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

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

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

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.

• What is $E_i$? Commented Apr 11, 2021 at 0:23
• $\uparrow$ See eq. (1). Commented Nov 3, 2022 at 20:12