Belt trick for spin 3/2
frank_k...@yahoo.co.uk
Dirac's belt trick models the behaviour of spin 1/2 particles.
When turning the end of the belt (the particle) by 4 pi,
the belt can be untwisted even while keeping the particle fixed.
Is there an analogous trick for spin 3/2? It seems that this would
imply rotations by 2 pi/3. What object would one have to take
(instead of the belt) to get a "spin 3/2 trick"?
Frank
Starbla...@excite.com
Could you explain this trick in more details? Do you also happen to
know the trick for spin 2 particles?
Thomas Mautsch
schrieb frank_k...@yahoo.co.uk <frank_k...@yahoo.co.uk>:
>
> Dirac's belt trick models the behaviour of spin 1/2 particles.
No, it doesn't.
> When turning the end of the belt (the particle) by 4 pi,
> the belt can be untwisted even while keeping the particle fixed.
>
> Is there an analogous trick for spin 3/2? It seems that this would
> imply rotations by 2 pi/3. What object would one have to take
> (instead of the belt) to get a "spin 3/2 trick"?
The belt trick is a visualization of the spin *group*
(see the thread "SU(2)" on sci.math, in particular
<news:e9n48i$ab3$1...@glue.ucr.edu> ),
and not of the spin 1/2 *representation* of this group,
as some people seem to think.
See my answer <news:45001d83$1...@news1.ethz.ch>
in the sci.math-thread "More tangles" for details.
Thomas Mautsch
frank_k...@yahoo.co.uk
>
> The belt trick is a visualization of the spin *group*
> as some people seem to think.
The question was whether there is something
analogous to the belt trick for spin 3/2.
Is there such a trick?
Frank
Thomas Mautsch
schrieb frank_k...@yahoo.co.uk <frank_k...@yahoo.co.uk>:
My point was that there is no "belt trick for spin 1/2",
there is only a "belt trick for visualising the spin group";
so the question whether there is an _analogous_ "belt trick for spin 3/2"
does not make sense.
Well, you _can_ visualize spin 1, but this is somehow trivial;
and I suppose there are also means to visualize the other
integer-spin representations, because they are representations of
the groups SO(3), resp. of SO(n).
The problem is that the only way I know
to obtain the spin 1/2 representation of the spin groups
is via embedding the spin groups into Clifford algebras
and using the identification of these Clifford algebras
with matrix algebras. (In particular Cliff(3) = IH in dim. 3).
But the picture of the spin group Spin(3) as shown by the belt trick
is more the topological definition of the spin group
as the equivalence class of closed paths in SO(3)
modulo homotopy, and I don't know how to _directly_ read off
from this picture that this group of paths is isomorphic to SU(2). -
So I see no way of visualizing any half-spin representation
with the belt trick.
frank_k...@yahoo.co.uk
an applet like the beautiful example at
http://gregegan.customer.netspace.net.au/APPLETS/21/21.html
but for spin 3/2 insteand of 1/2?
That one shows the spinor state, the path through SO(3), and more.
Frank
Cl.Massé
1161802444.7...@i3g2000cwc.googlegroups.com
> The question was whether there is something
> analogous to the belt trick for spin 3/2.
> Is there such a trick?
No, it's a conjuring trick. The fact that the two rotations are in the
inverse direction is concealed by an stealthy change of configuration, like
a conjuror would make a stealthy manipulation just at the time of a change
of motion. It is indeed impossible to have such a trick in a three
dimensional space with classical objects.
--
The Surprise Guest
Jonathan Scott
two orientation values (the ends of the belt) determines the twist of a
path between them modulo 4pi, where one's first instinct would be to
expect it to determine the twist modulo 2pi. This illustrates the fact
that if the orientation is described by a spinor quantity, which varies
as half of the rotation angle, a rotation of 2pi changes the sign of
the spinor and is hence distinct from the original state, but a
rotation of 4pi restores the original sign.
