classical mechanics - What is Maupertuis' principle good for?

The strength of Hamilton's principle is obvious to me and I see the advantage. Now, for conservative systems we also have Maupertuis' principle that says:

$$ \delta \int p dq =0$$

and I am not sure how to derive an equation of motion from this? Is this of any use in practical computations? So, can one apply this principle for example to the harmonic oscillator?- I have never seen anybody using it.

Further, I read in Goldstein's classical Mechanics that the variation in Maupertuis' principle is not the one in Hamilton's principle, since we have constant Hamiltonian and changing time, whereas Hamilton's principle has constant time and varying Hamiltonian (in general).

I am a little bit wondering about this, since you could easily get Maupertuis' principle from Hamilton's principle: $$ \delta \int L dt = \delta \int p \dot{q} - H dt = \delta \int p \dot{q} dt = \delta \int p dq =0,$$ if $H$ is constant. Can anybody here explain to me, why we have to use a different variation and how one can use this principle?

Answer

In fact, Maupertuis or Euler principles of least action are historically the first formulation of a least action principle, but one have to wait Lagrange and Hamilton to have a modern version, with the so-called Euler-Lagrange equations, which allow us to obtain the equations of movement.

If I am not mistaken, you cannot deduce directly the equations of movement from the Maupertuis/Euler principles. The problem I see is that you cannot know the dependence of the potential energy $V$ in $x$, in seeing only the kinetic energy $T$.

Now, as you state, but written differently, for a movement with conservative energy, one may see that the variation of the Maupertuis' action is equivalent to the variation of the Lagrange/Hamilton action, for instance, starting with Maupertuis' principle :

$$ \delta\int (2T) dt = 0$$

We have : $2T = T + (E - V)$, so Maupertuis' principle can be written :

$$ \delta\int (E + (T-V)) dt = 0$$

But $E$ being a constant, this is not useful in the variation, so finally, we have :

$$ \delta\int (T-V) dt = 0$$ which is the usual Lagrange/Hamilton action.

But, to really have the equations of movement, you have to use a functional, that is :

$$ \delta\int L(x,\dot x,t) dt = 0$$ and this gives you, thanks to the Euler/Lagrange equations, the equations of movement.