)$ lead to the same solutions for the Euler-Lagrange (EL) eqs. as the Lagrangian $f(L)$, where $f$ is a differentiable function?

   Let's calculate: $$\begin{aligned} E_i [f(L)] -f^{\prime}(L)~ E_i[L]~:=~& \left(\frac{\partial }{\partial q^i}- \frac{d}{dt} \frac{\partial }{\partial \dot{q}^i}\right)f(L) \cr ~-~& f^{\prime}(L)\left(\frac{\partial }{\partial q^i}- \frac{d}{dt} \frac{\partial }{\partial \dot{q}^i}\right)L \cr ~=~&-f^{\prime\prime}(L)\frac{dL}{dt} \frac{\partial L}{\partial \dot{q}^i}.\end{aligned}\tag{1} $$

   In eq. (1) $E_i[L]$ denotes the $i$th EL eq. for the Lagrangian $L$.

   Sufficient conditions are apparently:

   • $L$ is a constant of motion (COM), and $f^{\prime}(L)\neq 0$ is not zero.
   • $L(q,t)$ does not depend on velocities $\dot{q}$, and $f^{\prime}(L)\neq 0$ is not zero.
   • $f$ is a non-constant affine function.

2. But how can we guarantee that $L$ is a COM? Here is a common strategy.

   • If $L$ does not depend explicitly on $t$, then the energy $$h~:=~\left(\dot{q}^i\frac{\partial }{\partial\dot{q}^i} -1\right)L \tag{2}$$ is a COM, cf. Noether's theorem.

   • If moreover $L$ is homogeneous in the velocities $\dot{q}$ of weight $w\neq 1$, then $$L~\stackrel{(2)}{=}~\frac{h}{w-1}\tag{3}$$ is also a COM as well!

3. Main example. How does the theory from sections 1 & 2 apply to the square root Lagrangian $$L~:=~\sqrt{ g_{ij}(q) \dot{q}^i \dot{q}^j}\tag{4}$$ for geodesics? Well, it fails because of a tiny but important detail: The weight $w=1$ turns out to be exactly one! This is related to the fact that the solutions to the EL eqs. for $L$ in eq. (4) are all parametrized geodesics, but the solutions to the EL eqs. for $L^2$ are only all affinely parametrized geodesics. See e.g. this Phys.SE post for details.