This Week's Finds in Mathematical Physics (Week 181)
John Baez
Also available at http://math.ucr.edu/home/baez/week181.html
May 1, 2002
This Week's Finds in Mathematical Physics (Week 181)
John Baez
At the beginning of April I went up to Mathematical Sciences Research
Institute in Berkeley to a conference on n-categories and nonabelian
Hodge theory, which I should tell you about sometime... but the very
first thing I did is take a detour and give a talk at the University
of California at Santa Cruz.
U. C. Santa Cruz has a beautiful campus, with enormous rolling grassy
fields and groves of pine trees. And indeed it must be pretty idyllic
there, because the main thing the students used to complain about was
that the courses *aren't graded* - which makes it harder for them to get
jobs when they leave this paradise. I think grades are being phased in
now. Too bad!
Anyway, I wound up talking a lot to Richard Montgomery, who teaches in
the math department and works on the gravitational 3-body problem.
Except when one mass is much smaller than the other two - see my
discussion of Lagrange points in "week150" - this problem is still
packed with mysteries! Montgomery and other have turned their attention
to the case where all 3 masses are equal and found solutions with
amazing properties: for example, one where the total angular momentum is
zero and all 3 masses chase each other around a figure-8-shaped curve!
For details, see:
1) Alain Chenciner and Richard Montgomery, A remarkable periodic
solution of the three-body problem in the case of equal masses,
Ann. of Math. 152 (2000), 881-901. Also available as math.DS/0011268.
For a more popular account see:
2) Richard Montgomery, A new solution to the three-body problem,
AMS Notices 48 (May 2001), 471-481. Also available as
http://www.ams.org/notices/200105/fea-montgomery.pdf
and for Java applets illustrating this and other solutions based on
computer simulations by Carles Simo, try this:
3) Bill Casselman, A new solution to the three body problem - and more,
at http://www.ams.org/new-in-math/cover/orbits1.html
There are lots of other unsolved puzzles concerning point particles
interacting via Newtonian gravity. They're not very practical, but
they're lots of fun!
For example, can we find a periodic orbit where N particles move around
in the plane and trace out an arbitrary desired *braid*? For attractive
potentials stronger than the 1/r potential of gravity, very general
results like this are easily shown using variational methods - but
the question remains largely open for gravity. See:
4) Christopher Moore, Braids in classical gravity, Phys. Rev.
Lett. 70 (1993), 3675-3679.
5) Richard Montgomery, The N-body problem, the braid group, and
action-minimizing periodic solutions, Nonlinearity 11 (1998), 363-371.
There is also the issue of whether a particle can shoot off to infinity
in a finite amount of time. Of course this isn't possible in the real
world, but Newtonian physics has no "speed limit", and we're idealizing
our particles as points. So, if two or more of them get arbitrarily
close to each other, the potential energy they liberate could in
principle give another particle enough kinetic energy to zip off to
infinity! Then our solution becomes undefined after a finite amount
of time.
Zhihong Xia showed this can actually happen in the N-body problem
for N = 5 or bigger:
6) Zhihong Xia, The existence of non-collision singularities in Newtonian
systems, Ann. Math. 135 (1992), 411-468.
or for a more popular account:
7) Donald G. Saari and Zhihong Xia, Off to infinity in finite time,
AMS Notices (May 1995), 538-546. Also available at
http://www.ams.org/notices/199505/saari-2.pdf
As far as I know, the question is still open for N = 4. Another
question concerns how *likely* it is for our solution to become
undefined in a finite amount of time. If it's infinitely improbable, we
say we have "asymptotic completeness" for the N-body problem. I seem to
recall that the N-body problem has been shown asymptotically complete
for N = 3, but not higher N.
Now - back to my tale of Lie groups and geometry!
So far I've talked about how to any complex simple Lie group G we can
associate an "incidence geometry": a generalization of projective
geometry having G as its symmetry group. Each different type of
"figure" in this geometry corresponds to a dot in the Dynkin diagram
for G. For example, when G = SL(4,C) we have
o------o------o
points lines planes
For each dot, the space of all figures of the corresponding type is
called a "Grassmannian", and it's a manifold of the form G/P, where P
is a "maximal parabolic" subgroup of G.
More generally, any subset of dots in the Dynkin diagram corresponds to
a type of "flag". A flag is a collection of figures satisfying certain
incidence relations. For example, this subset:
x------o------x
points lines planes
corresponds to the type of flag consisting of a point lying on a plane.
The space of all flags of a particular type is called a "flag manifold",
and it's a manifold of the form G/P, where P is a "parabolic" subgroup
of G.
