Potential Energy tends to infinity on the N-Body Problem

I need help to solve this problem related with the N-Body problem, i dont understand quite well what I need to define or to express in order to solve it.

We assume a particular solution to the N-Body Problem, for all $t>0$, and $h>0$, where $h$ is the total energy of the N-Bodies, show that $U\rightarrow \infty$ as $t\rightarrow \infty$. This mean that the distance between a pair of particles goes to infinity? (No.)

In the N-Body problem $U$ is given by $U=\sum_{1\leq i< j\leq N}\frac{Gm_{i}m_j}{\left \| q_i-q_j \right \|}$, where $G$ is the gravitational constant. The Kinetic energy is $T=\sum_{i=1}^{N}\frac{\left \| p_i \right \|^2}{2m_i}=\frac{1}{2}\sum_{i=1}^{N}m_i{\left \| \dot{q}_i \right \|^2}$

The vector $q_{i}$ define the position vector of the $i$ particle. So Basically $U$ is like the sum of all the potential energies between all the $N$ particles. Also by the Lagrange Jacobi Formula, we have that $I$ is the moment of inertia, $T$ the kinetic energy so we can express:

$$\ddot{I}=2T-U=T+h\quad,$$

where $h$ is a conserved quantity.

I think that if $U\rightarrow \infty$, then $T\rightarrow \infty$ (because $h$ is constant), the problem is that the only way that i see to $U\rightarrow \infty$ is when the distance between all the particles $\left \| q_i-q_j \right \| \rightarrow 0$, but it means that it will be a collision, so if we have a collision then $t\rightarrow t_1$ and not to $\infty$, because a collision takes a finite amount of time (Sundmanns theorem of total collapse), as I said i dont know what i have to define to show that $U\rightarrow \infty$ as $t\rightarrow \infty$, or maybe i need to define a $q_i(t)$ that in some way that $\left \| q_i-q_j \right \|$ goes very near to zero, but never zero, so $t$ can $t\rightarrow \infty$?

Also, what about the question of a pair of particles going to infinity? It is clear that they should not go to $\infty$ because then $U\rightarrow 0$, and we are trying to prove the other case.

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Are you miss the -ve sign in front of the potential? I means $U=-GMm/r^2$ so that $U \to -\infty$. Also, the definition of the total energy $h$ should be $h = T+U$ not $T-U$ –  hwlau Dec 11 '12 at 5:06
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1 Answer

From virial theorem, stationary states are given by $2T=U$. The "particular solution" your teacher is assuming is a gravitational collapse where $U \gt 2T$ and therefore $U\rightarrow \infty$ as $t\rightarrow \infty$. Of course, the interparticle distance goes to zero in a collapse but this is not a collision: there is a lower bound in a collision and after collision particles increase their separation. In a collapse there is an asymptotic evolution towards a singularity.

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woow, interesting!, it blown my mind, Thank you now everything is clear –  JHughes Dec 11 '12 at 20:15
You are welcome and thanks by the vote. –  juanrga Dec 12 '12 at 9:42
But how do you solve the problem using only the hypothesis of the statement? $h>0$, the solution is defined for all $t$ therefore $U\rightarrow +\infty$ –  Janus Gowda yesterday
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