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I'm trying to put the 3 body problem mathematically. But I don't know how. I always get something reasonable, but I get something that is wrong.

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Not sure what you're struggling with. Could you show us your work and detail where you want to get with it? –  Raskolnikov Mar 18 '12 at 20:55
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It is just an initial-value problem. Wikipedia is pretty clear about it: "In its traditional sense the three-body problem is the problem of taking an initial set of data that specifies the positions, masses and velocities of three bodies for some particular point in time and then determining the motions of the three bodies, in accordance with the laws of classical mechanics: Newton's laws of motion and of universal gravitation." –  lhf Mar 18 '12 at 21:12
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Why not, "find $x_1,x_2,x_3:\mathbb{R}\to\mathbb{R}^3$ which solve $x_i'' = \sum_{j\neq i}\frac{m_j}{|x_i-x_j|^2}$, $i=1,2,3$, subject to the initial conditions $x_i(0) = x_{i,0}$, $x_i'(0)=v_{i,0}$, $i=1,2,3$"? –  Neal Mar 18 '12 at 22:00
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@Neal, you forgot the gravitational constant $G$ in the sum. Perhaps you could add your comment as an answer? –  lhf Mar 19 '12 at 12:46
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@Neal: The equations that you give are not entirely correct because they are scalar equations. We need vector equations. –  Haskell Curry Jan 3 '13 at 0:11

1 Answer 1

  • Suppose that we have three celestial bodies, called Bodies $ 1 $, $ 2 $ and $ 3 $.

  • Let $ \mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3} $ and $ m_{1},m_{2},m_{3} \in \mathbb{R}_{> 0} $ denote the displacement functions and masses of Bodies $ 1 $, $ 2 $ and $ 3 $ respectively.

  • Let $ \mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3} \in \mathbb{R}^{3} $ and $ \mathbf{x}_{1},\mathbf{x}_{2},\mathbf{x}_{3} \in \mathbb{R}^{3} $ denote the initial velocities and initial displacements of Bodies $ 1 $, $ 2 $ and $ 3 $ respectively.

  • According to Newton's Third Law of Motion and his Law of Universal Gravitation, the vector equation of motion for Body $ i $ is given as \begin{align} m_{i} \cdot \mathbf{X}_{i}'' &= \sum_{j \neq i} \frac{G m_{i} m_{j}}{\| \mathbf{X}_{j} - \mathbf{X}_{i} \|^{2}} \cdot \underbrace{\left[ \frac{1}{\| \mathbf{X}_{j} - \mathbf{X}_{i} \|} \cdot (\mathbf{X}_{j} - \mathbf{X}_{i}) \right]}_{\text{Unit vector in the direction $ \mathbf{X}_{j} - \mathbf{X}_{i} $}} \\ &= \sum_{j \neq i} \frac{G m_{i} m_{j}}{\| \mathbf{X}_{j} - \mathbf{X}_{i} \|^{3}} \cdot (\mathbf{X}_{j} - \mathbf{X}_{i}). \end{align} Dividing by $ m_{i} $ on both sides of the equation, we obtain $$ \mathbf{X}_{i}'' = \sum_{j \neq i} \frac{G m_{j}}{\| \mathbf{X}_{j} - \mathbf{X}_{i} \|^{3}} \cdot (\mathbf{X}_{j} - \mathbf{X}_{i}). $$

  • Therefore, the Three-Body Problem is mathematically expressed as the following set of nine equations:

    \begin{align} \forall i \in \{ 1,2,3 \}: \quad {\mathbf{X}_{i}''}(t) &= \sum_{j \neq i} \frac{G m_{j}}{\| {\mathbf{X}_{j}}(t) - {\mathbf{X}_{i}}(t) \|^{3}} \cdot [{\mathbf{X}_{j}}(t) - {\mathbf{X}_{i}}(t)], \\ {\mathbf{X}_{i}'}(0) &= \mathbf{v}_{i}, \\ {\mathbf{X}_{i}}(0) &= \mathbf{x}_{i}. \end{align}

Note: For certain initial-data sets $ (\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{x}_{1},\mathbf{x}_{2},\mathbf{x}_{3}) \in \mathbb{R}^{18} $, a global solution $ (\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3}): \mathbb{R} \to \mathbb{R}^{9} $ does not exist, due to singularities that result from binary or triple collisions among the bodies. However, the collection of all such initial-data sets has Lebesgue measure $ 0 $ in $ \mathbb{R}^{18} $, so these initial-data sets are not generic in the sense of measure.

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This isn't a definition of a three-body problem, it's a statement of one specific example of a three-body problem: the three-body problem for three pointlike particles in Newtonian mechanics, interacting gravitationally. –  Ben Crowell Jan 4 '13 at 1:47
    
@BenCrowell: the question is tagged classical-mechanics –  robjohn Feb 26 '13 at 8:01

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