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Suppose one has in-hand an accurate time-space trajectory in $\mathbb{R}^3$ of a (small) body, say an asteroid or satellite—effectively a point. To what extent does this trajectory determine the point masses that could gravitationally determine it (according to inverse-square gravitation)? Is this highly underdetermined, in that there are many point-mass distributions that would lead to the (exact) same trajectory, or does the trajectory essentially uniquely determine the masses? Perhaps this question only has a sharp answer with some assumptions on the size of the point masses, i.e., planetary or star-like, as opposed to spread-out asteroid belts or dust clouds...?
        IOP
        (Suggestive image from: "Spacetime symmetries and Kepler's third law," 2012, Class. Quantum Grav.: 29. 217002 (arXiv link)).


(Added 9Feb13). In light of Ben Crowell's incisive analysis, and the various comments and answers (by Joel, Abhinav, Brendan, Theo) which point to fundamental nonuniqueness, permit me to rephrase the question:

Given an accurate (space-time) trajectory of a point-mass (planet) in a fixed coordinate system, and given a number $n$, under the assumption that there are $n$ stationary/fixed point masses (stars) that might have caused the planet's trajectory solely due to inverse-square Newtonian gravitation, does the trajectory of the planet determine the fixed stars' masses and positions?

It might be useful to distinguish general positions of the stars vs. special arrangements. If this can be answered for each $n$, then one could explore successively larger values of $n$.

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If the trajectory lies in a plane, then couldn't one distribute masses outside of that plane, but symmetrically, in a variety of ways so as to realize the same trajectory? –  Joel David Hamkins Feb 9 '13 at 2:13
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Although the original question has been answered, I wonder: is it true that in $n$-dimensional space $n$ unit masses in general position (no $k$ of them lying on a common $k-2$-dimensional subspace, for each $k$) have a unique effect on any trajectory? –  Brendan McKay Feb 9 '13 at 5:49
    
@Brendan McKay: For the reasons given in my answer, I think your question needs to be refined by saying, basically, "relative to what?" If you have a finite number $j$ of masses, there's a well-defined center of mass. Let's say the trajectory is given relative to that center of mass. The masses have to interact, because otherwise you're not talking about a well-defined physical theory. So in an $n$-dimensional space, there are $2nj$ degrees of freedom, and I think infinitely precise knowledge of $2j+2$ points on the trajectory of a test particle should suffice. Why do you want $n=j$? –  Ben Crowell Feb 9 '13 at 13:56
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Two quibbles with the 9Feb13 restatement: "Accurate" should be "infinitely accurate," and "n stationary/fixed point masses" should be "n stationary/fixed point masses in general position." Without "infinitely," we can make differing and arbitrarily good explanations of the same trajectory by forming a regular-polytope shell centered on the test particle and expanding the shell to an arbitrary size. Without "in general position," we already have several counterexamples. –  Ben Crowell Feb 10 '13 at 3:24
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5 Answers 5

Both Newtonian gravity[1] and general relativity can be expressed in a geometric form such that the trajectory of a test particle is a geodesic. When the theory is expressed in this form, the question becomes whether or not we can recover detailed information about an entire semi-Riemannian space from knowledge of a single geodesic. Typically there is absolutely zero information contained in a geodesic; the information content comes from incidence relations between different geodesics. (You could have information about whether a geodesic is self-intersecting, but even in a spacetime with closed timelike curves we don't expect self-intersection for a geodesic in general position.)

The question really presupposes some additional machinery, such as a global Cartesian coordinate system for space plus a Newton-style absolute time. The existence of geometrical formulations of Newtonian gravity shows that this machinery isn't fundamental to the theory. From the point of view of Isaac Newton, we assume an omniscient observer O who can see the state of all matter in the universe. O can tell the difference between an object that is moving inertially and one that is accelerating uniformly because of a sheet of mass a billion light years away. Such an inertially moving object implies an inertial frame of reference K. Once we have K, we can observe the motion of a test particle and say whether it's straight or curved.

So the answer to the question is that the trajectory of a single test particle tells us nothing at all about gravitational sources. Only an omniscient O who already knows about all distant sources can even construct K and describe the trajectory as something other than a geodesic.

To extract information about the spacetime, we need incidence relations between geodesics. As an example, suppose that two test particles cross paths, and that their world-lines intersect not just at that event but at some other, later event as well. The two geodesics form a lune. This tells us something about the Ricci curvature, which implies that there is at least one gravitational source whose attraction brought the particles back together.