A closely related trick is to take a flexible closed band (of rubber,
tape or paper) and wrap it round something. You will find that it can
still be made to lie flat if the number of loops is changed by an even
number, but not if it is changed by an odd number. Such a band can
normally be wrapped around any odd number of times (corresponding to a
fermion). It is also possible to take a band with one whole twist in
it (not half, which makes a Moebius band), so that it looks like a
figure of 8 edge on, and try the same with that. Such a band can be
wrapped round any even number of times (corresponding to a boson).
I'm personally interested in another related effect. A spin-1/2 wave
function has the weird feature that the phase changes by pi along a
closed path which goes around the spin axis, which says that the value
changes sign by the time you get back to the original point in space
and time, making it at two-valued at any point in space, which seems
not only inconvenient but somewhat implausible if the wave function
might eventually have some underlying physical interpretation.
However, there are actually two forms of phase factor which can be
involved; geometric rotation (or twist, depending on how the rotation
is oriented) and complex scalar phase, and if these both independently
change phase by pi and hence change sign, then the product is
single-valued again. If the wave function has enough geometric
structure to support both of these forms of phase, then one of them can
be actually associated with the spin and the other with some other
conserved and quantised physical quantity (which will necessarily be
possessed by fermions in odd multiples of 1/2 and by bosons in even or
zero multiples of 1/2). I'm interested in pinning down exactly what
that physical quantity might be; my guess is that it is equal to or
closely related to the generalized extension of the I_3 component of
isospin (or weak isospin), Q+B/2-L/2, which is related to charge and
whether particles or antiparticles are involved.
Gerard Westendorp
objects.
So I was thinking, maybe it makes sense to think of N belts.
We could put these N belts in series. I am not sure if this
makes sense, but some aspects do fall into place.
Suppose we take an object with 2X spin 1/2.
If the 2 spins have opposite signs, then the composite object
has spin 0, if the 2 signs are equal, then we get spin 1.
Now if we rotate by 360 degrees, the belt will get a
360 degree twist for each spin 1/2 component. In the spin
0 case (opposite signs), the twists cancel, and the belt
will have no net twist. If the spins add up to a spin 1,
then we get a 720 degree twist, which is identity, like
we expect from a spin 1 object. For spin 3/2, we would
get a net 360 degree twist again.
A problem with this might be that we predict an identity
if a 3/2 spin object is rotated by 720*2/3=240 degrees. That
doesn't seem right. I guess we need to figure out how a spin
N/2 object is built up out of N X 1/2.
Gerard
Jonathan Scott
describes how things work in spinor space, but has no specific
connection with spin 1/2. You need to use spinor space to describe
spin 1/2, but you can describe any spin in spinor space.
Gerard Westendorp wrote:
> Spin N/2 objects can be seen as a composition of N spin 1/2
> objects.
>
> So I was thinking, maybe it makes sense to think of N belts.
> We could put these N belts in series. I am not sure if this
> makes sense, but some aspects do fall into place.
..
Jonathan Scott
Thomas Mautsch
> frank_k...@yahoo.co.uk <frank_k...@yahoo.co.uk> schrieb:
>> Thomas Mautsch wrote:
>>>
>>> The belt trick is a visualization of the spin *group*
>>> and not of the spin 1/2 *representation* of this group,
>>> as some people seem to think.
>>
>> The question was whether there is something
>> analogous to the belt trick for spin 3/2.
>> Is there such a trick?
>
> My point was that there is no "belt trick for spin 1/2",
> there is only a "belt trick for visualising the spin group";
> so the question whether there is an _analogous_ "belt trick for spin 3/2"
> does not make sense.
> The problem is that the only way I know
> to obtain the spin 1/2 representation of the spin groups
> is via embedding the spin groups into Clifford algebras
> and using the identification of these Clifford algebras
> with matrix algebras. (In particular Cliff(3) = IH in dim. 3).
>
> But the picture of the spin group Spin(3) as shown by the belt trick
> is more the topological definition of the spin group
This should read:
" as the set of equivalence classes of
all paths in SO(3) with fixed starting point"
heinrich...@yahoo.com
> My point was that there is no "belt trick for spin 1/2",
> there is only a "belt trick for visualising the spin group";
> so the question whether there is an _analogous_ "belt trick for spin 3/2"
> does not make sense.