I also said a bit about how we can quantize this entire story! This is
actually what got me interested in this whole business. In loop quantum
gravity we run around claiming that space is made of quantum triangles,
quantum tetrahedra and the like - see "week113" and "week134" if you
don't believe me. The whole theory emerges naturally from the way
Euclidean and Lorentzian geometry are related to representations of the
rotation and Lorentz groups, but it got me wondering how the story
would change if we changed the group to something fancier - as we might
in a theory that tried to unify gravity with other forces, for example.
So I started studying incidence geometry and group representations, and
wound up learning lots of math so beautiful that it has, so far, completely
sidetracked me from my original goal! I'll get back to it eventually....
Anyway, let me say more about this quantum aspect now. This is the
royal road to understanding representations of simple Lie groups. For
starters, fix a complex simple Lie group G and any parabolic subgroup
P. Since G and P are complex Lie groups, the flag manifold G/P is a
complex manifold. More precisely, it has a complex structure that is
invariant under the action of G.
On the other hand, we can write the flag manifold as K/L, where K is the
maximal compact subgroup of G, and L is the intersection of K and P -
L is called a "Levi subgroup". Since K is compact, we can take any
Riemannian metric on the flag manifold and average it with respect to
the action of K to get a Riemannian metric that is invariant under the
action of K.
So, the flag manifold has a complex structure and metric that are both
invariant under K!
If this doesn't thrill you, consider the simplest example:
G = SL(2,C)
K = SU(2)
P = {upper triangular matrices in SL(2,C)}
L = {diagonal matrices in SL(2,C)}
Here G/P = K/L is a 2-sphere, the complex structure is the usual way of
thinking of this as the Riemann sphere, and the metric can be any
multiple of the usual round metric on the sphere. The complex structure
is invariant under all of G = SL(2,C). That's why SL(2,C) is the double
cover of the group of conformal transformations of the Riemann sphere!
The metric is only invariant under K = SU(2). That's why SU(2) is the
double cover of the group of rotations of the sphere!
All this stuff is wonderfully important in physics - especially since
SL(2,C) is also the double cover of the Lorentz group, and the Riemann
sphere is also the "heavenly sphere" upon which we see the distant
stars. I have already lavished attention on this network of ideas
in "week162"... but what we're engaged in now is generalizing it to
*arbitrary* complex simple Lie groups!
Now, a basic principle of geometry is that any two of the following
structures on a manifold determine the third *if* they satisfy a
certain compatibility condition:
complex structure J
Riemannian metric g
symplectic structure w
and in this case we get a "Kaehler manifold": a manifold with a complex
structure J and a complex inner product on the tangent vectors whose
real part is g and whose imaginary part is w.
Furthermore, one of the big facts of quantization is that while the
phase space of a classical system is a symplectic manifold, we can only
quantize it and get a Hilbert space if we equip it with some extra
structure... for example, by making it into a Kaehler manifold! Once
the phase space is a Kaehler manifold, we can look for a complex line
bundle over it with a connection whose curvature is the symplectic
structure. If this bundle exists, it's essentially unique, and we can
take the space of its holomorphic sections to be the Hilbert space of
states of the *quantum* version of our system. For details, try my
webpage on geometric quantization, or these books, listed in rough order
of increasing difficulty and depth:
8) John Baez, Geometric quantization,
http://math.ucr.edu/home/baez/quantization.html
9) J. Snyatycki, Geometric Quantization and Quantum Mechanics,
Springer-Verlag, New York, 1980.
10) Nicholas Woodhouse, Geometric Quantization, Oxford U. Press, Oxford,
1992.
11) Norman E. Hurt, Geometric Quantization in Action: Applications of
Harmonic Analysis in Quantum Statistical Mechanics and Quantum Field
Theory, Kluwer, Boston, 1983.
In the beautiful situation I'm discussing now, the math gods are kind:
the complex structure and metric on the flag manifold fit together to
make it into a Kaehler manifold, so we can quantize it and get a Hilbert
space. And since everything in sight is invariant under the group K,
our Hilbert space becomes a unitary representation of K. This rep turns
out to be irreducible... and we get all the unitary irreps of compact
simple Lie groups this way!
By easy abstract nonsense, the unitary irreps of K are also all the
finite-dimensional irreps of G. So, we've just conquered a great deal
of territory in the land of group representations. You may have seen
other ways to get all the irreps of simple Lie groups: for example,
"heighest-weight representations" or "geometric quantization of
coadjoint orbits". In fact, all these tricks are secretly just
different ways of talking about the same thing. It took me years
to learn this secret, but it's yours for free!
However, there are some small subtleties we shouldn't sweep under the
rug. We've seen that our flag manifold has a god-given complex
structure, but it usually has *lots* of K-invariant metrics, since we
could take *any* metric and average it with respect to the action of K.
So, there are lots of K-invariant Kaehler structures on our flag manifold.
How many are there? Well, I said that we get a flag manifold from any
subset of the dots in the Dynkin diagram for G. It turns out that
K-invariant Kaehler structure on this flag manifold correspond to ways
of labelling the dots in this subset with positive real numbers. And
we can geometrically quantize the flag manifold to get an irrep of G
precisely when these numbers are *integers*!
