Three billiard balls collide simultaneously.
CobaltFjord
Can one know what will happen if three billiard balls
simultaneously collide? I was once informed that
that one could not predict the ball trajectories if this
occurred. I know the possibility of this actually
occuring is infinitely small. Is there any connection
between this problem and "The three body problem"?
Charles Francis
In message <20030126001734...@mb-cv.aol.com>, CobaltFjord
<cobal...@aol.com> writes
>
>Can one know what will happen if three billiard balls
>simultaneously collide? I was once informed that
>that one could not predict the ball trajectories if this
>occurred.
If A is to hit B and B is to hit C it is easy to see that the impacts
will be different depending on whether A hits B before B hits C or not.
In the general case of simultaneous collision there is not enough
information to predict motions.
> I know the possibility of this actually
>occuring is infinitely small.
Infinitely small is zero. But actually each impact takes a finite time
interval. There is then a finite probability that these finite time
intervals overlap.
>Is there any connection
>between this problem and "The three body problem"?
No. The three body problem generally refers to the motions of three
gravitating bodies in each others gravitational fields. This is
"insoluble" only because there are no defined mathematical functions
giving the solution. But it is soluble to given accuracy by means of
computer simulation. The billiard ball problem is insoluble for quite a
different reason, that there is not enough information for a solution.
If you knew the exact elastic properties of the balls, and could work
out what happens during the period of the impact, and if you knew
exactly when the other impact (or impacts) take place, then in
principle, you could work out the motion - it would of course require
remarkably accurate time keeping.
Regards
--
Charles Francis
Suresh __NoJunkMail kumar
cobal...@aol.com (CobaltFjord) wrote in message news:<20030126001734...@mb-cv.aol.com>...
> Is there any connection
> between this problem and "The three body problem"?
Although 'naively', interaction of 2 billard balls can be solved using
Netwon's law. So, there is no real "three body problem".
Although there is an implied one.
Even classically, the billbards do not really touch, each other while
colliding. When they come too close, they are repelled by each other's
electromagnetic forces. Rutherford, i think, understood this effect
when he was trying to study atomic structure using scattering
techniques.
I think then physics of this type of body interaction, could be
interpreted as a three body problem.
In terms of finding a solution, i think, you have to setup some type
of classical Lagragian and except the 3 bodies to travel the path the
minimizes action.
Then again, to get any meaningful result, you would have to model the
ball as if it was composed of not just 1 particle, but a myraid of
them.
-suresh
Alan Horowitz
> if three billiard balls simultaneously collide?
is there more than one way for 3 billiard balls to be each touching
the other two? this is an equilateral-triangle sort of formation, i
imagine. we are talking about a flatland case, correct? Videlicet,
there is motion in only two dimensions to begin with?
there certainly should be a solution for that case.
i'm just guessing, but if each has the same momentum at the moment of
impact, and if the theory of limits lets us say there _is_ a "moment"
of impact (vice a time interval) wherein the _epsilon_ is that we can
no longer observe a difference by presuming that the interval
decrements.... then all the kinetic energy is turned into heat.
If three vehicles really did this, who's insurance would have to pay?
CobaltFjord
> In the general case of simultaneous collision there is
> not enough information to predict motions.
I believe this is correct. Now, I am searching for the
proof. I would be greatly indebted if you could
explain why this is true, or provide some internet
link that explains why this true. Thanks.
>> Is there any connection between this problem
>> and "The three body problem"?
>
> No. The three body problem generally refers to the
> motions of three gravitating bodies in each others
> gravitational fields. This is "insoluble" only because
> there are no defined mathematical functions giving
> the solution. But it is soluble to given accuracy by
> means of computer simulation.
I am not sure this is true. I believe even to this
day, we cannot determine if our solar system is
stable or not. I think Henry Poincare proved that
such a system is extremely sensitive to initial
conditions, so sensitive that without knowing the
initial conditions with INFINITE precision, one
cannot predict the future. I believe the three body
problem is an example of deterministic chaos.
> The billiard ball problem is insoluble for quite a
> different reason, that there is not enough information
> for a solution.
This is what I am looking for.
> If you knew the exact elastic properties of the balls,
> and could work out what happens during the period
> of the impact, and if you knew exactly when the other
> impact (or impacts) take place, then in principle, you
> could work out the motion - it would of course require
> remarkably accurate time keeping.
The question is, is this a chaotic occurrence which
cannot be calculated because conditions need to
be known with infinite precision to make a legitimate
prediction, or can one make a legitimate prediction within
certain parameters simply by using better measurement
devices. Note that better measuring devices don't
necessarily give you a more accurate result if your
system is inherently chaotic like our solar system.
If you like, I can provide the references where I
read this. I think computer simulations can predict
our solar system only for a very limited time into
the future. That without knowing initial conditions
with infinite accuracy, all computer similuations
are fatally flawed. Computer models predicting
the weather with miniscule differences in input
but major differences in output led Lorenz (not
to be confused with Lorentz ;-) to the "butterfly
effect."
Do quantum effects play a major role in what will happen
when three billiard balls simultaneously collide?
