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2207.02582

DROPPING BODIES

Richard Montgomery

correctmedium confidence
Category
Not specified
Journal tier
Specialist/Solid
Processed
Sep 28, 2025, 12:56 AM

Audit review

The paper establishes, for periodic brake orbits whose two brake triangles are related by a symmetry F (a plane isometry possibly composed with an equal-mass label transposition), the half-period identity q(t+T/2)=F(q(t)) and the quarter-period fixed-point relation q(T/4)=F(q(T/4)). From these, it identifies the geometry at T/4 (reflection implies syzygy; F=R∘σ with R a 180° rotation implies an Euler configuration), and further derives F^2=Id hence R^2=Id, forcing R to be a 180° rotation. In the pure-rotation case F=R, the fixed-point condition at T/4 forces total collision, so no such periodic brake orbit exists for the gravitational N-body problem. These points are stated explicitly in the paper (see the derivation of q(t+T/2)=F(q(t)) and q(T/4)=F(q(T/4) and the associated geometric consequences, as well as the argument that R must be 180° and that pure rotation forces triple collision, which is inadmissible for classical gravitational solutions) . The candidate solution mirrors this line of reasoning, supplying standard ODE uniqueness/equivariance details and the equal-mass permutation symmetry noted in the paper’s discussion of F=R∘σ . Net: both are correct and essentially the same proof.

Referee report (LaTeX)

\textbf{Recommendation:} minor revisions

\textbf{Journal Tier:} specialist/solid

\textbf{Justification:}

The paper's argument is logically sound, compact, and insightful, tying endpoint congruences to global spatio-temporal symmetries and resolving the rotational paradox via permutations and fixed-point constraints at quarter period. Minor clarifications—explicit hypotheses (equal masses, collision-free domain), a lemma on commutation of rotations with permutations, and a brief remark delineating classical gravitational solutions from regularizable toy models—would improve precision without changing the results.