I've been discussing this question with my brother (a biophysicist): we've been gradually developing a clearer picture. A geometric model for n bodies in 1 dimension that takes into account the kinetic energy lost in each collision would be to start with a in R^n-1 (= the subspace center of mass = 0) with an accessible region that is the cone on a simplex (linearly equivalent to an orthant). At each collision, you shorten the velocity in the normal direction by a constant (called the coefficient of restitution, I believe). I.e. the two colliding particles travel more nearly in the same direction. This changes the geometry of reflections, so continuity around a 3-body collision never holds, no matter what the masses. For the idealized elastic case, I've checked out more about special cases that correspond to kaleidoscopic dynamics that are continuous (with help from my brother). If you have two particles (or objects) in a box of masses m1 and m2, say [0,1], then the dynamics is equivalent to a billiard ball in a right triangle with sides in the ratio sqrt(m1)/sqrt(m2). In particular, if the masses are equal, it's a 45-45-90 triangle. If the masses are in the ratio 1:3, then it's a 30-60-90 triangle. In these cases, the dynamics is continuous near the configuration where both particles collide simultaneously with each other and with the end; in all other configurations, the dynamics discontinuous. I think the 1:3 ratio should be particularly nice to make some kind of model (superballs on a track? magnets swinging?). If you start with just one of the masses moving, the dynamics is periodic (for the idealized dynamics). For three particles on a line with masses a, b and c at positions x,y,z, in the 2-dimensional plane center of mass = the origin, there are three lines x=y, y=z and x=z. If you give this plane a metric where vectors of kinetic energy 1 have length 1, then you can figure out the angles by drawing a hexagon with sides parallel to the lines and vertices on these lines. They are the same as the angles of the triangle with side lengths sqrt(a(b+c)), sqrt(b(a+c)) and sqrt(c(a+b)). By varying the masses, you can make any acute-angled triangle. The dynamics confines the arrangement to one particular order <--> one of the six chambers formed by these lines. If you add a 4th particle of mass d, then you get an arrangement of 6 planes in R^3, linearly equivalent to the set of planes through edges of a regular tetrahedron centered at the origin, or equivalently, the planes that go through a pair of opposite edges of a cube centered at the origin. Collisions confine the arrangement to one of the 24 = 4! chambers. The intersection of any two of these planes is either a 3 body collision or a pair of simultaneous collisions of 2 disjoint pairs of particles. In the latter case, the angles are 90 degrees, and in the former case, they are determined by the 3-body case above. Each chamber is the cone on a right-angled spherical angle whose other two angles are acute, and every such triangle can be attained by suitable choices of masses. For instance, to get the pi/2, pi/3, pi/5 Kaleidoscope (the cone on a triangle in the barycentric subdivision of an icosahedron), one solution for the masses is approximately a=55.6, b=13, c=5.8 and d=100. These numbers are imprecise because I got them by a Manipulate command in mathematica, where there are 4 sliders to adjust while you watch the angles change. The angles are 59.9969, 36.0558, and 90.) As the two outer weights become large, the sum of the angles tends toward 90 degrees, and it converges to the configuration of two particles bouncing between two walls. In a configuration with three bodies of masses 2, 1, 1, the angles of the chamber are to the 3-body collisions are 54.7356, 60, 90, ... , which confirms that the Newton's cradle configuration is discontinuous near the set of 3-body collisions where the first mass is double the other two. It's continuous near the particular trajectories where the heavy ball collides with a pair of stationary balls. Bill Thurston On Dec 12, 2010, at 1:11 AM, Gary Antonick wrote:
Bill-
I have a related question I think.
Kinetic energy and momentum for three bodies make a sphere and intersecting plane if plotted in 3D using the three body velocities (am not sure "inertial coordinates" mean but perhaps something similar. would be nice to see some animations of this geometry if you know of any. eg:
the first ball has twice the mass of the other two: sphere flattens (which you point out) the third ball always has mass zero (is never struck): sphere turns into cylinder first ball hits 2nd and 3rd ball simultaneously (guess you did this but I need to digest)
would be especially interesting to see how this geometry changes with totally inelastic collisions. kinetic energy is lost, so the sphere shrinks with each hit. momentum stays the same, so the plane stays where it is. and eventually all bodies are stuck together traveling at the same velocity at coordinate [v,v,v].
cheers,
Gary
On Sat, Dec 11, 2010 at 6:27 PM, Bill Thurston <wpthurston@me.com> wrote:
Here's a way to visualize the discontinuity near a 3-body collision of most combinations of masses in 1 dimension:
When there are just two bodies, if you use inertial coordinates, there is just one degree of freedom, and using as parameter the distance between the two bodies, the motion is that of a reflection.
When there are three bodies, then in an inertial coordinate system there are two degrees of freedom. The parameter space is an angle between two lines. There is a physically natural metric, ds^2 = kinetic energy of system and in this metric, the dynamics follows geodesics that reflect off the walls just like light reflects off a mirror.