In any event, what I need here are references, or at
least a convincing argument. I am having this discussion
with another person in a private Mensa LISTSERV.
Do quantum effects play a major role in Brownian
Motion, or is Brownian motion mostly a classical
phenomenon?
Warm regards,
michael
CobaltFjord
without doing any math. If three exactly equall billard
balls collide equidistant apart from the same distance
with the same force, then the only unique solution is
for each ball to bounce eactly backwards. It would
have to be this answer or the results would have to
entirely random. No other causal solution is possible
because any other solution would violate the symmetry
of the problem.
Thomas Sauvaget
>
> >Is there any connection
> >between this problem and "The three body problem"?
>
> No. The three body problem generally refers to the motions of three
> gravitating bodies in each others gravitational fields. This is
> "insoluble" only because there are no defined mathematical functions
> giving the solution. But it is soluble to given accuracy by means of
> computer simulation.
Some precisions are in order here. The 3-body problem does admit an
analytical solution, which is due to Sundman (even the n-body problem
in fact, due to Wang). It's just that they give no geometrical
information on the behaviour of the solutions. This has been described
recently by M.Henkel (in french) here:
<http://www.arxiv.org/abs/physics/0203001> [this is based in part on
D.Saari's paper (in english) in Am. Math. Monthly, 97: 105-119]
It is the geometrical study which is hard, and there is still a lot to
do there. In fact, one knows much more on the geometrical behaviour of
the orbits in some special cases (restricted problem, collinear
problem, isocele configuration...).
Computing accurate solutions is yet another issue; one has to use
relevant and powerful numerical methods (e.g. 'geometric
integrators'). But even in this case, one doesn't want to get close to
a 3-body collision, since already in the collinear problem it is known
that, after such a close encounter, one of the bodies will have an
arbitarily high kinetic energy. [Technically, the 3-body collision is
not block-regularizable. This implies a very wild behaviour for orbits
experiencing close 3-body approaches.]
The recent breakthrough deal with finding complicated periodic orbits.
See for instance on this page:
<http://www.maia.ub.es/dsg/2000/index.html> the preprints by
A.Chenciner, J.Gerver, R.Montgomery and C.SimÛ. There's been a nice
wide-audience paper on that by B. Casseleman, see
<http://www.ams.org/new-in-math/cover/orbits1.html>.
--
thomas.
Russell Blackadar
>
> > if three billiard balls simultaneously collide?
>
> is there more than one way for 3 billiard balls to be each touching
> the other two?
No, of course there's only one way. But the instantaneous
position of the balls is not a complete specification of the
problem -- you also have to know the instantaneous velocity
of each ball at the moment they come together. And since
those velocities won't necessarily be symmetrical, there's
an infinitude of possible cases.
this is an equilateral-triangle sort of formation, i
> imagine. we are talking about a flatland case, correct? Videlicet,
> there is motion in only two dimensions to begin with?
>
> there certainly should be a solution for that case.
Not one that ignores the detailed physical properties of
the balls, in general.
>
> i'm just guessing, but if each has the same momentum at the moment of
> impact, and if the theory of limits lets us say there _is_ a "moment"
> of impact (vice a time interval) wherein the _epsilon_ is that we can
> no longer observe a difference by presuming that the interval
> decrements.... then all the kinetic energy is turned into heat.
Why do you assume this happens with three balls, when it
doesn't happen with two? The only difference your argument
would take in that case would be that the symmetry would
be linear, rather than triangular; and we know your conclusion
is false in the linear case, at least for billiard balls that
aren't covered with sticky glue.
If we assume the three balls are perfectly elastic, are
arranged in an equilateral triangle, and all approach their
collective center of mass at the same speed, then they'll
rebound symmetrically and all move away at the same
speed after the collision. There are a number of other
scenarios where symmetry will dictate the solution in a
simple way; but the general case requires more info about
the balls.
By the way, you said "same momentum", but momentum is a
vector quantity; so if their momentum were really the same
there would be no collision at all -- the balls would just
continue to move in formation. In the example I gave, the
momentum of each ball has the same *magnitude* but the
directions differ by 120 degrees.
Charles Francis
In message <b1f1ot$70d$1...@panther.uwo.ca>, CobaltFjord
<cobal...@aol.com> writes:
It's an interesting argument, for the case that assumes symmetry in the
centre of mass frame. But it has the same flaw as arguing that if you
balance a pencil exactly on its tip it will not fall. In practice
symmetry is not that precise. In impact problems we assume negligible
time of impact, but when you look at simultaneous collision this
assumption becomes false. There is a finite duration for each impact
when the balls are in contact. Even if the duration of the three impacts
overlaps (which is as much as you can mean by "simultaneous") if one
impact is even infinitesimally earlier there will be a different result
as compared to a different impact being infinitesimally earlier.
Regards
--
Charles Francis
CobaltFjord
> In the general case of simultaneous collision there is
> not enough information to predict motions.
I believe this is correct. Now, I am searching for the
proof. I would be greatly indebted if you could
explain why this is true, or provide some internet
link that explains why this true. Thanks.
>> Is there any connection between this problem
>> and "The three body problem"?
