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Comment author: cousin_it 15 August 2017 12:09:59PM *  0 points [-]

Note that the case analysis is tricky because the function can be discontinuous. The interesting cases are when a ball's X velocity becomes zero due to a collision, or (more subtly) two balls with only Y velocity gain X velocity due to an off-center collision. But I think the statement about monotonicity still holds.

Comment author: Oscar_Cunningham 15 August 2017 01:14:34PM 0 points [-]

Yeah I thought about those two cases as well, but I agree that they are correct. Perhaps we could make the proof a bit simpler by picking the X direction to be one that the balls never travel perpendicular too (although in fact I can't even think of a proof that such a direction exists).

Comment author: Thomas 14 August 2017 07:32:28PM 0 points [-]

Doesn't matter how many collisions will happen, the momentum conservation will hold. Even if only two small balls of each gravity center will collide, the sum of momentums of that two gravity centers will remain the same. Doesn't matter which partition of balls we chose, only that is the same before and after the collision.

Comment author: Oscar_Cunningham 14 August 2017 09:21:49PM 0 points [-]

After some collisions have happened and the two parts are heading away from each other the two parts could still overlap and then some more of their balls could collide. This could lead to the two parts heading back together.

Comment author: Thomas 14 August 2017 03:57:51PM 0 points [-]

Then, both gravity centers travel with a certain velocity each and will collide. How can they return back here? If they will reverse both velocities after the collision. Then, they can return. But with the opposite velocities, and therefore this will not be the same state.

Since they always move in lines, there will be no another collision. And therefore no return to the present state.

Comment author: Oscar_Cunningham 14 August 2017 07:17:10PM 0 points [-]

What do you mean by "the" collision? If each part has several balls then there will be multiple collisions.

Comment author: Thomas 14 August 2017 02:36:04PM 0 points [-]

Do we agree, that if there will be no collision, and both gravity centers move, that they will never return to the present position?

Comment author: Oscar_Cunningham 14 August 2017 02:44:27PM 0 points [-]

Yes.

Comment author: Thomas 14 August 2017 01:23:41PM 0 points [-]

Sure.

If the center of gravity moves, it moves with the velocity v. So it will be in the position r+v*t after some time t. Now it's at the position r. A different position of gravity (mass) center means different position. For the whatever finite t.

In the case when the gravity center doesn't move, you can divide the composition into two sub-compositions, where both gravity centers do move. If only one had moved, then the combined gravity center would move and we would have the solved case above.

But if both gravity centers move, they can either move apart and never collide - in which case they will both have different position vectors lately - or they will collide. In that case, they will reverse their directions after the elastic collision and we have a solved case then.

Well, that's an approximate proof.

Comment author: Oscar_Cunningham 14 August 2017 01:50:57PM 0 points [-]

But if both gravity centers move, they can either move apart and never collide - in which case they will both have different position vectors lately - or they will collide. In that case, they will reverse their directions after the elastic collision and we have a solved case then.

I'm not convinced by this bit. Usually we can calculate the results of an elastic collision by using both conservation of energy and conservation of momentum. But we can't know the energy of the sub-compositions based just on the velocity of their centres of mass. They will also have some internal energy. So we can't calculate the results of the collision.

Comment author: cousin_it 14 August 2017 11:55:17AM *  3 points [-]

1) If only positions should match, then the answer is yes. Just roll two identical balls at each other.

2) If both positions and velocities should match, then the answer is no. Here's a sketch of a proof:

Assume without loss of generality that some balls have nonzero velocity along the X axis. Let's define the following function of time: take all balls having nonzero X velocity, and take the lowest X coordinate of their centers. A case analysis shows that the function can change from increasing to decreasing, but never the other way. Therefore it cannot be periodic. But since it's determined from the configuration, that means the configuration can't be periodic, QED.

Comment author: Oscar_Cunningham 14 August 2017 12:13:39PM 0 points [-]

Yep, that looks pretty airtight to me. Well done!