How can this be? It is mentioned in "Gravitation" by MSW, by Feynman,
etc.
And the website shown in one of the previous answers even shows
how the components of the spinor depend on the orientation of the
belt's end.
So the belt's end does behave like a spin 1/2 particle.
Why should all these famous people be wrong, especially when the
end of the belt really behaves like a spin 1/2 spinor?
Is this rejection of the belt trick a minority view, or did it
turn out that Wheeler and Feynman are wrong?
Heinz
CarlB
> function has the weird feature that the phase changes by pi along a
> closed path which goes around the spin axis, which says that the value
> changes sign by the time you get back to the original point in space
> and time, making it at two-valued at any point in space, which seems
> not only inconvenient but somewhat implausible if the wave function
> might eventually have some underlying physical interpretation.
When a spin-1/2 state is written in spinor form the above happens, but
when
it is written in density matrix form there is no arbitrary complex
phase and
there is no factor of -1 from closed paths (uh, rotations).
See the section starting page 10, "Naught spinor behavior" of this
introduction
to density matrices:
http://www.brannenworks.com/dmaa.pdf
Carl
frank_k...@yahoo.co.uk
> Spin N/2 objects can be seen as a composition of N spin 1/2
> objects.
>
> So I was thinking, maybe it makes sense to think of N belts.
> We could put these N belts in series. I am not sure if this
> makes sense, but some aspects do fall into place.
One could also try to take 3 belts in a star formation, or
three belts in a triangle. Each belt end would act like a spin
1/2 particle.
What seems less clear is in which case the three spins
add up to 3/2, and in which case they add up to spin 1/2.
(The same question appears when one connects TWO spin 1/2
by a belt: is that a spin zero or a spin 1 configuration?
And how does one model the other case?)
Frank
frank_k...@yahoo.co.uk
On Nov 1, 12:36 am, Jonathan Scott <jonathan_sc...@vnet.ibm.com>
wrote:
> Sorry, Gerard, but you've missed an earlier point. The belt trick
> describes how things work in spinor space, but has no specific
> connection with spin 1/2. You need to use spinor space to describe
> spin 1/2, but you can describe any spin in spinor space.
But IF this is correct, the question remains:
How can you describe spin 3/2?
Spin 3/2 means that a structure comes back to itself after
2/3rd of a turn. If this is possible to describe with the
belt trick, how does one do it?
Frank
Thomas Mautsch
schrieb frank_k...@yahoo.co.uk <frank_k...@yahoo.co.uk>:
^^^^^^^^ !!!
I can't get the applet to run on my computer,
but from the reading the description on the page
I get the impression that this is not the usual belt trick:
Here we have a belt (fixed at one end)
_together_ with _two complex numbers_.
These numbers should change values
whenever the buckle of the belt is moved,
but not, when both ends of the belt are kept fixed
and only its middle part is moved. Is that correct?
In this case, the analogue construction for spin 3/2
would consist of a belt - like for the spin 1/2 construction -
and *four* complex numbers, and these complex numbers should
behave similarly as described above, only corresponding
to a spin 3/2 representation instead of spin 1/2.
(_Four_ complex numbers instead of two,
because the spin 3/2 representation is _four_-dimensional.)
To explain why this is so, consider the following
completely analogous example of representations of SO(3):
Any _vector_ v in R^3 can be represented by
a choice of an oriented orthonormal frame (e1,e2,e3)
and three real numbers (a1,a2,a3):
v = a1 e1 + a2 e2 + a3 e3.
The frame corresponds to the belt,
the real numbers correspond to the two complex numbers
in the applet above.
When we change, i.e. rotate, the orthonormal frame
by a matrix A in SO(3)
- which corresponds to moving the belt
and in particular its buckle in the picture above -
then the triple of real numbers is transformed by the transpose of A.