>
> No. The three body problem generally refers to the
> motions of three gravitating bodies in each others
> gravitational fields. This is "insoluble" only because
> there are no defined mathematical functions giving
> the solution. But it is soluble to given accuracy by
> means of computer simulation.
I am not sure this is true. I believe even to this
day, we cannot determine if our solar system is
stable or not. I think Henry Poincare proved that
such a system is extremely sensitive to initial
conditions, so sensitive that without knowing the
initial conditions with INFINITE precision, one
cannot predict the future. I believe the three body
problem is an example of deterministic chaos.
[Moderator's note: Poincare's first name was Henri. - jb]
> The billiard ball problem is insoluble for quite a
> different reason, that there is not enough information
> for a solution.
This is what I am looking for.
> If you knew the exact elastic properties of the balls,
> and could work out what happens during the period
> of the impact, and if you knew exactly when the other
> impact (or impacts) take place, then in principle, you
> could work out the motion - it would of course require
> remarkably accurate time keeping.
The question is, is this a chaotic occurrence which
cannot be calculated because conditions need to
be known with infinite precision to make a legitimate
prediction, or can one make a legitimate prediction within
certain parameters simply by using better measurement
devices. Note that better measuring devices don't
necessarily give you a more accurate result if your
system is inherently chaotic.
Do quantum effects play a major role in what will happen
when three billiard balls simultaneously collide?
In any event, what I need here is references, or
a convincing argument. I am having this discussion
with another person in a private Mensa LISTSERV.
Warm regards,
michael
Charles Francis
<cobal...@aol.com> writes:
>Some poor uncited soul wrote:
>> In the general case of simultaneous collision there is
>> not enough information to predict motions.
>I believe this is correct. Now, I am searching for the
>proof.
I did, but you snipped it. Think about the order of collisions, If A is
to hit B and B is to hit C it is easy to see that the impacts will be
different depending on whether A hits B before B hits C or not. It
follows that you cannot know, from this information, what happens in
simultaneous collision.
>>> Is there any connection between this problem
>>> and "The three body problem"?
>>
>> No. The three body problem generally refers to the
>> motions of three gravitating bodies in each others
>> gravitational fields. This is "insoluble" only because
>> there are no defined mathematical functions giving
>> the solution. But it is soluble to given accuracy by
>> means of computer simulation.
>I am not sure this is true.
If it were not true, the moderator would never have let me say it.
[Moderator's note: that's not true. - jb]
>I believe even to this
>day, we cannot determine if our solar system is
>stable or not. I think Henry Poincare proved that
>such a system is extremely sensitive to initial
>conditions, so sensitive that without knowing the
>initial conditions with INFINITE precision, one
>cannot predict the future. I believe the three body
>problem is an example of deterministic chaos.
Perhaps I should not have simply said given accuracy. Nonetheless if the
initial condition is known to a level of accuracy, the first part of the
subsequent motion can also be calculated, until such point as cumulative
errors predominate. In the case of the solar system this is certainly
thousands of years.
>> The billiard ball problem is insoluble for quite a
>> different reason, that there is not enough information
>> for a solution.
>This is what I am looking for.
>> If you knew the exact elastic properties of the balls,
>> and could work out what happens during the period
>> of the impact, and if you knew exactly when the other
>> impact (or impacts) take place, then in principle, you
>> could work out the motion - it would of course require
>> remarkably accurate time keeping.
>The question is, is this a chaotic occurrence which
>cannot be calculated because conditions need to
>be known with infinite precision to make a legitimate
>prediction,
Yes.
>or can one make a legitimate prediction within
>certain parameters simply by using better measurement
>devices.
In principle, yes. In practice I don't actually know whether such
accurate measurements can be done.
>Note that better measuring devices don't
>necessarily give you a more accurate result if your
>system is inherently chaotic like our solar system.
They do. The more accurately you know the initial condition, and the
more accurately you calculate by digital simulation, the longer the
computer model will mimic real behaviour.
>If you like, I can provide the references where I
>read this. I think computer simulations can predict
>our solar system only for a very limited time into
>the future.
A limited time. Certainly many years.
> That without knowing initial conditions
>with infinite accuracy, all computer similuations
>are fatally flawed. Computer models predicting
>the weather with miniscule differences in input
>but major differences in output led Lorenz (not
>to be confused with Lorentz ;-) to the "butterfly
>effect."
The weather is very much more complicated than the solar system.
>
>Do quantum effects play a major role in what will happen
>when three billiard balls simultaneously collide?
No.
>In any event, what I need here are references, or at
>least a convincing argument. I am having this discussion
>with another person in a private Mensa LISTSERV.
That was why I suggested you think about the effect of collision between
balls A, B, C. A different order of impact has quite different results.
>Do quantum effects play a major role in Brownian
>Motion, or is Brownian motion mostly a classical
>phenomenon?
Mostly classical.
Regards
--
Charles Francis
Oz
>I think then physics of this type of body interaction, could be
>interpreted as a three body problem.
>
>In terms of finding a solution, i think, you have to setup some type
>of classical Lagragian and except the 3 bodies to travel the path the
>minimizes action.