This seems related to the Einstein hole argument.

[1] Giulini, "Some remarks on the notions of general covariance and background independence," http://arxiv.org/abs/gr-qc/0603087

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For a bounded trajectory, then one can imagine placing the entire region inside a uniform spherical shell, which has no net gravitational effect on what happens inside (see wikipedia, also hyperphysics). Thus, no bounded trajectory can uniquely determine the ambient mass distribution.

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I think the question is restricting to point masses (as opposed to spread-out regions, like shells). Otherwise one could also use the same argument to replace any point mass by a ball of the same mass (as long as the trajectory does not intersect it). –  Abhinav Kumar Feb 9 '13 at 1:59
    
Yes, so let's not take this answer as the end of the story. –  Joel David Hamkins Feb 9 '13 at 2:10
    
Or take a Newtonian spacetime with countably many identical point masses distributed on an infinite cubical lattice, all initially at rest relative to their neighbors. By symmetry, no information is recoverable from observing the motion of one such mass relative to the others. We can replace one of the masses with a test particle of negligible mass, and the test particle will still never accelerate relative to the average of the others. If the space has the topology of a torus, the number of masses can be made finite. –  Ben Crowell Feb 9 '13 at 5:11
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In $\mathbb{R}^1$, an arbitrarily short part of a nonconstant trajectory determines the locations and sizes of a finite collection of distinct point masses. From the trajectory (in space-time, not space), you can find the acceleration on an interval.

Suppose $\sum_i m_i/(x-x_i)^2 = \sum_j m_j/(x-x_j)^2$ on a nonempty interval. Then their difference is a meromorphic function which is $0$ on an interval. Since the zeros are not isolated, the function is identically $0$. So, the sets of singularities $\lbrace x_i \rbrace$ and $\lbrace x_j \rbrace$ have to be the same, with the same coefficients.

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Nice, Douglas! (Yes, I meant space-time trajectory---Now clarified.) –  Joseph O'Rourke Feb 10 '13 at 2:30
    
This is a nice, rigorous demonstration of two of the three features that I think are generally required: (1) general position (here, nonconstant trajectory), and (2) a finite number of masses. The other general issue is (3) what are you measuring relative to? (Here it could be the center of mass of the finite collection of masses.) –  Ben Crowell Feb 10 '13 at 3:11
    
@Ben Crowell: Finiteness can be weakened in this proof, say to collections of masses with no limit points so that the sum converges everywhere except for the masses. –  Douglas Zare Feb 10 '13 at 19:11
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Taking a cue from Joel's comment: it's easy to consider a trajectory going along the z-axis. A system of unit masses at $(1,0,0)$, $(-1,0,0)$, $(0,1,0)$ and $(0,-1,0)$ will give the same $(0,0,z)-t$ trajectory as for the system of unit masses at $(1,0,0)$, $(-1,0,0)$, $(\cos \theta, \sin \theta, 0)$ and $(-\cos \theta, -\sin \theta, 0)$.

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I wonder if such examples are very rare or very common among all trajectories. –  Theo Johnson-Freyd Feb 9 '13 at 4:12
    
This example only works if the four masses don't interact among themselves and respond to each other's forces by accelerating. –  Ben Crowell Feb 9 '13 at 5:16
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A similar but dynamically valid example would be a test particle moving along the longitudinal axis of a gravitational rosette such as a Klemperer rosette. There are many different rosettes that could produce the same motion. –  Ben Crowell Feb 9 '13 at 5:28
    
Along the lines of Theo's and Brendan's questions, I wonder if some general-position assumption would lead to uniqueness... –  Joseph O'Rourke Feb 9 '13 at 16:51
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Although this question has been answered many times, I would like to add the following simple way to generate examples. It is clear that in the plane any central force, no matter how complicated, generates circular orbits as special cases. There are many ways to distribute matter in three space to generate such a force field. For example, any distribution of mass (which can, of course, consist of point masses) along the $z$-axis will produce such a field in the $x,y$-plane. As a side remark, such configurations occur in practice---e.g., as the gravitational field generated by a wire.

I might add for those interested in the theme of trajectories that there is a considerable body of work on the question of which families of curves can occur as the trajectories of force fields, including a monograph "Differential-geometric aspects of dynamics", by Edward Kasner and his associates, much of which is easly accessible online. This is, however, not directly relevant to the query.

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@jbc: Nice observation re circle trajectories! –  Joseph O'Rourke Feb 13 '13 at 12:22
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