But vectors are not the only representations of SO(3)!
If instead of a vector, we consider a _bilinear form_ q on R^3
- a combination of different representations of SO(3) -
then this bilinear form can be represented by
a choice of an oriented orthonormal frame (e1,e2,e3) - again! -
and - now - nine real numbers (a11,a12,a13,a21,a22,a23,a33)
instead of three:
aij := q(ei,ej) for all i, j in {1,2,3}.
And again, when we rotate the orthonormal frame
by a matrix A in SO(3),
the nine-tuple will transform by a certain 9x9 matrix
connected to A.
Can you see my point that the belt itself has
nothing to do with the representation of SU(2) -
whether it is spin 1/2, or spin 3/2, or whatever -
just like the choice of orthonormal frame alone
says nothing about whether you describe
a vector, or a bilinear form, or whatever with it?!
Thomas Mautsch
schrieb heinrich...@yahoo.com <heinrich...@yahoo.com>:
> Thomas Mautsch wrote:
>> My point was that there is no "belt trick for spin 1/2",
>> there is only a "belt trick for visualising the spin group";
>> so the question whether there is an _analogous_ "belt trick for spin 3/2"
>> does not make sense.
>
> How can this be? It is mentioned in "Gravitation" by MSW, by Feynman,
> etc.
Could you please post a citation of a statement
that stands in contradiction to what I wrote! -
This would make it easier for me to try to defend my point of view.
> And the website shown in one of the previous answers even shows
> how the components of the spinor depend on the orientation of the
> belt's end.
Yes, but the components of the spinor are usually not part
of the belt trick as I understand it - and without giving
these components the "belt trick for spin 1/2" and the
"belt trick for spin 3/2" will be identical - and that is what I wrote.
The belt trick as I understand it just shows
that if you double a certain closed path with values in SO(3),
this doubled path becomes contractible -
nothing to do with _representations_ of Spin(3).
> So the belt's end does behave like a spin 1/2 particle.
^^
"So"?!
If you replace the two components of the spin 1/2 representation
in the applet by four components of a spin 3/2 representation,
then the same argument could be used in order to show that
the belt's end does also behave like a spin 3/2 particle
(and likewise for any other representation of Spin(3) -
well, maybe not the trivial representation...).
> Why should all these famous people be wrong, especially when the
^^^^
> end of the belt really behaves like a spin 1/2 spinor?
This "when" is based on your inconclusive "so"
from the previous sentence.
> Is this rejection of the belt trick a minority view,
> or did it turn out that Wheeler and Feynman are wrong?
What's this?! - Proof by authority?! :-)
Wheeler and Feynman may not be wrong on the belt trick,
but there certainly _are_ mistakes in MTW's "Gravity" book
(e.g. in the geometric picture of parallel transport).
Noone is infallible, I think.
Thomas Mautsch
schrieb frank_k...@yahoo.co.uk <frank_k...@yahoo.co.uk>:
> Gerard Westendorp wrote:
As far as I can see, this is complete and utter nonsense!
You and Gerard Westendorp are using
two different ways of "composition" without even noticing:
"Composition" in the first line of Gerard Westendorp's message
has to do with taking the _tensor product_ of representations;
"composition" in the remainder has to do with _concatenating paths_
(paths with values in SO(3),
or better, values in an SO(3)-torsor - but never mind this).
Both of these compositions are VASTLY different from each other,
have NOTHING much in common,
and thus leave no room for such idle speculations...
Thomas Mautsch
schrieb Cl.Massé <ret...@contactprospect.com>:
><frank_k...@yahoo.co.uk> a écrit:
>
>> The question was whether there is something
>> analogous to the belt trick for spin 3/2.
>> Is there such a trick?
>
> No, it's a conjuring trick. The fact that the two rotations are in the
> inverse direction is concealed by an stealthy change of configuration, like
> a conjuror would make a stealthy manipulation just at the time of a change
> of motion. It is indeed impossible to have such a trick in a three
> dimensional space with classical objects.
