2

I am trying to develop intuition about Why happen to be true the Hamilton's principle of stationary action, and after seen this video I have a few questions that are more related to maths than physics.

I know beforehand than in general, searching for a stationary points of a function $F(t,\vec{x})$ will lead to different answers than looking for stationary points of the function $|F(t,\vec{x})|$ (it is under the absolute value function).

Now, since the Lagrangian of a system is defined as $\textit{L}=E_k - U$ where $E_k$ is the kinetic energy and $U$ is the potential energy, so the action will be $S=\int\limits_{t_0}^{t_f} \textit{L}\, dt$, the statinonary path will lead to the Euler-Lagrange equations.

Now, there are other external assumptions that will affect how the equations will look like, like having a time-invariant potential will make the forces equal to the negative of its gradient, or the abscence of dissipative forces will make the time derivatives of the lagrangian equal the derivatives of the momentum.

Now for the main question:

  • What will happen if instead of solving for the classic action, I will start looking for the stationary points of the following? $$S^*=\int\limits_{t_0}^{t_f} |E_k-U|\, dt$$
  • Under which assumptions $S^*$ will have the same stationary points that $S$?
  • or they will never happen to be equivalent?
Joako
  • 1,957
  • 1
    they will be equivalent as far $E_k\geqslant U$ all the time, or when $E_k\leqslant U$ all the time, that is, when $|E_k-U|=E_k-U$ or $|E_k-U|=U-E_k$ all the time –  Aug 17 '22 at 19:24
  • @Masacroso since total energy is conserved $E=E_k+U=\text{constant}$ I can say, as example when $U$ is time independent, that $E-2U=E_k-U$ so somehow now $E_k-U$ is time independent in this case and is going to be positive/negative as long $E-2U$ is positive/negative on each section of space... right?... intuitively this doesn't make sense (and is probably wrong), but maybe there are some assumptions of the system that will made them equivalent... this is what I am trying to understand. – Joako Aug 17 '22 at 19:53
  • @Joako The action you wrote might be hard to minimize analytically because the integrand is not smooth. Check out the geodesic equation for an example of when we want to minimize such a nonsmooth action. In that setting we replace the norm by the square norm, and under the condition that the curve is unit speed, the critical points of the new action are the same as those of the old action. – Mason Aug 18 '22 at 00:28

1 Answers1

1

OP ponders:

When does the Lagrangian $L(x,v,t)$ lead to the same solutions for the Euler-Lagrange (EL) eqs. $$ \frac{\partial L}{\partial q^j} ~=~\frac{d}{dt}\frac{\partial L}{\partial v^j}\tag{1}$$ as the absolute valued Lagrangian $|L|$?

Well, let's see. The new EL eqs. are $$ \begin{align} {\rm sgn}(L)\frac{\partial L}{\partial q^j} ~=~&\frac{\partial |L|}{\partial q^j}\cr ~\stackrel{\text{EL eq.}}{=}&\frac{d}{dt}\frac{\partial |L|}{\partial v^j}\cr ~=~&\frac{d}{dt}\left({\rm sgn}(L)\frac{\partial L}{\partial v^j}\right)\cr ~=~&{\rm sgn}(L)\frac{d}{dt}\frac{\partial L}{\partial v^j} + \delta(L)\frac{dL}{dt}\frac{\partial L}{\partial v^j},\tag{2} \end{align}$$

A sufficient condition is apparently:

  • The Lagrangian $L\neq 0$ is never zero.

Related Math.SE post: When can the Lagrangian be squared without changing the stationary path?

Qmechanic
  • 13,259