The simplest situation is when our flag manifold is a Grassmannian.
This corresponds to a single dot in the Dynkin diagram. If we label
this dot with the number 1, we get a so-called "fundamental
representation" of our group. I sketched in "week180" how to get all
the other irreps from these.
Now let me illustrate all this stuff by going through all the classical
series of simple Lie groups and seeing what we get.
A_n: Here are the Grassmannians for some of the A_n series, that is, the
groups SL(n+1,C). I've drawn the Dynkin diagrams with each dot labelled
by the corresponding type of geometrical figure and the dimension of the
Grassmannian of all figures of this type. We can think of these figures
as vector subspaces of C^{n+1}. We can also think of them as spaces of
one less dimension in CP^n. Either way, we are talking about *projective*
geometry:
A1 1d subspaces
SL(2,C) o
1
A2 1d subspaces 2d subspaces
SL(3,C) o---------------o
2 2
A3 1d subspaces 2d subspaces 3d subspaces
SL(4,C) o------------o---------------o
3 4 3
A4 1d subspaces 2d subspaces 3d subspaces 4d subspaces
SL(5,C) o------------o---------------o------------o
4 6 6 4
Recognize the numbers labelling the Dynkin diagram dots? It's a weird
modified version of Pascal's triangle - but can you figure out the pattern?
No? I claim you learned this table of numbers when you were in grade
school: just tilt your head 45 degrees and you'll recognize it!
Next, here's what we get from quantizing these Grassmannians. I've
labelled each dot by the name of the corresponding fundamental
representation and its dimension. All these reps are exterior powers
of the obvious rep of SL(n+1,C) on C^{n+1}. We call elements of the pth
exterior power "p-vectors", or "multivectors" in general:
A1 vectors
SL(2,C) o
2
A2 vectors bivectors
SL(3,C) o---------------o
3 3
A3 vectors bivectors 3-vectors
SL(4,C) o-------------o--------------o
4 6 4
A4 vectors bivectors 3-vectors 4-vectors
SL(5,C) o-------------o--------------o------------o
5 10 10 5
Here the numbers labelling the dots form Pascal's triangle! So we
see that Pascal's triangle is a quantized version of the multiplication
table. (That was the answer to the previous puzzle, by the way - our
triangle was just the multiplication table viewed from a funny angle.)
B_n: Next let's look at the B_n series. B_n is another name for the
complexified rotation group SO(2n+1,C), or if you prefer, its double
cover Spin(2n+1,C). A Grassmannian for this group is a space consisting
of all p-dimensional "isotropic" subspaces of C^{2n+1} - that is,
subspaces on which a nondegenerate symmetric bilinear form vanishes.
As I explained in "week180", these Grassmannians show up when we study
relativity in odd-dimensional Minkowski spacetime, especially when we
complexify and compactify. Another way to put it is that this is all
about *conformal* geometry in odd dimensions! We've already seen that
conformal geometry in even dimensions is very different, and we'll get
to that later.
Here are the Grassmannians and their dimensions:
isotropic
B1 1d subspaces
Spin(3,C) o
1
isotropic isotropic
B2 1d subspaces 2d subspaces
Spin(5,C) o=======>=======o
3 3
isotropic isotropic isotropic
B3 1d subspaces 2d subspaces 3d subspaces
Spin(7,C) o--------------o=======>=======o
5 7 6
isotropic isotropic isotropic isotropic
B4 1d subspaces 2d subspaces 3d subspaces 4d subspaces
Spin(9,C) o-------------o---------------o=======>======o
7 11 12 10
I'm sure these are well-known, but James Dolan and I had a lot of
fun one evening working these out, using a lot of numerology that we
eventually justified by a method I'll explain later.
Here's a bigger chart of these dimensions:
A1 1
A2 3 3
A3 5 7 6
A4 7 11 12 10
A5 9 15 18 18 15
A6 11 19 24 26 25 21
A7 13 23 30 34 35 33 28
A8 15 27 36 42 45 45 42 36
I leave it as an easy puzzle to figure out the pattern, and a harder
puzzle to prove it's true. Don't be overly distracted by the symmetry
lurking in rows 2, 5, and 8 - every third row has this symmetry, but
it's a bit of a red herring!
If we quantize these Grassmannians we get these fundamental reps of
Spin(2n+1,C):
B1 spinors
Spin(3,C) o
2
B2 vectors spinors
Spin(5,C) o=======>=======o
5 4
B3 vectors bivectors spinors
Spin(7,C) o--------------o=======>=======o
7 21 8
B4 vectors bivectors 3-vectors spinors
Spin(9,C) o-------------o---------------o=======>======o
9 36 84 16
As before, the dimension of the space of p-vectors in q-dimensional
space comes straight from Pascal's triangle: it's q choose p. But now
we also have spinor reps; the dimensions of these are powers of 2.
