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andyandy
19th November 2006, 03:15 AM
I was reminded of this by Tez's post here (http://forums.randi.org/showthread.php?postid=2108720#post2108720).......

why is it so difficult to predict what will happen when more than 2 moving objects collide? is it likely to be something that can be predicted in the future? How (un)accurately can it be done at the moment?

Zep
19th November 2006, 04:45 AM
A collision of three balls was not what Tez was talking about. Read again more carefully.

Brian Jackson
19th November 2006, 05:22 AM
3 balls collide....tell me what happens!

I got the wrong prostitute:jaw-dropp

ponderingturtle
19th November 2006, 06:28 AM
I was reminded of this by Tez's post here (http://forums.randi.org/showthread.php?postid=2108720#post2108720).......

why is it so difficult to predict what will happen when more than 2 moving objects collide? is it likely to be something that can be predicted in the future? How (un)accurately can it be done at the moment?

That is not about collisions but the three body problem, and the three body problem has no analytic solution. There are no general ways to describe three massive bodies interacting with gravity, you have to solve such thing numerically.

Physics is full of such problems, there are actually only a handful of problems that can be solved in mechanics, the rest have no analytic solution

Dr. Imago
19th November 2006, 08:07 AM
why is it so difficult to predict what will happen when more than 2 moving objects collide?

Just one suggestion: don't play pool for money.

-Dr. Imago

andyandy
19th November 2006, 08:10 AM
A collision of three balls was not what Tez was talking about. Read again more carefully.

yes, Tez said

we cannot actually simulate three massive particles moving under the influence of Newtonian gravity very well at all, and no one expects we ever will be able to.

and that reminded me of the problem in mechanics in modelling what happens when 3 moving balls collide. Maybe you should read my OP again more carefully :)

Modified
19th November 2006, 09:08 AM
I was reminded of this by Tez's post here (http://forums.randi.org/showthread.php?postid=2108720#post2108720).......

why is it so difficult to predict what will happen when more than 2 moving objects collide? is it likely to be something that can be predicted in the future? How (un)accurately can it be done at the moment?

I would assume it's because any margins of error in the initial conditions or the calculations are compounded with each collision.

Iamme
19th November 2006, 09:29 AM
Originally Posted by andyandy
why is it so difficult to predict what will happen when more than 2 moving objects collide?

Response by Dr. Imago:
Just one suggestion: don't play pool for money.

-Dr. Imago

Excellent point, doctor. Consider trick-shot artists who can arrange every ball...not just 3...and have ALL balls go into predicted pockets.

andyandy
19th November 2006, 09:34 AM
Excellent point, doctor. Consider trick-shot artists who can arrange every ball...not just 3...and have ALL balls go into predicted pockets.

the 3 balls (or more) are not all moving when they collide......

say you have 3 balls A,B,C

all with mass 1kg and velocity 1m/s,

A travelling due south
B travelling due north
C travelling due east

such that after 1 second, if unimpeded, all would have travelled exactly 1 metre and be on point X

then what is the best prediction that can be made about resultant direction and speed of the balls post-collision?

Zep
19th November 2006, 06:11 PM
Assuming no other forces involved, simple vector arithmetic applies.

clarsct
19th November 2006, 07:04 PM
How big are the balls...(She said! *cue rimshot*)

If you're talking pool balls, and we can assume non-elastic collisions, then it is fairly simple.

If you're talking basketballs, where the collisions will deform the balls, thus creating another force at play, then not so easy.

Add balls the size and mass of dwarf stars, and you have a whole new set of calculations.

19th November 2006, 07:48 PM
I have shot a few racks of snooker and I can tell you that billiard game ball collisions can be very predictable.
I think the question refers to perfect spheres meeting instantaneaously on exactly plotted courses.
i do not see why the resultant carrom should not be predictable.
In billiard games there is almost always a succession of collissions influenced by deliberate spin applied to the spheres by the shooter and even when the queball is directed at the midpoint between two close or touching balls it is a case of one ball hitting two others rather than three balls colliding simultaeneously.
The relevant mathematics is far beyond me. I find the question interesting and believe the unpredictability to be a result of imperfect spheres and unequal momentum and trajectory ( if thats the right term).
How does the math play out?

19th November 2006, 07:52 PM
Say the balls are carbide, to minimize if not eliminate the compression factor, and approach from the points of an equilateral triangle with no spin and at equal speed.

wollery
19th November 2006, 07:54 PM
How big are the balls...(She said! *cue rimshot*)

If you're talking pool balls, and we can assume non-elastic collisions, then it is fairly simple.

If you're talking basketballs, where the collisions will deform the balls, thus creating another force at play, then not so easy.

Add balls the size and mass of dwarf stars, and you have a whole new set of calculations.[Pedantic git]
Pool ball collisions are about as close to perfectly elastic as we can get in a real life situation, due to the fact that the deformation of the balls is almost zero and friction between the balls is almost zero. An object such as a basketball, by virtue of the fact that it deforms, does not undergo elastic collision. The deformation causes energy losses from heating of the material as does friction between the elastic surfaces of the balls if the collision is not head on.
[/Pedantic git]

clarsct
19th November 2006, 07:59 PM
I knew I had a term confused there...

I meant what Wollery said...switch non-elastic for elastic.

Thank you.

Zep
19th November 2006, 10:24 PM
Booooooooiiiiing!

wollery
19th November 2006, 10:37 PM
I'm trying to work out whether that's an extended 'boing' or an extended 'booing'.

andyandy
20th November 2006, 08:34 AM
Assuming no other forces involved, simple vector arithmetic applies.

really?

so 3 balls A,B,C

all with mass 1kg and velocity 1m/s,

A travelling due south [0,-1]
B travelling due north [0,1]
C travelling due east [1,0]

such that after 1 second, if unimpeded, all would have travelled exactly 1 metre and be on point X

say you assume the three balls were zero size points for simplicity, assuming total elastic collision, ignoring friction and gravity....

then at point X, ball A would be simultaneously exerting force on both balls B and C, and having a force exerted upon it by B and C. The same situation applies to the other two balls...
how would you model that with vector arithmetic?
Say collision was T=1 second, at T=2 seconds where would the balls be?

