6

So I am doing a Presentation on the $N$-Body Problem for my Physics final (12th grade) and the main content of the Presentation is going to be the comparison between an analytical and a numerical approach to solve the problem.

I chose the "$N$-Body by Taylor series" method as an analytical solution (since the Sundman method seemed too advanced). However I have not been able to find out if the Taylor series converges or not. I have read that the Taylor series solution too the problems has no real life applications because it normally only works in small time frames. Is it because the function doesnt converge or just because its unpractical working with infinitely many terms?

Keep in mind that I am not an expert on the subject of differential equations and Taylor Series

Edit: So the differential equation I have to solve is: $\begin{align} \frac{d^2}{dt^2}\vec{r_i}(t)=G \sum_{k=1}^{n} \frac {m_k(\vec{r}_k(t)-\vec{r}_i(t))} {\lvert\vec{r}_k(t)-\vec{r}_i(t)\rvert^3} \end{align}$

where: $\vec{r_i}(0)$ and $\frac{d}{dt}\vec{r_i}(0)$ are given therefore $\frac{d^2}{dt^2}\vec{r_i}(0)$ is also known

so the taylor series would be: $\begin{align} \sum_{n=0}^{\infty} \frac{1}{n!} \cdot \frac{d^n}{dt^n}\vec{r_i}(0) \cdot t^n \end{align}$ (correct me if I'm wrong)

Eli
  • 11,878
SIMONE ESPOSITO
  • 71
  • 3
  • 1
    Can you give some more details? Such as what N-body problem you're studying, and exactly what the Taylor series for that problem is. Perhaps give a link to what you've been reading so that people can understand what your problem is better. – Dan Pollard May 12 '21 at 13:17
  • @DanPollard i have made an edit I hope its more clear now – SIMONE ESPOSITO May 12 '21 at 13:30
  • Possible duplicates: https://physics.stackexchange.com/q/1235/2451, https://physics.stackexchange.com/q/293245/2451 and links therein. – Qmechanic May 12 '21 at 13:56
  • 1
    @Qmechanic thank you for the suggestions. However the answers there refer to the solution proposed by Wang Qui-Dong which is too advanced for my scope. I am specifically referring tho the simple Taylor series i have described. – SIMONE ESPOSITO May 12 '21 at 14:21
  • Just an editorial comment, I would call this an advanced topic for 12th grade. Keep up the good work! – RC_23 Jul 30 '22 at 00:14
  • Technically the series for the positions can not converge for long time scales because the n-body problem ejects its bodies over the long term, so at least some of them will eventually have coordinates that go towards infinity. At most the velocities or momenta can converge. – FlatterMann Nov 02 '22 at 02:30

1 Answers1

-1

This answer may be a bit late for your final project, but it was fun to research.

Spacetime is a smooth manifold. This is not enough to guarantee analyticity. But it just so happens that spacetime also possesses an analytic structure (see Smooth vs analytic spacetimes), so it is an analytic manifold, so the taylor series of $\vec{r}(t)$ is equal to its Taylor Series for all time $t \geq 0$.

So the Taylor Series is equal to $\vec{r}(t)$, but its convergence is very inefficient. Using the Taylor Series, only short time scales can be practically implemented.

Hypernova
  • 91
  • -1. Spacetime structure is not what is at stake here. The question is whether the solution to the differential equation is analytic or not. In principle, given an ODE system of second order, we only have reason to believe it to be twice continuously differentiable (and it might even fail to be so, as exemplified by any situation in which two bodies cross the same point). Furthermore, notice that the problem is being considered in Newtonian gravity, not General Relativity (spacetime is still a manifold, but your answer gave me the impression you could be thinking of GR) – Níckolas Alves Jul 30 '22 at 01:58
  • From a physical viewpoint, even in Newtonian mechanics, position has to be be a smooth function (no jumps, discontinuities, finite accelerations, etc.), so position is analytic, so it equals its taylor series no matter the situation. Also, physically two bodies won't be able to cross the same point. – Hypernova Jul 30 '22 at 03:18
  • @Níckolas Alves In general, for a 2nd order ODE system, twice differentiability is all that is provided, but doesn't the physical nature of the underlying space and motion (nature of the position vectors which are solutions to the ODE) give the needed additional information about the higher order derivatives? Or am I leaping over a disconnect between mathematical rigor and physical intuition? – Hypernova Jul 30 '22 at 03:28
  • Position doesn't have to be smooth. It does need to be continuous (no teleportation), but it isn't absurd to consider a discontinuous velocity (e.g., for a simple model of an elastic ball bouncing off the ground). Also, even if it was smooth, smoothness of a function doesn't imply that it is continuous. Furthermore, notice that if two bodies of OP's problem pass through the same spot (as it happens if $N=2$ and both start at rest), acceleration becomes infinite at an instant as a consequence of the ODE. That might not happen physically, but the model does admit these problems – Níckolas Alves Jul 30 '22 at 04:01
  • Notice then that there is a difference between the physical world and the model: if you consider that the bodies have finite extension, no two bodies will be at the same place ever, but the problem is way more complex. If you treat them like points, it is easier, but they can get in the same point. Anyway, that was just an immediate example of how the solution might fail to be analytic. There could be other sources of issues – Níckolas Alves Jul 30 '22 at 04:03
  • As for your second comment, I don't agree that the spacetime structure leads to the higher order derivatives. Notice that would imply pretty much that any differential equation has an analytic solution, provided someone can come up with a potential that leads to that ODE, and that is certainly false. Also, even if it did imply smoothness, analyticity is a stronger requirement. – Níckolas Alves Jul 30 '22 at 04:08
  • Yeah, I don't think this is a great answer! Even if it is analytic in t around a point, its radius of convergence is a separate question! I would bet it's analytic (but I don't know about existence theorems in this context) with a finite radius of convergence. – David Nov 02 '22 at 02:09