[Moderator's note: Comment about moderation policy deleted. -P.H.]
Dirac's belt trick is _not_ a conjuring "trick", as the name might
suggest, but a proof of the fact that given a path of rotations around a
fixed axis with rotation angle going from 0 degrees to 360 degrees to
720 degrees, this path can be homotopically deformed to become constant
without moving its endpoints.
Look at Berger's "Geometry" book, volume 1, figure 8.10.3.1,
if you don't believe it!
Gerard Westendorp
[..]
> As far as I can see, this is complete and utter nonsense!
>
> You and Gerard Westendorp are using
> two different ways of "composition" without even noticing:
>
> "Composition" in the first line of Gerard Westendorp's message
> has to do with taking the _tensor product_ of representations;
> "composition" in the remainder has to do with _concatenating paths_
> (paths with values in SO(3),
> or better, values in an SO(3)-torsor - but never mind this).
Well, I already said myself in my first post that this probably
wouldn't work. Maybe I was being lazy to post a first guess.
I agree that the belt trick demonstrates something about rotations
in general: The SO(3) is not simply connected.
But I am not convinced that the belt does not correspond to
a spin(1/2) object in an interesting way.
btw, We had a discussion on this already in 2000:
http://www.lepp.cornell.edu/spr/2000-09/msg0028333.html
The belt buckle's orientation can be represented by a normalized
vector with 3 components. (ie. 2 independent real numbers)
A spin (1/2) spinor is given by 2 complex components, which
can be written out as 4 real ones. This is again normalized,
so there are 3 independent real numbers.
But part of the spinor is a phase factor that does not influence
its dimensional orientation. So that leaves only 2 real
independent numbers. If we try to make a correspondence with
the the 2 independent real numbers that describe the buckle orientation,
we find a 2-to-1 correspondence because of the double cover stuff.
But that is where the untwistability comes in. That gives you
the extra information to create a 1-to-1 correspondence.
With higher spins, such a correspondence becomes more troublesome.
For example, the 2 (3/2) spinors:
( 1 ) ( 0 )
( 0 ) ( 1 )
( 0 ) ( 0 )
( 0 ) ( 0 )
both have spin in the z-direciton only. But they differ by more
than just a phase factor. So a single belt cannot encode a
(3/2) spinor, while a single belt can, upto a phase factor,
encode a (1/2) spinor.
I 'm a bit confused about the spin 1 case. The spinor
( 0 )
( a )
( 0 )
gives expectation value zero for x,y and z directions when
you do the math.
So in the spin 1 case, there is an entire subspace of the
spinor that has no orientation at all. If we consider only
spinors orthogonal to this subspace, we again get 3 independent
numbers that describe the state.
hmm, I'm still tempted not to give up on the multi-belt idea,
but I might wait until I've checked the details, that is,
if I have time.
Gerard
Cl.Massé
4549...@news1.ethz.ch
> Dirac's belt trick is _not_ a conjuring "trick", as the name might
> suggest, but a proof of the fact that given a path of rotations around a
> fixed axis with rotation angle going from 0 degrees to 360 degrees to
> 720 degrees, this path can be homotopically deformed to become constant
> without moving its endpoints.
Like the conjuring trick of the cord that seems to be winded two times, but
that is made loose because the second time was in the other direction.
The first time, the buckle is passed beneath, the second time it is passed
above. All what have been done is reversing the direction of rotation with
respect to the belt, stealthily because it is supple. The axis isn't fixed,
it is stealthily changed by a small amount, that has a big topological
consequence because the belt is very thin. To see that, it suffices to make
a continuous deformation, such that the buckle remains in a plane
perpendicular to the axis of the belt, which is kept as straight as
possible.
That is a quite rigorous reasoning, based on attentive observation. The
"belt trick" has never been intended to be wholly mathematically equivalent,
but to give a rough picture to lay persons. A true mathematical proof
exists, and it's all what is needed by a physician.
[Moderator's note: Should be "physicist". In French, a physicist is a
physicien, and a physician is a medicin. As Steve Martin said, these
French have their own word for EVERYTHING. -P.H.]