Crazycowbob
20th November 2006, 09:17 AM
Say the balls are carbide, to minimize if not eliminate the compression factor, and approach from the points of an equilateral triangle with no spin and at equal speed.

Then, barring any resistances, and assuming also for a perfectly round ball, after the collusion they would all be traveling with their original velocities, only traveling directly towards their origins (reversed directions). Now change an angle or two, or the velocity of a ball or two, or the roundness of a ball or two, and there you've got some major figuring to do.

boooeee
20th November 2006, 09:34 AM
Actually, I think andyandy is right. In the idealized situation he described, the resulting trajectories of the balls is indeterminate.

There are three unknowns.

From symmetry, you know that ball C will travel in the east-west direction.

Also from symmetry, you know that the east-west velocity component of balls A and B will have the same magnitude, and that the north south component of their velocity will have the same magnitude, but opposite direction.

Applying momentum and energy conservation only gets you two equations for these three unknowns, meaning that there's an infinite number of possibilities.

Does this mean that Newtonian physics is indeterminate? No.

It just means you haven't specified enough information about the balls' interaction with eachother. It's like asking what the orbit of a planet around a star will be without knowing what the gravitational constant is.

Any realistic interaction between the balls will have a finite range. You need to specify the nature of that interaction at all distances. You can then solve for the trajectory of the balls for any desired degree of precision (although a closed form solution would probably be impossible).

Crossbow
20th November 2006, 09:41 AM
the 3 balls (or more) are not all moving when they collide......

say you have 3 balls A,B,C

all with mass 1kg and velocity 1m/s,

A travelling due south
B travelling due north
C travelling due east

such that after 1 second, if unimpeded, all would have travelled exactly 1 metre and be on point X

then what is the best prediction that can be made about resultant direction and speed of the balls post-collision?

Thanks for the extra data, however without knowing exactly where the balls are when they impact, it is impossible to say which ball will be doing what after the impact.

For example,
IF all three balls impacted each other at exactly the same moment by forming a straight line between their centers with ball 'A' north, ball 'C' in the middle, and ball 'C' south, and provided that one ignores things like inelastic collisions and spin problems,
THEN ball 'A' would turn around and travel north at 1 m/s, while ball 'C' would continue would continue east at 1 m/s, and ball 'B' would turn around and travel south at 1 m/s.

However, it is quite unlikely that such an ideal collision would occur between three separate objects, therefore, all that one can say is that the momentum of the system will be conserved after the collision so that the three balls will continue to have 1 kg*m/s of momentum in a direction to the east.

Beleth
20th November 2006, 10:33 AM
For example,
IF all three balls impacted each other at exactly the same moment by forming a straight line between their centers with ball 'A' north, ball 'C' in the middle, and ball 'C' south, and provided that one ignores things like inelastic collisions and spin problems,
THEN ball 'A' would turn around and travel north at 1 m/s, while ball 'C' would continue would continue east at 1 m/s, and ball 'B' would turn around and travel south at 1 m/s.
I don't think that's possible given the constraints of the original problem:

such that after 1 second, if unimpeded, all would have travelled exactly 1 metre and be on point X
If they formed a straight line like that, ball C would be on point X, and the other two balls would not be.

A quick calculation shows that C will simultaneously collide with A and B before A and B collide with each other. The three balls (pre-collision) form an isosceles right triangle with angle ACB being the right angle.

boooeee
20th November 2006, 11:00 AM
I don't think that's possible given the constraints of the original problem:

If they formed a straight line like that, ball C would be on point X, and the other two balls would not be.

A quick calculation shows that C will simultaneously collide with A and B before A and B collide with each other. The three balls (pre-collision) form an isosceles right triangle with angle ACB being the right angle.
He specifies later on to assume that the balls have zero size. Under that condition, the collision between all three would be simultaneous.

Unnamed
20th November 2006, 11:08 AM
He specifies later on to assume that the balls have zero size. Under that condition, the collision between all three would be simultaneous.
Take Beleth's explanation to the limit where the ball size goes to zero. It remains valid all the way.

Beleth
20th November 2006, 11:13 AM
He specifies later on to assume that the balls have zero size. Under that condition, the collision between all three would be simultaneous.
Doh! I see that now.

He also said to assume that "for simplicity"... but IMO it actually makes the problem harder. Indeed, it sounds like it makes it impossible.

69dodge
20th November 2006, 11:22 AM
really?

so 3 balls A,B,C

all with mass 1kg and velocity 1m/s,

A travelling due south [0,-1]
B travelling due north [0,1]
C travelling due east [1,0]

such that after 1 second, if unimpeded, all would have travelled exactly 1 metre and be on point X

say you assume the three balls were zero size points for simplicity, assuming total elastic collision, ignoring friction and gravity....

then at point X, ball A would be simultaneously exerting force on both balls B and C, and having a force exerted upon it by B and C. The same situation applies to the other two balls...
how would you model that with vector arithmetic?
Say collision was T=1 second, at T=2 seconds where would the balls be?Does assuming zero size really make things simpler? I think that, on the contrary, it's what's causing the trouble. [ETA: as I now see Beleth has already said]

If you assume zero size, there's already trouble even with only two balls. For example, forget about ball C. When A and B collide, will they reverse directions, and end up moving, respectively, north and south? Or will they perhaps glance off each other, and end up moving east and west? Each of these outcomes (as well as infinitely many others) conserves both momentum and energy.

If we make the reasonable assumption that all collisions between points are direct hits, and not glancing blows, then the original three-point problem is also solvable: C bounces back due west at 1/3 m/s, and A and B reverse the directions of their 1 m/s north-south velocity components while also each gaining an eastward component of 2/3 m/s.

(If a pair of large balls approach each other, you can figure out how they'll rebound by looking at their angular momentum, which also needs to be conserved: zero means a direct hit, nonzero means a glancing blow. Points are so small that they won't collide at all unless they're moving directly toward each other, but if they are, their zero angular momentum is conserved regardless of the directions in which they rebound.)

Crazycowbob
20th November 2006, 11:24 AM
Hmm, actually, if you take them down to zero size, then you don't have to deal with the angle that their sides impact, and can instead use vector addition/subtraction to get the resulting ball movements. Given that situation, the result would be ball A traveling north at 1m/s, east at 1/2m/s, ball B traveling south 1m/s, east 1/2m/s, and ball C stopping at the point of impact.