C_n: Next let's look at the Grassmannians for the C_n series, that is,
the symplectic groups Sp(2n,C). This is the only series of classical
groups I haven't touched yet! Just as the A_n series are symmetry
groups of projective geometry and the B_n and D_n series are symmetry
groups of conformal geometry, the C_n series are symmetry groups of
"projective symplectic" geometry. Unfortunately I don't know much
about this subject - at least not consciously. It should be important
in physics, but I'm not sure where!
Anyway, Sp(2n,C) is the group of linear transformations of C^{2n} that
preserve a symplectic form: that is, a nondegenerate *antisymmetric*
bilinear form. A Grassmannian for this group again consists of all
p-dimensional isotropic subspaces of C^{2n}, where now a subspace is
"isotropic" if the symplectic form vanishes on it.
Here's a little table of these Grassmannians:
isotropic
C1 1d subspaces
Sp(2,C) o
1
isotropic isotropic
C2 1d subspaces 2d subspaces
Sp(4,C) o=======<=======o
3 3
isotropic isotropic isotropic
C3 1d subspaces 2d subspaces 3d subspaces
Sp(6,C) o--------------o=======<=======o
5 7 6
isotropic isotropic isotropic isotropic
C4 1d subspaces 2d subspaces 3d subspaces 4d subspaces
Sp(8,C) o-------------o---------------o=======<======o
7 11 12 10
You'll notice the dimensions are the same as in the B_n case! That's
because their Dynkin diagrams are almost the same: for reasons I may
someday explain, dimensions of flag manifolds don't care which way the
little arrows on the Dynkin diagrams point, since they depend only on
the *reflection group* associated to this diagram (see "week62").
However, the dimensions of the fundamental representations are different
from the B_n case - and I don't even know what they are! The basic idea
is this: the space of p-vectors is no longer an irrep for Sp(2n,C), but
contracting with the symplectic form maps p-vectors to (p-2)-vectors,
and the kernel of this map is the pth fundamental rep of Sp(2n). Let's
call these guys "irreducible p-vectors".
Oh heck, I can *guess* the dimensions of these guys from this... I guess
they're just the dimension of the p-vectors minus the dimension of the
(p-2)-vectors. Here's a table of these guesses:
C1 vectors
Sp(2,C) o
2
irreducible
C2 vectors bivectors
Sp(4,C) o=======<=======o
4 5
irreducible irreducible
C3 vectors bivectors 3-vectors
Sp(6,C) o--------------o=======<=======o
6 14 14
irreducible irreducible irreducible
C4 vectors bivectors 3-vectors 4-vectors
Sp(8,C) o-------------o---------------o=======<======o
8 27 28 14
Maybe someone can tell if they're right.
D_n: Finally, D_n is another name for the complexified rotation group
SO(2n,C) or its double cover Spin(2n,C). The pth Grassmannian for
this group consists of all p-dimensional isotropic subspaces of the
space C^{2n} equipped with a nondegenerate symmetric bilinear form -
*except* for the top-dimensional Grassmannians, as I explained
last week. These consist of self-dual or anti-self-dual subspaces.
Self-duality is the special feature of conformal geometry in *even*
dimensions!
Here are the Grassmannians and their dimensions:
1
o self-dual 2d subspaces
D2
Spin(4,C)
o anti-self-dual 2d subspaces
1
3
o self-dual 3d subspaces
/
/
D3 /
Spin(6,C) 4 o isotropic 1d subspaces
\
\
\
o anti-self-dual 3d subspaces
3
6
o self-dual 4d subspaces
/
isotropic /
D4 1d subspaces /
Spin(8,C) o-------o isotropic 2d subspaces
6 9\
\
\
o anti-self-dual 4d subspaces
6
10
o self-dual 5d subspaces
/
/
D5 1d subspaces 2d subspaces /
Spin(10,C) o-----------o---------o isotropic 3d subspaces
8 13 15\
\
\
o anti-self-dual 5d subspaces
10
You'll notice that the numbers on the "fishtails" are triangular
numbers: 1, 3, 6, 10... I'll say more later about how to calculate
the rest of these numbers.
As explained last week, the fundamental reps of the D_n consist of
p-vectors, except for those at the fishtails, which are left- and
right-handed spinor reps:
2
o left-handed spinors
D2
Spin(4,C)
o right-handed spinors
2
4
o left-handed spinors
/
/
D3 /
Spin(6,C) 6 o vectors
\
\
\
o right-handed spinors
4
8
o left-handed spinors
/
/
vectors /
Spin(8,C) o-------o bivectors
8 28\
\
\
o right-handed spinors
8
16
o left-handed spinors
/
/
D5 vectors bivectors /
Spin(10,C) o-----------o---------o 3-vectors
10 45 120\
\
\
o right-handed spinors
16
Again the dimension of the space of p-vectors in q-dimensional space
comes from Pascal's triangle, while the dimensions of the spinor reps
are powers of 2.
