# what this Lagrangian stands for?

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

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 taking V as a potential, m as a mass, then each term is J^2... no idea what it is physically though! – Nic Nov 24 '11 at 20:43

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).

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thanks but i can't understand how comes [L]= energy and [T]= energy and L ~ T^2??? – 0x90 Nov 24 '11 at 21:22
The Lagrangian does not have to have dimension of energy in classical mechanics. For instance, if one scales a Lagrangian with a dimensionfull (non-zero) constant, the equations of motion will stay the same. – Qmechanic Nov 24 '11 at 21:29
but L =T-U, i can't get it. – 0x90 Nov 24 '11 at 21:30
No,L=T-U is only one possible Lagrangian, namely the (simplest) one that describes Newtonian mechanics. In general the Lagrangian can be anything as long as it is a function of the proper variables (and is dimensionally consistent). – David Zaslavsky Nov 25 '11 at 5:28
@ZoZo123 Qmechanic's point is the $\mathcal{L}' = \mathcal{L} \times 10\text{ m} = (T-U)\times 10\text{ m}$ has different dimensions, but can still be used to get the same equations of motion, so clearly there is no requirement that a Lagrangian have units of energy. – dmckee Nov 26 '11 at 19:35

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.

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... this does not help the question at all; you just redefined "Lagrangian". – Chris Gerig Nov 25 '11 at 12:45