Crazycowbob
20th November 2006, 11:26 AM
Does assuming zero size really make things simpler? I think that, on the contrary, it's what's causing the trouble. [ETA: as I now see Beleth has already said]

If you assume zero size, there's already trouble even with only two balls. For example, forget about ball C. When A and B collide, will they reverse directions, and end up moving, respectively, north and south? Or will they perhaps glance off each other, and end up moving east and west? Each of these outcomes (as well as infinitely many others) conserves both momentum and energy.

If we make the reasonable assumption that all collisions between points are direct hits, and not glancing blows, then the original three-point problem is also solvable: C bounces back due west at 1/3 m/s, and A and B reverse the directions of their 1 m/s north-south velocity components while also each gaining an eastward component of 2/3 m/s.

(If a pair of large balls approach each other, you can figure out how they'll rebound by looking at their angular momentum, which also needs to be conserved: zero means a direct hit, nonzero means a glancing blow. Points are so small that they won't collide at all unless they're moving directly toward each other, but if they are, their zero angular momentum is conserved regardless of the directions in which they rebound.)

You forgot that ball C has no opposing force on it, ideally it should transfer all it's momentum to balls A and B, like a ball in pool stopping when it hit's another ball square on.

69dodge
20th November 2006, 11:39 AM
You forgot that ball C has no opposing force on it, ideally it should transfer all it's momentum to balls A and B, like a ball in pool stopping when it hit's another ball square on.But C hits A and B simultaneously, and together A and B weigh more than C. So it's as if a light ball hit a heavy one, and in that case it bounces back instead of stopping.

Crazycowbob
20th November 2006, 11:50 AM
ooops, you're entirely correct, me and my showboat physics LOL...

Ben Tilly
20th November 2006, 11:53 AM
That is not about collisions but the three body problem, and the three body problem has no analytic solution. There are no general ways to describe three massive bodies interacting with gravity, you have to solve such thing numerically.

Sorry, that's not quite right.

There do exist analytic solutions of the 3 body problem. And in the 90's a solution was found for the general n body problem. However the analytic solutions converge very slowly so they are computationally useless.

Cheers,
Ben

Ben Tilly
20th November 2006, 12:24 PM
really?

so 3 balls A,B,C

all with mass 1kg and velocity 1m/s,

A travelling due south [0,-1]
B travelling due north [0,1]
C travelling due east [1,0]

such that after 1 second, if unimpeded, all would have travelled exactly 1 metre and be on point X

say you assume the three balls were zero size points for simplicity, assuming total elastic collision, ignoring friction and gravity....

then at point X, ball A would be simultaneously exerting force on both balls B and C, and having a force exerted upon it by B and C. The same situation applies to the other two balls...
how would you model that with vector arithmetic?
Say collision was T=1 second, at T=2 seconds where would the balls be?

Zero point size is a complication. Let's make them have finite size so we can calculate angles.

Let's simplify further by supposing that, all three hit at the same time, forming an equilateral triangle.

Now we still have an indeterminate solution. The problem is that there are many possible solutions. For instance consider the nearby situation where A hits C, then C hits B. In that case most of C's eastward momentum goes to A. But we also have the nearby situation where B hits C, then C hits A. In that case most of C's eastward momentum goes to B. The limits of both situations provide two very different yet perfectly valid solutions to the mathematics of the three way collision. (There are others) Therefore the problem is indeterminate.

For a simpler version of the same problem, try throwing a bouncy ball at a corner. Depending on which wall the ball hits first, the ball will wind up bouncing off in one of two different ways. What happens if you hit the corner dead on? Well both solutions are valid, so simple Newtonian mechanics is indeterminate. (I invite people to try it. While it is theoretically possible, I would be very surprised if anyone can get it to hit squarely enough to get anything other than one of the two limiting solutions that I just outlined.)

Cheers,
Ben

Crazycowbob
20th November 2006, 01:31 PM
Zero point size is a complication. Let's make them have finite size so we can calculate angles.

Let's simplify further by supposing that, all three hit at the same time, forming an equilateral triangle.

Now we still have an indeterminate solution. The problem is that there are many possible solutions. For instance consider the nearby situation where A hits C, then C hits B. In that case most of C's eastward momentum goes to A. But we also have the nearby situation where B hits C, then C hits A. In that case most of C's eastward momentum goes to B. The limits of both situations provide two very different yet perfectly valid solutions to the mathematics of the three way collision. (There are others) Therefore the problem is indeterminate.

For a simpler version of the same problem, try throwing a bouncy ball at a corner. Depending on which wall the ball hits first, the ball will wind up bouncing off in one of two different ways. What happens if you hit the corner dead on? Well both solutions are valid, so simple Newtonian mechanics is indeterminate. (I invite people to try it. While it is theoretically possible, I would be very surprised if anyone can get it to hit squarely enough to get anything other than one of the two limiting solutions that I just outlined.)

Cheers,
Ben

But what you're describing is them hitting at different (albiet miniscule) times. The ideal that we were looking at has all three striking at precisely the same instant, it would be like the bouncing ball hitting the corner at exactly the center, in which the vector forces of all the walls would act on it equally, sending the ball right back to it's source, which is the solution the limit should resolve to as the position of the ball reaches the center of the corner.

Aepervius
20th November 2006, 01:44 PM
Quote from wikipedia "As it has been shown by Siegel, that collisions which involve more than 2 bodies cannot be regularised analytically, [hence Sundman's regularization cannot be generalised.]".

I know you are not trying to find a regular analitycal solution, but IMO since you are speaking of a simultaneous 3 body problem, I fear your hope of agreeing or finding a solution to it is, let us say, thin.

Ben Tilly
20th November 2006, 02:45 PM
But what you're describing is them hitting at different (albiet miniscule) times. The ideal that we were looking at has all three striking at precisely the same instant, it would be like the bouncing ball hitting the corner at exactly the center, in which the vector forces of all the walls would act on it equally, sending the ball right back to it's source, which is the solution the limit should resolve to as the position of the ball reaches the center of the corner.

Sorry, the limits don't resolve to that situation.

The limits resolve to 2 very different solutions, as you approach the limit from different directions. The actual question under consideration has an infinite number of possible mathematical solutions, including the two limiting solutions. Therefore the problem is indeterminate.