--
The Surprise Guest
Bossavit
in the sense of "cheating", in Dirac's belt trick.
Yet, one may legitimately feel some frustration with
it, for though it "proves" its point (that SO3 is not
simply connected), it does so in a needlessly involved
way.
"Proves", here, actually means, "help visualize".
(Mathematical proof is another issue.) As
illustrated in Greg Egan's display, such visualization
is easy without losing one's pants: Think of a solid
ball in 3D space, centered at some point, dubbed "the
origin" from now on, with pairs of antipodal points
identified. Topologically, this is SO3, the 3D rotations
Lie group. The point to make is that there are two
classes of origin-centered loops in this space, those of
the first class being shrinkable to 0, while the others
are not: Any loop not crossing the surface belongs in
the first class. To get a representative of the second
class, go straight from 0 to the boundary (say, at point
A), reenter from the opposite point (A', say), return
straight to 0. This is a closed loop in SO3, obviously
not reducible to 0 by continuous deformation within SO3.
On the other hand, run along the previous path twice (or
any even number of times), and the resulting loop can
now be shrunk to 0 (provided one correctly interprets
the meaning of "loop": one is supposed to return to 0
at the end of the journey, but the mid-trip passage
through 0 is not mandatory--actually it *must* be avoided
as the loop is deformed).
The latter loop is thus from 0 to A, then jump to A',
go from A' to A, jump to A' again, and return to 0.
How to shrink it to 0, is what Greg Egan's
animation shows. It involves changing direction the
second time around, progressively: Instead of going
from A' to A, go straight from A' to some B
close to A, on the surface (do *not* go through 0
this time), jump to B' opposite to B, then back
home. This new loop, 0AA'BB'0, is "close", in
an obvious sense, to 0AA'AA'0. Now, keeping the first
leg of the journey (from 0 to A aka A') unchanged,
move B (and hence, B', the same point in SO3,
actually) towards A', on the surface. The
deformed loop is now 0AA'A'A0, that is to say, 0A0,
which nicely retracts to 0.
[All this is very easy to draw, though not in ASCII
unfortunately. I tend to think that such a series of
drawings, with comments, may be better as a teaching
aid than an animation, but that's a side issue.
(I did enjoy this animation, as well as others proposed
by Greg Egan, but I always had to first "see things with
the eyes of the mind", as Poirot is suppposed to say
in French translations, before being able to appreciate
them.)]
Now, the belt. Fasten the buckle in some fixed
position (a door knob may do), hold firm the
other end. Pause to show the belt is flat,
i.e., untwisted. Now give it two full (360o) turns
(around some fixed spatial axis, for neatness
--the vertical, say). You used both hands to do
that, presumably. Pause, keep loose end in one
hand, bring attention to the double twist in
the belt. Now, *always keeping this loose end of the
belt (approximately) vertical and not rotating
(not by more than a few degrees)*, pass it behind the
rest of the belt, then back to where it was (notice
you *must change hands* [or perhaps jump above
the belt... never tried that one] to achieve
that; maybe this is what evokes a "conjuring
trick" to some--we'll return to that). See the
belt, miraculously, flatten back to its original
configuration. Make the point that
nothing of this kind could be done after only
*one* full turn (or three, for that matter, if
the belt could stand it). Demonstrate also that
the loose end can be kept very close to the knob-
and-buckle end in this process, with its own frame
almost parallel, always, to the buckle's frame.
What has thus been "proved"? With respect to the
knob and buckle, each segment t of the belt, t
going from 0 to 1, has a translation vector v(t)
in 3D vector space V3, and a rotation s(t), element
of SO3, with respect to the buckle's frame. So
the belt materializes a path, t --> {v(t, s(t)},
in the product space V3 x SO3, i.e., in the Lie group
D3 of (direct) 3D displacements, the unit element
of which corresponds to the {knob, buckle's frame}
pair. [The algebraic structure of D3, as a
semi-direct product, is ignored here--perhaps
not a good thing.] Since buckle and loose end can
be kept very close and almost parallel,
this path is an honest approximation to a *loop*,
a loop in D3. Flat belt and doubly twisted belt
materialize two such loops, which are shown to be
homotopic by performing the trick. So what is
addressed, really, is the non-trivial homotopic
structure *of D3*, not SO3.