Let me conclude by listing the dimensions of Grassmannians for the
exceptional groups, as computed by James Dolan. I strongly doubt he's
the first to have computed these - at this stage we're mainly learning
and reinventing known stuff - but he did it using a nice trick I'd like
to mention. I was shocked at how unfamiliar these numbers were to me,
because all these Grassmannians should be definable using the octonions:
-------
G2: 5 --->--- 5
-------
F4: 43------48===>===48------43
20
|
|
|
E6: 15------24------28-----24------15
35
|
|
|
E7: 26------40------46------43------35------21
56
|
|
|
E8: 42------62------70------68------61------48------28
You can calculate dimensions of these and all the other Grassmannians
for simple Lie groups the following simple trick. Given the Dynkin
diagram for G and a chosen dot in it, remove this dot to get one or more
Dynkin diagrams for groups G_i. Work out the dimension of the space
of maximal flags for G, and subtract all the dimensions of the spaces
of maximal flags for the G_i. Voila! You get the dimensions of the
Grassmannian corresponding to the ith dot.
The dimensions of maximal flag manifold for G is easy to compute, in
turn, because it's just dim(G) - dim(B), where B is the Borel. And
dimension of the Borel is in turn easy to compute because... oh, let's
quit here! - if you've read this far, you probably know already.
-----------------------------------------------------------------------
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mathematics and physics, as well as some of my research papers, can be
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http://math.ucr.edu/home/baez/
For a table of contents of all the issues of This Week's Finds, try
http://math.ucr.edu/home/baez/twf.html
A simple jumping-off point to the old issues is available at
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If you just want the latest issue, go to
John Baez
>Let me conclude by listing the dimensions of Grassmannians for the
>exceptional groups, as computed by James Dolan.
>
> 20
> |
> |
> |
>E6: 15------24------28-----24------15
>
>
>
>
> 35
> |
> |
> |
>E7: 26------40------46------43------35------21
>
>
>
>
> 56
> |
> |
> |
>E8: 42------62------70------68------61------48------28
I'm sorry: here I copied down the results of an older, erroneous
calculation!
Our current theory is:
21
|
|
|
E6 16------25------29------25------16
42
|
|
|
E7 33------47------53------50------42------27
92
|
|
|
E8 90------98-----106-----104------99------83------57
The "16" at the tips of the E6 diagram makes sense, because this
is the dimensions of the octonionic projective plane, of which E6
acts as the collineation group.
Squark
news:<aapt9e$lk1$1...@glue.ucr.edu>...
> There is also the issue of whether a particle can shoot off to infinity
> in a finite amount of time. Of course this isn't possible in the real
> world, but Newtonian physics has no "speed limit", and we're idealizing
> our particles as points. So, if two or more of them get arbitrarily
> close to each other, the potential energy they liberate could in
> principle give another particle enough kinetic energy to zip off to
> infinity! Then our solution becomes undefined after a finite amount
> of time.
Hmm, something here sounds strange... For this to happen, the particle
which "shoots off" should develop infinite momentum. However, for this to
happen, some other particle in the bunch should develop infinite momentum
in the opposite direction. This should make them come apart and stop
feeling the gravitational influence of one another which would spoil the
whole trick, because we don't know of such a scenario for less than 5
particles, and even if we had, the argument could be continued by
induction (and a particle certainly can't run away to infinity all on its
own). What am I missing here?
Best regards,
Squark
------------------------------------------------------------------
Write to me using the following e-mail:
Skvark_N...@excite.exe
(just spell the particle name correctly and use "com" rather than
"exe")
Gordon D. Pusch
> ba...@math.ucr.edu (John Baez) wrote in message
> news:<aapt9e$lk1$1...@glue.ucr.edu>...
>
>> There is also the issue of whether a particle can shoot off to infinity
>> in a finite amount of time. Of course this isn't possible in the real
>> world, but Newtonian physics has no "speed limit", and we're idealizing
>> our particles as points. So, if two or more of them get arbitrarily
>> close to each other, the potential energy they liberate could in
>> principle give another particle enough kinetic energy to zip off to
>> infinity! Then our solution becomes undefined after a finite amount
>> of time.
>
> Hmm, something here sounds strange... For this to happen, the particle
> which "shoots off" should develop infinite momentum. However, for this to
> happen, some other particle in the bunch should develop infinite momentum
> in the opposite direction. This should make them come apart and stop
> feeling the gravitational influence of one another which would spoil the
> whole trick, because we don't know of such a scenario for less than 5
> particles, and even if we had, the argument could be continued by
> induction (and a particle certainly can't run away to infinity all on its
> own). What am I missing here?