The point of bringing up the limiting solutions is that it is easy to see, without having to write a bunch of equations down, that they are possible yet very different solutions.

Cheers,
Ben

Ben Tilly
20th November 2006, 03:00 PM
Quote from wikipedia "As it has been shown by Siegel, that collisions which involve more than 2 bodies cannot be regularised analytically, [hence Sundman's regularization cannot be generalised.]".

I know you are not trying to find a regular analitycal solution, but IMO since you are speaking of a simultaneous 3 body problem, I fear your hope of agreeing or finding a solution to it is, let us say, thin.

Read wikipedia more closely. The simultaneous 3 body problem has been solved analytically for almost all (1) cases. As has the general n body problem.

The analytic solutions fail in the case of a collision. Guess what? Numerical solutions also fail in the case of a collision! And in fact the simple theory of gravity doesn't handle the case of a collision either!

In my books, that's solved.

Cheers,
Ben

(1) "Almost all" has a precise definition, which I am using. That definition is, "The set of exceptions has Lesbegue measure 0." A technical explanation of that takes a while, but in the case of a line, a set of measure zero is a set of points without any length. In the case of a plane, a set of measure zero might be a set of lines without any area. In the case of 3 dimensions, a set of measure zero might be a set of planes without any volume. And so on.

69dodge
20th November 2006, 03:12 PM
Sorry, the limits don't resolve to that situation.

The limits resolve to 2 very different solutions, as you approach the limit from different directions. The actual question under consideration has an infinite number of possible mathematical solutions, including the two limiting solutions. Therefore the problem is indeterminate.

The point of bringing up the limiting solutions is that it is easy to see, without having to write a bunch of equations down, that they are possible yet very different solutions.I see that if the balls collide two at a time, then what happens depends on the order in which they collide. And infinitesimal differences in their initial positions can change that order. So their behaviour after colliding is a discontinuous function of their previous behaviour. But I don't see why there are an infinite number of mathematical solutions in the idealized case where the three collide simultaneously. Doesn't the symmetry of the problem require that the solution also be symmetric?

Beleth
20th November 2006, 03:32 PM
But I don't see why there are an infinite number of mathematical solutions in the idealized case where the three collide simultaneously. Doesn't the symmetry of the problem require that the solution also be symmetric?
I tend to agree. In the idealized, non-point-source case, the momentum and location of impact of each of the three balls should be easily discernable. Then movement on each of the two axes in the plane of collision could be figured out separately and added together at the end.

Especially if the problem were changed just slightly so that instead of them all reaching point X at the same time, they all collide at the same time.

Then again, you know what they say about problems that you don't have to figure out yourself...

Ben Tilly
20th November 2006, 03:58 PM
I see that if the balls collide two at a time, then what happens depends on the order in which they collide. And infinitesimal differences in their initial positions can change that order. So their behaviour after colliding is a discontinuous function of their previous behaviour. But I don't see why there are an infinite number of mathematical solutions in the idealized case where the three collide simultaneously. Doesn't the symmetry of the problem require that the solution also be symmetric?

There are an infinite number of solutions to the mathematical constraints of Newtonian mechanics. The math is indeterminate, it does not uniquely specify what will happen.

You're right that there is only one solution if one adds the additional rule that symmetry must be maintained. But that is not a rule from Newtonian mechanics. That is an extra constraint. And furthermore when you try to run the experiment with, say, hockey pucks, you'll never actually see anything that looks like that solution.

Now what happens if we tweak the problem so the weights and masses differ by .0001%? There are still an infinite number of solutions to the mathematical constraints of Newtonian mechanics, but we no longer have any symmetry to help us decide which solution that we prefer. (And the "preferred" solution, not being one of the limiting solutions, will never actually be observed in practice anyways...)

Cheers,
Ben

andyandy
20th November 2006, 11:25 PM
Zero point size is a complication. Let's make them have finite size so we can calculate angles.

Let's simplify further by supposing that, all three hit at the same time, forming an equilateral triangle.

Ben

yes...i put zero point in to ensure that in my example the balls would all collide simultaneously.....

as has been pointed out, this would not be the case with [1,0] [0,1] [0,-1]

[0,-1] [-1/2, sin120] [1/2, sin120]

would ensure, i think (dusting off some trig......:) ) that they do......

Crossbow
21st November 2006, 05:30 AM
yes...i put zero point in to ensure that in my example the balls would all collide simultaneously.....

as has been pointed out, this would not be the case with [1,0] [0,1] [0,-1]

[0,-1] [-1/2, sin120] [1/2, sin120]

would ensure, i think (dusting off some trig......:) ) that they do......

The problem I see is that while you do specify all three of the balls impact simultaneously, you do not specify how they impact simultaneously.

The post-collision veolicities of each of the balls can be readily calculated using conservation of momentum provided that you know exactly where the balls are relative to each other when they impact.

This the is fact that I tried to point in my earlier post.

Ben Tilly
21st November 2006, 10:48 AM
The problem I see is that while you do specify all three of the balls impact simultaneously, you do not specify how they impact simultaneously.

The post-collision veolicities of each of the balls can be readily calculated using conservation of momentum provided that you know exactly where the balls are relative to each other when they impact.

This the is fact that I tried to point in my earlier post.

This "fact" is not.

Don't believe me? Try to calculate it!

Since everyone has been lazy so far, I'll show you how to do it. And I'll show you why the calculation does not lead to a unique solution.

Take the case where three rigid balls A, B, and C impact at once in an elastic collision. All three have the same mass, which we will say is 1. A has velocity (0, -1) (ie due south), B has velocity (0,1) (ie due north) and C has velocity (1, 0) (ie due east). At the point of impact A and B hit dead on vertically, and C hits forming an equilateral triangle with the other two. In other words if the radius of each ball is 1, then at the moment of impact A is at (0, 1), B is at (0, -1) and C is at (sqrt(3)/2, 0).

Now we have 6 sets of momentum transfers. x1 = transfer from A to B, x2 = transfer from B to A, x3 = transfer from A to C, x4 = transfer from C to A, x5 = transfer from B to C, and x6 = transfer from C to B. Conservation of momentum is easy to satisfy, we must have x1 = - x2, x3 = -x4, and x5 = - x6. So we just need to figure out x1, x3 and x5.

After the collision the velocity of A is going to be (0, -1) + (0, x2) + (sqrt(3)*x4/2, x4/2). Which is (-sqrt(3)*x3/2, -1 - x1 - x3/2).

Similarly we can work out the velocity of B and C.

Now we specified an elastic collision. That means that there is one more constraint, namely that energy is conserved. (Inelastic solutions include things like all three hit, and fuse into a lump moving at (1/3, 0). Elastic is a stronger constraint.) Which means that the sum of the squares of the velocities has to remain the same.

This gives us one second-order equation in 3 variables. We can discard the solutions that result in balls continuing to move towards each other as being physically impossible. We still wind up with 1 equation and 3 variables, and as a result there is an infinite number of solutions. (In fact one could graph this in 3 dimensions and the set of possible solutions would form a 2-dimensional surface.)

In point of physical fact, any attempt to run this experiment is very likely to see one of only 6 solutions come out of it. And those 6 are the limiting cases of 3 sets of pairwise collisions very close together.

Are there any questions about this explanation? If not then I hope we can put this to rest. The solution is indeterminate. Really.

Note that if you apply this reasoning in the 2 ball case you get a very different outcome. Rather than 6 variables to solve for initially, we have 2. Conservation of momentum gets rid of one. Then conservation of energy gives us a second order equation in one variable. This has 2 solutions. One solution is that the balls continue in their original way. We discard that because the balls have collided and cannot travel into each other. The other solution is our answer.

Cheers,
Ben

boooeee
21st November 2006, 01:06 PM
Not to speak for Crossbow, but I imagine that he is assuming perfectly rigid billiard balls. Knowing that and knowing the radius of the balls leads to a unique solution.

Conservation of momentum and energy can only get you so far (as you showed above). You also need to know the specifics of the interaction.

Formally, the balls interaction with eachother can be modelled with a delta function of the balls' mutual distance. The delta function would spike at 2R, where R is the ball's radius.

Ben Tilly
21st November 2006, 02:24 PM
Not to speak for Crossbow, but I imagine that he is assuming perfectly rigid billiard balls. Knowing that and knowing the radius of the balls leads to a unique solution.

Conservation of momentum and energy can only get you so far (as you showed above). You also need to know the specifics of the interaction.

Formally, the balls interaction with eachother can be modelled with a delta function of the balls' mutual distance. The delta function would spike at 2R, where R is the ball's radius.

Please demonstrate how to get from "perfectly rigid billiard balls" to the exact delta functions that fully model the interactions.

When you try, you should find that your delta functions just turn out to be a way of restating what I just said with x1, x3 and x5. However the assumption of rigidity does not specify the delta functions and does not specify the interaction.

If you disagree, don't wave your hands and say it can be done. Instead get out a piece of paper and attempt to do the calculation.

Incidentally your bringing up delta functions reminds me of an old math joke. A novice was learning math at a Buddhist monastery. He goes up to the monk after class and says, "Master, I do not understand the Dirac delta." The monk slapped him and the novice was enlightened.

Cheers,
Ben

Ben Tilly
21st November 2006, 02:27 PM
You're right that there is only one solution if one adds the additional rule that symmetry must be maintained. But that is not a rule from Newtonian mechanics. That is an extra constraint. And furthermore when you try to run the experiment with, say, hockey pucks, you'll never actually see anything that looks like that solution.

And I was, of course, wrong. Even if you insist on symmetry there remain an infinite number of solutions. For one example, C can fly back west at a rate of 1 while both A and B wind up going east at a rate of 1. For another, C can stop, while A and B wind up travelling east at a rate of 1/2, and north and south at sqrt(5)/2.

Cheers,
Ben

andyandy
21st November 2006, 02:29 PM
Please demonstrate how to get from "perfectly rigid billiard balls" to the exact delta functions that fully model the interactions.

If you disagree, don't wave your hands and say it can be done. Instead get out a piece of paper and attempt to do the calculation.

Incidentally your bringing up delta functions reminds me of an old math joke. A novice was learning math at a Buddhist monastery. He goes up to the monk after class and says, "Master, I do not understand the Dirac delta." The monk slapped him and the novice was enlightened.

Cheers,
Ben

I've been interested to read everyone's posts on this.....so there's no need to be quite so combative - we have the politics forum for that :D

Ben Tilly
21st November 2006, 02:51 PM
I've been interested to read everyone's posts on this.....so there's no need to be quite so combative - we have the politics forum for that :D

My apologies. I didn't think I was being that combative. But I must admit that I am getting tired of this continuing with "is", "isn't", "is", "isn't" with some of my isn'ts accompanied by painfully long explanations.

boooeee
21st November 2006, 05:25 PM
Ben - Are you saying that the delta function does not accurately represent the interaction?

Or, are you saying that even if you assume a delta function interaction, the answer is still indeterminate?

Loved the delta function joke, by the way.

Ben Tilly
21st November 2006, 06:03 PM
Ben - Are you saying that the delta function does not accurately represent the interaction?

Or, are you saying that even if you assume a delta function interaction, the answer is still indeterminate?

I'm saying that knowledge of the ball's rigidity and radius does not tell you the right delta functions to use. Of course knowing the right delta functions to use will tell you exactly what will happen. In fact you only need to know 2 of the delta functions - the third can be worked out using conservation of energy.

But if you do the calculation, you'll find that it is looking eerily familiar. In fact you're going to be reproducing what I did with x1, x3 and x5 above. Why? Because delta functions are just a representation of transferring momentum in an impact. So talking about the right delta function to use is just another way of talking about how much momentum you wish to transfer, and you're just restating what I said in a different language.

Loved the delta function joke, by the way.

That looked like encouragement. You don't want to encourage me. Just trust me on this. :-)

Cheers,
Ben

ObTriva: A delta function is not actually a function.

Soapy Sam
22nd November 2006, 05:04 PM
Side issue.
Can a zero dimensional point move?
How would we know?

Art Vandelay
22nd November 2006, 06:12 PM
Read wikipedia more closely. The simultaneous 3 body problem has been solved analytically for almost all (1) cases. As has the general n body problem.Of course, three balls, in "almost all" cases, don't hit silmultaneously. Which means that in any real situation, we need to find a solution for the balls colliding almost simultaneously, which means that we're probabably going to take the limit, which is a problem if the limit is different depending on how we approach.

But I don't see why there are an infinite number of mathematical solutions in the idealized case where the three collide simultaneously. Doesn't the symmetry of the problem require that the solution also be symmetric?If there is a deterministic answer, it must preserve symmtery. But then you're making a circular argument: if it's deterministic, then there can't be an infinite number of answers.

Side issue.
Can a zero dimensional point move?
How would we know?Well, all points are zero dimensional. And they can't really "move"; a point is a particular location in space. If a point could move, then it wouldn't be a particular location. Of course, while all points have zero dimensions, not everything with zero dimensions is a point. And I think that what you're asking is whether something with zero dimensions can move. And the answer is that it is just as capable of moving as anything else. As for whether we'd know, we'd know it moved just as much as we know that it exists in the first place.

wollery
23rd November 2006, 05:54 PM
Photons are massless point particles, i.e. they are dimensionless, and we know that they exist and move.

23rd November 2006, 09:50 PM
Ok, look, I cant do the math and it looks as though no one else can.
I can tell you this much, thre balls comming at equal speed with no spin from the points of an equilateral triangle would be rebounded back towards their points of origin.

Your E. S. and W. conunundrum can only be based on the uncertainty originating from your inability to measure ie: you cant do the math. I can, and I do not need a pen and paper, only my eyes and brain.
yf
cm

23rd November 2006, 10:02 PM
In your , mathematically unapproachable scenario, that being E.S.N. APPROACHES OF EQUAL spherical bodies with no spin, the result of collision would be; reversal of the E. sphere directly back to its point of origin. the N. and S. spheres would 'kiss off' the E. sphere at 45 degree angles subsequently meet and ... may i leave the rest to the significant logic/science oriented people who attend this site?
I know what happens next, do you?

23rd November 2006, 10:04 PM
I saw lots of allusion to math, I did not seee any math applied to the question. If you are so smart, do the math.

23rd November 2006, 10:12 PM
My brother Michel or either one of his sons could address this problem mathematically, showing their work, inside five minutes. Why cant you?

Art Vandelay
23rd November 2006, 10:29 PM
Photons are massless point particles, i.e. they are dimensionless, and we know that they exist and move. In what sense are they dimensionless? They have a wavefunction that is spread over a three dimensional volume of space. And they lack "mass" in the sense of rest mass, but they have mass in the sense of energy. So if they are restricted to a point, doesn't that imply infinite density (and that the photon is smaller than its Schartzchild radius, and is thus a black hole)?

wollery
23rd November 2006, 10:59 PM
In what sense are they dimensionless? They have a wavefunction that is spread over a three dimensional volume of space. And they lack "mass" in the sense of rest mass, but they have mass in the sense of energy. So if they are restricted to a point, doesn't that imply infinite density (and that the photon is smaller than its Schartzchild radius, and is thus a black hole)?The wavefunction is a measure of the probability of interaction between the photon and anything else as a function of position, but the wavefunction is not the photon, and although it has mass equivalence it does not have mass.

MortFurd
24th November 2006, 02:10 AM
I got the wrong prostitute:jaw-dropp
You've only got one ball? :jaw-dropp

andyandy
24th November 2006, 03:40 AM
My brother Michel or either one of his sons could address this problem mathematically, showing their work, inside five minutes. Why cant you?

you yourself say "Ok, look, I cant do the math" and then ask "why can't you?"

when you say that your brother and his kids could model this problem.....do you mean to impress with your genetic heritage or belittle the problem by showing its childlike simplicity?

what a strange mixed up troll you are......

Cuddles
24th November 2006, 04:44 AM
I saw lots of allusion to math, I did not seee any math applied to the question. If you are so smart, do the math.

I suggest you read post #42.

Also, if you have nothing productive to add, why bother?

69dodge
24th November 2006, 10:29 AM
If there is a deterministic answer, it must preserve symmtery. But then you're making a circular argument: if it's deterministic, then there can't be an infinite number of answers.I don't get it. Why must we already know that there's only one answer, in order to conclude that any answer must preserve the symmetry of the question?

69dodge
24th November 2006, 10:42 AM
And I was, of course, wrong. Even if you insist on symmetry there remain an infinite number of solutions. For one example, C can fly back west at a rate of 1 while both A and B wind up going east at a rate of 1. For another, C can stop, while A and B wind up travelling east at a rate of 1/2, and north and south at sqrt(5)/2.I was talking about the case symmetric in all three balls, I think.

I see your point here, where A and B are equivalent but C is different.

Art Vandelay
25th November 2006, 01:25 PM
I don't get it. Why must we already know that there's only one answer, in order to conclude that any answer must preserve the symmetry of the question?If there's more than one answer, then the possible answers taken together can preserve the symmetry, without individual answers preserving it.

69dodge
25th November 2006, 09:57 PM
"Symmetry" can mean all sorts of things. Are you sure you aren't changing meaning midstream?

Three identical balls are in identical situations. How can they behave differently from each other?

I guess I do assume that each ball behaves determistically. But I don't assume that the system of the three of them does; I conclude from symmetry that it does.

More precisely: I assume that each ball behaves in some deterministic way, but I make no assumptions, beyond conservation of momentum and energy, about which way that is. And then I conclude from this assumption, together with the symmetry of the situation, that the three-ball system behaves in one particular deterministic way, namely, that each ball reverses its velocity.

Art Vandelay
26th November 2006, 02:46 PM
Three identical balls are in identical situations. How can they behave differently from each other?How can two coins in identical situations act differently? Two photons?

Crossbow
27th November 2006, 06:25 AM
This "fact" is not.

Don't believe me? Try to calculate it!

Since everyone has been lazy so far, I'll show you how to do it. And I'll show you why the calculation does not lead to a unique solution.

Take the case where three rigid balls A, B, and C impact at once in an elastic collision. All three have the same mass, which we will say is 1. A has velocity (0, -1) (ie due south), B has velocity (0,1) (ie due north) and C has velocity (1, 0) (ie due east). At the point of impact A and B hit dead on vertically, and C hits forming an equilateral triangle with the other two. In other words if the radius of each ball is 1, then at the moment of impact A is at (0, 1), B is at (0, -1) and C is at (sqrt(3)/2, 0).