From which, of course, the same about SO3 is
easily *derived*. But let's not be fooled. The
Dirac's trick is an involved one, which achieves
more than was bargained for. Too much, perhaps.
Arguably, it takes too much classroom time,
if compared to the sequence-of-drawings process.
But also, and more of a concern, it tends to elicit
further questions which the teacher may not feel
ready to address.
For example, we noticed that the path in D3 could
be considered, for what it's worth, as a loop. So
why not make it a real, physical loop? Take the belt,
give it two full turns, now *fasten the belt*. Any
way to disentangle it? You bet. The inspired remark
that "if the belt's material was this kind of magic
stuff that can intersect itself, then the belt could
untwisted (after 2n turns, and not after 2n + 1)"
may not win applause.
And yet, this is it, for a part: D3 can be
considered as a fibered space, fiber SO3, base V3
(or rather, more accurately, base A3, the affine
3D space), and while homotopies in D3 do
generate homotopies in SO3, they don't project
to homotopies in A3. This is why the belt must
intersect itself to untwist, or be left unbuckled
in order to leave room for the manipulation (which
by the way could not be achieved if the loose end
was always kept in the same hand). The suspicion
of a "conjuring trick", though unfair, can thus be
understood, if not endorsed.
The "plate trick", though its analysis is involved
too, looks a little more transparent, but discussing
this would take too long. More interestingly perhaps,
it would be nice if topologists could explain to us,
more thoroughly and accurately than suggested here,
the mathematics behind the belt trick, beyond the
basic fact that SO3 is doubly connected.
Cl.Massé
4550B26B...@lgep.supelec.fr
> "Proves", here, actually means, "help visualize".
With a diagram (or an equation, or symbols, in short a full-fledged
mathematical proof). With a material object, it's impossible.
> Now, the belt. Fasten the buckle in some fixed
> position (a door knob may do), hold firm the
> other end. Pause to show the belt is flat,
> i.e., untwisted. Now give it two full (360o) turns
> (around some fixed spatial axis, for neatness
> --the vertical, say). You used both hands to do
> that, presumably. Pause, keep loose end in one
> hand, bring attention to the double twist in
> the belt. Now, *always keeping this loose end of the
> belt (approximately) vertical and not rotating
> (not by more than a few degrees)*, pass it behind the
> rest of the belt, then back to where it was (notice
> you *must change hands* [or perhaps jump above
> the belt... never tried that one] to achieve
> that; maybe this is what evokes a "conjuring
> trick" to some--we'll return to that). See the
> belt, miraculously, flatten back to its original
> configuration. Make the point that
> nothing of this kind could be done after only
> *one* full turn (or three, for that matter, if
> the belt could stand it). Demonstrate also that
> the loose end can be kept very close to the knob-
> and-buckle end in this process, with its own frame
> almost parallel, always, to the buckle's frame.
The second operation, however intricate it may be, is but a stealthy twist
in the other direction, concealed behind a 360° rotation. Indeed, if the
loose end is bend 90° to the right, passed behind, and unbended, the twist
is obvious. If during the operation, the bended end is kept parallel to
itself, there is no untwist. Houdini presumably knew that.
> What has thus been "proved"? With respect to the
> knob and buckle, each segment t of the belt, t
> going from 0 to 1, has a translation vector v(t)
> in 3D vector space V3, and a rotation s(t), element
> of SO3, with respect to the buckle's frame. So
> the belt materializes a path, t --> {v(t, s(t)},
> in the product space V3 x SO3, i.e., in the Lie group
> D3 of (direct) 3D displacements, the unit element
> of which corresponds to the {knob, buckle's frame}
> pair. [The algebraic structure of D3, as a
> semi-direct product, is ignored here--perhaps
> not a good thing.] Since buckle and loose end can
> be kept very close and almost parallel,
> this path is an honest approximation to a *loop*,
> a loop in D3. Flat belt and doubly twisted belt
> materialize two such loops, which are shown to be
> homotopic by performing the trick. So what is
> addressed, really, is the non-trivial homotopic
> structure *of D3*, not SO3.