Well, I have drunk several glasses of wine.
However it seems to me that this problem may be one in which there may
be no definable minimum. That is, many paths have an equal minimum. In
this case there is no unique 'path of minimum action'.
Which is much the same as saying 'not enough information for a
solution', or alternatively 'many solutions'.
One suspects that the result might show quantumlike behaviour.
<hic>
<sorry>
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Note: soon (maybe already) only posts via despammed.com will be accepted.
Stephen Riley
<cobal...@aol.com> writes
>Hi Charles,
>
>> In the general case of simultaneous collision there is
>> not enough information to predict motions.
>
>I believe this is correct. Now, I am searching for the
>proof. I would be greatly indebted if you could
>explain why this is true, or provide some internet
>link that explains why this true. Thanks.
>
I'm not sure if I agree (did you change your mind?), so I would be
interested in this too. Especially since your example (three equally
sized and massed billiard balls colliding simultaneously) appears to me
to be simpler case than a more general problem. My reasons for this are
expanded on below, but don't assume I'm correct, since like you, I've
found next to no information to verify or contradict from the web.
If I have this imagined right, immediately prior to collision the three
balls lie on a plane connecting their centres. Since the balls are equal
size, each ball centre is at 60 degrees to another, forming an
equilateral triangle. We might avoid unwanted complications of friction
and spin so that only components of velocity in directions connecting
the balls centres (and so also restricted to that plane) play a part in
the collision. And indeed any changes in velocities produced by the
collision will only be on that plane and in those directions. These
directions (the three lines connecting the balls centres) can be called
normals.
>From now, I'll assume equal massed balls of equal radius, no friction
and no spin, and no energy losses. I don't think you were interested in
these originally, they complicate descriptions, the maths, are the same
for a 2-body case, detract from the essence of the question, and could
be added later (easier said than done I imagine:).
In which case collisions are central, and post-collision changes in
velocities will be along normals connecting the balls centres. The balls
therefore move apart after a single 3-body collision, without further
impacts. Any component of velocity not along a normal will not affect
the collision in any way, so can be forgotten about as far as the
collision is concerned. A pre-collision state might therefore be viewed
simply as a set of three pared-down velocities whose only components
worth knowing are the velocities projected onto these normals, which are
60 degrees apart - and might even held as scalar values if being sloppy
:) Put in other terms, the dot product of any two unit normals equal 0.5
(cosine 60). I imagine the symmetries would allow a solution without
need of solving long tedious equations, but I won't pursue that.
In any case this setup appears simpler than it could have been since the
balls separate without later collision between themselves after the
first 3-way collision. Else I imagine this would preclude a single
closed form equation.
Easier for visualisation is to have two of the balls initially
stationary and only one ball moving, which I guess is what you
originally had in mind. In which case (or rather with a drawing, or a
few coins) it can be seen that the moving ball would need to be moving
*exactly* [1] toward the midpoint of the other two balls centres for a
simultaneous collision to occur. Anything else would mean one ball being
hit first, or only one ball being hit at all. So it's an extremely
unlikely occurrence. Possibly impossibly unlikely for certain
definitions of simultaneous, which may need to be defined in an
appropriate way, which is possibly why the phrase "not enough
information is provided" were used previously, I'm not sure and what has
provoked my post.
One final simplification: while this is a 2D/3D problem I'm nevertheless
going to be working in one dimension, to start with.
I was going to copy a solution I had worked, but I can't be bothered
with the ASCII. I've posted my solution and how I arrived at it at :
While the result appears to work for me, I don't know if it is right,
and even if it is, I think I could likely find it in a neater way than
this brutish first principles approach that I used there. But it is all
simple algebra.
Remaining with the setup mentioned, of two balls initially stationary
and one ball moving, the moving ball rebounds with 1/3 of its original
*speed* while the other two rebound with 2/3 of that original speed. The
two initially stationary balls move along their normals 60 degrees
apart, while the originally moving ball moves back in the direction it
came from, since it is midway between *two* normals involved in its
collision (a diagram would be appropriate here I think :) [There would
be two normals for the other two balls too, but for the fact that in
this case one of them adds nothing to their velocities because they are
initially stationary].
Taking direction into account, the sum of the speeds are -1/3 + 2/3 +
2/3 = 1, the original speed. Speed isn't a conserved quantity, but since
we have equal massed balls, we'd expect the sum of the velocities before
and after to be equal, in this special case.
I did a search on the net on multiple simultaneous collisions, and came
across very little, but there was an abstract from a Masters Thesis that
looked as if it could be interesting...
http://www.dgp.toronto.edu/~funge/mscthesis.html
but unfortunately no link to the actual document itself.
[1] Barring a suitable definition of simultaneous.
I just tried a 3 way collision using 3 pennies, about 20 times (two of
them initially stationary). Not once did I get anything close to a good
result, i.e. one where the separating speeds looked equal with angles
equal (friction will alter angles and speeds a little, but symmetry
should remain). Maybe a bit of practise, bigger pennies, and a smoother
table would help :) I wonder how often a snooker player could hit two
stationary balls and have them separate with equal speeds and angle,
even at a level of precision he would regard them to have been hit
equally?