Now we have 6 sets of momentum transfers. x1 = transfer from A to B, x2 = transfer from B to A, x3 = transfer from A to C, x4 = transfer from C to A, x5 = transfer from B to C, and x6 = transfer from C to B. Conservation of momentum is easy to satisfy, we must have x1 = - x2, x3 = -x4, and x5 = - x6. So we just need to figure out x1, x3 and x5.

After the collision the velocity of A is going to be (0, -1) + (0, x2) + (sqrt(3)*x4/2, x4/2). Which is (-sqrt(3)*x3/2, -1 - x1 - x3/2).

Similarly we can work out the velocity of B and C.

Now we specified an elastic collision. That means that there is one more constraint, namely that energy is conserved. (Inelastic solutions include things like all three hit, and fuse into a lump moving at (1/3, 0). Elastic is a stronger constraint.) Which means that the sum of the squares of the velocities has to remain the same.

This gives us one second-order equation in 3 variables. We can discard the solutions that result in balls continuing to move towards each other as being physically impossible. We still wind up with 1 equation and 3 variables, and as a result there is an infinite number of solutions. (In fact one could graph this in 3 dimensions and the set of possible solutions would form a 2-dimensional surface.)

In point of physical fact, any attempt to run this experiment is very likely to see one of only 6 solutions come out of it. And those 6 are the limiting cases of 3 sets of pairwise collisions very close together.

Are there any questions about this explanation? If not then I hope we can put this to rest. The solution is indeterminate. Really.

Note that if you apply this reasoning in the 2 ball case you get a very different outcome. Rather than 6 variables to solve for initially, we have 2. Conservation of momentum gets rid of one. Then conservation of energy gives us a second order equation in one variable. This has 2 solutions. One solution is that the balls continue in their original way. We discard that because the balls have collided and cannot travel into each other. The other solution is our answer.

Cheers,
Ben

Sorry about the delayed reply, I have been away from the Internet over the holiday.

Also, I did what 'Ben Tilly' suggested and tried to calculate the post-collision momentum vectors of the three balls (assuming perfectly elastic collisions, 1 kg mass, etc.) and found out that there is simply not enough data to determine what would happen to each of the three balls.

While one can accurately work out the pre-collision momentum of each individual ball as well as the total momentum of the system of the three balls, however there is simply not enough information furnished that would enable one to determine the post-collision momentum of each individual ball.

Originally, I was thinking that provided one knew the pre-collision momentum of each ball and that one knew the position of each ball at the point of impact, then one could calculate the post-collision velocity of each ball. However, I did do some drawings, did some calculations, and I did not get the results I expected. So, I did still more drawings and more calculations, but still did not get the results I expected. Finally, I realized that the problem was not my math skills, it is that one simply does not provide enough data to determine the post-collision of each individual ball.

Sorry about that! I was wrong and 'Ben Tilly' is quite correct.

PS: Thanks for the correction 'Ben Tilly'! You are way cool.

boooeee
27th November 2006, 08:53 AM
Well, I'm going to probably make an ass out of myself here, but I think the problem is solvable, assuming perfectly elastic collisions and perfectly rigid balls with finite radii.

It is true that conservation of momentum and energy doesn't lead to a single solution. However, we know more than that. We also know that the force exerted by the balls on eachother can only point radially outward. It is this added knowledge that makes the problem tractable.

If you don't believe me, then consider the collision of two perfectly rigid balls. We have four unknowns (the two momentum components of the two balls), but we only have three equations (1 for conservation of energy and two for conservation of momentum). Is the two ball problem intractable? Of course not. Knowing that the force exerted by the balls has to point radially outward provides the additional constraint.

So, in regards to the original problem, one must be clear on what the initial conditions are. I will use andyandy's formulation:

so 3 balls A,B,C

all with mass 1kg and velocity 1m/s,

A travelling due south [0,-1]
B travelling due north [0,1]
C travelling due east [1,0]

such that after 1 second, if unimpeded, all would have travelled exactly 1 metre and be on point X
As Beleth pointed out, if the balls have finite radii, this will not lead to a simultaneous collision, but rather, C impacts A and B at the same time, but A and B do not touch.

The resulting trajectories of the balls are as follows:

A travels due east at 1 m/s.
B travels due east at 1 m/s.
C travels due west at 1 m/s.

Note that you will get a different answer if the balls meet simultaneously (i.e. in an equilateral triangle). I'm still working on this one because the trigonometry is a little trickier.

Art Vandelay
27th November 2006, 10:43 AM
While one can accurately work out the pre-collision momentum of each individual ball as well as the total momentum of the system of the three balls, however there is simply not enough information furnished that would enable one to determine the post-collision momentum of each individual ball.Either the "while" or the "however" is superfluous.

boooeee
Note that you will get a different answer if the balls meet simultaneously (i.e. in an equilateral triangle). I'm still working on this one because the trigonometry is a little trickier.I think that in that case, it truly is unresolvable, because there are now three directions in which force can be exerted. Consider a collision that leaves ball A stationary, while B and C travel away from each other. It's possible to choose the velocities of B and C such that momentum and energy are conserved. And the center-to-center restriction doesn't eliminate this possibility, either. By symmetry, there are then two other solutions, as well as all of their linear solutions (one of which is the one in which they just bounce back symmetrically).

Ben Tilly
27th November 2006, 12:00 PM
Well, I'm going to probably make an ass out of myself here, but I think the problem is solvable, assuming perfectly elastic collisions and perfectly rigid balls with finite radii.

It is true that conservation of momentum and energy doesn't lead to a single solution. However, we know more than that. We also know that the force exerted by the balls on each other can only point radially outward. It is this added knowledge that makes the problem tractable.

It is not generally sufficient. The 3-ball collision that I described had an infinite number of solutions with all forces being exerted radially outward.

So, in regards to the original problem, one must be clear on what the initial conditions are. I will use andyandy's formulation:

[...]

As Beleth pointed out, if the balls have finite radii, this will not lead to a simultaneous collision, but rather, C impacts A and B at the same time, but A and B do not touch.

The resulting trajectories of the balls are as follows:

A travels due east at 1 m/s.
B travels due east at 1 m/s.
C travels due west at 1 m/s.

Note that you will get a different answer if the balls meet simultaneously (i.e. in an equilateral triangle). I'm still working on this one because the trigonometry is a little trickier.

Interesting.

In this case C hits A and B at 45 degree angles, and A does not touch B. The limiting solutions of C hitting A then B, or C hitting B then A all work out to be the same solution. (If C hits A, then A heads due east, and C due south. Then C hits B and B heads due east and C due west. Similarly the other way around.) Therefore in this case the limiting solutions are the same, and there is only one physically likely solution.

However there are other mathematical solutions. They are all asymmetric in some way. For instance C-A can hit slightly harder than in yours, and C-B slightly weaker. The numbers get ugly, but you satisfy momentum and energy. (They will not come up in experiments though.)

Cheers,
Ben

Crossbow
27th November 2006, 12:53 PM
Either the "while" or the "however" is superfluous.

...

Possibly so, but the I do believe that the text used does serve to get the point across.

boooeee
27th November 2006, 02:05 PM
Okay, I think I understand now.

A collision in which all three balls touch at the same time and form an equilateral triangle is indeterminate. It is indeterminate because a slight change to the parameters (e.g. changing Ball C's starting point by nanometers) can result in a discontinuous change in the resulting trajectories.

However, a simultaneous collision in which all three balls do not form an equilateral triangle is not indeterminate and should be solvable. For example, A strikes B at the same time C strikes B, but A and C do not touch. Changing the starting conditions slightly should only change the resulting trajectories slightly.

Is the equilateral triangle collision the only indeterminate one, or are there others?

jsfisher
27th November 2006, 02:19 PM
A collision in which all three balls touch at the same time and form an equilateral triangle is indeterminate. It is indeterminate because....
It is indeterminate (period). It is indeterminate because Newtonian mechanics provides insufficient constraints to lead to a unique solution, not because variations from "simultaneous" lead to differing solutions.

However, a simultaneous collision in which all three balls do not form an equilateral triangle is not indeterminate and should be solvable. For example, A strikes B at the same time C strikes B, but A and C do not touch.
Are you sure? Consider a straight-line collision. A, B, and C all of the same mass; A and C traveling in opposite directions but at the same speed to strike a stationary B. The sum of the final velocities must equal 0 (conservation of momentum), and the sum of the squares of the velocities must equal some constant (= the sum of the squares of the velocities before the collision, conservation of energy). Three unknowns (the final velocities of A, B, and C), but only two equations.

ETA:
If the initial velocities of A, B, and C along the straight line are (1, 0, -1), then the "common sense" result would be (-1, 0, 1). That is, A and C rebound off B equally. However, nothing in the laws of physics requires that solution. Another, equally feasible result would be (-1/2 sqrt(4/3), -1/2 sqrt(4/3), sqrt(4/3)).

boooeee
27th November 2006, 03:53 PM
It is indeterminate (period). It is indeterminate because Newtonian mechanics provides insufficient constraints to lead to a unique solution, not because variations from "simultaneous" lead to differing solutions.

Are you sure? Consider a straight-line collision. A, B, and C all of the same mass; A and C traveling in opposite directions but at the same speed to strike a stationary B. The sum of the final velocities must equal 0 (conservation of momentum), and the sum of the squares of the velocities must equal some constant (= the sum of the squares of the velocities before the collision, conservation of energy). Three unknowns (the final velocities of A, B, and C), but only two equations.

ETA:
If the initial velocities of A, B, and C along the straight line are (1, 0, -1), then the "common sense" result would be (-1, 0, 1). That is, A and C rebound off B equally. However, nothing in the laws of physics requires that solution. Another, equally feasible result would be (-1/2 sqrt(4/3), -1/2 sqrt(4/3), sqrt(4/3)).In this case, I am pretty sure that the common sense answer is also the correct answer. I think the key is to imagine that the collisions actually occur a nanosecond apart.

I can almost guarantee that if you duplicated the situation with billiard balls, you would end up with the (-1,0,1) result.

jsfisher
27th November 2006, 05:48 PM
In this case, I am pretty sure that the common sense answer is also the correct answer. I think the key is to imagine that the collisions actually occur a nanosecond apart.
You can image all you like. The problem is there is no basis in physics for an answer, neither in a Newtonian nor relativistic model.

I can almost guarantee that if you duplicated the situation with billiard balls, you would end up with the (-1,0,1) result.
Unfortunately, "simultaneous" is a bit elusive and not likely obtainable under experimental conditions.

jsfisher
27th November 2006, 06:11 PM
This whole discussion has driven me to my old college text bookshelf. I now hold in my hand Physics: Part I by Resnick and Halliday.

I hate you all.

69dodge
27th November 2006, 08:42 PM
Consider a straight-line collision. A, B, and C all of the same mass; A and C traveling in opposite directions but at the same speed to strike a stationary B. [...] If the initial velocities of A, B, and C along the straight line are (1, 0, -1), then the "common sense" result would be [...]. However, nothing in the laws of physics requires that solution. Another, equally feasible result would be (-1/2 sqrt(4/3), -1/2 sqrt(4/3), sqrt(4/3)).That's weird. A and B exert forces on each other as they approach each other; but then somehow stop exerting forces on each other, even though they're still close, so that they don't move apart again?

(I'm thinking of the balls as being very stiff, but not quite infinitely stiff.)

boooeee
27th November 2006, 10:38 PM
Unfortunately, "simultaneous" is a bit elusive and not likely obtainable under experimental conditions.

Exactly. We're doing physics here, not philosophy or mathematics.

Let's say that somebody tried to duplicate your experiment. Obviously, they aren't going to get a simultaneous collision, but they could probably get it pretty close. Are you contending that all velocities are possible as long as momentum and energy are conserved?

If somebody tried to run this experiment (producing a simultaneous collision of A, B, and C), I guarantee that (-1,0,1) will be the result, within experimental tolerances.

The point here is that the resulting trajectories approach a limit as we get closer and closer to simultaneity. That limit is (-1,0,1). Since we're doing physics, that's all we care about.

This is different from the equilateral triangle collision because there is no limit. If our experimenter is one nanometer off to the left, balls A and C may never collide and we have one set of trajectories. If she is off one nanometer to the right, balls A and B may never collide and we have a different set of trajectories. I can't predict the outcome of this experiment with any accuracy. I can predict the outcome of your scenario.