what this Lagrangian stands for?

Question

i saw this Lagrangian in notes i have printed:

$$ L(x,dx/dt) = (m^2(dx/dt)^4)/12 + m(dx/dt)^2*V(x) -V^2(x) $$

what is it? is it physical? it seems like it doesn't have a right units of energy,

thanks

Answer 1

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.

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$^1$ 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).

Answer 2

The Lagrangian was first developed in classical mechanics and can be viewed as a mathematical tool used to simplify and generalize the understanding and development of a system's dynamics.

It first appeared as the function that gives a physical meaning to the action defined as

$$ S(L) = \int_{t_{0}}^{t_1} Ldt $$

in a way that if the action is stationary (i.e. minimized, maximized), we get the Euler-Lagrange equations that describe the dynamics of the system in cause.

This is due to the analogy between the newton equations and the principle of least action.

More formally, Lagrangian is strictly related to the Hamiltonian by a Legendre transformation: the dynamics is known by either knowing the Lagrangian and solve the Euler-Lagrange equations, or by knowing the Hamiltonian and solving the Hamilton equations.