--
Stephen Riley
thief#37
all-embracive, that is excluding all other possibilities. Surely quantum
possibilities must include an "infinity" of all possible variations whether
these variations be plausible/feasible or not? Thus the set of quantum
possibilities DOES include one where a truly simultaneous collision occurs?
One only indeed, just one, but the one-only must be there. So the
questioner is entitled to have that variant considered? Bruce Main-Smith +++
Mark Biggar
As the three points of contact are spacially separated, by special
relativity the very concept of simultaneous collision is ill defined.
--
Mark Biggar
mark.a...@attbi.com
[Moderator's note: we can eliminate this ambiguity by working
in the center of mass frame. - jb]
Stephen Riley
<St...@rileys.demon.co.uk> writes
<snipped>
I don't know what I was thinking of in that post of mine (probably not
much!), but on reflection I'd ask the original poster to ignore it.
First (and my foremost reason for replying to myself), on checking my
result, it doesn't work, it's rubbish. And second, on reading the
initial post, a symmetrical collision wasn't mentioned anyway, the balls
could collide in various other ways.
I'm not yet convinced that it's not solvable though, or that the result
is undefined/random. Another possibility that comes to mind (if
instantaneous impulse doesn't fit the bill) might be to transfer
momentum between the colliding objects in stages, this might be better
than some kind of repulsive/restoring force for conserving
energy/momentum, though how well this would work in practise I've no
idea. Since this event is so unlikely, I've never really given it
thought. Interesting problem though.
For what it's worth, I'll post a result I do think is right (done in a
conventional way, which I should have used to check my original posting,
and now have done) for a symmetrical 3 way collision with two balls
initially stationary.
It's probably unreadable in this ASCII form though.
Kinetic Energy
0.5*M1*V1f^2 + 0.5*M2*V2f^2 + 0.5*M3*V3f^2 = 0.5*M1*V1i^2 +
0.5*M2*V2i^2 + 0.5*M3*V3i^2
Equal massed balls, two initially stationary
V1f^2 + V2f^2 + V3f^2 = V1i^2
By symmetry ball2 final speed = ball3 final speed
V1f^2 + 2*V2f^2 = V1i^2
Since x axis chosen coincident with Ball 1's velocity vector:
V1i = V1xi and V1f = V1xf
V1xf^2 + 2*V2f^2 = V1xi^2 (1)
Momentum
V1f + V2f + V3f = V1i
Along x axis :
V1xf + 2*V2xf = V1xi (2)
From (1) and (2)
(V1xi - 2*V2xf )^2 = V1xi^2 - 2*V2f^2
Since tan30 = V2yf/V2xf
so V2yf = V2xf/tan30 = V2xf/3^0.5
V2f = (V2xf^2 + V2yf^2)^0.5 = (V2xf^2 + (V2xf^2)/3)
V2f = (V2fx^2 + V2fy^2) = (V2fx^2 + (V2fx^2)/3)
Therefore
V2xf = (V1xi + (V1xi^2 - 2*V2f^2)^0.5 )/2
V2xf = 0.6*V1xi
*************************************************************
V1xf = -0.2*V1xi
V1xf = 0
V2xf = 0.6*V1xi
V2yf = ((3^0.5)/5)*V1xi
V3xf = 0.6*V1xi
V2yf = -((3^0.5)/5)*V1xi
*************************************************************
Check:
------
E.g. ball 1 initial velocity = (10,0), hitting stationary balls
Outcome Ball1 velocity = (-2, 0) Magnitude = -2
Ball2 velocity = ( 6, 3.464) Magnitude = 6.928
Ball3 velocity = ( 6, -3.464) Magnitude = 6.928
Before:
Kinetic energy (unit masses for convenience)
0.5 * 1 * 10^2 + 0 + 0 = 50
Momentum
1 * 10 + 0 + 0 = 10
After:
Kinetic energy
0.5 * 1 * (-2)^2 + 0.5 * 1 * 6.928^2 + 0.5 * 1 * 6.928^2
= 2 + 24 + 24 = 50
Momentum (along x axis)
1 * (-2) + 1 * 6 + 1 * 6 = 10
--
Stephen Riley
Charles Francis
<sez...@hotmail.com> writes
>Thus the set of quantum
>possibilities DOES include one where a truly simultaneous collision occurs?
>One only indeed, just one, but the one-only must be there. So the
>questioner is entitled to have that variant considered? Bruce Main-Smith +++
We only need classical mechanics to model billiard ball collisions. The
point is that this can only be an approximation, even without
considering quantum effects. It is not legitimate to consider two
precisely simultaneous and instantaneous collisions when the physics of
the situation dictates that in practice we have collisions of finite
duration such that one collision inevitable starts before the other.
Regards
--
Charles Francis
Oz
>We only need classical mechanics to model billiard ball collisions. The point is
>that this can only be an approximation, even without considering quantum
>effects. It is not legitimate to consider two precisely simultaneous and
>instantaneous collisions when the physics of the situation dictates that in
>practice we have collisions of finite duration such that one collision
>inevitable starts before the other.
