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My question is quite specific as it refers to this article but I hope that someone here could help me. I cite the relevant part of the article:

... The second example consists of gravitational attraction of a configuration of material points on a particular material point. For convenience, we place the particular material point, whose motion is observed, at the center of a system of spherical coordinates $(r, \theta, \varphi)$. The configuration $a$ in $A$ is a function that assigns to each triple $(r, \theta, \varphi)$ either zero or a mass point $a(r, \theta, \varphi)$. If the function $a$ is nonzero at $n-1$ points, then the total system, involving the configuration and the observed point, is an $n$-body system. If $a$ consists of $n-1$ points and $b$ of $m$ points, then $a\oplus b$, defined in the obvious way, consists of $n+m-1$ points, yielding and $(n+m)$-body system. The exception occurs when $a$ and $b$ are nonzero at some common coordinate $(r, \theta, \varphi)$; we then need a method whereby two material points combine to yield a third [...] In this example, if $A$ is taken to be the set of configurations actually occuring in some empirically realizable celestial system, then of course, $\oplus$ and $*$ cannot be defined.

Why could the "addition" (i.e. superposition) of two configurations not defined? Maybe because two point masses (or bodies) could "physically" not be at the same position in space. But then we could simply define the "addition" as a configuration where at the common point of two point masses a new point is put with the mass of the two other point masses. This would be a valid configuration in my opinion, so where is the problem?

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He is saying that if $A$ is the set of configurations of a binary star system, then the sum $\oplus: A \times A \to A$ can be not defined, because if $a_1$ is the configuration at some time and $a_2$ is the configuration some time later, then $a_1 \oplus a_2$ will not be a configuration of a binary star system, since it will have four stars. In other words, $A$ is not closed under $\oplus$.

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  • $\begingroup$ In the quoted paragraph the sum of two configurations, one an $n$-body systen the other an $m$-body system, is cleary defined to be an $(n+m)$-body system. So $A = \{ A' : A \mbox{ is an $n$-body system for some $n$ } \}$ is the set of all $n$-body configurations for every $n$. So this could not be the problem. $\endgroup$ – StefanH Feb 19 '14 at 12:33
  • $\begingroup$ I don't know what you mean by $A=\{ A':A \textrm{ is an } n \textrm{-body system for some }n \}$. Do you mean $A=\{ a:a \textrm{ is an } n \textrm{-body system for some }n \}$? Anyway the version of the question that shows up for me clearly says, "... $A$ is taken to be the set of configurations actually occuring in some empirically realizable celestial system, ..." Since celestial bodies don't pop in and out of existence, each element of $A$ has the same number of "bodies" $n$. $\endgroup$ – Brian Moths Feb 19 '14 at 14:35
  • $\begingroup$ Yes, this is a typo. I meant: $A = \{ A' : A' \mbox{ is an $n$-body system for some $n$} \} = \{ a : a \ldots \}$. Here they can "pop up" into existence. The paper seeks for an algebraic description, and of course if I have an configuration I can put in other bodies, or if I have two I can combine them, or if I have one I can decompose it if more than one body is present. Further the author has an equilibrium axiom stating that for every configuration $a$ there is one $b$ which puts $a$ and $b$ (i.e. $a \oplus b$) in equilibrium. $\endgroup$ – StefanH Feb 19 '14 at 20:28
  • $\begingroup$ It may be helpful to look at more of the quote to give additional context: "... $A$ is taken to be the set of configurations actually occuring in some empirically realizable celestial system, then of couse $\oplus$ and $*$ cannot be defined. If $a$ and $b$ are configurations in $A$, then $a\oplus b$ is a physically possible configuration, but since it is never observed in the system under study, it is not in $A$." In the last sentence he is explicity stating that $A$ is not closed under $\oplus$ the reason for this is that you never see a system of two starts turn into a system of four stars. $\endgroup$ – Brian Moths Feb 19 '14 at 20:43

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