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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 the right units of energy,

thanks

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It should be noted that this Lagrangian appears in Goldstein's first chapter excercises. –  jinawee Feb 1 at 23:13

1 Answer 1

up vote 11 down vote accepted

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
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The Lagrangian does not have to have dimension of energy in classical mechanics. For instance, if one scales a Lagrangian with a dimensionful (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
2  
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 Z Nov 25 '11 at 5:28
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@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

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