You tell me at one point that it is a solution, if that is the type of solution

you had in mind I have found it. But it is not a solution to the question you

asked.

RonL

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- Jul 27th 2007, 08:34 PMCaptainBlack
- Jul 28th 2007, 03:37 AMtopsquark
(sigh) Of course I am aware that $\displaystyle \sum_{n = 1}^{\infty} \left ( \frac{1}{2} \right ) ^n = 1$. The point is that if any of the masses are non-zero then they will be attracted to the rest of the collection and start accelerating. Thus the situation is not static.

Perhaps because we can't think of a way to do it?

I have grave doubts about the possibility of this problem having a solution as stated. I have (vaguely) heard of Ulam, but have never studied any of his books so I can't say for sure, but certainly within the limits of my knowledge of Physics and my creativity I can't think of a way to construct such a series of non-zero masses.

Until someone can post a workable solution I'm going to have to say that I think it is impossible.

Perhaps you could try posting your question here. If they do manage to come up with a solution I'd be interested in hearing about it.

-Dan - Jul 28th 2007, 05:22 AMtopsquark
I suppose there

*is*a possibility. I thought of it more in terms of an electrodynamics problem, rather than gravity.

If we had, say two point masses (m = 1/2 each), one positioned at + infinity and one at - infinity. This would do the trick. We can, in fact, place any single mass at the origin (or any other point in between.)

I must add that this isn't really a constructable problem, but as we are already considering point masses at infinity...

-Dan

Edit: No, this isn't going to work either, because the masses need to be constrained to [0, 1] on some line, not out at infinity. Ah well. - Jul 29th 2007, 12:36 PMCaptainBlack
- Jul 29th 2007, 04:04 PMtopsquark
I knew about neither. As for Monte-Carlo I've always viewed that method with a bit of suspicion. Not that I think it doesn't work (though I've heard of solutions that it's missed) but because you pretty much have to know what the solution is going to be before you start in order for it to work well. (I could be wrong about this; I've never studied how to do it, just quickly read an article about its use once.)

-Dan - Aug 1st 2007, 04:53 AMCaptainBlack
It is essentially a method for doing very high dimensional integrals, as such

it is just a question of how much precision you want rather than a question

of it working or not working (at least that is if you can specify the sample

space and distributions).

It is essentially a method of last resort, and as we know these occur almost

all the time in scientific work.

RonL - Aug 1st 2007, 04:57 AMtopsquark