Um, the possiblity that one or more _other_ particles could leave
the system with infinite momentum, such that the momenta of all the
``transcendent particles'' sums to zero?
(Granted, such situations are ``infinitely unlikely'' and rather contrived,
but AFAIK, all known solutions that exhibit this pathological behavior have
been carefully constructed by hand, and highly symmetrical in geometry...)
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
t...@rosencrantz.stcloudstate.edu
Squark <fii...@yahoo.com> wrote:
>ba...@math.ucr.edu (John Baez) wrote in message
>news:<aapt9e$lk1$1...@glue.ucr.edu>...
>
>> There is also the issue of whether a particle can shoot off to infinity
>> in a finite amount of time. Of course this isn't possible in the real
>> world, but Newtonian physics has no "speed limit", and we're idealizing
>> our particles as points. So, if two or more of them get arbitrarily
>> close to each other, the potential energy they liberate could in
>> principle give another particle enough kinetic energy to zip off to
>> infinity! Then our solution becomes undefined after a finite amount
>> of time.
>
>Hmm, something here sounds strange... For this to happen, the particle
>which "shoots off" should develop infinite momentum. However, for this to
>happen, some other particle in the bunch should develop infinite momentum
>in the opposite direction.
My understanding of the five-particle solution is that it involves two
pairs of orbiting particles with a single particle shuttling back and
forth between them. Each time the shuttle particle encounters one of
the orbiting pairs, it extracts energy from it, sending the orbiting
particles into a closer orbit and thus gaining kinetic energy for both
itself and the orbiting pair. It then zips over to the other pair and
does the same thing. Surprisingly, it's possible to cook things up so
that both orbiting pairs shoot off to infinity in finite time.
>This should make them come apart and stop
>feeling the gravitational influence of one another which would spoil the
>whole trick, because we don't know of such a scenario for less than 5
>particles, and even if we had, the argument could be continued by
>induction (and a particle certainly can't run away to infinity all on its
>own). What am I missing here?
I can't begin to understand your induction argument. From your conclusion
("and a particle certainly can't run away to infinity all on its own"),
it sounds like you're trying to run the induction downwards, to get from
an N-particle solution to an (N-1)-particle solution. I can't imagine
how such an argument would work.
-Ted
Russell Easterly
"Squark" <fii...@yahoo.com> wrote in message
news:939044f.02050...@posting.google.com...
> ba...@math.ucr.edu (John Baez) wrote in message
> news:<aapt9e$lk1$1...@glue.ucr.edu>...
>
> > There is also the issue of whether a particle can shoot off to infinity
> > in a finite amount of time. Of course this isn't possible in the real
> > world, but Newtonian physics has no "speed limit", and we're idealizing
> > our particles as points. So, if two or more of them get arbitrarily
> > close to each other, the potential energy they liberate could in
> > principle give another particle enough kinetic energy to zip off to
> > infinity! Then our solution becomes undefined after a finite amount
> > of time.
>
I assume we are talking about particles held together by gravity.
A particle would not have to have infinite momentum to "zip off to
infinity".
All it needs is enough energy to achieve escape velocity.
Escape velocity would depend on the mass of the other particles.
I've written some simple multi-body simulations.
Its easy to show that most 3 body configurations are unstable.
Two of the bodies quickly conspire to launch the third "off to infinity".
Its usually the smallest body that gets expelled, but not always.
[Sci.physics.research moderator's note: the situation people were
discussing here was much stranger than that. This discussion was
about situations in which particles actually get to infinity *in
finite time*. For that, the momentum of the particle certainly does
have to go to infinity. And, of course, since momentum is conserved,
something else has to have momentum that goes to infinity in the
opposite direction. -TB]
> Hmm, something here sounds strange... For this to happen, the particle
> which "shoots off" should develop infinite momentum. However, for this to
> happen, some other particle in the bunch should develop infinite momentum
> in the opposite direction. This should make them come apart and stop
> feeling the gravitational influence of one another which would spoil the
> whole trick, because we don't know of such a scenario for less than 5
> particles, and even if we had, the argument could be continued by
> induction (and a particle certainly can't run away to infinity all on its
> own). What am I missing here?
>
> Best regards,
> Squark
>
My simulations did have the problem described above.
The distance between two paricles goes to 0.
If particles can come arbitrarily close to one another
then "infinite" acceleration can occur.
Both particles would be accelerated.
In my simulations they would essentially cease to exist.
They would move so fast that they didn't have time
to interact with the other particles.
Something that I've often wondered about.
If the second derivative of a function has a limit
does this mean that the third derivative also has a limit?
I think of the speed of light as a limit on the velocity of an object.
Is there a "maximum" acceleration as well?
