5
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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 $T-U$ $\endgroup$
    – unsym
    Commented Dec 11, 2012 at 5:06

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ woow, interesting!, it blown my mind, Thank you now everything is clear $\endgroup$
    – JHughes
    Commented Dec 11, 2012 at 20:15
  • $\begingroup$ But how do you solve the problem using only the hypothesis of the statement? $h>0$, the solution is defined for all $t$ therefore $U\rightarrow +\infty$ $\endgroup$ Commented Dec 4, 2013 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.