Comment author: Thomas 14 August 2017 11:17:23AM 0 points [-]

Two balls which are orbiting around its common mass center, they do return to a previous position. If there is no gravity, a finite bunch of rolling balls will never again return to the present state. Never again.

Comment author: Oscar_Cunningham 14 August 2017 11:27:52AM 0 points [-]

I'm assuming no gravity (and that at least one ball is moving). Do you have a proof for your assertion?

Comment author: Thomas 14 August 2017 07:51:03AM 2 points [-]

Another week, another open thread, another problem, too!

Comment author: Oscar_Cunningham 14 August 2017 10:59:39AM *  1 point [-]

EDIT: Clarified some things.

Suppose we have a bunch of spherical billiard balls rolling around on an infinite plane. Suppose there is no friction and the collisions are elastic. They don't feel the influence of gravity or any other force except the collisions. At least one ball is moving. Can they ever return to their original positions and velocities?

Comment author: MrMind 07 August 2017 12:25:51PM 0 points [-]

The intuitive answer seems to me to be: the last one. It's the tallest, so it witness exactly one billion towers. Am I misinterpreting something?

Comment author: Oscar_Cunningham 07 August 2017 12:43:44PM 0 points [-]

I guess some of the towers might block each other from view.

Comment author: Manfred 31 July 2017 10:24:30PM 0 points [-]

Cool insight. We'll just pretend constant density of 3M/4r^3.

This kind of integral shows up all the time in E and M, so I'll give it a shot to keep in practice.

You simplify it by using the law of cosines, to turn the vector subtraction 1/|r-r'|^2 into 1/(|r|^2+|r'|^2+2|r||r'|cos(θ)). And this looks like you still have to worry about integrating two things, but actually you can just call r' due north during the integral over r without loss of generality.

So now we need to integrate 1/(r^2+|r'|^2+2r|r'|cos(θ)) r^2 sin(θ) dr dφ dθ. First take your free 2π from φ. Cosine is the derivative of sine, so substitution makes it obvious that the θ integral gives you a log of cosine. So now we integrate 2πr (ln(r^2+|r'|^2+2r|r'|) - ln(r^2+|r'|^2-2r|r'|)) / 2|r'| dr from 0 to R. Which mathematica says is some nasty inverse-tangent-containing thing.

Okay, maybe I don't actually want to do this integral that much :P

Comment author: Oscar_Cunningham 02 August 2017 07:50:20PM *  1 point [-]

EDIT: On second thoughts most of the following is bullshit. In particular, the answer clearly can't depend logarithmically on R.

I had a long train journey today so I did the integral! And it's more interesting than I expected because it diverges! I got the answer (GM^2/R^2)(9/4)(log(2)-43/12-log(0)). Of course I might have made a numerical mistake somewhere, in particular the number 43/12 looks a bit strange. But the interesting bit is the log(0). The divergence arises because we've modelled matter as a continuum, with parts of it getting arbitrarily close to other parts.

To get an exact answer we would have to look at how atoms are actually arranged in matter, but we can get a rough answer by replacing the 0 in log(0) by r_min/R, where r_min is the average distance between atoms. In most molecules the bond spacing is somewhere around 100 nm. So r_min ~ 10^-10, and R = 6.37*10^6 so log(r_min/R) ~ -38.7, which is more significant than the log(2)-43/12 = -2.89. So we can say that the total is about 38.7*9/4*GM^2/R^2 which is 87GM^2/R^2 or 5.1*10^27.

But after working this out I suddenly got worried that some atoms get even closer than that. Maybe when a cosmic ray hits the earth it does so with such energy that it gets really really close to another nucleus, and then the gravitational force between them dominates the rest of the planet put together. Well the strongest cosmic ray on record is the [Oh-My-God particle with mass 48J. So it would have produced a spacing of about h_bar*c/48, which is about 6.6*10^-28. But the mass of a proton is about 10^-27, so Gm^2/r^2 is about G, and this isn't as significant as I feared.]

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