Russell
- 2 many 2 count
Ted Shoemaker
Russell Easterly wrote:
>
> If the second derivative of a function has a limit
> does this mean that the third derivative also has a limit?
All derivatives (including the second and third derivatives)
can be defined in terms of limits, at least in standard analysis.
Perhaps you meant:
If a second derivative is bounded, then is the third derivative bounded?
In this case, the answer is no.
As a counterexample, consider a function whose second derivative is
x*log(x) on the open interval (0,1). The third derivative has
unbounded values as x --> 0.
Ted Shoemaker
shoema...@yahoo.com
Ilja Schmelzer
> ba...@math.ucr.edu (John Baez) wrote in message
> news:<aapt9e$lk1$1...@glue.ucr.edu>...
> > There is also the issue of whether a particle can shoot off to infinity
> > in a finite amount of time. Of course this isn't possible in the real
> > world, but Newtonian physics has no "speed limit", and we're idealizing
> > our particles as points. So, if two or more of them get arbitrarily
> > close to each other, the potential energy they liberate could in
> > principle give another particle enough kinetic energy to zip off to
> > infinity! Then our solution becomes undefined after a finite amount
> > of time.
> Hmm, something here sounds strange... For this to happen, the particle
> which "shoots off" should develop infinite momentum. However, for this to
> happen, some other particle in the bunch should develop infinite momentum
> in the opposite direction. This should make them come apart and stop
> feeling the gravitational influence of one another which would spoil the
> whole trick, because we don't know of such a scenario for less than 5
> particles, and even if we had, the argument could be continued by
> induction (and a particle certainly can't run away to infinity all on its
> own). What am I missing here?
Oscillations with increasing frequency and amplitude.
One pair of particles close to each other (therefore able to spend
arbitrary amounts of energy) runs to +oo, another pair to -oo, and a
last particle oscillates between them. Thenever the oscillating
particle reaches one of the pairs, the pair gets closer, this
increases kinetic energy by increasing the speed of the pair by some
factor and (to conserve momentum) the other particle bounces back with
much higher speed, high enough to reach the other end.
Then the numbers can be made up to give infinities in finite time.
Ilja
--
I. Schmelzer, <il...@ilja-schmelzer.net>, http://ilja-schmelzer.net
Squark
news:<abcl1l$6gk$1...@sue.its.caltech.edu>...
> My understanding of the five-particle solution is that it involves two
> pairs of orbiting particles with a single particle shuttling back and
> forth between them. Each time the shuttle particle encounters one of
> the orbiting pairs, it extracts energy from it, sending the orbiting
> particles into a closer orbit and thus gaining kinetic energy for both
> itself and the orbiting pair. It then zips over to the other pair and
> does the same thing. Surprisingly, it's possible to cook things up so
> that both orbiting pairs shoot off to infinity in finite time.
> ...
> I can't begin to understand your induction argument. From your conclusion
> ("and a particle certainly can't run away to infinity all on its own"),
> it sounds like you're trying to run the induction downwards, to get from
> an N-particle solution to an (N-1)-particle solution. I can't imagine
> how such an argument would work.
It's alright, you have just explained the fault of the arguement: I assumed
all of the particles have momentum which runs off to infinity _in a certain
direction_, while in the actual solution there's the "shuttle" particle for
which the momentum oscillates and the directions flips an infinity of times.
Without this, the particles would separate into far away groups (because of
momentum conservation) and this would yield the downward induction.
Actually, I have seen this hole in my arguement, but considered it minor
because I couldn't believe the assumption wouldn't hold somehow. Your
explanation makes it clear, though, that it does fail, and shows how it can
fail.
Tony Smith
I am baffled by your calculations of Grassmannians in week 181.
According to Representations of Compact Lie Groups
by Brocker and tom Dieck (Springer 1985) at page 38:
"... The manifold of all k-dimensional linear subspaces of R^n
is called the Grassmann manifold Gk(R^n) ...
Gk(R^n) = O(n) / O(k)xO(n-k)
...
Gk(C^n) = U(n) / U(k)xU(n-k)
...
Gk(H^n) = Sp(n) / Sp(k)x\Sp(n-k) ...". [here H denotes quaternions]
When I try to compare that with your week 181,
using U(n) for SL(n,C) and O(n) for Spin(n,C)
(maybe that messes up in some way by a discrete Z2, but even
trying to take that into account I still don't understand)
I seem to get the following (with your week 181 stated,
followed by my attempt at calculations).
What am I doing wrong? Why are my results different?
Could there be two so-very-different definitions of "Grassmannian"
floating around in the literature ?
============================================================
week 181:
A4 1d subspaces 2d subspaces 3d subspaces 4d subspaces
SL(5,C) o------------o---------------o------------o
4 6 6 4
My calculation for Grassmannian with k-dim subspaces:
U(5) / U(k)xU(5-k)
k = 1 dim = 25 - 1 - 16 = 8
k = 2 dim = 25 - 4 - 9 = 12
k = 3 dim = 25 - 9 - 4 = 12
k = 4 dim = 25 - 16 - 1 = 8
Here I note that my calculation gives exactly twice yours,
so maybe I have messed up by a factor of 2 somehow?
