# What does a Lagrangian of the form $L=m^2\dot x^4 +U(x)\dot x^2 -W(x)$ represent?

I saw this Lagrangian in notes I have printed:

$$L\left(x,\frac{dx}{dt}\right) = \frac{m^2}{12}\left(\frac{dx}{dt}\right)^4 + m\left(\frac{dx}{dt}\right)^2\times V(x) -V^2(x).$$ (It appears in the exercises in the first chapter of Goldstein.)

What is it? Is it even physical? It seems like it doesn't have the right units of energy.

Lagrangian:

$$L~=~\frac{1}{3}T^2+2TV-V^2, \qquad T~:=~\frac{m}{2}\dot{x}^2.$$

Lagrange equation:

$$2(T-V)V^{\prime}~=~\frac{\partial L}{\partial x} ~=~ \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}}\right) ~=~ \frac{d}{dt} \left[\left(\frac{2}{3}T +2V\right)m\dot{x}\right]$$ $$~=~ \left(\frac{2}{3}T +2V\right)m\ddot{x} + \left(\frac{2}{3}m\dot{x}\ddot{x} +2V^{\prime}\dot{x}\right)m\dot{x} ~=~ 2(T+V)m\ddot{x} +4TV^{\prime},$$ or,

$$- 2(T+V)V^{\prime}~=~ 2(T+V)m\ddot{x}.$$

In other words, one gets Newton's second law$^1$

$$m\ddot{x}~=~-V^{\prime}. \qquad\qquad\qquad(N2)$$

So the Lagrangian $L$ is equivalent to the usual $T-V$ at the classical level.

--

$^1$ One may wonder about the second branch $T+V=0$, but since $T+V={\rm const}$ is a first integral to (N2), the second branch is already included in the first branch (N2).