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  1. On p.464 of Spivak's mechanics book, the author proves the equivalence of Maupertuis' principle and stationary action principle by considering variation of some path $c(t)$, such that other paths in the variation have the same energy as $c(t)$ and the same end points, but are not necessarily defined on the same time interval. His proof becomes sketchy towards the end and refers to a lemma in the addendum (p.473). However that lemma and its corollary seem to deal with variation with integral constraint, and assume all maps are defined on the same interval. How is that useful in our situation, where "energy is constant" seems like a non-integral constraint (and moreover maps are defined on different intervals)?

  2. I am trying to apply the theorem about variation with non-integral constraint and see what I can get (here I am assuming all paths are defined on the same interval, so this probably will not give what I want, but it shouldn't give anything wrong). We are considering the action defined by $\int Adt$, where $$A=\frac{\partial L}{\partial\dot{x}}\cdot\dot{x}.$$ The energy $E$ is equal to $A-L$. For a path to be stationary for this action among all paths with the same endpoints and same energy $E=C$, it has to satisfy the Euler-Lagrange equations for $$A+\lambda(t)(A-L-C),$$ where $\lambda(t)$ is some unknown function. The equations reduce to the following if my calculation is right:

$$(1+\lambda(t))\left[\frac{d}{dt}\frac{\partial A}{\partial\dot{x}}-\frac{\partial A}{\partial x}\right]-\lambda(t)\left[\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}\right]+\dot{\lambda}(t)\left(\frac{\partial A}{\partial\dot{x}}-\frac{\partial L}{\partial\dot{x}}\right)=0.$$

But this doesn't seem right because if $c(t)$ is a true trajectory, then the first two terms should vanish, which implies the last term is also zero, but there is no reason why $A-L$ being constant should imply $\frac{\partial A}{\partial\dot{x}}-\frac{\partial L}{\partial\dot{x}}=0$. Where did I make a mistake?

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  • $\begingroup$ The energy conservation constraint is integral when considering the trajectory in phase space (not just in configuration space). You don't need to enforce energy conservation by a Lagrange multiplier as it is automatically satisfied. $\endgroup$
    – LPZ
    Commented Dec 19, 2022 at 9:07
  • $\begingroup$ @lpz I forgot to add that we assume the Lagrangian doesn't explicitly depend on time, so energy is always conserved. The point is all paths in the variation must have the same energy. $\endgroup$
    – Lxm
    Commented Dec 19, 2022 at 15:59

1 Answer 1

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  • OP seems correct that the Addendum 13A on p. 471 for integral constraints is irrelevant for the proof of Maupertuis' principle at the bottom of p. 465.

  • The (off-shell) abbreviated action functional $A[q, E]$ can be defined as a energy $\leftrightarrow$ time Legendre-type transformation $$ A[q; E, t_f, t_i] ~=~ I[q; t_f,t_i] + E (t_f-t_i) \tag{1} $$ of the (off-shell) action functional $I[q; t_f,t_i]$. In Maupertuis' principle we restrict to virtual paths with constant and same fixed energy $E$ but with free endpoint times $t_i$ and $t_f$. Formula (1) then becomes equal to $$\begin{align} A[q; E, t_f, t_i]~=~& \int_{t_i}^{t_f} p\dot{q} ~\mathrm{d}t, \cr p~:=~&\frac{\partial L}{\partial \dot{q}},\end{align}\tag{2} $$ cf. my Phys.SE answer here.

  • Returning to OP's point 2, one could try to impose the non-integral condition $$\forall t\in[t_i,t_f]:~~ h~=~E \tag{3}$$ for the Maupertuis' principle with a Lagrange multiplier $\lambda$, where $$h~:=~p_j\dot{q}^j-L \tag{4}$$ is the energy function. However, we cannot use the stationary action principle as OP seems to suggest, as we need to allow for variations of the endpoints $t_i$ and $t_f$ as well. In fact the previous sentence is the whole point of Maupertuis' principle.

References:

  1. M. Spivak, Physics for mathematicians, Mechanics I, 2010; chapter 13.
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  • $\begingroup$ My point 2 was that suppose the path $c(t)$ is critical w.r.t. any "Maupertuis variation", then it is also critical w.r.t. to any variation where all paths have the same $t_i$ and $t_f$, so applying stationary action principle should at least gives me a partial result, but it seems to give me something wrong $\endgroup$
    – Lxm
    Commented Dec 20, 2022 at 16:53
  • $\begingroup$ @Lxm it is not critical wrt variations that explore different total energies. $\endgroup$
    – timur
    Commented Nov 16, 2023 at 23:45

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