=============================================================
week 181:
isotropic isotropic isotropic isotropic
B4 1d subspaces 2d subspaces 3d subspaces 4d subspaces
Spin(9,C) o-------------o---------------o=======>======o
7 11 12 10
My calculation for Grassmannian with k-dim subspaces:
O(9) / O(k)xO(9-k)
k = 1 dim = 36 - 0 - 28 = 8
k = 2 dim = 36 - 1 - 21 = 14
k = 3 dim = 36 - 3 - 15 = 18
k = 4 dim = 36 - 6 - 10 = 20
k = 5 dim = 36 - 10 - 6 = 20
k = 6 dim = 36 - 15 - 3 = 18
k = 5 dim = 36 - 21 - 1 = 14
k = 6 dim = 36 - 28 - 0 = 8
Here, however, my calculation ( 8 14 18 20 ) seems to me to
have no clear relation to yours ( 7 11 12 10 ),
except that my 20 is twice your 10.
=============================================================
week 181:
10
o self-dual 5d subspaces
/
/
D5 1d subspaces 2d subspaces /
Spin(10,C) o-----------o---------o isotropic 3d subspaces
8 13 15\
\
\
o anti-self-dual 5d subspaces
10
My calculation for Grassmannian with k-dim subspaces:
O(10) / O(k)xO(10-k)
k = 1 dim = 45 - 0 - 36 = 9
k = 2 dim = 45 - 1 - 28 = 16
k = 3 dim = 45 - 3 - 21 = 21
k = 4 dim = 45 - 6 - 15 = 24
k = 5 dim = 45 - 10 - 10 = 25
k = 6 dim = 45 - 15 - 6 = 24
k = 7 dim = 45 - 21 - 3 = 21
k = 8 dim = 45 - 28 - 1 = 16
k = 9 dim = 45 - 36 - 0 = 9
Here, again, my calculation ( 9 16 21 24 25 ) seems to me to
have no clear relation to yours ( 8 13 15 10 and 10 ),
and also I am not clear on why you have no 4d subspaces?
Maybe 5 as the square root of my 25 could correspond to
self-dual and anti-self-dual 5-dim subspaces.
If so, then my 5 might be half your 10.
=============================================================
I am even more confused because when I calculate fundamental representation
dimensions, I get the same numbers that you get in week 181.
Tony 11 May 2002
John Baez
>I am baffled by your calculations of Grassmannians in week 181.
>
>According to Representations of Compact Lie Groups
>by Brocker and tom Dieck (Springer 1985) at page 38:
>"... The manifold of all k-dimensional linear subspaces of R^n
>is called the Grassmann manifold Gk(R^n) ...
>Gk(R^n) = O(n) / O(k)xO(n-k)
>...
>Gk(C^n) = U(n) / U(k)xU(n-k)
>...
>Gk(H^n) = Sp(n) / Sp(k)x\Sp(n-k) ...". [here H denotes quaternions]
These are the Grassmannians of *all* k-dimensional subspaces in a
n-dimensional real, complex, or quaternionic vector space. If you
reread "week181", you'll see that I'm talking about some other
Grassmannians, which are more natural from the viewpoint of simple
Lie group theory.
For the A series I am indeed talking about Grassmannians consisting
of k-dimensional subspaces of C^n; if your answers here are twice
mine it's probably because my Grassmannians are always complex manifolds,
so I'm stating their complex dimension, not their real dimension.
For the B and D series I am talking about Grassmanians consisting of
k-dimensional *isotropic* subspaces of C^n, where C^n has been equipped
with a nondegenerate symmetric bilinear form - also known as an
orthogonal structure.
Similarly, for the B and D series I am talking about Grassmanians
consisting of k-dimensional *isotropic* subspaces of C^{2n}, where now
C^{2n} has been equipped with a nondegenerate antisymmetric bilinear
form - also known as a symplectic structure.
Thus, in the B, C, and D cases your calculations are completely unrelated
to mine.
If you're wondering what "isotropic" means and why this is so important,
try rereading "week180", especially the brand new addendum where I talk
more about this in the B and D cases.
One last warning: when I use the letter n above, this is not the
same number n as in A_n, B_n, C_n, and D_n.
>Could there be two so-very-different definitions of "Grassmannian"
>floating around in the literature ?
Yes. Your Grassmannians are defined whenever one has a vector space
over a field - possibly a skew field. My Grassmannians are defined
whenever one has a complex simple Lie algebra. They are defined in a
systematic way using the concept of "maximal parabolic subgroup", and
they even make sense for the exceptional Lie algebras!