Does a Lagrangian of a system multiplied by an arbitrary constant still work? If if I apply the Euler-Lagrange equations, do they still guarantee that the action is extremal? I arrived to the following Lagrangian in a homework problem: L=$\frac{1}{2}l^2m\frac{d\theta}{dt}^2 + mgl\cos\theta$. The solution gives the exact same expression, but without the $m$ (mass) constant. Can I just discard it?

Note: This is the lagrangian for a pendulum with a streachable string, so $l$, its length, depends on $t$

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    $\begingroup$ Why don't you write down yours and the other solution /eom here. $\endgroup$ – Your Majesty Mar 31 '16 at 1:50
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    $\begingroup$ EL eqs. are linear in $L$, so yes. $\endgroup$ – Qmechanic Mar 31 '16 at 7:12

First of all, the Lagrangian is not unique, Multiplying by a constant will give the right Equations of motion (EOM's) when there are no constraint forces.

In case of when there are constraints (holonomic) the variation of action would be

$$\delta S=\delta \int_{t_1}^{t_2}L+\lambda f \space \mathrm dt = 0$$

and the new Lagrangian $L'$ would be

$$L'=L+\lambda f$$ Now, multiplying $L'$ with a constant would still give the right EOM's, but multiplying just $L$ with a constant will give the wrong EOM's

This will be true for the non-holonomic case too but now the equations are in the differential form

$$\mathrm dL'=\mathrm dL+\lambda\, \mathrm df$$


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