0
$\begingroup$

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 $t$

$\endgroup$
  • 1
    $\begingroup$ Why don't you write down yours and the other solution /eom here. $\endgroup$ – Your Majesty Mar 31 '16 at 1:50
  • 1
    $\begingroup$ EL eqs. are linear in $L$, so yes. $\endgroup$ – Qmechanic Mar 31 '16 at 7:12
2
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.