Maupertuis principle says that if we know the initial and final coordinates but not time, the total energy and the fact that energy is conserved, we can choose the "right" path from all mathematically possible paths (those that have same energy E and follow the conservation of energy) by minimizing the action integral which is given by

$S = \int p\mathrm{d}q$ taken from initial coordinates to final.

So consider two paths that are mathematically possible in the sense that they follow law of conservation of energy. Then for both paths, the Hamiltonian is time-independent. In section 44 in Landau & Lifshitz, they say

" Thus the abbreviated action has a minimum with respect to all paths which satisfy the law of conservation of energy and pass through the final point at any instant".

This statement suggests that the action is minimum for all mathematically possible paths because for all of these virtual paths, the energy is conserved.

So i feel it contradicts the fact that abbreviated action is minimized only for the correct path. Please help me what i am thinking wrong.


1 Answer 1

  1. Consider the set $\Gamma(E,q_i,q_f)$ of all virtual paths $\gamma:[t_i,t_f]\to \mathbb{R}^n$ for arbitrary $t_i\leq t_f$ with energy $E$ and that satisfy the boundary conditions $$\gamma(t_i)~=~q_i\qquad\text{and}\qquad \gamma(t_f)~=~q_f.$$

  2. OP already know that not all virtual paths $\gamma\in\Gamma(E,q_i,q_f)$ are stationary path for the abbreviated action functional.

  3. The quoted sentence is marred by conditional cases, which makes it difficult to read, but Landau & Lifshitz do not mean what OP suggests. The energies of the allowed virtual paths are not arbitrary; they all have to be constant and equal to the same given energy $E$.

  4. Concerning OP's title question, see also my Phys.SE answer here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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