Hmmm.
Perhaps you could give your balls some amount of deformability.
Or, failing that, approach from each end of one impact being ever so
slightly before the other. Do the solutions, that is:
A hits B, then after a short time t hits C
and
A hits C then B
converge to a unique solution as t->0?
Given
masses and radii can be st A=/=B=/=C ?
Somehow, though, I suspect not.
Yes, yes, I know I could so it .....
Nicolaas Vroom
Nick
***********************************************************
Charles Francis wrote:
>
> In message <20030128202150...@mb-cv.aol.com>, CobaltFjord
> <cobal...@aol.com> writes:
>
> >I believe even to this
> >day, we cannot determine if our solar system is
> >stable or not. I think Henry Poincare proved that
> >such a system is extremely sensitive to initial
> >conditions, so sensitive that without knowing the
> >initial conditions with INFINITE precision, one
> >cannot predict the future. I believe the three body
> >problem is an example of deterministic chaos.
>
> Perhaps I should not have simply said given accuracy. Nonetheless if the
> initial condition is known to a level of accuracy, the first part of the
> subsequent motion can also be calculated, until such point as cumulative
> errors predominate. In the case of the solar system this is certainly
> thousands of years.
IMO there are two issues.
First of all our Solar System behaves as it is, evolves as it does.
You can not prove such a behaviour.
For example. As a result of continuous bombardment of asteroids
the average position of the planets and the moon slowly changes
over the ages.
Secondly you can try to simulate this behaviour.
Two issues are very important:
a) all the measurements on which those simulations are based.
b) the model used.
A general a model consists of differential equations.
Such a model of two variables looks like:
dx1/dt = a11*x1 + a12*x2
dx2/dt = a21*x1 + a22*x2
In general dx/dt = A * x
You can also rewrite your model as:
v = dx/dt = a11*x + a12*v
a = dv/dt = a21*x + a22*v
And then it looks more like Newton's Law.
The most important question to answer is:
What are the parameters a11,a12,a21 and a22.
In general what is the matrix A.
To answer that question you need a) measurements
over a long range of time, b) a starting date for your
simulation, c) initial values for A and d) initial conditions
for each variable at the starting date.
With the items b,c, and d you can solve your equations (model)
and you can compare the results (calculate an error)
with your measurements.
You do that for different sets of initial values and initial
conditions and the set with the smallest errors (temporary) wins.
And you continue...
What I want to point out is that your final set of matrix A values
and initial conditions of your variables is a function of
all the measurements. The more and accurate measurements the
better will be your matrix and initial conditions.
It is important to remark that in principle your starting date
can be any value, also one of which no actual value of all the
objects, subject of your simulation, is available.
For example: your starting value can be 1 jan 1800 12:00:00.
As already indicated above you can use Newton's Law as your
model. Newton's Law requires a parameter mass for each object
involved. The mass values are embedded in your matrix parameters.
Again the more and accurate measurements the better you can
calculate those mass values.
With Newton's Law as your model you cannot correctly simulate
the movement of the planet Mercury.
You need GR and inturn a new matrix A and new initial conditions.
But for GR as a model the same implies:
The more and accurate measurements the better will be your matrix
and initial conditions in order to predict future behaviour.
> >> If you knew the exact elastic properties of the balls,
> >> and could work out what happens during the period
> >> of the impact, and if you knew exactly when the other
> >> impact (or impacts) take place, then in principle, you
> >> could work out the motion - it would of course require
> >> remarkably accurate time keeping.
>
> >The question is, is this a chaotic occurrence which
> >cannot be calculated because conditions need to
> >be known with infinite precision to make a legitimate
> >prediction,
>
> Yes.
The original question is based on 3 billiard balls.
Let me raise a slightly different question:
Consider a billiardtable with has a grid of
100*100 = 10000 "squares" with two bbs:
bb1 near the side and bb2 in the centre.
When you hit bb1 is it possible to predict in which
"square" each bb will come to rest ?
First you do trial shot and you mark the outcome.
This is your prediction.
The better you can repeat this trial shot the better is your
prediction.
Maybe this is not what you are looking for.
Maybe you will ask me: is it possible to calculate and or to
simulate the outcome of one shot.
The answer is the same as previously: You need measurements
the more the better (Specific when you hit the first bb)
a model (including friction, side effects ?), the matrix A
and initial conditions.
Using the measurements you can calculate the matrix parameters.
Again the more accurate those measurements and the better your matrix
then the better you can predict the outcome of a particular shot
assuming you have measured how you hit the first billiard ball (bb1)
Your next exercise should be to predict what happens
when the second bb (bb2) is also moving. This is more complex.
To call the behaviour of bb's chaotic does not help you in
understanding this exercise.
(What does it chaos mean compared with deterministic chaos ?)
Nick
http://users.pandora.be/nicvroom/
Ed Keane III
Please feel free to correct both my terminology and my
assumptions. I think that way to much is being read into
this, especially any suggestion of quantum or relativistic
effects.
Three balls can hit simutaneously. If you play you know
the sound and the fine points about which one *really*
hit first does not change the physics used in the
analysis of the resulting motion of the balls.
The simplest way to think about this is one ball
colliding with two stationary balls already in contact
with each other in a stationary frame. I think the
following point holds true in any frame. This system
is not chaotic or complex. The addition of the
velocities is linear and the angles are set by boundary
conditions.
The angle at which the first ball hits the second does
not change the angle at which the second ball hits the
third. As long as you hit any spot on the side of the
ball facing away from the other the other ball moves in
the same direction every time. Am I correct in saying
that the non-linear angular dynamics of the balls do not
couple and that this is not a chaotic system? Small
initial changes in the velocity of the balls will not
result in large changes or unpredictable velocities.
Am I correct in thinking that a complex system is not
necessarily a chaotic one?
-Ed Keane III
Dr Tim
I suspect that there are three different problems here,
depending on how you interpret the question.
(I assume that we can ignore the spin on the balls.)
Interpretation 1:
Three point-like particles with known mass and velocity
reach a simultaneous elastic collision.
Interpretation 2:
Three small, perfectly hard spherical particles
with known mass and velocity reach
a simultaneous elastic collision.
At the moment of impact, the particles are arranged
in a known configuration.
Interpretation 3:
As in interpretation 2, but the balls have known composition
and finite elasticity.
In all cases, we want to know three momenta,
which is nine unknowns in three dimensions.
In interpreation 1, we know the total momentum and total energy.
Four equations in nine unknowns: not enough.
In interpretation 2, we know the total momentum, total energy,
and total angular momentum.
Seven equations in nine unknowns: not enough.
But in interpretation 3, we can use continuum mechanics
to track the deformations of the balls and determine the result.
Oz
>Your next exercise should be to predict what happens
>when the second bb (bb2) is also moving. This is more complex.
>
>To call the behaviour of bb's chaotic does not help you in
>understanding this exercise.
>(What does it chaos mean compared with deterministic chaos ?)
[Warning: I am totally ignorant]
I was under the impression that with chaotic systems, small changes in
any of the initial parameters gave radically different 'results'. That
is in general pertubation techniques don't work because they do not
converge on a given answer. In a way you could describe the level of
chaos as the largest step you can take and still get a convergent
result.
This is used for giving some confidence to UK weather forecasting, and
has been very successful in that when they give detailed five-day
forecasts on TV, they are usually pretty accurate for a day or two
ahead. When they just give tomorrow and handwave after that, do NOT rely
on the forecast.
When people say a process is chaotic I assume they mean that they do not
know the individual parameters accurately enough (and that includes the
model) to make a prediction. Often they may not know the parameters
accurately enough by many orders of magnitude.
I can see no reason why some systems should not exhibit this on very
tiny scales as do the chaotic regions in mandelbrot sets, or regions of
newton's approximations. I trust everyone here has played with fractint
for many hours (days, even), if not you are in for a stonking delight:
download it (it's free!).
Stephen Riley
<timro...@paradise.net.nz> writes
>
>I suspect that there are three different problems here,
>depending on how you interpret the question.
...
>
>But in interpretation 3, we can use continuum mechanics
>to track the deformations of the balls and determine the result.
>
Hmm, "continuum mechanics" is a new one on me (but then again I'm not a
physicist). I agree that the problem leaves room for interpretation, and
that some of those interpretations are unphysical.
I have a few observations and questions, partly spurred by your post and
partly because yours is the last I've read - so don't feel obliged to
reply :)
Consider the following 1-Dimensional case, a line of billiard balls
immediately prior to a 3 ball collision :
A --> B C
The above is meant to represent the velocities of three balls, with ball
A initially moving toward B, and where balls B and C are initially
stationary. All 3 balls are touching at that instant in time.
I think with the billiard balls in that configuration, C and A exchange
velocities (since they are equal masses), while B remains in place. I
also have reason to think, from a quick simulation I have done just now,
that this also works with C moving with arbitrary velocity, and
remarkably (if my simulation is correct) as well with all 3 balls moving
with arbitrary velocity. In the latter case, with all 3 moving, A and B
exchange velocities while B carries on on its merry way unaffected by
the collision. This is the same result as that obtained if both
collisions are done separately, one after the other, in any order. I
hope I got that right, since it does sounds a bit way-out. At least I
was surprised.
It's interesting since the unknowns you mention, and finite /
overlapping collision times / integration's, etc. aren't required. Of
all the combinations of velocities available on that 1 dimensional line
(as least those allowable by conservation of energy and momentum alone,
which leaves a lot) and it's left as simple as a two body collision. For
this 3-body elastic collision of equal masses in one dimension at least.
--
Stephen Riley
Nicolaas Vroom
>
> Nicolaas Vroom <nicolaa...@pandora.be> writes
>
> >Your next exercise should be to predict what happens
> >when the second bb (bb2) is also moving. This is more complex.
> >
> >To call the behaviour of bb's chaotic does not help you in
> >understanding this exercise.
> >(What does it chaos mean compared with deterministic chaos ?)
>
> any of the initial parameters gave radically different 'results'.
Different initial conditions will give different solutions of
differential
equations. The same for computer simulations using difference eqautions.
Protostars are formed out of gas clouds. Stars are formed out of
protostars,
Different gas clouds (size) give rise to different classes of Stars.
Approximate the same, you can say related to galaxies.
To call those processes chaotic (because the results are different)
will not improve our understanding.
> When people say a process is chaotic I assume they mean that they do not
> know the individual parameters accurately enough (and that includes the
> model) to make a prediction.
If "you" do not know the parameters accurately than "you" should say
and include an error analysis.
That is no reason to call the process chaotic.
My point is that in order to do any simulation you need variables, a
model
(relations between those variables), parameters of this model (matrix A)
and initial conditions.
However what is the most important you need observations.
Using the observations you have to calculate the initial conditions,
initially based on estimates.
My point is that you can not use a random set of initial conditions
to predict future behaviour of a particular system.
For example: To simulate our solar system there exists only one
specific set of initial conditions for each particular date and time.
To make the whole exercise easier it is important that all observations
should be done under the same conditions i.e. with the same tools.
Stephen Riley
>I suspect that there are three different problems here,
>depending on how you interpret the question.
...
>But in interpretation 3, we can use continuum mechanics
>to track the deformations of the balls and determine the result.
[Apologies, there was a typo in my post of a few minutes ago. This is a
correction of that. Hopefully only one will be posted].
Hmm, "continuum mechanics" is a new one on me (but then again I'm not a
physicist). I agree that the problem leaves room for interpretation, and
that some of those interpretations are unphysical.
I have a few observations and questions, partly spurred by your post and
partly because yours is the last I've read - so don't feel obliged to
reply :)
Consider the following 1-Dimensional case, a line of billiard balls
immediately prior to a 3 ball collision :
A --> B C
The above is meant to represent the velocities of three balls, with ball
A initially moving toward B, and where balls B and C are initially
stationary. All 3 balls are touching at that instant in time.
I think with the billiard balls in that configuration, C and A exchange
velocities (since they are equal masses), while B remains in place. I
also have reason to think, from a quick simulation I have done just now,
that this also works with C moving with arbitrary velocity, and
remarkably (if my simulation is correct) as well with all 3 balls moving
with arbitrary velocity. In the latter case, with all 3 moving, A and C
Oz
>Oz wrote:
>> I was under the impression that with chaotic systems, small changes in
>> any of the initial parameters gave radically different 'results'.
>Different initial conditions will give different solutions of
>differential equations. The same for computer simulations using
>difference eqautions.
>
>To call those processes chaotic (because the results are different)
>will not improve our understanding.
No, I said *some* were chaotic. That is very small, even infinitesimal,
changes result in radically different solutions.
For example try iterating
x_1 = R x_0 (1 - x_0), 0 < x_0 < 1.
With R =(say) 3.8
>> When people say a process is chaotic I assume they mean that they do not
>> know the individual parameters accurately enough (and that includes the
>> model) to make a prediction.
>If "you" do not know the parameters accurately than "you" should say
>and include an error analysis.
>That is no reason to call the process chaotic.
It will be chaotic if the results deviate radically for small changes in
the input data. That is, with (say) x=1 your 'answer' is 10, but for
x=1.000010 your answer is 10,000 and with x=1.000011 your answer is .001
and your results continue to deviate radically no matter how small are
the changes you make. That is, the results do not converge.
Even the very simple iteration I quote above has regions of stability
interspersed with regions of chaos depending on the parameter R.
>My point is that in order to do any simulation you need variables, a
>model (relations between those variables), parameters of this model
>(matrix A) and initial conditions.
Indeed.
>However what is the most important you need observations.
>Using the observations you have to calculate the initial conditions,
>initially based on estimates.
>My point is that you can not use a random set of initial conditions
>to predict future behaviour of a particular system.
Of course not (well you can, but it would be pointless).
>For example: To simulate our solar system there exists only one
>specific set of initial conditions for each particular date and time.
To the quantum level, yes.
>To make the whole exercise easier it is important that all observations
>should be done under the same conditions i.e. with the same tools.
You assume a model converges to a single solution.
There is no reason to believe this will always be the case, and some
known physical systems, like the weather, are known not to be convergent
on some scales.
Stephen Riley
writes
>It will be chaotic if the results deviate radically for small changes in
>the input data. That is, with (say) x=1 your 'answer' is 10, but for
>x=1.000010 your answer is 10,000 and with x=1.000011 your answer is .001
>and your results continue to deviate radically no matter how small are
>the changes you make. That is, the results do not converge.
I'm not sure whether you're talking about billiard balls, the three body
problem or chaos in general. But going with billiard balls, I've written
an applet that shows results for small changes in input data, which
might be of interest. The slight shortcoming of this applet however is
that collisions are done one at a time, so the radical deviation under
interest may be lost (though a limiting range for any radicalness is
apparent, I think). To handle multiple simultaneous collisions would
have been more work, though having done some previously in one
dimension, I didn't notice much there in the way of chaos. I'll leave it
up to you to interpret the results (or validity) of the following
applet, if you're interested.
http://www.sriley.co.uk/chaotic
--
Stephen Riley