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Disclaimer: I have asked this question on Physics SE a week ago, but got no answers. I know that some MO users are interested in the $n$-body problem, so I decided to cross post here as well.

Consider the $n$-body problem where we are interested in describing the time evolution of $n$ masses interacting through a potential $U$. Let $D$ be the matrix containing all pairwise distances between our masses; i.e. $D_{ij} = \sqrt{(x_i - x_j)^2 + (y_i - y_j)^2 + (z_i - z_j)^2}$ where $x_i, y_i, z_i$ are the Cartesian coordinates of the $i$'th mass. Assume that $U$ only depends on $D$, and not individual positions. [1]

I am interested in finding a differential equation that describes the time evolution of $D$, without individual positions appearing in it. Is it possible to formulate such an equation in classical mechanics?

[1] This is indeed true for Newton's Law of Gravitation, but it need not hold for a generic potential.

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  • $\begingroup$ If you want the equations to just depend on $D$ and $U$, then no. With just $D$ and $U$ you cannot distinguish between 2 bodies orbiting each other (constant $D$) and 2 bodies crashing into each other. $\endgroup$ May 1, 2016 at 16:33
  • $\begingroup$ Basically: you can think of this in terms of symmetry reductions. The "general" system has $6n$ degrees of freedom (position and velocity of each mass). If you demand that they interact only through relative displacement and not from absolute position, then you have translation invariance which can kill 3 degrees. rotational invariance (in the Newtonian case) kills another 3. Galilean invariance (linear change of reference frames) kills yet 3 more. So the equations should number $6n-9$. $\endgroup$ May 1, 2016 at 16:38
  • $\begingroup$ The number of pairwise distances is $n(n-1)/2$. Including the pairwise speeds you have $n(n-1)$ degrees of freedom captured by just $D$. (Note, the actual number should be smaller since when $n > 4$ there are non-trivial relationships between the $D_{ij}$ to be admissible as actual distances between masses. ) For $n \leq 5$ we have $6n-9 > n(n-1)$ so you definitely will be missing degrees of freedom: your equations would not fully capture the classical mechanics. For $n \geq 6$ I suspect after factoring in the aforementioned nontrivial relationships you would still be short, but I am not 100%. $\endgroup$ May 1, 2016 at 16:50

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