When the particles all have equal mass, the angle is 60 degrees: two mirrors at a 60 degree angle form a continuous image across the corner, and the dynamics is continuous in a neighborhood of the triple collision.
As the mass of the particles varies, the angle between the lines changes. If the mass of the central object is small compared to the other two, it bounces back and forth many times before they all separate: the angle is very small. If the mass in the middle is very large, the angle is almost 90 degrees.
Just as for two mirrors, the dynamics is continuous if and only if the angle has the form pi/n.
If there are more masses, the geometric figure is wedge formed between n hyperplanes in R^n, linearly equivalent to an orthant (whose coordinates are the distances between masses) but metrically of a variable shape. The angle between two faces that corresponding to collisions of disjoint pairs of particles are at right angles. The dynamics is continuous if and only if it is continuous for the collision of each triple of consecutive particles, i.e. if these angles have the form pi/n. It should be fun to make a Newton's cradle with some of the variants. --- I figure any of the Euclidean Coxeter groups should be doable. Bill Thurston
On Dec 11, 2010, at 3:34 PM, Andy Latto wrote:
On Sat, Dec 11, 2010 at 1:45 PM, Robert Munafo <mrob27@gmail.com> wrote:
How ill-defined?
For example, take the case where all three have equal mass, the central body starts out motionless and the outer two are approaching at the same speed (from opposite directions). The midpoint between the positions of the outer two is X, and isn't changing until the first impact because of the initial conditions I just stated.
You happen to have chosen a case where the order doesn't make a difference. But let's try a slightly different example.
One ball of mass 2 is stationary very near the origin, but to the left of the origin. Two balls of mass 1, one at -X and one at +X, approach it at speed V.
First the ball on the left hits the stationary ball. Now the ball on the left is going left at speed V/3, while the center ball is going right at speed 2V/3.
An instant later, the center ball hits the ball on the right. Now the ball on the right is going right with speed 11V/9, while the center ball is moving left with speed 4V/9
Of course, if the large ball started out slightly to the right of the origin, the collision order would be reversed, and we'd end up with the left ball moving left with speed 11V/9, the center ball moving right with speed 4V/9, and the right-hand ball moving right with velocity V/3.
So what happens when the center ball is exactly at the origin, and the two collisions happen simultaneously? One might argue from symmetry that the center ball would stay stationary, and the other two would rebound and end moving away from the center at speed V. But experimentally, you can't place the ball *exactly* at the origin, so wouldn't you instead expect one of the two asymmetric outcomes above?
And what if the ball coming from the left has mass 1, the center ball has mass 3, and the left hand ball has mass 2, and again, the outer balls are approaching the origin with speed V from equal distances? I know how to calculate what happens if the left-hand collision happens first, and what happens if the right-hand collision happens first. But is there a third "simultaneous collisions", possibility, and if so, in the absence of symmetry, how do you calculate what happens? You have 2 equations (conservation of momentum and energy) in 3 unknowns, so there are infinitely many solutions, and I have no idea how to choose which solution "between" the two non-simultaneous-collision answers is the correct answer!
So the answer to the original question of "why don't physicists get the problem of Newton's Cradle with a double-mass ball hitting the stationary balls" may be "It's an ill-posed problem"! If we model it as a series of non-simultaneous collisions, then after the big ball hits the second ball, it's still moving in the same direction, at speed V/3. So it will be involved in more collisions. All the collisions are nearly simultaneous, and the answer will depend on what order they occur in, so I don't see any way to give a unique answer.
Andy
If the central body starts out at position "X+epsilon", I get two non-simultaneous collisions after which the central body ends up at position "X-epsilon". In the limit as epsilon goes to zero, I get two simultaneous collisions, the central body remains motionless and the outer two simultaneously bounce off it.
If I decompose in the other order (starting with a negative "epsilon" so the other impact happens first) the limit still gives me the same answer.
Why is the limit ill-defined? Because the velocity of either particle is not defined (and discontinuous) at the moment of impact? Wouldn't that make all individual collision events ill-defined?
(Of course this is all assuming the bodies are ideal perfectly rigid, i.e. the speed of sound in their material is infinite and similar idealizations)
- Robert
On Sat, Dec 11, 2010 at 07:55, Fred lunnon <fred.lunnon@gmail.com> wrote:
[...] I've never been convinced that these problems are even well-defined, for the same reason that a simultaneous collision between three bodies in the plane is ill-defined. In that case, it's immediately obvious that you get completely different results according to how you decompose into a sequence of 2-body collisions.
The resemblance to the preceding discussion is striking! WFL
-- Robert Munafo -- mrob.com Follow me at: mrob27.wordpress.com - twitter.com/mrob_27 - youtube.com/user/mrob143 - rilybot.blogspot.com
math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun