rolling ball argument
John Baez
Here's a fun question that a friend and I got to arguing about today.
Suppose we have two balls of the same radius. I'll call them
the "rolling ball" and the "fixed ball".
The fixed ball is not allowed to move.
The rolling ball must touch the fixed ball - and it's allowed to roll
without slipping or twisting on the surface of the fixed ball.
Now:
1) Start with the rolling ball touching the north pole of the fixed ball.
2) Roll it down to the equator along a great circle.
3) Then roll it along the equator for an arbitrary distance.
4) Then roll it back up to the north pole along a great circle.
The question is: when we carry out this process, can the rolling
ball come back *rotated* relative to its original orientation... or not?
One of us thinks the rolling ball comes back unrotated at the end of this
process. The other thinks it rotates by a certain angle that depends
on the distance we rolled it along the equator.
I won't say who thinks what, or who this anonymous friend is.
What do you think?
Of course, anyone who knows some math can solve this by doing lots
of calculations. So, it's more fun if you try to solve it by sheer
thought, or visualization.
Daniel L
answers before you come up with your own, you should skip this thread
for now. -- KS]
> One of us thinks the rolling ball comes back unrotated at
> the end of this
> process. The other thinks it rotates by a certain angle
> that depends
> on the distance we rolled it along the equator.
>
> I won't say who thinks what, or who this anonymous friend is.
>
> What do you think?
They come back unrotated.
> Of course, anyone who knows some math can solve this by doing lots
> of calculations. So, it's more fun if you try to solve it by sheer
> thought, or visualization.
If one marks the path followed by the common point of the spheres on
both spheres, one gets - obviously - two identical spherical triangles.
Imagine these triangles drawn on the spheres after they reached the
final position. Since the last generated triangular edge on both spheres
was done simultaneously, these two edges are both in a vertical plane
(containing the centres of the spheres). Since the triangles are
identical, it means that the firstly generated edges are - at the final
position - also in a vertical plane. Therefore, the spheres are
unrotated.
Cheers,
Daniel L.
--
Posted via http://web2news.com
To contact in private, remove nno-6spp5amm
Ken S. Tucker
ba...@galaxy.ucr.edu (John Baez) wrote in message news:<b89ths$ma3$1...@glue.ucr.edu>...
IMO, No...it won't be rotated.
Regards Ken S. Tucker
PS: Snippable...
I confess, I used a pair of dice to confirm my posting,
(disqualified for 1st) :(
Each sphere rotates 90 degrees relative to the point of
contact moving from the pole to the equator, or, by setting
one sphere to a fixed reference, then the 'rolling' sphere
rotates 180 degrees relative.
Set the rolling sphere to R0 =(x,y,z) and then rolling first
around x reverses y,z to give R(1) =(x,-y,-z). Then roll R(1)
around y reverses x and z giving R(2) = (-x,-y,z), and then
rolling about R(3) = (x,y,z).
A sharp aerobatic pilot can do all three at once, and still
continue headed in the same direction, afterward.
KST
Barry
John Baez wrote:
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or not?
When the ball is touching the equator, the point "x" that is originally
touching the North Pole must be "pointing" North. This does not change
as the ball moves around the equator. So, when the ball returns to the
North Pole, X just "retraces" its track - it "unrotates".
Correction to my first post, what I should have written was no net
rotation about the North-South axis.
This can be seen more clearly by "flatenning out" the N. hemisphere into
a circle and seeing that the rotations are either along a radius or
along the perimeter of the circle. All axes of rotation are parallel to
the plane of the circle, with no orthogonal component.
Hence the ball must always come back unrotated.
Barry
Richard D. Saam
John Baez wrote:
>Here's a fun question that a friend and I got to arguing about today.
>
>Suppose we have two balls of the same radius. I'll call them
>the "rolling ball" and the "fixed ball".
>
>The fixed ball is not allowed to move.
>
>The rolling ball must touch the fixed ball - and it's allowed to roll
>without slipping or twisting on the surface of the fixed ball.
>
>Now:
>
>1) Start with the rolling ball touching the north pole of the fixed ball.
>
>2) Roll it down to the equator along a great circle.
>
>3) Then roll it along the equator for an arbitrary distance.
>
>4) Then roll it back up to the north pole along a great circle.
>
>The question is: when we carry out this process, can the rolling
>ball come back *rotated* relative to its original orientation... or not?
>
>One of us thinks the rolling ball comes back unrotated at the end of this
>process. The other thinks it rotates by a certain angle that depends
>on the distance we rolled it along the equator.
>
>I won't say who thinks what, or who this anonymous friend is.
>
>What do you think?
>
The answer lies in the fact that it will take two revolutions for the
rolling ball to roll its way around to its original position.
The rolling ball will make 1/2 revolution on rolling 1/4 way around the
fixed ball to the equatorial position (step 2)
The answer is necessarily quantized.
Unrotated rolling sphere orientation after Step 4 can result when
original rolling sphere is rolled on great circle to north pole from two
places on the equator (0 and 180 degree)
Richard Saam
Willem H. de Boer
It's quite hard to visualise, but I think a bit of real-life
experimentation would give the definite answer once and forever.
Here is my gedankenexperiment:
Start with the two balls in their initial position. Suppose we can
give the point of contact between the two balls a certain colour,
with an imaginary marker that colours the point of contact at a
certain point in time on _both_ the balls.
Now let's start with a green imaginary marker.
Start rolling the ball along a great arc towards the equator, making
sure that at every point on this geodesic, we mark it with our green
marker on both the balls.
Now, take a blue marker, roll the ball along the equator, marking all
the points on this geodesic blue on both the balls.
For our final step, let's take a yellow marker, and roll the ball
upwards until it reaches the north pole of the fixed ball, marking
every point on the geodesic yellow on both balls.
Now what? Well, if the moving ball comes back to its original spot
unrotated, we could do the entire path again, without tracing out
any additional lines with our markers (if you see what I mean), or
without overriding the original path with a _different coloured_ marker.
Errrm, it's a little hard to visualise, so I can't really give you a
clear cut answer. But this would be the experiment I would use to test
it..
Cheers,
--
Willem H. de Boer
http://www.whdeboer.com
groups_at_whdeboer.com
To email me, replace _at_ with @
Courtney Mewton
On Fri, 25 Apr 2003, John Baez wrote:
>
> Here's a fun question that a friend and I got to arguing about today.
>
> Suppose we have two balls of the same radius. I'll call them
> the "rolling ball" and the "fixed ball".
>
> The fixed ball is not allowed to move.
>
> The rolling ball must touch the fixed ball - and it's allowed to roll
> without slipping or twisting on the surface of the fixed ball.
>
> Now:
>
> 1) Start with the rolling ball touching the north pole of the fixed ball.
>
> 2) Roll it down to the equator along a great circle.
>
> 3) Then roll it along the equator for an arbitrary distance.
>
> 4) Then roll it back up to the north pole along a great circle.
>
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or not?
>
> One of us thinks the rolling ball comes back unrotated at the end of this
> process. The other thinks it rotates by a certain angle that depends
> on the distance we rolled it along the equator.
>
> I won't say who thinks what, or who this anonymous friend is.
>
> What do you think?
I would say that the ball comes back unrotated.
Although this isn't how I originally visualised it, but imagine squashing
the whole system vertically, so the two balls become two pancakes stacked
on top of one another (with the rolling ball/pancake on top). The act of
rolling one ball down the equator is now replaced by flipping the "rolling"
pancake upside down, but such that it maintains constant contact at only
one point on its circumference with the fixed pancake. Mark the point
where they contact with a dot of ice cream on each pancake for later
reference.
Rolling the ball about the equator is replaced by rotating the pancake.
Note that all times, the system is symmetric about the point of contact.
Call the new point of contact between the two contacts "J".
Now roll the ball back to the equator, or flip the moving pancake over,
maintaining contact at point J. If done carefully, the two balls of ice
cream will bash into each other, showing that the pancake, and hence the
ball, was not rotated over all.
I realised halfway into typing this that I could have used "disk" instead
of "pancake". I think my subconscious is trying to tell me something...
That was fun.
Regards,
CJM
Ralph Hartley
John Baez wrote:
> One of us thinks the rolling ball comes back unrotated at the end of this
> process. The other thinks it rotates by a certain angle that depends
> on the distance we rolled it along the equator.
In fact I've just about convinced myself that the ball comes back unrotated
after following *any* path that brings it back to the north pole. All the
business with equator is a red herring.
Instead of fixing one ball and letting the other roll, let them both roll
on opposite sides of a plain. If the path comes back to the north pole, the
contact point must come back to the same point on the plane, so the only
possible rotation is about the vertical axis.
Also the two spheres end up rotated by the same amount (it is harder to see
that they rotate in the *same* direction, think of the plain as a mirror).
If you rotate them both to bring the "fixed" sphere back to its original
position, the other sphere will end up unrotated as well.
If this is right, it should be possible to build some sort of "spherical
gear", that meshes in such a way that the spheres can *only* roll without
slipping. That wouldn't be possible if rolling around could rotate one of
them in place.
Ralph Hartley
Willem H. de Boer
Actually, if we simplify the case a bit, it's easier to see what's going on.
Suppose instead of rolling the ball, we slide the ball. Bear with me for
a minute:
Start with the two balls at the initial position. Now move the rolling ball
down to the equator along a great arc, but instead of rolling it, slide it.
We can take the initial tangent vector of the great arc at the north pole
of the fixed ball as our z-axis of the sliding ball. By sliding the ball, we
rotate this z-axis in such a way that it will always point in the same
direction
as the tangent vector of the great arc. When it reaches the equator
the z-axis will have rotated 90 degrees (1).
Moving an arbitrary distance along the equator, will rotate the x-axis
of the sliding ball by an arbitrary angle (2).
Moving back to the north-pole the z-axis of the sliding ball will rotate
90 degrees in the opposite direction as (1).
From this we can clearly see that the orientation of the sliding ball
has changed. Its final orientation will depend on how much we moved
the ball along the equator.
Now back to our _rolling_ ball. As some people have pointed out
this actually corresponds to rotating the axes of the rolling ball by
twice the amount as in the last paragraph.
So from this I conclude that the orientation _does_ depend on the
amount we have rolled the ball along the equator.
Willem H. de Boer
Willem H. de Boer
Hmmm, forget the sliding ball example. The ball will always come back
unrotated.
Willem H. de Boer
Dave Hilbert
>1) Start with the rolling ball touching the north
pole of the fixed ball.
>
>2) Roll it down to the equator along a great circle.
>
>3) Then roll it along the equator for an arbitrary
distance.
>
>4) Then roll it back up to the north pole along a
great circle.
Change frames of reference. Move to a frame in
which the centers of each ball remain at fixed
coordinates. In this frame the point of contact
also remains fixed. The ball's perform mirror
image actions of each other:
1) Call the initial point of contact the north
pole for both balls.
2) Roll the balls 90 degrees in opposite
directions along an axis parallel to their
tangent plane (in opposite directions, so no
slipping occurs). Now, both north-south axes are
parallel.
3) Roll the balls (again, opposite directions so they
roll and don't slip) along their north-south axes an
arbitrary amount.
4) Reverse 2)
Mirror images can't rotate relative to each
other. :)
So what highly mathematical question was reduced
to this simple query?
Dave (?)
Mad Scientist
news:b89ths$ma3$1...@glue.ucr.edu...
>Suppose we have two balls of the same radius. I'll call them
>the "rolling ball" and the "fixed ball".
>
>The fixed ball is not allowed to move.
>
>The rolling ball must touch the fixed ball - and it's allowed to roll
>without slipping or twisting on the surface of the fixed ball.
>
>Now:
>
>1) Start with the rolling ball touching the north pole of the fixed >
>ball.
>
>2) Roll it down to the equator along a great circle.
>
>3) Then roll it along the equator for an arbitrary distance.
>
>4) Then roll it back up to the north pole along a great circle.
>
>The question is: when we carry out this process, can the rolling
>ball come back *rotated* relative to its original orientation... or
>not?
My instict tells me that the ball would come back rotated through some
angle as a function of the distance it travels along the equator. I am
imagining the case where the ball simply starts at the north pole,
rotates down to the equator, and then rotates right back to the north
pole. Obviously, in this case, the ball would return unrotated,
because it simply makes the same trip back and forth. Add some travel
time around the equator, though, and you will pick up a rotation that
the trip back to the north pole will not correct.
Just my knee-jerk reaction,
Jeff
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Ken S. Tucker
> John Baez wrote:
> > What do you think?
> They come back unrotated.
> > Of course, anyone who knows some math can solve this by doing lots
> > of calculations. So, it's more fun if you try to solve it by sheer
> > thought, or visualization.
> If one marks the path followed by the common point of the spheres on
> both spheres, one gets - obviously - two identical spherical triangles.
> Imagine these triangles drawn on the spheres after they reached the
> final position. Since the last generated triangular edge on both spheres
> was done simultaneously, these two edges are both in a vertical plane
> (containing the centres of the spheres). Since the triangles are
> identical, it means that the firstly generated edges are - at the final
> position - also in a vertical plane. Therefore, the spheres are
> unrotated.
Agreed, that suggests a symmetry...
An axis throught the centers of the two spheres of course intersects
the point of contact, and each ball would be a mirror image of the
other along this axis, assuming the balls do not relatively rotate on
the point of contact then this 'mirror symmetry' will be preserved.
A dot is placed on a mirror, and the sphere is placed there initially,
no matter how the sphere is rolled on the mirror, when it is returned
to that point, it will be a mirror image of the other, and no relative
rotation would have occurred... (I think?).
This would allow any path of rolling.
Regards
Ken S. Tucker
PS: Hope a second posting is allowed.
Doug Goncz
the obverse and reverse shared an axis of symmetry.
I found that rolling to the equator is a lot like flipping the coin
over. And rolling any distance along the equator, then flipping over
along the tangent to the point of contact, produces the exact same
result as no procedure at all.
Gimbal lock.
Yours,
Doug Goncz, Replikon Research, Seven Corners, VA
http://users.aol.com/DGoncz
If a computer won't do what needs to be done, lie to it.
Don't try this trick on people.
Bryan Erickson
I vote for no rotation, as best as I could tell. I started by
visualizing two vectors internal to each of the two spheres from the
center of a sphere to a point on the surface. These included a vector
inside each sphere ending on the point of contact between the spheres
in their initial configuration, making them a north pole vector for
the fixed sphere and a south pole vector for the moving sphere. I'll
call them p(f) (polar vector of fixed sphere) and p(m) (polar vector
of moving sphere). The second vector in each sphere ends on the
equator in the direction of the initial rolling, call it zero degrees
longitude, so they are parallel to each other initially. I'll call
them e(f) and e(m), for equatorial vector of fixed/moving sphere.
At the end of the first rolling, the moving sphere has rotated 180
degrees and put p(f) and p(m) parallel, while e(f) and e(m) now
contact each other at the point of contact between the two spheres at
their equators.
Afer the second arbitrary-angle rolling along the mutual equators,
p(f) & p(m) remain parallel, while e(f) remains oriented to zero
degrees longitude and e(m) has been rotated by an arbitrary angle
2*phi. I imagined a tangential plane between the two spheres
intersecting each sphere only at their new point of contact, so each
sphere with defined vectors was the mirror image of the other through
the plane. Then I defined two new vectors t(f) and t(m), one in each
sphere that is parallel to the tangential plane and ending on the
equator. The angle between t(f) and e(f) is phi, while the angle
between t(m) and e(m) is also phi, in the opposite direction.
Then the final rolling takes place. p(m) will rotate 180 degrees to
become antiparallel to p(f) and in contact with it on their endpoints.
t(m) and t(f) will remain parallel to each other throughout the final
rolling, since they are also parallel to the mirror plane. e(f)
remains oriented to zero degrees longitude throughout the whole
process of course.
As for e(m), it rotates 180 degrees about t(m), so its final
orientation is equivalent to a mirror translation through t(m). It
remains at the angle phi from t(m), but now in the opposite direction
as it was originally, which is also the same direction that e(f) is
set from t(f). Since e(f) always pointed toward zero longitude, and is
phi from t(f), and t(m) is parallel to t(f), and e(m) is phi from t(m)
in the same direction, e(m) now also points toward zero longitude, as
it did before any rolling. Since two separate vectors in the moving
sphere, e(m) and p(m), are both in their original orientation, the
moving sphere is in its original orientation with no net rotation.
This remains true regardless of the value of 2phi, the arbitrary angle
of rotation.
That's my story anyway, hopefully not too far off. What is the correct
answer?
Chase Boulware
> Suppose we have two balls of the same radius. I'll call them
> the "rolling ball" and the "fixed ball".
>
> The fixed ball is not allowed to move.
>
> The rolling ball must touch the fixed ball - and it's allowed to roll
> without slipping or twisting on the surface of the fixed ball.
>
> Now:
>
> 1) Start with the rolling ball touching the north pole of the fixed ball.
>
> 2) Roll it down to the equator along a great circle.
>
> 3) Then roll it along the equator for an arbitrary distance.
>
> 4) Then roll it back up to the north pole along a great circle.
>
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or not?
I think the rolling ball will not be rotated, and here's why I think
that. I'll build on the idea that the original contact point is the
north pole of the fixed ball. Let's make that contact the south pole
of the rolling ball and put meridians on each ball (the name is
meridians, yes?, by which I mean half of a great circle connecting the
poles). The original configuration has corresponding meridians on
each ball in the same plane.
We now roll down the Prime Meridians, the price of fame becoming
manifest as a squished Greenwich on the fixed ball, and get to the
equators touching at the Prime Meridian-equator intersection on both
balls. Then we roll around with equators in contact and we see that
corresponding meridian-equator intersections always match up. It then
makes sense that when we roll back up some arbtitrary meridian, the
same meridian on each ball is matched up all the way and we get back
to our original configuration.
If I really knew what I was talking about, I could say something about
what happens for different paths. My guess is that without sliding or
twisting, we always move on great circles on each ball, and we can't
ever get back to our original contacts with a nonzero rotation of the
rolling ball.
...Chase
Eric A. Forgy
I like this question :)
I have no solution, but I'd like to extend the problem with another
challenge.
Instead of rolling the ball without slipping or twisting while it
traverses the right spherical triangle, what if it moves about an
arbitrary closed loop? Does it return rotated for a general closed
loop? Can you phrase the answer in terms of curvature and parallel
transport (as I imagine the original motivation for the question)?
Eric
Uncle Al
> Suppose we have two balls of the same radius. I'll call them
> the "rolling ball" and the "fixed ball".
>
> The fixed ball is not allowed to move.
>
> The rolling ball must touch the fixed ball - and it's allowed to roll
> without slipping or twisting on the surface of the fixed ball.
>
> Now:
>
> 1) Start with the rolling ball touching the north pole of the fixed ball.
>
> 2) Roll it down to the equator along a great circle.
>
> 3) Then roll it along the equator for an arbitrary distance.
>
> 4) Then roll it back up to the north pole along a great circle.
>
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or not?
The rolling ball returns apparently unrotated. Being a chemist and
thereby knowing nothing, Uncle Al claims full empowering social
diversity. Objective qualification is discrimination!
The lockstep motion, absent translational and angular slippage, along
*any* path restores the initial state whenever the initial pair of
kissing points occurs. The changing point of contact anchors a mirror
plane tangent to both surfaces - a more useful point of view. A
reflection can only lose synchronization if rotations are not in
lockstep (chiral ordering and parity). Translations are irrelevant.
So much for congruent Euclidean ellipsoids. What about "balls"
embedded in the other seven simply-connected geometric 3-manifolds
with compact quotients? What about other minimal surfaces of constant
curvature, especially concave ones? Imagine what a little practical
chemistry will do to resolve 400 years of Equivalence Principle
elegance in physics. Test mass symmetry is defining, not test mass
composition. "8^>)
--
Uncle Al
http://www.mazepath.com/uncleal/eotvos.htm
(Do something naughty to physics)
jacques Fric
Answer : no rotation
Very simple demo:
You need to decorate each of the two ball, in starting position with:
Equator line (horizontal great circle)
North and south poles
Meridian lines (great circles giving longitude, oriented in the same way),
including Zero meridians initialy in the same vertical plane (starting
position), all meridian of corresponding longitude of the two balls in
"symetrical vertical planes" all intersecting along vertical symetry axis.
Step one: Roll zero meridian of rolling ball along Zero meridian up to
equator of fixed ball:
Zero meridians of the two balls still in contact, rolling ball upside down.
Step two. Roll over the equator from meridian zero up to longitude "a":
As rolling ball is upside down, when rolling, meridian "longitude x" of
rolling ball always in contact with corresponding meridian "longitude x" of
fixed ball.
Step three.
Roll back to pole along meridian "longitude a".
Final position: Meridian "longitude a" of rolling ball in same
"symmetrical vertical plane" of corresponding meridian "longitude a"
of fixed ball, north pole up again.
Conclusion: No turn
Jacques
"John Baez" <ba...@galaxy.ucr.edu> a écrit dans le message de news:
b89ths$ma3$1...@glue.ucr.edu...
Pete
"John Baez" <ba...@galaxy.ucr.edu> wrote in message
news:b89ths$ma3$1...@glue.ucr.edu...
> Suppose we have two balls of the same radius. I'll call them
> the "rolling ball" and the "fixed ball".
>
> The fixed ball is not allowed to move.
>
> The rolling ball must touch the fixed ball - and it's allowed to roll
> without slipping or twisting on the surface of the fixed ball.
>
> Now:
>
> 1) Start with the rolling ball touching the north pole of the fixed ball.
>
> 2) Roll it down to the equator along a great circle.
>
> 3) Then roll it along the equator for an arbitrary distance.
>
> 4) Then roll it back up to the north pole along a great circle.
>
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or not?
Seems to me that if you mark the orientation of the rolling ball by
one point on its north pole and one at some arbitrary point on its
equator, then after the described manoeuvre, the first point will
return to its original point but the second will be rotated in the
horizontal plane by an amount equal to your arbitrary roll along the
equator (of the fixed ball).
But then I'm probably suffering from the same confusion as your friend.
Pete
[Moderator's note: don't assume my friend was the confused one! - jb]
Timo Nieminen
On Fri, 25 Apr 2003, John Baez wrote:
> Suppose we have two balls of the same radius. I'll call them
> the "rolling ball" and the "fixed ball".
[cut]
> What do you think?
Rolling ball returned to original position must be in original
orientation.
Easily shown experimentally by playing with my spherical coordinate
system tennis balls next to my desk.
> Of course, anyone who knows some math can solve this by doing lots
> of calculations. So, it's more fun if you try to solve it by sheer
> thought, or visualization.
But the maths is easy!
Just stick a little arrow on the uppermost point of the rolling
ball. Rolling to the equator reverses the direction of this if rolled
parallel to arrow, no change if rolled transverse. Change is (pi-2*phi)
where phi is angle between arrow and rolling direction. Do intitial roll
at phi=0 for simplicity. Rolling around the equator changes the angle of
the arrow by 2*phi, since the ball must spin twice in one full roll around
the equator. Roll up back to N pole, noting that angle between original
arrow position and rolling angle is -phi, 2*phi terms cancel.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
John Polasek
Baez) wrote:
>Suppose we have two balls of the same radius. I'll call them
>the "rolling ball" and the "fixed ball".
>
>The fixed ball is not allowed to move.
>
>The rolling ball must touch the fixed ball - and it's allowed to roll
>without slipping or twisting on the surface of the fixed ball.
>
>Now:
>
>1) Start with the rolling ball touching the north pole of the fixed ball.
>
>2) Roll it down to the equator along a great circle.
>
>3) Then roll it along the equator for an arbitrary distance.
>
>4) Then roll it back up to the north pole along a great circle.
>
>The question is: when we carry out this process, can the rolling
>ball come back *rotated* relative to its original orientation... or not?
It's fairly simple. The path is a spherical triangle. There is a
series of angular transformations, call them A down, and B across and
-A up again. To be accurate we must use 3x3 matrices which are 2d rank
tensors.
Let the ball roll down one meridian to the equator. Call the rolldown
axis x (now parallel to the equator)and call the rotation matrix Ax.
Name the axis lying along the meridian to the north y.
Now rotate along the equator for angle b, about axis y, and call this
B_y.
So far, using row vector notation we have A_x*B_y.
Now if b were 0, we could again roll by -A_x, but the x axis is
displaced now by angle b. So we use a similarity transform to make the
-Ax matrix fit in the tilted by b axis:
Tr(-Ax)= -By*-Ax*By
to "cure" rotation -Ax, making the total trip:
Ax*By*Tr(-Ax) = Ax*By*-By*-Az*By = By
The ball is turned by the matrix By!
John C. Polasek
Mr. Dual Space
If you have something to say, write an equation.
If you have nothing to say, write an essay.
jacques Fric
- key point is when ball is rolling over the equator
It looks like there is a rotation of "rolling ball" in regard to reference
frame defined by fixed ball.
It's an artifact, there is no "relative rotation", no wonder the rolling
return unrotated relatively to the fixed ball !
This is especially obvious if you consider the two ball as two copies of the
Earth (with its coordinate system).
At start of roll, take a Mercator Map of both, GMT meridian over GMT
meridian (one is upside down).
Roll the ball, you can check at any point on the path that if you take
Mercator Map, relative position of two Maps are unchanged (you may follow
the path of the contact point on the maps, same on both)
Starting and Going back to north pole in Mercator map will just reverse the
direction of the map (easy to check according to Mercator Transform).
Good exercise for keeping neurons alive.
Jacques
"John Baez" <ba...@galaxy.ucr.edu> a écrit dans le message de news:
b89ths$ma3$1...@glue.ucr.edu...
> Suppose we have two balls of the same radius. I'll call them
Ross Smith
> Suppose we have two balls of the same radius. I'll call them
> the "rolling ball" and the "fixed ball".
>
> The fixed ball is not allowed to move.
>
> The rolling ball must touch the fixed ball - and it's allowed to roll
> without slipping or twisting on the surface of the fixed ball.
>
> Now:
>
> 1) Start with the rolling ball touching the north pole of the fixed
> ball.
>
> 2) Roll it down to the equator along a great circle.
>
> 3) Then roll it along the equator for an arbitrary distance.
>
> 4) Then roll it back up to the north pole along a great circle.
>
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or
> not?
If you visualise it in a slightly different way, the answer becomes
obvious.
Paint a dot on the equator of each ball, lined up one above the other.
When you perform step 2, roll the moving ball along the arc containing
the fixed ball's dot, so that you end up with the balls touching at the
dots.
Now, look down on the system from above, and let your point of view
rotate as the moving ball rolls in step 3, following the centre of the
moving ball, so that the line through the centres of the two balls
continues to run left-to-right. From your point of view, the balls are
rotating in opposite directions. The two dots both started out at the
point of contact, so the two balls remain mirror images of each other
-- reflections in the vertical line between them, with the dots in
corresponding positions.
Now, when we perform step 4, from your (rotated) point of view, what
we're doing is exactly that: reflecting the moving ball in the vertical
axis. So it ends up superimposed on the fixed ball, with the dots lined
up, i.e. in exactly the position it started in.
--
Ross Smith ......... r-s...@ihug.co.nz ......... Auckland, New Zealand
Our landscape raped by different armies
Soldier-slaves who have no faces
Control our ways and lives completely
Our minds are torn, time left its traces
-- L'Ame Immortelle
Richard D. Saam
John Baez wrote:
>Here's a fun question that a friend and I got to arguing about today.
>
>Suppose we have two balls of the same radius. I'll call them
>the "rolling ball" and the "fixed ball".
>
>The fixed ball is not allowed to move.
>
>The rolling ball must touch the fixed ball - and it's allowed to roll
>without slipping or twisting on the surface of the fixed ball.
>
>Now:
>
>1) Start with the rolling ball touching the north pole of the fixed ball.
>
>2) Roll it down to the equator along a great circle.
>
>3) Then roll it along the equator for an arbitrary distance.
>
>4) Then roll it back up to the north pole along a great circle.
>
>The question is: when we carry out this process, can the rolling
>ball come back *rotated* relative to its original orientation... or not?
>
>One of us thinks the rolling ball comes back unrotated at the end of this
>process. The other thinks it rotates by a certain angle that depends
>on the distance we rolled it along the equator.
>
>I won't say who thinks what, or who this anonymous friend is.
>
>What do you think?
>
>
It will be unrotated at end of track if:
pi D1 / 4 = pi D2
This condition does not exist for your equal radius balls so the ball
will be rotated at end of track.
This is similar to a problem posed by von Savant in parade magazine in
January 1994.
Richard Saam
>
>
Brian Catlin
I believe the ball's rotation will be a function of the distance it rolled
around the equator. Consider the rolling ball starting on the North pole.
Imagine a dot painted on the top of the rolling ball. The ball makes N
rotations down to the equator, and the dot is now on the rolling ball's equator
(at the rolling ball's prime meridian). The rolling ball then travels around
the fixed ball's equator for M rotations, with the dot now on the East side of
the rolling ball. Rotating the rolling ball back up to the North pole, would
leave the dot still on the side of the rolling ball, and not at the top (North
pole), where it started. The only way in which the rolling ball would not end
up rotated, is if M were always a whole number of rotations.
-Brian
Willem H. de Boer
> Set the rolling sphere to R0 =(x,y,z) and then rolling first
> around x reverses y,z to give R(1) =(x,-y,-z). Then roll R(1)
> around y reverses x and z giving R(2) = (-x,-y,z), and then
> rolling about R(3) = (x,y,z).
You're assuming that you always move 90 degrees along the
equator, whereas the original question said that you can
move an arbitrary distance along the equator.
Willem H. de Boer
Scott Stephens
news:b89ths$ma3$1...@glue.ucr.edu...
> Suppose we have two balls of the same radius. I'll call them
> the "rolling ball" and the "fixed ball".
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or not?
> One of us thinks the rolling ball comes back unrotated at the end of this
> process. The other thinks it rotates by a certain angle that depends
> on the distance we rolled it along the equator.
First I thought about it, and considered a rolling ball of <larger> radius
than the fixed ball, and concluded it would be rotated. Considering balls of
equal radii, I find there not rotated (after I wimped out and tested it on
some real balls). This being the case, perhaps you're both right, if you
express the angle of rotation and distances in terms of ratios of ball
radii.
Scott
--
*********************************************
Nature is reciprocal. Those paid with love of kindness repay with love and
kindness,
those paid with hate and meanness repay with hate and meaness,
those paid with hate and meanness wise guys call love and kindness,
repay with the hate and meanness they believe is love and kindness.
*********************************************
man_mars
Baez) wrote:
>
>Here's a fun question that a friend and I got to arguing about today.
>
>Suppose we have two balls of the same radius. I'll call them
>the "rolling ball" and the "fixed ball".
>
>The fixed ball is not allowed to move.
>
>The rolling ball must touch the fixed ball - and it's allowed to roll
>without slipping or twisting on the surface of the fixed ball.
>
>Now:
>
>1) Start with the rolling ball touching the north pole of the fixed ball.
>
>2) Roll it down to the equator along a great circle.
>
>3) Then roll it along the equator for an arbitrary distance.
>
>4) Then roll it back up to the north pole along a great circle.
>
>The question is: when we carry out this process, can the rolling
>ball come back *rotated* relative to its original orientation... or not?
>
>One of us thinks the rolling ball comes back unrotated at the end of this
>process. The other thinks it rotates by a certain angle that depends
>on the distance we rolled it along the equator.
>
>I won't say who thinks what, or who this anonymous friend is.
>
>What do you think?
>
>Of course, anyone who knows some math can solve this by doing lots
>of calculations. So, it's more fun if you try to solve it by sheer
>thought, or visualization.
>
Unrotated. Visualize as two cubes.
>
Ralph Hartley
> 1) Start with the rolling ball touching the north pole of the fixed ball.
Let's call the axis through the north pole z, the axis of rotation for step
2 x, and the remaining axis y.
> 2) Roll it down to the equator along a great circle.
This rotates the rolling ball by 180 degrees around its x axis, so its x
axis is unchanged and y and z axes are each reversed.
It is touching the end of the y axis of the fixed sphere.
> 3) Then roll it along the equator for an arbitrary distance.
To make the visualization clear make it 90 degrees. The sphere rotates 180
degrees around the z axis reversing the x and y axes.
The combined moves reverse the z and x axes, and it is touching the end of
the x axis of the fixed sphere
> 4) Then roll it back up to the north pole along a great circle.
This reverses the z and x axis, binging it back to it's original position.
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or not?
That wasn't the result I expected.
Maybe 90 degrees was a bad choice.
Lets see, for an arbitrary angle, the x axis rotates twice as far as the
line about which it will be rotated by 180 degrees, so it will always end
up back were it stared.
The z axis isn't affected either, so there is no rotation.
> One of us thinks the rolling ball comes back unrotated at the end of this
> process.
That seems to be correct, though it wasn't what I expected.
> Of course, anyone who knows some math can solve this by doing lots
> of calculations. So, it's more fun if you try to solve it by sheer
> thought, or visualization.
Well, I didn't use any paper, but what I did was closer to a mental
calculation than to a visualization, and the answer *still* doesn't *seem*
right, even though I cheated a little.
Trickier puzzle:
For what class of paths does the sphere come back to its original
orientation? Original orientation and position?
Someone should be able to explain it all in terms of curvature on a fiber
bundle. But I can't do that all in my head.
The obvious bundle has the points of contact as the base space, and the
orientation of the rotating sphere as the fiber.
Ralph Hartley
Tim S
>
> Here's a fun question that a friend and I got to arguing about today.
>
> Suppose we have two balls of the same radius. I'll call them
> the "rolling ball" and the "fixed ball".
>
> The fixed ball is not allowed to move.
>
> The rolling ball must touch the fixed ball - and it's allowed to roll
> without slipping or twisting on the surface of the fixed ball.
>
> Now:
>
> 1) Start with the rolling ball touching the north pole of the fixed ball.
>
> 2) Roll it down to the equator along a great circle.
>
> 3) Then roll it along the equator for an arbitrary distance.
>
> 4) Then roll it back up to the north pole along a great circle.
>
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or not?
>
> One of us thinks the rolling ball comes back unrotated at the end of this
> process. The other thinks it rotates by a certain angle that depends
> on the distance we rolled it along the equator.
>
<snip>
Hmm.
Take out your pen with the infinitely thin nib.
On each ball, draw the equator. On each ball, draw two great circles which
meet the equator at a right angle and which meet each other at the poles at
angle phi. On each ball, the intersections of these circles with the equator
are separated by a distance D (= phi R, where R is the radius of the balls).
Place the balls together with their north poles touching and the initial
segments of the great circles lining up there.
Roll the rolling ball down to the equator along the first great circle of
the fixed ball. This will also pull the point of contact along the first
great circle of the rolling ball (by symmetry, among other reasons). Now
roll distance D along the equator of the fixed ball. This will also take the
point of contact a distance D along the equator of the rolling ball.
Finally, roll back up the second great circle of the fixed ball until the
north poles are in contact again. The tangents to the second great circles
at the north poles of the two balls will still line up.
So no rotation.
I think.
Tim
Bill Vajk
> John Baez wrote:
>> One of us thinks the rolling ball comes back unrotated at the end
>> of this process. The other thinks it rotates by a certain angle
>> that depends on the distance we rolled it along the equator.
> In fact I've just about convinced myself that the ball comes
> back unrotated after following *any* path that brings it
> back to the north pole. All the business with equator
> is a red herring.
Using a second sphere is actually another red herring.
The solution is obvious using a point of origin on a plane
surface. The same solution holds true for any continuous
surface shape one places the ball on and any path so long
as there's no slippage.
Conservation at its most evident.
William J. Vajk
Techny, Illinois
Tom Snyder
"Ralph Hartley" <har...@aic.nrl.navy.mil> wrote in message
news:3EA9A37B...@aic.nrl.navy.mil...
The introduction of a plane which may be thought of as a mirror separating
the spheres is a neat way to think about it.
Experiment: Use your hands to make two fists. Place them together so that
they appear as mirror-images of one another. Then arbitrarily roll one fist
over the other (no slipping) and you will soon convince yourself that there
will be no net rotation when the rolling fist returns to its initial point
of contact with the other fist.
So, the objects need not be spheres. They only need to be mirror images of
one another. Cool.
Tom S.
Barry
John Baez wrote:
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or not?
When the ball is touching the equator, the point "x" that is originally
touching the North Pole must be "pointing" North. This does not change
as the ball moves around the equator. So, when the ball returns to the
North Pole, X just "retraces" its track - it "unrotates".
Considering the point "y" that is originally on the Eastern edge of the
ball, and the point "z" the westernmost point. None of the moves
involves a rotation around this original "x-y", East- West axis.
Therefor "x" and "y" must also return to their original positions.
Hence the ball must always come back unrotated.
Barry
Richard D. Saam
John Baez wrote:
>Here's a fun question that a friend and I got to arguing about today.
>
>Suppose we have two balls of the same radius. I'll call them
>the "rolling ball" and the "fixed ball".
>
>The fixed ball is not allowed to move.
>
>The rolling ball must touch the fixed ball - and it's allowed to roll
>without slipping or twisting on the surface of the fixed ball.
>
>Now:
>
>1) Start with the rolling ball touching the north pole of the fixed ball.
>
>2) Roll it down to the equator along a great circle.
>
>3) Then roll it along the equator for an arbitrary distance.
>
>4) Then roll it back up to the north pole along a great circle.
>
>The question is: when we carry out this process, can the rolling
>ball come back *rotated* relative to its original orientation... or not?
>
>One of us thinks the rolling ball comes back unrotated at the end of this
>process. The other thinks it rotates by a certain angle that depends
>on the distance we rolled it along the equator.
>
>I won't say who thinks what, or who this anonymous friend is.
>
>What do you think?
>
>
The answer lies in the fact that it will take two revolutions for the
rolling ball to roll its way around the fixed ball on a great circle
path to its original position.
In general a ball of arbitrary radius will rotate around a ball of equal
or larger radius by the larger balls circumference divided by the
smaller balls radius + 1.
also:
In general a ball of arbitrary radius will rotate inside a ball of equal
or larger radius by the larger balls circumference divided by the
smaller balls radius - 1.
Therefore in terms of the equal sized balls, the rolling ball will make
1/2 revolution on rolling 1/4 way around the fixed ball to the
equatorial position (step 2)
The answer is necessarily quantized.
Unrotated rolling sphere orientation after Step 4 will result when
original rolling sphere is rolled on great circle from equator to north
pole from two places on the equator (0 and 180 degrees).
Rotated rolling sphere orientation after Step 4 will result when
original rolling sphere is rolled on great circle from equator to north
pole from arbitrary places on the equator (0 < > 180 degrees).
In conclusion, John Baez and friend are both right.
Richard Saam
>
>
John Polasek
Baez) wrote:
>Suppose we have two balls of the same radius. I'll call them
>the "rolling ball" and the "fixed ball".
>
>The fixed ball is not allowed to move.
>
>The rolling ball must touch the fixed ball - and it's allowed to roll
>without slipping or twisting on the surface of the fixed ball.
>
>Now:
>
>1) Start with the rolling ball touching the north pole of the fixed ball.
>
>2) Roll it down to the equator along a great circle.
>
>3) Then roll it along the equator for an arbitrary distance.
>
>4) Then roll it back up to the north pole along a great circle.
>
>The question is: when we carry out this process, can the rolling
>ball come back *rotated* relative to its original orientation... or not?
My note showed that by using 3x3 rotation matrices we finally had the
transformation
AB(-B-AB) = B
where the expression in the brackets is the last leg, -A, but
similarity-transformed by the tilt angle B. The xyz set is rotated by
B regardless of the values of a or b.
OK, this may convince you. In the aircraft and inertial guidance
business we speak of gimbal order, of which there are 6 possible sorts
of 2 axis gimbals. This is obviously a pitch-over-yaw gimbal problem.
Let the letter "Y" be our gimbal yoke, with the vertical a yaw
bearing. Put an axle across the open end and this will be the pitch
bearing.
Now lock the yaw and pitch axles. Put an xyz triad on top, y going
vertical and x horizontal.
Now unlock the pitch bearing and pull y down to any angle "p". Lock
the axle.
Now unlock the yaw bearing and drag y to the west and lock the angle y
for yaw.
Now unlock the pitch bearing and pull y up till p = 0.
This has performed all the moves needed and it's apparent the xyz is
rotated by y.
It has been rotated by the value of angular trip around the equator,
or any parallel of latitude.
Ralph Hartley
Ralph Hartley wrote:
> Instead of fixing one ball and letting the other roll, let them both roll
> on opposite sides of a plain. If the path comes back to the north pole, the
> contact point must come back to the same point on the plane
That isn't right (it can end up anywhere on the plane), but that doesn't
matter.
Picture a ball rolling without slipping on a mirror. It would also be
rolling without slipping with its image in the mirror.
If it starts and ends with the same point touching the mirror, then it
can only have rotated about the axis that passes through that point. The
other ball is its image in the mirror, which is perpendicular to the
axis, so it rotates by the same angle in the same direction.
So when two spheres (or *any* convex smooth solid for that matter, if
they start out with mirror symmetry about their contact plane) roll on
each other without slipping, and they come back to their original
contact point, there is *never* any relative rotation. It doesn't matter
if they follow the path given in the statement of the problem.
ba...@galaxy.ucr.edu (John Baez) wrote:
>>Of course, anyone who knows some math can solve this by doing lots
>>of calculations. So, it's more fun if you try to solve it by sheer
>>thought, or visualization.
Ok. I've gotten it boiled down to an easy visualization.
I wonder if anyone uses toothed spherical (or ellipsoidal etc.) gears
for anything.
Ralph Hartley
Volker Braun
Here is a way to do it without looking at spherical triangles:
Instead of balls, roll spheres. Since "rolling" is a local property of
the contact point, you might just as well roll one sphere *inside* the
other.
As the radii are the same it is now obvious that the rolling ball
comes back unrotated when the concact point runs around a closed loop.
Best regards,
Volker
Eugen Winkler
"John Baez" <ba...@galaxy.ucr.edu> wrote
>
> Here's a fun question that a friend and I got to arguing about today.
>
Nice question.
After reading 8 responses I just wonder why the words 'geometric phase' and
'symmetry' are missing.
Let's make it a little more complicated:
Let there be a third ball b3 between the two balls b1 and b2 such, that the
three centers are always on a straight line. Again the rolling balls must
touch - and it's allowed to roll without slipping or twisting on the
surface.
0) Let the third ball b3 be arbitrary small compared to the two balls b1 and
b2 having the same radius.
1) Start with b3 touching the north pole of the fixed ball b1.
2) Take b2 and roll b3 down to the equator along a great circle.
3) Then roll it along the equator for an arbitrary distance.
4) Then roll it back up to the north pole along a great circle.
The question is: when we carry out this process, can b2 come back *rotated*
relative to its original orientation... or not?
Eugen Winkler
Greg Egan
geodesic from A to B on the fixed sphere? If A and B are described by
unit vectors, then it's the rotation in the common plane of those
vectors, through an angle equal to twice the angle between them.
But that rotation happens to be equal to the product of the reflection in
the plane normal to A, followed by the reflection in the plane normal to
B.
If we write this:
rot(A,B) = ref(B) ref(A)
It then follows that:
rot(B,C) rot(A,B) = ref(C) ref(B) ref(B) ref(A)
= ref(C) ref(A)
= rot(A,C)
So it doesn't matter how many intermediate points you go through as you
roll the ball, it's exactly the same as going along a geodesic from the
starting point to the end point. And if the starting point *is* the end
point, the net result must be the identity.
backdoorstudent
news.verizon.net
John Baez wrote:
> Here's a fun question that a friend and I got to arguing about today.
>
>
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or
> not?
To help visualisation, let's imagine both balls as having a north and south
pole and an equatorial band showing degrees.
In the first move, we roll the ball along the 0° meridian. When the equators
are parallel, the north pole on the rolling ball is now pointing south.
In the next move, we roll the ball around the equator. Assume that we roll
it through 20° in a clockwise direction seen from above the fixed ball (i.e.
looking down on the north pole of the fixed ball and the south pole of the
rolling ball). The rolling ball now has the 340° point on its equator
opposite the 20° point on the fixed ball's equator.
We finally roll the ball back to its original vertical orientation.
The rolling ball has rotated through 40° in the horizontal plane WRT the
fixed ball.
(Subject to gedankenexperimental error ...)
Russell Blackadar
[snip]
> It's fairly simple. The path is a spherical triangle. There is a
> series of angular transformations, call them A down, and B across and
> -A up again. To be accurate we must use 3x3 matrices which are 2d rank
> tensors.
>
> Let the ball roll down one meridian to the equator. Call the rolldown
> axis x (now parallel to the equator)and call the rotation matrix Ax.
> Name the axis lying along the meridian to the north y.
>
> Now rotate along the equator for angle b, about axis y, and call this
> B_y.
> So far, using row vector notation we have A_x*B_y.
Ok....
>
> Now if b were 0, we could again roll by -A_x, but the x axis is
> displaced now by angle b.
Woops, no. The axis has rotated only half the angle that
the ball has rotated. Call that matrix C_y, where we have
C_y*C_y = B_y.
So we use a similarity transform to make the
> -Ax matrix fit in the tilted by b axis:
> Tr(-Ax)= -By*-Ax*By
Should change those B's to C's.
> to "cure" rotation -Ax, making the total trip:
> Ax*By*Tr(-Ax) = Ax*By*-By*-Az*By = By
Actually Ax*By*(-Cy*-Ax*Cy) = Ax*Cy*-Ax*Cy
and since Ax rotates the y-axis 180 degrees, we have
Ax*Cy*-Ax = -Cy.
> The ball is turned by the matrix By!
Actually by -Cy*Cy = I.
Ken S. Tucker
Thank you Willem, kindly accept my explanation.
Yes I did make the assumption of a 90 degree rotation
about the equator, as you say. I did indeed over simplify
Dr, Baez's question in that regard.
(I had just been reading Baez's article about parallel
transport from North pole to equator, then shifting 90
along the equator and up again to the North pole,
as recommended by Ted Bunn to me in thread "falling balls",
so I suppose I had a preconceived geometry ).
May I add, the rules of the OP (jb) was to use
visualisation, and my original post used a limited amount
of math to support the argument. After all, 90 degrees
was my arbituary choice as I moved along the equator
- by the rules, and I did admit a cheat by checking this
solution with the relativity of dice.
Studying other postings, it seemed that a symmetry is
involved, so I posted a generalization on 04-28, about
mirror symmetry, for perusal.
Regards Ken S. Tucker
Paul Colby
>
> Here's a fun question that a friend and I got to arguing about today.
>
> Suppose we have two balls of the same radius. I'll call them
> the "rolling ball" and the "fixed ball".
>
Consider a slightly more general problem.
Consider any closed path, P1, on the fixed sphere that has a tangent
direction at every point. We may roll the movable ball along this
path. As we roll it the point of contact between the spheres maps
out a path, P2, on the moving sphere. For spheres of equal radius the
path P1 is the same shape and all as P2 since the tangent vectors of
each curve agree at every point because no slipping or twisting
is allowed. Therefore P2 is closed if P1 is closed. If we choose
a path that has a different starting tangent than ending tangent
then we've shown the rolling ball has the same angular location
before and after transversing the path since these two independent
directions agree before and after.
--
Regards
Paul Colby
Ken S. Tucker
news:<b8gc9j$1hu$1...@ngspool-d02.news.aol.com>...
>Nice question.
>After reading 8 responses I just wonder why the words 'geometric phase' and
>'symmetry' are missing.
>
>Let's make it a little more complicated:
>
>Let there be a third ball b3 between the two balls b1 and b2 such, that the
>three centers are always on a straight line. Again the rolling balls must
>touch - and it's allowed to roll without slipping or twisting on the
>surface.
>
>0) Let the third ball b3 be arbitrary small compared to the two balls b1 and
>b2 having the same radius.
>
>1) Start with b3 touching the north pole of the fixed ball b1.
>
>2) Take b2 and roll b3 down to the equator along a great circle.
>
>3) Then roll it along the equator for an arbitrary distance.
>
>4) Then roll it back up to the north pole along a great circle.
>
>The question is: when we carry out this process, can b2 come back *rotated*
>relative to its original orientation... or not?
Yup, Baez threw a curve ball...
I hesitated to respond to this question, (hoping for some
guidance from other posters)...
Let me redesignate the spheres, A, M, B, with radii A = B and
M=Middle sphere of any radius, b3 = M for simplicity of notation.
It's seems obvious that sphere A and B will sustain parallel axes,
(conguence might be the correct term here) and thus return
unrotated to their original orientation, since M creates the
equivalent to sliding, and this allows A and B to sustain the
same orientation.
Correct me if I'm wrong, but couldn't we substitute many
middle balls that convey rotation between sphere's A and B,
ie, M1, M2....Mn, and, provided we use this alignment/contact
rule (your -1 rule, ie. above rule 0) we can convey rotational
and relative orientation?
I think so, there would be no difference in conveying
orientational information using the middle sphere whether it
be sphere M1 or N many spheres M1....Mn.
Why I think that's important in physics, is because Winkler's
introduction of the Middle sphere mediates a relation between
spheres A and B, and no matter how many spheres M separate
spheres A and B, both A and B synchronize by touching
the nearest sphere M.
Regards and thanks, comments are warmly welcomed...
Ken S. Tucker
Larry D. Sumner
As the ball rolls down from the pole to the equator it is rotating about
an axis running in an east-west direction. When the ball reaches the
equator as it rotates along the equator the original axis of rotation
also rotates about an axis running in a north south direction. The
ball also travels along the equator. Unless the angle, through which
the original axis is rotated, is rotated by an exact multiple of 2 pi
when it returns to the original point of contact with the equator it
will be rotated if rolled northward to the poll.
Therefore the ball will always be rotated unless it reaches the
original point of contact with the equator in an integer number of
rotations about the equator.
--
Nature intended me for the tranquil pursuits of Science by rendering them
my supreme
Delight -- Thomas Jefferson
backdoorstudent
work it out rigorously.
It reminded me of some interesting psychological experiments that I
read about which demonstrated how limited the mind is when it comes to
intuition in visualizing rotations. The brain is easily confused by
rotations (especially compounded rotations). This makes evolutionary
sense since predators and other natural hazards do not come at us
rotating.
I hate being retarded.
Andy Ruina
ba...@galaxy.ucr.edu (John Baez) wrote in message news:<b89ths$ma3$1...@glue.ucr.edu>...
> Here's a fun question that a friend and I got to arguing about today.
>
I posed this question to two friends, Mark Levi and Jim Papadopoulos.
Independently they both pointed out that if you take any pair of
convex objects that are mirror images of each other, that when two
points return to contact so does the whole relative orientation
return to previous.
Both used the idea of rolling on two sides of a plane. And the symmetry
argument depends of course on there being no spin about the axis normal
to the contact plane.
This applies to eggs, and if you stay away from the collapsed-to-the-plane
configuration, also to torii. And funny assymetric things too if the
pair are mirror images of each other.
Bill Vajk
Eugen Winkler wrote:
> Nice question.
> After reading 8 responses I just wonder why the words 'geometric phase' and
> 'symmetry' are missing.
Great observation!
> Let's make it a little more complicated:
> Let there be a third ball b3 between the two balls b1 and b2 such, that the
> three centers are always on a straight line. Again the rolling balls must
> touch - and it's allowed to roll without slipping or twisting on the
> surface.
> 0) Let the third ball b3 be arbitrary small compared to the two balls b1 and
> b2 having the same radius.
> 1) Start with b3 touching the north pole of the fixed ball b1.
> 2) Take b2 and roll b3 down to the equator along a great circle.
> 3) Then roll it along the equator for an arbitrary distance.
> 4) Then roll it back up to the north pole along a great circle.
> The question is: when we carry out this process, can b2 come back *rotated*
> relative to its original orientation... or not?
The interloper merely alters the direction of rotation of the outer
ball without changing the virtual distance(s) traveled. A cm traveled
on one side of a ball equals a cm traveled on the other side of the
ball.
Consider the case of a single ball, any diameter so long as there's
no slippage, and an infinitely large plane with a marked start point
and N-S axis marked on the ball, aligned normal at the start point.
Roll the ball around on the plane on any path(s) you chose, and
eventually return to the point of origin. The alignment is conserved.
Then contemplate the shape of the infinitely large plane. What you see
of that plane is, after all, just a segment of a huge sphere (ball.)
The rest of the problem is merely a matter of scale, and the problem
does scale nicely.
backdoorstudent
John Polasek <jpol...@cfl.rr.com> wrote in message news:<6o6ravkabiuvjgj7j...@4ax.com>...
> The ball is turned by the matrix By!
>
> John C. Polasek
> Mr. Dual Space
>
> If you have something to say, write an equation.
> If you have nothing to say, write an essay.
You forgot "draw a picture."
My first intuition agrees with your conclusion.
John Baez
In article <b8c4rl$166$1...@hercules.btinternet.com>,
Right. This comment also applies to the person who studied
this issue by taking two cubical dice and rolling one around
the other. The above analysis is fine in the special case
of rolling the ball 90 degrees along the equator - but the
really interesting question concerns rolling the ball for
an arbitrary angle.
Russell Blackadar
...snip...
> Consider the case of a single ball, any diameter so long as there's
> no slippage, and an infinitely large plane with a marked start point
> and N-S axis marked on the ball, aligned normal at the start point.
> Roll the ball around on the plane on any path(s) you chose, and
> eventually return to the point of origin. The alignment is conserved.
But that's wrong.
(This is essentially a duplicate of something I submitted
last Friday, which may yet appear on s.p.r.)
In the first place, not every closed path on the plane will
return the ball to its original-axis-normal-to-the-plane
orientation. Indeed, by requiring this, you are putting
a severe constraint on the set of allowed paths.
In the second place, even if the path meets your requirement,
in general the ball will end up twisted.
Here's an example:
Place the ball on the origin of your plane, and draw a tiny arrow
on its top, pointed in the +x direction. Now roll the ball 1/2
turn in that same +x direction -- let's call that East. The arrow
is now on the bottom of the ball, pointed West. Next, without
twisting or sliding the ball, roll it 1/2 turn in the NE direction.
The arrow is now back on top of the ball, and it is pointing North.
Next, roll the ball 1/2 turn West; the arrow is now on the bottom
of the ball and still pointing N. Finally, roll the ball SW back
to its starting point, closing the path (a rhombus). The arrow is
now on top of the ball -- and it is pointing W. The axis of the
ball is exactly in its original position, but the ball has twisted
180 degrees about that axis.
> Then contemplate the shape of the infinitely large plane. What you see
> of that plane is, after all, just a segment of a huge sphere (ball.)
You have "generalized" beyond what is valid. The curvature of
the rolled-on surface is crucial to this problem; you can't
neglect it.
But note, my closed path was a rhombus, not a square. It turns
out that regular polygonal paths (of 1/2 turn on a side, at least --
but I think maybe in general) *do* preserve the ball's orientation,
assuming that your requirement is met.
jacques Fric
in terms of invariance (symmetry) when rolling ball is moving over its
closed path (the "transformations").
I developed a similar argument in my post dated 29/04/03
Jacques
"Ken S. Tucker" <dyna...@vianet.on.ca> a écrit dans le message de news:
2202379a.03042...@posting.google.com...
> "Daniel L" <dlapadatu.n...@web2news.net> wrote in message
news:<2667...@web2news.com>...
> > If one marks the path followed by the common point of the spheres on
> > both spheres, one gets - obviously - two identical spherical triangles.
> > Imagine these triangles drawn on the spheres after they reached the
> > final position. Since the last generated triangular edge on both spheres
> > was done simultaneously, these two edges are both in a vertical plane
> > (containing the centres of the spheres). Since the triangles are
> > identical, it means that the firstly generated edges are - at the final
> > position - also in a vertical plane. Therefore, the spheres are
> > unrotated.
> Agreed, that suggests a symmetry...
>
> An axis throught the centers of the two spheres of course intersects
> the point of contact, and each ball would be a mirror image of the
> other along this axis, assuming the balls do not relatively rotate on
> the point of contact then this 'mirror symmetry' will be preserved.
>
> A dot is placed on a mirror, and the sphere is placed there initially,
> no matter how the sphere is rolled on the mirror, when it is returned
> to that point, it will be a mirror image of the other, and no relative
> rotation would have occurred... (I think?).
>
> This would allow any path of rolling.
David M. Palmer
wrote:
> 1) Start with the rolling ball touching the north pole of the fixed ball.
> 2) Roll it down to the equator along a great circle.
> 3) Then roll it along the equator for an arbitrary distance.
> 4) Then roll it back up to the north pole along a great circle.
>
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or
> not?
No, it will return to the same orientation.
Take two globes of Earth. (Or if you have a matter duplicator, take
Earth and a duplicate). Stack them so that the Rolling Earth's
Antarctica is sitting on fixed Earth's north pole, and that their
Greenwich meridians are aligned.
Without loss of generality, roll the top Earth down the Greenwich
meridian. England, France etc. on the fixed Earth get splashed into
the South Atlantic of the rolling Earth. When you are done, the
(lat,lon) = (0,0) points (South of Benin) are touching and the North
poles are pointing in opposite directions.
Now, when you roll the rolling Earth around the equator, the point of
contact has the same longitude on both Earths. Amazon meets Amazon
mouth to mouth, Galapagos turtles can cross over, etc. When you stop,
the longitude of the contact point is the same for both.
Then you roll the Earth back up along the meridian, and you are left
with that (and every other) meridian aligned on the two Earths.
So there is no rotation.
--
David M. Palmer dmpa...@email.com (formerly @clark.net, @ematic.com)
Alex Boeglin
This puzzle needs not be visualized as a 3D problem at all:
picture two discs, one on top the other. "Open" them flat on
a table like a "book". Move one the disc along the perimeter
of the other and close the "book" again using the new contact
point as a hinge!
To convince oneself that there will be no resulting change in
orientation of the disc that's been moved, one can even use
old fashioned geometry with triangles defined by radii of the
circles and the tangent to the circles where they meet.
alternatively, one can also just fold a piece of paper with
three circles drawn on it (with salient points and lines
penciled in with a heavy hand...). Of course this amounts
to a bit of cheating since you weren't supposed to fold the
sheet of paper in geometry: the greeks would have solved it
drawing their figures in the sand...
--
AJB
Russell Blackadar
...
> Using a second sphere is actually another red herring.
No.
>
> The solution is obvious using a point of origin on a plane
> surface. The same solution holds true for any continuous
> surface shape one places the ball on and any path so long
> as there's no slippage.
Your claim is false whether you require the north pole of the
sphere to return to its original position on the plane, or not.
In the latter case (plane path not closed) counterexamples
abound -- three quarter-rolls at right angles, for example --
but I found the closed-plane-path case a bit tricky. Easiest
way to satisfy the hypothesis is to roll along a polygon each
of whose sides is a multiple of pi*r. (If the total is an
odd multiple you have to go around twice.) The tricky part
is that if the polygon is regular, it's not a counterexample!
(I haven't proved this, but it's true for n=3, 4, and 6 at
least.)
However, a nonsquare rhombus of half-rotations *is* a
counterexample. Place a ball on the xy plane and draw
a little arrow on its top, in the +x direction. Roll it
a half turn in the x direction; the arrow is now on the
bottom and pointing toward -x. Now roll 1/2 turn at
45 degrees; the arrow is now on top and pointing toward
+y. Complete the rhombus and you will return to the
origin with the arrow on top and pointing toward -x.
news.verizon.net
> John Baez wrote:
>> Here's a fun question that a friend and I got to arguing about today.
> [snipped]
> To help visualisation, let's imagine both balls as having a north and
> south pole and an equatorial band showing degrees.
> In the first move, we roll the ball along the 0° meridian. When the
> equators are parallel, the north pole on the rolling ball is now
> pointing south. In the next move, we roll the ball around the
> equator. Assume that we roll it through 20° in a clockwise direction
> seen from above the fixed ball (i.e. looking down on the north pole
> of the fixed ball and the south pole of the rolling ball). The
> rolling ball now has the 340° point on its equator opposite the 20°
> point on the fixed ball's equator. We finally roll the ball back to
> its original vertical orientation. The rolling ball has rotated
> through 40° in the horizontal plane WRT the fixed ball.
>
> (Subject to gedankenexperimental error ...)
.... and the gedankenexperimental error is in failing to visualise the
effect of the inversion involved in rolling the ball "back to its
original vertical orientation", which moves the 20°-340° points on the
rolling ball's equator back in alignment with the same points on the
fixed ball's equator.
Gavin Collings
> Of course, anyone who knows some math can solve this by doing lots
> of calculations. So, it's more fun if you try to solve it by sheer
> thought, or visualization.
If you parameterize the rotation of the moving ball in terms of the
rotation of a vector that joins the centres, you find that the moving
ball rotates twice as much as the vector. The vector is an element of
SO(3), the orientation of the moving ball must be one of the
corresponding elements of SU(2). The exact double cover between the
two groups means that the moving ball will have exactly the same
orientation at both N and S poles. i.e. no relative rotation.
Gavin Collings
Tim S
on 4/5/03 1:26 pm, Tim S at T...@timsilverman.demon.co.uk wrote:
> on 29/4/03 11:51 pm, Tim S at T...@timsilverman.demon.co.uk wrote:
>
>> on 25/4/03 7:15 am, John Baez at ba...@galaxy.ucr.edu wrote:
>>
>>>
>>> Here's a fun question that a friend and I got to arguing about today.
>>>
>>> Suppose we have two balls of the same radius. I'll call them
>>> the "rolling ball" and the "fixed ball".
>>>
>>> The fixed ball is not allowed to move.
>>>
>>> The rolling ball must touch the fixed ball - and it's allowed to roll
>>> without slipping or twisting on the surface of the fixed ball.
>>>
>>> Now:
>>>
>>> 1) Start with the rolling ball touching the north pole of the fixed ball.
>>>
>>> 2) Roll it down to the equator along a great circle.
>>>
>>> 3) Then roll it along the equator for an arbitrary distance.
>>>
>>> 4) Then roll it back up to the north pole along a great circle.
>>>
>>> The question is: when we carry out this process, can the rolling
>>> ball come back *rotated* relative to its original orientation... or not?
>>>
>>> One of us thinks the rolling ball comes back unrotated at the end of this
>>> process. The other thinks it rotates by a certain angle that depends
>>> on the distance we rolled it along the equator.
>>>
>> <snip>
>>
>> Hmm.
>>
>> Take out your pen with the infinitely thin nib.
>>
>> On each ball, draw the equator. On each ball, draw two great circles which
>> meet the equator at a right angle and which meet each other at the poles at
>> angle phi. On each ball, the intersections of these circles with the equator
>> are separated by a distance D (= phi R, where R is the radius of the balls).
>>
>> Place the balls together with their north poles touching and the initial
>> segments of the great circles lining up there.
>>
>> Roll the rolling ball down to the equator along the first great circle of
>> the fixed ball. This will also pull the point of contact along the first
>> great circle of the rolling ball (by symmetry, among other reasons). Now
>> roll distance D along the equator of the fixed ball. This will also take the
>> point of contact a distance D along the equator of the rolling ball.
>>
>> Finally, roll back up the second great circle of the fixed ball until the
>> north poles are in contact again. The tangents to the second great circles
>> at the north poles of the two balls will still line up.
>>
>> So no rotation.
>>
>
> In fact, come to think of it, this same argument leads straightforwardly to a
> much more general result:
>
> Take any two smooth surfaces whatever, call them A and B, and let them
> initially be in contact with point p_A in A touching point p_B in B.
>
> Now roll one of the surfaces over the other without slipping or twisting along
> any path at all such that eventually p_A ends up in contact with p_B again.
> (We assume that if the surfaces are non-convex, they can pass through each
> other when necessary anywhere away from the point of contact).
>
> The point of contact traces out a path on each surface. At any given moment,
> the tangents along these paths at the point of contact are parallel ('no
> twisting'). This holds all the way around the paths, ergo is still true when
> p_A again touches p_B. So the rolling surface must come back unrotated
> relative to the fixed surface.
Oops! No sooner had I sent this than I realised it assumes what it was
supposed to prove - that the angle between the tangents to the two ends of
the loop at p_A is the same as that at p_B. Symmetry is used to prove this
in the original argument with spheres, but doesn't apply to unrelated
surfaces.
But it's not so easy to construct blindingly obviously counterexamples...
Tim
Nicolaas Vroom
John Baez wrote:
>
> Here's a fun question that a friend and I got to arguing about today.
>
> The question is: when we carry out this process, can the rolling
> ball come back *rotated* relative to its original orientation... or not?
>
> One of us thinks the rolling ball comes back unrotated at the end of this
> process. The other thinks it rotates by a certain angle that depends
> on the distance we rolled it along the equator.
>
Observing the results of this thread as of 6 May:
There are 48 messages in this thread.
35 people have participated.
Assuming each one has one vote/opinion then this is the result:
1 Question: 1
2 Unrotated: 21
3 Rotated: 4
4 Rotated and Unrotated: 3
5 No answer: 2
6 New problem: 4
Category 4 means that in general the moving ball returns rotated
but in a special case the ball returns unrotated.
Category 5 means that the answer is very mathematical and at the
end very difficult to decipher what they mean.
Category 6 means that people do not answer the original problem
but immediate start with there own problem.
In summary 21 people think that the moving ball returns unrotated
and 7 people that the moving ball returns rotated.
The question is what is the right answer.
How do we decide ?
Does the majority wins ?
Pete
>
How about a Geometric Algebra approach.....
Let the balls be of unit radius with the fixed ball centred on the
origin. An orbit of the point of contact from point p1 to p2 is now
defined by sqrt(p1*p2) where * is the geometric product. If this
orbital rotation is accompanied by a spin of the moving ball by a
similar amount in the same plane (ie non slipping roll) then this
gives a total rotation of the rolling ball of sqrt(p1*p2)*sqrt(p1*p2)
= p1*p2.
A further roll to point p3 on the fixed ball would give a further
total rotation of the rolling ball of p2*p3. Combining these rotations
gives :-
(p1*p2)*(p2*p3) = p1*(p2*p2)*p3 = p1*(1)*p3 = p1*p3.
And by induction any non-slipping roll is equivalent (in terms of
rotation of the rolling ball) to a single non-slipping roll from the
start point to the end point. So any roll about a closed path roduced
no rotation.
Now, does this mean that the electron rolls without slipping about the
nucleus in the hydrogen atom?
Pete
John Baez
Ralph Hartley <har...@aic.nrl.navy.mil> almost wrote:
>John Baez wrote:
>> One of us thinks the rolling ball comes back unrotated at the end of this
>> process. The other thinks it rotates by a certain angle that depends
>> on the distance we rolled it along the equator.
>In fact I've just about convinced myself that the ball comes back unrotated
>after following *any* path that brings it back to the north pole. All the
>business with equator is a red herring.
You're right!
>Instead of fixing one ball and letting the other roll, let them both roll
>on opposite sides of a plane. If the path comes back to the north pole, the
>contact point must come back to the same point on the plane, so the only
>possible rotation is about the vertical axis.
>
>Also the two spheres end up rotated by the same amount (it is harder to see
>that they rotate in the *same* direction, think of the plane as a mirror).
>If you rotate them both to bring the "fixed" sphere back to its original
>position, the other sphere will end up unrotated as well.
I like this a lot. Let me try to say it a different way.
Imagine a plane between the rolling ball and the fixed ball.
Imagine moving this plane around in such a way that neither ball
slips or twists on the plane as the rolling ball moves around.
Both balls do exactly what they would do if the plane were not
there.
Now, switch to a moving frame of reference in which the plane is
fixed.
The rolling ball rolls around on this plane... and so does the
fixed ball! Since they both touch the plane at the same point
at each moment, and both are rolling without slipping, they
execute mirror-image versions of the same motion.
So, without losing any information, we can turn the plane into
a mirror, and replace the fixed ball with the mirror image of
the rolling ball!
The rolling ball starts out with some point marked X touching
the mirror. We roll it however we like until the X touches the
mirror again. When this happens, the X on the mirror image also
touches the mirror - and of course, the X on the mirror image
is not twisted relative to the X on the rolling ball: it's still
just a mirror-image X.
Translating back to the original setup...
... this means that when the rolling ball's point of contact
with the fixed ball returns to its original position, the
rolling ball has not come back twisted relative to its original
orientation!
So:
My friend James Dolan was right, and I was wrong. I knew
very well that a ball rolling without slipping or twisting
on a plane can come back twisted relative to its original
orientation. This is a textbook example of a "nonholonomic
constraint", where a physical system can move around while
satisfying some constraint, and achieve a motion that seems
superficially to be ruled out by the constraint. I got
interested in this rolling ball problem because of its
apparent relation to the split octonions, the exceptional
group G2, and Cartan's work on nonholonomic constraints:
http://groups.google.com/groups?selm=ao6rg8%246ap%241x%40glue.ucr.edu
So, I never dreamed that a ball rolling on a ball of equal
radius could be *holonomic*! But apparently it is.
(It seems to be crucial that that balls have equal radius.)
Jim and I are very confused about what this means for my
conjecture about how the rolling ball is related to the
split octonions and G2. But we'll eventually straighten it
out. In math, the truth is never bad. So, thanks! And
thanks to everyone else who tried this problem!
John Baez
Eric A. Forgy <fo...@uiuc.edu> wrote:
>I like this question :)
Thanks! It is rather cool.
>I have no solution, but I'd like to extend the problem with another
>challenge.
>
>Instead of rolling the ball without slipping or twisting while it
>traverses the right spherical triangle, what if it moves about an
>arbitrary closed loop? Does it return rotated for a general closed
>loop?
Apparently it returns unrotated for an arbitrary closed loop!
>Can you phrase the answer in terms of curvature and parallel
>transport (as I imagine the original motivation for the question)?
Actually that wasn't the original motivation. The original
motivation was to find a 5-dimensional manifold equipped with
an action of the split real form of the exceptional Lie group G2.
See:
http://groups.google.com/groups?selm=ao473v%24ren%241x%40glue.ucr.edu
for details.
But you're right: if we pick any smooth convex surface
S in R^3 and pick any number r > 0, we get a connection
on the trivial SO(3)-bundle over S for which the holonomy
along any path is just the rotation that a ball of radius r
would execute as it rolled along this path without slipping
or twisting.
If S is a plane and r > 0 is arbitrary, it's a famous
fact that this connection has nonzero curvature - look
in Goldstein's "Classical Mechanics" under "nonholonomic
constraints".
But apparently if S is itself a sphere of radius r, this
connection is flat!
If you're still wondering why, see my recent post in which
I expand on the argument that Ralph Hartley, Uncle Al and
others gave.
John Baez
Gavin Collings <gavin.c...@airbus.com> wrote:
It's not right to say "the vector is an element of SO(3)";
the vector is really just an element of the unit sphere,
which is not the same as SO(3). But I think you're onto
something interesting ... so I'd like to hear someone
make this idea precise.
The best I can do so far is this:
The point of contact between the rolling ball and the fixed
ball defines an element of the unit sphere in R^3. Let's think
of this 2-sphere as the unit imaginary quaternions:
{ai + bj + ck: a^2 + b^2 + c^2 = 1}
This is a subset of the unit quaternions:
{ai + bj + ck + d1: a^2 + b^2 + c^2 + d^2 = 1}
which is isomorphic to SU(2), which in turn maps
in a two-to-one and onto way to the rotation group SO(3).
So, we get a map from "points of contact" to "rotations",
and I believe we can use this to calculate how much the rolling
ball has rotated given its point of contact.
The only irksome thing is that in this description, no point of
contact corresponds to the identity rotation, since the quaternion
1 is not a unit *imaginary* quaternion. Instead, every point of
contact corresponds to the 180 degree rotation around the axis
defined by that point of contact... since unit imaginary quaternions
get mapped to 180 degree rotations.
John Baez
Dave Hilbert <wannado...@yahoo.com> wrote:
>Mirror images can't rotate relative to each
>other. :)
Yes, that appears to be the key!
>So what highly mathematical question was reduced
>to this simple query?
We were wondering if the configuration space of the
rolling ball was (double covered by) one of the
Grassmannians for the split real form of the exceptional
Lie group G2, with the "lines" in this Grassmannian
corresponding to paths where the ball rolls along a great
circle.
And in fact we're still wondering this, because the
answer you and others gave - "no rotation" - seems
to contradict other properties the lines on this
Grassmannian should have! But, we have other evidence
that this space really *is* double covered by the desired
Grassmannian. So, maybe our notion of "line" is screwed up.
We've been thinking about this for about a week.
I append some more of the background behind this problem.
...............................................................
From: ba...@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: octonions and the rolling ball
Date: Thu, 10 Oct 2002 18:32:13 +0000 (UTC)
Organization: UCR
I've been reading Richard Montgomery's book "A Tour of
Subriemannian Geometries, Their Geodeisics and Applications"
and I came across an interesting puzzle.
I don't have the energy to explain what this means -
(see pages 87-88 in his book if you're curious - but
in an 1910 paper Elie Cartan worked out an invariant
of "(2,3,5) distributions" on a manifold, and when this
invariant vanishes he showed the exceptional Lie group
G2 acts on the manifold in question.
Montgomery shows that an example of this situation
arises when we consider the phase space of a ball
rolling on a table. This manifold is 5-dimensional:
it takes 2 numbers to say where the center of the ball is,
and 3 more to describe the rotational degrees of freedom
of the ball. There is a 3-dimensional sub-bundle of
the tangent bundle which describes the ways the ball
can roll without slipping, and a 2-dimensional sub-bundle
which describes the ways the ball can roll without
slipping or "spinning", i.e. rotating about the vertical
axis. Perhaps this enough for you to guess why this
gives a "(2,3,5) distribution" in Cartan's terminology.
So, Montgomery poses the following open question:
Find an explicit geometric or physical description of the
G2 action on the ball-table system.
This is interesting, but it's especially interesting
to me because G2 is the automorphism group of the octonions!
Answering Montgomery's question would for the first time
give an example of the octonions appearing in real-world
physics (as opposed to unproven theories of particle physics,
like string theory).
But it's also puzzling, because the lowest-dimensional
example I know of a manifold on which G2 acts nontrivially
is 6-dimensional: the unit sphere in the imaginary octonions.
Of course, if you can answer Montgomery's question I'd
be delighted. But I'd settle for any example of a 5-dimensional
manifold on which G2 acts in a nontrivial way.
...............................................................
From: ba...@galaxy.ucr.edu
Newsgroups: sci.physics.research
Subject: Re: octonions and the rolling ball
Date: 14 Oct 2002 01:55:57 GMT
Organization: UCR
In article <ao5gj6$mq5$1...@news.fas.harvard.edu>,
Noam D. Elkies <elkie...@h.harvard.edu> wrote:
>For the compact real form of G2, I can believe that SU(3) is
>the proper subgroup of maximal dimension. For the split form,
>or for G2(C), the group has maximal parabolics P of dimension 9,
>so codimension 14-9=5. Fulton and Harris discuss these maximal
>parabolics around page 391 of their book _Representation Theory:
>A First Course_. For one of them, G/P is a quadric surface in P^6,
>which is presumably the octonions o such that o^2=0 (that is,
>Tr(o)=Norm(o)=0), modulo scaling.
Just to clarify: as you mentioned in your email to me, these
"octonions" are really the split octonions if we're working
with the split form of G2, or the complexified octonions
(aka bioctonions) if we're working with the complex form G2(C).
>This seems like a reasonable guess for John Baez's 5-manifold.
>Fulton-Harris indicates that the other G/P is harder to get
>one's hands on.
Naively I'd guess that for the split octonions, the
the equation Norm(o) = 0 amounts to something like
a^2 + b^2 + c^2 + d^2 - e^2 - f^2 - g^2 - h^2 = 0
where (a,b,c,d,e,f,g,h) are an 8-tuple of real numbers.
The condition Tr(o) = 0 says the "real part" of our split
octonion vanishes, and naively I'd guess this gives us
an equation like
b^2 + c^2 + d^2 - e^2 - f^2 - g^2 - h^2 = 0
in 7 variables. If we projectivize this we get a
manifold diffeomorphic to (S^2 x S^3) / Z_2 . It
seems a bit odd that this is the same projective
quadric we get from G/P where G = SO(4,4) and P is
a maximal parabolic, so maybe I'm screwing up, or
maybe this is one of those coincidences like how
S^7 is a homogenous space of both SO(8) and the
compact real form of G2.
Anyway, this space (S^2 x S^3) / Z_2 is a bit different,
but not *drastically* different, from the phase space
of a ball rolling on the plane - namely R^2 x (S^3 / Z_2).
So, maybe we're pretty close!
I'll work through it more carefully sometime and
straighten it out. Thanks a million!
Ken S. Tucker
news:<bd0ff8f.03050...@posting.google.com>...
>ba...@galaxy.ucr.edu (John Baez) wrote in message
news:<b89ths$ma3$1...@glue.ucr.edu>...
>> Here's a fun question that a friend and I got to arguing about today.
>How about a Geometric Algebra approach.....
>
>Let the balls be of unit radius with the fixed ball centred on the
>origin. An orbit of the point of contact from point p1 to p2 is now
>defined by sqrt(p1*p2) where * is the geometric product. If this
>orbital rotation is accompanied by a spin of the moving ball by a
>similar amount in the same plane (ie non slipping roll) then this
>gives a total rotation of the rolling ball of sqrt(p1*p2)*sqrt(p1*p2)
>= p1*p2.
>
>A further roll to point p3 on the fixed ball would give a further
>total rotation of the rolling ball of p2*p3. Combining these rotations
>gives:
>
>(p1*p2)*(p2*p3) = p1*(p2*p2)*p3 = p1*(1)*p3 = p1*p3.
>
>And by induction any non-slipping roll is equivalent (in terms of
>rotation of the rolling ball) to a single non-slipping roll from the
>start point to the end point. So any roll about a closed path roduced
>no rotation.
>
>Now, does this mean that the electron rolls without slipping about the
>nucleus in the hydrogen atom?
That's interesting....
Using John Baez's anology of placing an X on a mirror, then
rolling the ball - by any path - to another point P, will result in
in an orientation completely path independant, although not the
same as that originally at X. So two balls rolled from X to P by
any paths arrive at P with the same orientation.(IMO?).
Some quantization of orientation may be realized though.
A polar CS with R as radius and Theta as angle originates
from point X. Rolling along the Theta =0 axis, in direction R,
a ball of unit radius, will succesively have equal orientations at
R_n = 2pi, 4pi ... 2pi*n with n=a whole number.
At any n, along R, a roll can occur on Theta with R constant.
The circumference about X at these R_n are
C_n = 2pi*(R_n=(2pi)*n) = (2pi)^2*n
The circumference C of unit ball is 2pi, so that,
C_n/C = 2pi*n also allows the rolling to be equally
orientated.
This was only meant to be a baby step in addressing
Pete's question on Hydrogen atoms.
Regards
Ken S. Tucker
PS: I'm not sure if the Pauli Exclusion principle applies
to an H atom, but usually does in multi-electron atoms.(?)
Mention this thinking Pete's question might be for He...
KST
Eric A. Forgy
Do to a quirk in the way my google newsreader is set up, I was deep
into a reply to your response to David's post before I realized you
had replied to mine. Doh!
ba...@galaxy.ucr.edu (John Baez) wrote:
> Eric A. Forgy <fo...@uiuc.edu> wrote:
> >Instead of rolling the ball without slipping or twisting while it
> >traverses the right spherical triangle, what if it moves about an
> >arbitrary closed loop? Does it return rotated for a general closed
> >loop?
> Apparently it returns unrotated for an arbitrary closed loop!
Yes, the mirror/plane trick is very clever. My hat is off to those who
thought of it :)
> >Can you phrase the answer in terms of curvature and parallel
> >transport (as I imagine the original motivation for the question)?
> Actually that wasn't the original motivation.
Too bad :) Rolling the second ball without "slipping" or "twisting"
sounds suspiciously like the classic example (which suspiciously
involves the same path) of parallel transport on a sphere while
keeping the tangent vector "as straight as possible." When I catch a
few seconds here and there I've been trying to think of the rolling
ball problem in terms of parallel transport as well.
> The original
> motivation was to find a 5-dimensional manifold equipped with
> an action of the split real form of the exceptional Lie group G2.
> See:
>
> http://groups.google.com/groups?selm=ao473v%24ren%241x%40glue.ucr.edu
>
> for details.
I have no clue what this means :)
> But you're right: if we pick any smooth convex surface
> S in R^3 and pick any number r > 0, we get a connection
> on the trivial SO(3)-bundle over S for which the holonomy
> along any path is just the rotation that a ball of radius r
> would execute as it rolled along this path without slipping
> or twisting.
Cool :)
> If S is a plane and r > 0 is arbitrary, it's a famous
> fact that this connection has nonzero curvature - look
> in Goldstein's "Classical Mechanics" under "nonholonomic
> constraints".
>
> But apparently if S is itself a sphere of radius r, this
> connection is flat!
Exactly! Isn't that somewhat mystifying?
In the case of the standard example of parallel transport on a sphere
of radius a, the original tangent vector comes back rotated by an
amount epsilon, which is proportional to the area Aloop enclosed by
the loop and the curvature 1/a^2, i.e.
epsilon = Aloop/a^2
In the case of a spherical triangle, it is interesting (to me anyway)
that epsilon turns out to be the excess angle, i.e.
epsilon = "sum of the angles of the spherical triangle" - pi
and
Ast
= "Area of a spherical triangle"
= espilon*a^2
> If you're still wondering why, see my recent post in which
> I expand on the argument that Ralph Hartley, Uncle Al and
> others gave.
The mirror/plane argument makes it clear why it returns unrotated,
which implies the connection is flat, but I'd be interested in a more
satisfying explanation for why the connection is flat. In particular,
if you were to let S be a sphere of radius a != r, then the holonomy
would be an element of SO(3), apparently. Or maybe SU(2) would be more
appropriate (thinking of your response to Gavin)?
In the case of two nonidentical spheres, i.e. a != r, then I would
expect that the holonomy would involve the area of the loop Aloop in
an essential way.
I'm not very good at this stuff, but let me try a guess...
In the case of the standard example of parallel transport on a sphere,
the holonomies are elements of SO(2) (I hope I got at least this
correct), which is equivalent to U(1), i.e. elements of the form
exp(i*phi), phi in R. Therefore, for this case we have the vector
rotated by
exp(i*Aloop/a^2).
With such little information at my disposal, I am willing to speculate
that the holonomy for the case of the rolling ball problem is going to
be something like
exp[(i*Aloop/a^2)*n^]
where n^ is a unit vector and I am considering holonomies in SU(2).
Anyway, I know this can't be right because I can't see how to get the
identity for any closed loop from this when a = r, but this is along
the lines I am thinking.
Does anyone know the correct expression for holonomies for this
rolling ball problem for the slightly generalized case of a != r? I
think it would be instructive to then see how it reduces to a flat
connection for the case a = r.
Eric
Russell Blackadar
[...speculation re arbitrary surfaces...]
> Oops! No sooner had I sent this than I realised it assumes what it was
> supposed to prove - that the angle between the tangents to the two ends of
> the loop at p_A is the same as that at p_B. Symmetry is used to prove this
> in the original argument with spheres, but doesn't apply to unrelated
> surfaces.
Also, you might want to check your assumption that p_A in
fact touches p_B at the end. In effect you are assuming
that a closed path on one surface is always congruent to
a closed path on the other. 'Taint so, in general.
>
> But it's not so easy to construct blindingly obviously counterexamples...
For the assumption I just mentioned, obvious counterexamples
are much easier to find than examples, once you start looking
at all. Just roll a ball on the plane, pole-to-equator, then
along the equator, then back to the pole. Is the path on the
plane closed? If it is, then you have a planar triangle with
two right angles!
John Baez
>>How about a Geometric Algebra approach.....
>(p1*p2)*(p2*p3) = p1*(p2*p2)*p3 = p1*(1)*p3 = p1*p3.
It's worth noting that this equation is also hidden in
Greg Egan's post:
>What's the rotation induced in the rolling ball by rolling it along a
>geodesic from A to B on the fixed sphere? If A and B are described by
>unit vectors, then it's the rotation in the common plane of those
>vectors, through an angle equal to twice the angle between them.
>
>But that rotation happens to be equal to the product of the reflection in
>the plane normal to A, followed by the reflection in the plane normal to
>B.
>
>If we write this:
>
> rot(A,B) = ref(B) ref(A)
>
>It then follows that:
>
> rot(B,C) rot(A,B) = ref(C) ref(B) ref(B) ref(A)
> = ref(C) ref(A)
> = rot(A,C)
Geometric algebra, aka the theory of Clifford algebras,
heavily exploits the fact that two reflections gives a rotation.
The Clifford algebra generated by 3 square roots of 1 is
C tensor H, the complexified quaternions, or "biquaternions" -
an algebra isomorphic to the 2x2 complex matrices. We can
think of these 3 square roots of 1 as corresponding to
reflections about the x,y, and z axes. If you do a reflection
twice you get back where you started, so reflections become square
roots of 1 in the Clifford algebra.
The even part of this Clifford algebra, i.e. all linear combinations
of *even* numbers of generators, is just H, the quaternions.
The double cover of the rotation group lives in here, since an
even number of reflections gives a rotation.
This lies behind my use of the quaternions in another post on this
thread.
Pete
> >nucleus in the hydrogen atom?
>
>
> That's interesting....
> Using John Baez's anology of placing an X on a mirror, then
> rolling the ball - by any path - to another point P, will result in
> in an orientation completely path independant, although not the
> same as that originally at X. So two balls rolled from X to P by
> any paths arrive at P with the same orientation.(IMO?).
> Some quantization of orientation may be realized though.
> A polar CS with R as radius and Theta as angle originates
> from point X. Rolling along the Theta =0 axis, in direction R,
> a ball of unit radius, will succesively have equal orientations at
> R_n = 2pi, 4pi ... 2pi*n with n=a whole number.
> At any n, along R, a roll can occur on Theta with R constant.
> The circumference about X at these R_n are
> C_n = 2pi*(R_n=(2pi)*n) = (2pi)^2*n
> The circumference C of unit ball is 2pi, so that,
> C_n/C = 2pi*n also allows the rolling to be equally
> orientated.
>
> This was only meant to be a baby step in addressing
> Pete's question on Hydrogen atoms.
>
I must say that this thread has made me realise more than ever how suprising
aand subtle rotations in 3D can be. The various approaches presented in
response have opened up a veritable Aladins Cave of conceptual tools.
A while back we had a thread about spinning ball electrons as a way of
developing a classical model of the hydrogenic atom which was consistent
with some of the strange behaviour of spin 1/2 particles. I'm wondering now
whether the rolling ball electron might do an even better job?
It seems like such a model offers a natural way in which the electron needs
to turn twice to regain its original orientation. Plus we can see how a
single surface (the sphere in this case) can support two electrons in
anti-phase. But maybe I'm getting carried away...
Pete
Douglas B Sweetser
John Baez wrote:
> So, I never dreamed that a ball rolling on a ball of equal
> radius could be *holonomic*! But apparently it is.
> (It seems to be crucial that that balls have equal radius.)
The equal radius only applies if the balls are in contact with a mirror.
Hold a ball infront of your face, away from a distant mirror, and rotate
it. The ball in the mirror will be smaller, yet both rotate together and
end up at the not rotated. The "point of contact" changes, since it
requires a line to link between points on the two solids (they just need to
be three D and identical - oops, any chirality must be reversed - when at
zero distance from the mirror).
doug
quaternions.com
Gavin Collings
"John Baez" <ba...@galaxy.ucr.edu> wrote in message
news:b9hoj8$ivg$1...@glue.ucr.edu...
The mapping I had in mind was something like the rotation that takes the
moving ball from its original position at the N pole to its final
orientation is a member of SO(3), or at least a subset of it - corresponding
to no twists. And that this corresponds to the half-angle rotation that
moves the point of contact from the N pole so the point of contact rotation
is a member of the corresponding subset of SU(2).
If you start with the end result, it doesn't matter how you move the moving
sphere to its final position. So the easiest way is just one rotation:
rolling it down the relevant meridian. That means the orientation of the
moving sphere is always described by a rotation with only x and y
components. In SU(2), this is a unit quaternion with d = 0. Say the
rotation (of the moving sphere) is 2 * alpha about an axis rotated by beta
about the N pole. This gives the following quaternion:
a = cos( alpha )
b = sin( alpha ) * cos( beta )
c = cos( alpha ) * cos( beta )
d = 0
As you say, this subset of SU(2) can be mapped on to the 2-sphere, and if
you use the parameters alpha and beta as colatitude and longitude you map
directly to the point of contact.
So, going the other way, could I still use my double cover argument on this
subset?
Gavin Collings
Tim S
on 11/5/03 8:17 am, Russell Blackadar at rus...@mdli.com wrote:
> Tim S wrote:
>
> [...speculation re arbitrary surfaces...]
>
>> Oops! No sooner had I sent this than I realised it assumes what it was
>> supposed to prove - that the angle between the tangents to the two ends of
>> the loop at p_A is the same as that at p_B. Symmetry is used to prove this
>> in the original argument with spheres, but doesn't apply to unrelated
>> surfaces.
>
> Also, you might want to check your assumption that p_A in
> fact touches p_B at the end. In effect you are assuming
> that a closed path on one surface is always congruent to
> a closed path on the other. 'Taint so, in general.
Oh, yeah, I was was aware of that. I think I included the closed-ness of
both paths in the hypotheses.
>> But it's not so easy to construct blindingly obviously counterexamples...
>
> For the assumption I just mentioned, obvious counterexamples
> are much easier to find than examples, once you start looking
> at all. Just roll a ball on the plane, pole-to-equator, then
> along the equator, then back to the pole. Is the path on the
> plane closed? If it is, then you have a planar triangle with
> two right angles!
Yes -- the constraint that both paths shall be closed is a very strong one.
In fact, that's what's made it difficult for me to think of counterexamples
-- trying to incorporate that constraint. As you say, it's easy to construct
counterexamples where one of the paths isn't closed.
Tim
Greg Egan
It's possible to make a link between this problem and SU(2), but it's a
bit skewed.
If you nominate a particular starting point A on the fixed sphere, e.g.
the north pole, then every other point B on the fixed sphere specifies a
rotation, namely the rotation induced when you roll the moving sphere
from A to B. The net rotation is path-independent, but it's easiest to
see what it is when you imagine rolling along a geodesic from A to B,
when it's obvious that it's the rotation around the axis A cross B
(indentifying points on the sphere with unit vectors in R^3), by an angle
equal to twice the angle between A and B.
Half the sphere gives you all the possible rotations around axes normal
to A, and the whole sphere double covers that same subset of SO(3), so we
can look for a natural way to embed the whole sphere in SU(2). Varying A
should then give different embeddings of the sphere in SU(2).
Think of SU(2) as the unit quaternions, and identify the imaginary
quaternions with R^3. An arbitrary unit quaternion can be written as:
q = cos C + v sin C
where v is a unit imaginary quaternion. The action of the unit
quaternions on an arbitrary unit quaternion w is:
q(w) = q w q*
which, for q as defined above, is:
q(w) = (cos C + v sin C) w (cos C - v sin C)
and, identifying the imaginary quaternions with R^3, this is a rotation
around the vector corresponding to v, by an angle of 2C.
q(v) = (cos C + v sin C) v (cos C- v sin C)
= v (cos C)^2 - v^3 (sin C)^2
= v [(cos C)^2 - v^2 (sin C)^2]
= v [(cos C)^2 + (sin C)^2]
= v
So if we identify points A and B on the fixed sphere with unit vectors,
and if we identify all of R^3 with the imaginary quaternions, then if C
is the angle between A and B, the rotation induced by rolling from A to B
corresponds to the unit quaternion:
cos C + (unit vector parallel to A cross B) sin C
= A.B + (A cross B)
For a given A, and an arbitrary B, this embeds the whole fixed sphere in
the 3-sphere of the unit quaternions, aka SU(2). As A moves from the
north pole to the south pole along any meridian, these various embeddings
sweep over all of SU(2).
Tim S
Hmmm, this is along the right lines, but the sphere of unit imaginary
quaternions is sort of orthogonal to what you want. That is, you do want
a 'great sphere' (a maximal-size S^2) in S^3 (a.k.a. SU(2)). But you
want the one which corresponds to all rotations about axes
_perpendicular_ to a given axis, the axis in question being the
north-south axis of the fixed sphere (or whichever axis has the initial
position of the point of contact as one of its endpoints). Hence, in the
unit quaternions, the axis you suppress to get your great sphere is
never the real one; in fact, it's always purely imaginary. E.g. if the
north-south axis is the z-axis, we're talking about unit quaternions of
the form a + bi + cj.
To see why this is, note that, since the final state of the rolling ball
doesn't depend on what path the point of contact followed, we might as
well have it following a great circle. Then the orientation of the
rolling ball will undergo a smooth rotation about a fixed axis. The
fixed axis will always be perpendicular to the north-south axis of the
fixed ball.
Neatly, the fact that the angle of the rolling ball changes twice as
fast as the latitude of the point of contact corresponds precisely to
the way that SU(2) (a subspace of which is identified with the fixed
ball) doubly covers SO(3) (a subspace of which is identified with the
rotations of the rolling ball).
Hope this makes things clearer.
Tim
punk floyd
Greg Egan <greg...@netspace.zebra.net.au> wrote in message
news:<200305120741....@cumulus.netspace.net.au>...
>So if we identify points A and B on the fixed sphere with unit
>vectors, and if we identify all of R^3 with the imaginary
>quaternions, then if C is the angle between A and B, the rotation
>induced by rolling from A to B corresponds to the unit quaternion:
> cos C + (unit vector parallel to A cross B) sin C
> = A.B + (A cross B)
...which is
= B A^-1 (as quaternions)
So this actually means that the orientation of the rotating ball
(expressed as a unit quaternion) is proportional to the point of
contact on the fixed sphere (expressed as a unit imaginary
quaternion). The coefficient of proportionality is given by the
starting position and orientation.
This expression can itself be used to derive the independency of
orientation from the path. If we start with orientation q_1 and roll
the ball along a path a_1 a_2 a_3 .. a_n where each segment is a
geodesic, the resulting orientation is:
q_n = (a_n a_n-1^-1)...(a_3 a_2^-1)(a_2 a_1^-1)q_1 = a_n a_1^-1 q_1
So (after taking a limit) we conclude the result is path-independent.
Best,
Oleg
Ken S. Tucker
"Pete" <Pe...@removebtclick.com> wrote in message
news:<b9nrad$s1g$1...@hercules.btinternet.com>...
>"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
>news:2202379a.03051...@posting.google.com...
[unnecessary quoted text deleted by moderator]
>I must say that this thread has made me realise more than ever how
>suprising and subtle rotations in 3D can be. The various approaches
>presented in response have opened up a veritable Aladin's Cave of
>conceptual tools.
Yes, and imagining this beats counting sheep for an insomniac!
>A while back we had a thread about spinning ball electrons as a way
>of developing a classical model of the hydrogenic atom which was
>consistent with some of the strange behaviour of spin 1/2
>particles. I'm wondering now whether the rolling ball electron might
>do an even better job?
If you could provide the name of the thread, I'd like to read it.
(please).
>It seems like such a model offers a natural way in which the electron needs
>to turn twice to regain its original orientation.
Well the first rotation (through 2pi) of the rolling ball has it arrive
at the south pole of the fixed sphere, in an orientation equal to the
initial orientation. And then another rotation (total 4pi) has it
arrive at north pole, of course in it's original location and
orientation.
Would this be like a Mobius strip except opposite (with orientation
and location interchanged)?
On Mobius, a rotation of 2pi produces the same location but of
opposite orientation, while the rolling spheres produces the same
orientation but opposite locations, (being at the south pole after
rotating 2pi).
(I got as far as tensors, I need to study John Baez's quaternions...
prior to trying an intelligent comment please pardon my ignorance).
Would it make sense to add
locations and orientations of (colloquially expressed),
Mobius + Spheres = zero?,
naturally curious about more symmetry, I'm even wondering
if Baez's Sphere problem and Mobius have the same dimensionality.
>It seems like such a model offers a natural way in which the electron needs
>to turn twice to regain its original orientation. Plus we can see how a
>single surface (the sphere in this case) can support two electrons in
>anti-phase. But maybe I'm getting carried away...
Well, to paraphrase the meaning of anti-phase, would this be two
spheres s and s' rolling on the fixed sphere S, initially with opposite
locations relative to S with points marked on the rolling spheres so
that they are orientated relatively opposite, ie, x=-x' , y=-y' ,
z = -z', and described by locations relative to S as
X =-X', Y=-Y', Z=-Z'
so that the sum of orientation and location is zero?
Regards
Ken S. Tucker
Uncle Al
One would hope so.
If the spheres have identical radii (if convex shapes are mirror
images) then the Gedankenexperiment is indistinguishable from one
sphere (shape) rolling on a mirror tangent to the point of contact.
How does an object dephase from its reflection? (Concave/negative
curvature shapes are left as an exercise for the alert reader.)
Reflected rotations can certainly go awry - parity nonconservation in
Weak Interactions, axial vs. polar vectors - but those have added
degrees of freedom. Non-congruent shapes (spheres with unlike radii)
are different stories.
--
Uncle Al
http://www.mazepath.com/uncleal/eotvos.htm
(Do something naughty to physics)
Pete
news:2202379a.03051...@posting.google.com...
>
>
> >A while back we had a thread about spinning ball electrons as a way
> >of developing a classical model of the hydrogenic atom which was
> >consistent with some of the strange behaviour of spin 1/2
> >particles. I'm wondering now whether the rolling ball electron might
> >do an even better job?
>
> If you could provide the name of the thread, I'd like to read it.
> (please).
"Stern Gerlach and spiining ball electrons "
Note that at some point someone corrected the spelling of "spinning" so you
may need to Google on both spellings to explore the full thread.
I'm going to pass on the more difficult questions you raise since I don't
feel like I have anything useful to add.
Pete
Eric A. Forgy
> I'm not very good at this stuff, but let me try a guess...
>
> In the case of the standard example of parallel transport on a sphere,
> the holonomies are elements of SO(2) (I hope I got at least this
> correct), which is equivalent to U(1), i.e. elements of the form
> exp(i*phi), phi in R. Therefore, for this case we have the vector
> rotated by
>
> exp(i*Aloop/a^2).
>
> With such little information at my disposal, I am willing to speculate
> that the holonomy for the case of the rolling ball problem is going to
> be something like
>
> exp[(i*Aloop/a^2)*n^]
>
> where n^ is a unit vector and I am considering holonomies in SU(2).
>
> Anyway, I know this can't be right because I can't see how to get the
> identity for any closed loop from this when a = r, but this is along
> the lines I am thinking.
After spending another 60 seconds thinking about this, I have another
guess.
Let the two spheres be denoted a and b. Now let ra> denote the vector
pointing from the point of contact of the two spheres to the center of
sphere a. Diddo for rb> and sphere b.
My new guess is that when the rolling ball returns to its original
position it's holonomy will be of the form
exp[(i*Aloopa/|a|^2)*(ra> + rb>)],
where Aloopa is the area of the loop as measured on sphere a, and |a|
is the radius of sphere a.
Note,
Aloopa/|a|^2 = Aloopb/|b|^2
so the holonomy above is symmetric under replacing a <-> b, as it
should be.
Note also that when ra> = -rb>, as in the original problem, then we've
got
exp[i*0] = 1.
Furthermore, if rb> = 0, then we get
exp[(i*Aloopa/|a|^2)*ra>]
which agrees with the standard example of parallel transport (so does
that mean we can think of a tangent vector on the sphere as the
configuration of a sphere of radius 0? :))
Anyway, in the time it took me to write this post, I could have worked
out the details of this problem myself, but this is more fun :)
Eric
PS: I'm not giving up until I see the correct answer :)
punk floyd
news:<3EC2AACE...@hate.spam.net>...
> punk floyd wrote:
> > q_n = (a_n a_n-1^-1)...(a_3 a_2^-1)(a_2 a_1^-1)q_1 = a_n a_1^-1 q_1
> >
> > So (after taking a limit) we conclude the result is path-independent.
> One would hope so.
...
> Reflected rotations can certainly go awry - parity nonconservation in
> Weak Interactions, axial vs. polar vectors - but those have added
> degrees of freedom. Non-congruent shapes (spheres with unlike radii)
> are different stories.
I am not quite sure what you are getting at, but in case of spheres of
different radii the relationship between the rotation of the sphere
and the change of the point of contact is no longer a simple
proportion. If we roll the ball from point A to point B (expressed as
unit imaginary quats), then the additional rotation will be (if I
didn't miss something)
(B A^-1)^(fixed ball radius/moving ball radius)
Now, if you perform multiple rotations in a sequence, the terms will
not cancel thanks to non-commutativity of quaternions. The proof no
longer goes through and the rotation turns out to be path-dependent.
Best,
Oleg
[Moderator's note: I think that's what Uncle Al meant by
"sphere with unlike radii are different stories" - jb]
Tim S
<snip>
Some back-of-the-envelope scribblings seem to indicate that the holonomy is
something like
exp[i * Aloopa * (1/|a|^2 - 1/|b|^2) * SomeRadiusIndependentVector].
We clearly do get flatness when |a| = |b|, and when |a| goes to infinity, we
get a vaguely sensible-looking result for rolling on the plane.
I'm not sure how this relates to your suggestion. I'm a bit concerned about
Aloopa/|a|^2 = Aloopb/|b|^2
because, in general, (if |a| is not equal to |b|) rolling round a loop on
sphere a won't result in a complete loop being drawn on sphere b, so I don't
see how Aloopb can be defined.
Tim
Ralph E. Frost
Nicolaas Vroom <nicolaa...@pandora.be> wrote in message
news:3EB7A82B...@pandora.be...
My attempt (which apparently evaporated along the line) was of the variety
rotated except for set revolutions. But, after I wrote it, I think it
turned out I was wrong.
I "placed" a red dot on the top of the roller ball to start and then rolled
it to the equator. First time through, I had it visualized as NOT at the
radius height, but higher. Don't ask me why I did that. Away, rotating
that (wrong) visualization makes it so the red dot only resumed the right
place when returned to the north pole on certain revolutions (1/2 and 1...)
along the equator.
Now I realize that when it rolls to the equator first, along and then back
again, it would be unrotated. But I think it is correct that every place
else, short or long of to the equator except for 1/2.. spins along the
parallel, it comes back rotated.
So, the answer to the specific question (to equator and back) is unrotated.
Eric A. Forgy
Thanks a lot for taking the time to work that out.
Tim S <T...@timsilverman.demon.co.uk> wrote:
> Some back-of-the-envelope scribblings seem to indicate that the holonomy is
> something like
>
> exp[i * Aloopa * (1/|a|^2 - 1/|b|^2) * SomeRadiusIndependentVector].
>
> We clearly do get flatness when |a| = |b|, and when |a| goes to infinity, we
> get a vaguely sensible-looking result for rolling on the plane.
Great! :)
I'm really happy to see the term Aloopa/|a|^2 in there. Whatever the
curvature actually is, I'd be REALLY surprised if it weren't at least
a constant. For constant curvature, I expected the holonomy to have a
term Aloopa/|a|^2 in it somewhere. My suggestion was just a lame
attempt at guessing an answer when I really have no intuition about
this stuff.
> I'm not sure how this relates to your suggestion. I'm a bit concerned about
>
> Aloopa/|a|^2 = Aloopb/|b|^2
>
> because, in general, (if |a| is not equal to |b|) rolling round a loop on
> sphere a won't result in a complete loop being drawn on sphere b, so I don't
> see how Aloopb can be defined.
Thanks. For some reason (lack of brain cells), I was assuming that a
closed loop on one sphere implied a closed loop on the other. If that
were true, then thinking of spherical triangles, Aloopa/|a|^2 is just
the excess angle of the spherical triangle on a and Aloopb/|b|^2 is
the excess angle on sphere b. Since both triangles had the same
angles, they would have the same excess. Intuition failed me again.
I wonder if I'd be taking the problem too far to wonder if there were
any "meaning" we can distill from that formula for the holonomy. My
gut tells me there is still more story to tell.
Eric
Tim S
> Hi Tim! :)
>
> Thanks a lot for taking the time to work that out.
Hey, what is time for, if not working things out?
>
> Tim S <T...@timsilverman.demon.co.uk> wrote:
>
>> Some back-of-the-envelope scribblings seem to indicate that the holonomy is
>> something like
>>
>> exp[i * Aloopa * (1/|a|^2 - 1/|b|^2) * SomeRadiusIndependentVector].
>>
>> We clearly do get flatness when |a| = |b|, and when |a| goes to infinity, we
>> get a vaguely sensible-looking result for rolling on the plane.
>
> Great! :)
>
> I'm really happy to see the term Aloopa/|a|^2 in there. Whatever the
> curvature actually is, I'd be REALLY surprised if it weren't at least
> a constant. For constant curvature, I expected the holonomy to have a
> term Aloopa/|a|^2 in it somewhere. My suggestion was just a lame
> attempt at guessing an answer when I really have no intuition about
> this stuff.
I wouldn't have guessed it either. In fact, I was guessing it involved
(1/|a| - 1/|b|). I'm quite good at visualising curvy lines on surfaces
(despite the occasional moment of idiocy), but when it comes to the
curvature of connections on principal bundles, my powers of visualisation
leave something to be desired...
It's quite interesting (though not so surprising in retrospect) the way the
curvature sort of changes sign as the radius of sphere b grows and passes
through that of sphere a.
>
>> I'm not sure how this relates to your suggestion. I'm a bit concerned about
>>
>> Aloopa/|a|^2 = Aloopb/|b|^2
>>
>> because, in general, (if |a| is not equal to |b|) rolling round a loop on
>> sphere a won't result in a complete loop being drawn on sphere b, so I don't
>> see how Aloopb can be defined.
>
> Thanks. For some reason (lack of brain cells), I was assuming that a
> closed loop on one sphere implied a closed loop on the other. If that
> were true, then thinking of spherical triangles, Aloopa/|a|^2 is just
> the excess angle of the spherical triangle on a and Aloopb/|b|^2 is
> the excess angle on sphere b. Since both triangles had the same
> angles, they would have the same excess. Intuition failed me again.
Heh! And there was I thinking you'd done something tremendously clever which
I wasn't following.
>
> I wonder if I'd be taking the problem too far to wonder if there were
> any "meaning" we can distill from that formula for the holonomy. My
> gut tells me there is still more story to tell.
I'd like to hear more about how jb and James Dolan are getting on with their
grassmannians and G2 actions.
Tim
John Baez
Tim S <T...@timsilverman.demon.co.uk> wrote:
This is very nice except that we get a great sphere in SU(2)
which depends on the initial point of contact between the rolling
sphere and the fixed one. I.e., we decree the rolling sphere to
be "unrotated" when its point of contact is some chosen point P,
and then by rolling it around we get a map from S^2 to SU(2) whose
image is a great sphere in SU(2). This map and its image depend on P.
I was trying to eliminate the dependence on P by the following sneaky
trick: we decree the rolling sphere to be "rotated 180 decrees around
the P axis" when its point of contact is P. This is a funny convention,
but I hope it has the following nice effect: by rolling the ball around,
we get a map from S^2 to SU(2) which is just the inclusion of the unit
imaginary quaternions in the unit quaternions - a map that does not
depend on the choice of P! The idea is that by starting out with this
funny convention, after we roll the sphere so that its point of contact
is Q it will turn out to be be "rotated 180 degrees around the Q axis" -
so the convention doesn't depend on the choice of P. I hope that's true.
In another posted, you wondered how James Dolan and I are doing on
relating the configuration space of this rolling ball to the
Grassmannian for the exceptional Lie group G2. The short answer
is that we are still quite puzzled about some things, but we've
learned a lot about how the split real form of G2 sits inside SO(3,4)
and how their Grassmannians are related, so someday we'll use this
to figure everything out. And then I'd like to explain it!
Greg Egan
punk floyd <punk...@rambler.ru> wrote:
>A quick note on Gred Egan's argument:
>
>Greg Egan <greg...@netspace.zebra.net.au> wrote in message
>news:<200305120741....@cumulus.netspace.net.au>...
>
>>So if we identify points A and B on the fixed sphere with unit
>>vectors, and if we identify all of R^3 with the imaginary
>>quaternions, then if C is the angle between A and B, the rotation
>>induced by rolling from A to B corresponds to the unit quaternion:
>
>> cos C + (unit vector parallel to A cross B) sin C
>> = A.B + (A cross B)
>
>...which is
>= B A^-1 (as quaternions)
That's a much more elegant way of putting it. It also makes it easy to
see that as well as thinking of the rolling ball's net rotation as it
goes from A to B as being:
* the reflection in the plane normal to A
followed by the reflection in the plane normal to B
you can think of it as:
* a 180-degree rotation around A^{-1} (which is the same
as around A), followed by a 180-degree rotation around B.
because the rotation corresponding to any purely imaginary unit
quaternion q is a 180-degree rotation around q as a vector in R^3.
Tim S
> In article <BAE6FF00.1A44B%T...@timsilverman.demon.co.uk>,
> Tim S <T...@timsilverman.demon.co.uk> wrote:
>
<Relating position of the point of contact on the fixed sphere to action of
SO(3) on the rolling sphere...>
OK, I see what you are trying to do (I think). I somehow got the impression
you didn't like the approach you were taking. As to your hope: yes, it is
true.
Suppose the initial point of contact is p on the fixed ball, P on the
rolling ball. And suppose the final point of contact is q on the fixed ball,
Q on the rolling ball. And let the centre of the fixed ball be o, and the
centre of the rolling ball be O.
What we'll do is start with a completely unrotated ball and get it to be 180
degrees rotated around the oq axis in two different ways. In the first way,
we'll start it off at p, rotate 180 degrees around the op axis, then roll it
around the fixed ball to q. In the second way, we'll start it ofg at q, and
rotate it 180 degrees around the oq axis. The final state of the ball will
be identical in the two cases. I hope this is what you wanted.
First method:
Begin with an unrotated rolling ball, contacting at p & P. Below is a
drawing of the position of the points:
| O |
\ /
Q Q'
\_P_/
-p-
/ \q
/ \
| o |
(I apologise for the sheer ghastliness of the ASCII diagram. Those are
supposed to be circles.)
Call angle QOP t. Then angle qop is also t. So is the angle Q'OP.
We now rotate the rolling ball 180 degrees around the axis op. This leaves P
and its diametrically opposite point unchanged, but all other points on the
sphere are reflected through the OP axis. E.g. Q changes places with its
reflection Q'.
| O |
\ /
Q' Q
\_P_/
-p-
/ \q
/ \
| o |
We now roll the ball down a great circle so that Q contacts q. P has now
rolled up to a new position.
Q'
/
| O
P
-p- \
/ \q Q
/ \ \
| o |
I'm not sure these drawings aren't doing more harm than good, but...
Second method:
We start with the same initial position. We _slide_ the rolling ball so that
Q contacts q -- i.e. no rotation of the rolling ball. We now rotate it 180
degrees about the oq axis.
The position of Q is unchanged by this rotation, while P clearly ends up
above Q and lying in the plane of the paper, and since all the angles are
the same and q is in contact with Q, we end up with the same state as I have
feebly attempted to show in the third diagram above. The only actual
rotation of the rolling ball was the 180 switch about the oq axis.
More generally, if we 'pre-twist' the rolling sphere about the axis through
the point of contact by an angle t, then the points of the S^3 of unit
quaternions accessible from that state by rolling without twisting consist
of the great sphere S^2 orthogonal to the quaternion axis which passes
through (sin(t/2) + cos(t/2) (ai + bj + ck)), where (ai + bj + ck) is in the
direction of the axis through the point of contact. So for t = 0, we simply
have all rotations about axes perpendicular to the one through the point of
contact, while for larger t, there is more overlap, and for t = pi, every
point of contact picks out the same great sphere, viz. the one perpendicular
to the real axis, or in other words the imaginary unit quaternions.
After all that, I really hope that I've understood you this time!
>
> In another posted, you wondered how James Dolan and I are doing on
> relating the configuration space of this rolling ball to the
> Grassmannian for the exceptional Lie group G2. The short answer
> is that we are still quite puzzled about some things, but we've
> learned a lot about how the split real form of G2 sits inside SO(3,4)
> and how their Grassmannians are related, so someday we'll use this
> to figure everything out. And then I'd like to explain it!
I'm looking forward to it!
Tim
Ken S. Tucker
Alex Boeglin <al...@f1.u-strasbg.fr> wrote in message news:<3EB90090...@f1.u-strasbg.fr>...
> John Baez wrote:
>
> > Willem H. de Boer <nos...@mbx5.ucsd.edu> wrote:
>
> > >"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
>
> > >> around x reverses y,z to give R(1) =(x,-y,-z). Then roll R(1)
> > >> around y reverses x and z giving R(2) = (-x,-y,z), and then
> > >> rolling about R(3) = (x,y,z).
>
> > >You're assuming that you always move 90 degrees along the
> > >equator, whereas the original question said that you can
> > >move an arbitrary distance along the equator.
>
> > Right. This comment also applies to the person who studied
> > this issue by taking two cubical dice and rolling one around
> > the other. The above analysis is fine in the special case
> > of rolling the ball 90 degrees along the equator - but the
> > really interesting question concerns rolling the ball for
> > an arbitrary angle.
> This puzzle needs not be visualized as a 3D problem at all:
> picture two discs, one on top the other. "Open" them flat on
> a table like a "book". Move one the disc along the perimeter
> of the other and close the "book" again using the new contact
> point as a hinge!
>
> To convince oneself that there will be no resulting change in
> orientation of the disc that's been moved, one can even use
> old fashioned geometry with triangles defined by radii of the
> circles and the tangent to the circles where they meet.
>
> alternatively, one can also just fold a piece of paper with
> three circles drawn on it (with salient points and lines
> penciled in with a heavy hand...). Of course this amounts
> to a bit of cheating since you weren't supposed to fold the
> sheet of paper in geometry: the greeks would have solved it
> drawing their figures in the sand...
The criticism of rolling dice are certainly true,
but may have applications...
The idea that spheres need to used might not be
necessary in every case. If there're quantized,
a small cube can roll on a large cube. For example,
roll a cube on a chess board from square to square,
with the chess board being the surface of a very
large cube. You could use a dice for the small cube.
This is meant only as a math model, for example
draw out on a sheet this,(pardon my ascii)...
____________________________________
| | | | | |
| 1,1 | 1,2 | 1,3 | 1,4 | 1,5 |
|______|___ ___|______|______|_____|_
| | | | | |
| 2,1 | 2,2 | 2,3 | 2,4 | 2,5 |
|______|___ ___|______|______|_____|_
| | | | | |
| 3,1 | 3,2 | 3,3 | 3,4 | 3,5 | Right
|______|___ ___|______|______|_____|_
| | | | | |
| 4,1 | 4,2 | 4,3 | 4,4 | 4,5 |
|______|___ ___|______|______|_____|_
| | | | | |
| 5,1 | 5,2 | 5,3 | 5,4 | 5,5 |
|______|___ ___|______|______|_____|_
Front
Since dice are usually available, lets define a die as a
small rolling cube,
x= 1 -x =6
y= 2 -y =5
z= 3 -z =4
being the axes corresponding to the numbers on the die.
The die can be set in 1,1 with Front =1, Right=2 and Top=3,
the *starting position* is then at 1,1 orientated as,
Front Right Top
x y z
relative to the board.
To move the die from 1,1 to 1,2 the die is rotated 90 degrees
around x, and produces an orientation at 1,2 to be
Front Right Top... shows on the die as 1,3,5 corresponding to,
x z -y
after rolling to 1,2 for example.
Call Loop1: Rolling from 1,1=>1,2=>2,2=>2,1=>1,1
The results are as follows (orientations are given at 1,1)
Front Right Top
1st Loop1 -z x -y
2nd Loop1 y -z -x
3rd Loop1 x y z
The 1st Loop1 does not return the ball to it's original
orientation. Three loops around Loop1 do so.
Call Loop2: Rolling from 1,1 =>1,2=>1,3=>2,3=>3,3
=>3,2=>3,1=>2,1=>1,1 produces,
Front Right Top
1st Loop2 x y z
does return the ball to it's original orientation.
Some concluding questions...
I'm assuming a sphere rotated by rolling 90 degrees on a
mirror is equivalent to the cubic model suggested above?
Given Baez's conditions, (two equal non slipping spheres),
Rolling from a beginning point on a mirror will follow
some arbituary path and arrive at that initial point with
an orientation differing from the starting orientation
(as demonstrated by Loop1), but the spheres are not
relatively rotated along the connecting axis.
A rotation will occur relative to the mirror, as
demonstrated by Loop1?
This little cubic experiment suggests no rotation occurs
about the axis connecting the centers of Baez's sphere's
but a relative rotation will occur relative to axes
perpendicular to the connecting axis.
Regards
Ken S. Tucker
Patrick Massot
splipping or twisting. I've just discovered the book /Differential
geometry/ by R. W. Sharpe, Springer GTM 166. One of the anpendices
deals with the general notion of rolling one submanifold on another in
R^n and answer many questions concerning generalized rolling-balls
(excluding the connection with G_2). I think book is very nice and
currently it is cheap thanks to Springer Yellow Sales.