The belt trick is intended to illustrate SU2, not SO3. For SO3, it's merely
a tautology, a rotation of 360° is equivalent to the identity. It's not
like in SU2 where a rotation of 360° isn't equivalent to the identity
(that's why a 360° *twist* is used), while a rotation of 720° is. That can
be seen with any rigid object. In that case, the proof can be made with a
material model, but only because the space is modelled by a three
dimensional Euclidean space. As on the earth surface it is actually
modelled by a Riemannian space, the proof can't be made, e basta.
Sorry for not having made the Grandes Ecoles.
--
The Surprise Guest
Bossavit
>in the other direction, concealed behind a 360° rotation. Indeed, if >the
>loose end is bend 90° to the right, passed behind, and unbended, the
>twist
>is obvious.
Which "twist" do you mean? As the loose end is moved
around, its own frame doesn't rotate. As you say
>during the operation, the bended end is kept
>parallel to itself
and the belt flatten out.
>The belt trick is intended to illustrate SU2,
>not SO3
It "demonstrates" (with still the same caveat about what the
word means) that SO3 is doubly connected. Hence
it has a simply connected double cover, which one easily
proves (mathematically speaking, this time) is SU2.
>a rotation of 360° is equivalent to the identity.
No doubt.
>in SU2 [], a rotation of 360° isn't equivalent
>to the identity
But what justifies calling an element of SU2 a "rotation"?
The group SU2 is acting on a different space than good
old 3D space (where belts, plates, etc., live). The
"demonstration", being a classical experiment in
this space, tells something about SO3. I don't
object to the statement that "the belt trick is
intended to illustrate SU2, not SO3". Most
of those who write about the subject seem to
intend that. But as the trick is about SO3,
really, and only indrectly about SU2, readers
get confused (and rightly so).
Thomas Mautsch
> <heinrich...@yahoo.com> schrieb:
>> Thomas Mautsch wrote:
>>> My point was that there is no "belt trick for spin 1/2",
>>> there is only a "belt trick for visualising the spin group";
>>> so the question whether there is an _analogous_ "belt trick for spin 3/2"
>>> does not make sense.
>>
>> How can this be?
>> And the website shown in one of the previous answers even shows
>> how the components of the spinor depend on the orientation of the
>> belt's end.
>
> Yes, but the components of the spinor are usually not part
> of the belt trick as I understand it - and without giving
> these components the "belt trick for spin 1/2" and the
> "belt trick for spin 3/2" will be identical - and that is what I wrote.
>
> The belt trick as I understand it just shows
> that if you double a certain closed path with values in SO(3),
> this doubled path becomes contractible -
> nothing to do with _representations_ of Spin(3).
>
>> So the belt's end does behave like a spin 1/2 particle.
>
> If you replace the two components of the spin 1/2 representation
> in the applet by four components of a spin 3/2 representation,
> then the same argument could be used in order to show that
> the belt's end does also behave like a spin 3/2 particle
> (and likewise for any other representation of Spin(3) -
> well, maybe not the trivial representation...).
I should have written here "and likewise
for any other _non-integer-spin_ representation of SU(2),"
because I meant to say that
only these representations of SU(2) are _faithful_ -
no two elements of SU(2) act in the same way on these representation
spaces.
Unfortunately, I found a message by John Baez
<news:60991l$9...@agate.berkeley.edu>
from 1997/09/23 in the sci.physics.research thread
"Weyl system, observable algebra, Hilbert space",
where he writes:
"Whoops, the spin-3/2 rep isn't faithful: when you rotate a spin-3/2
particle by 4pi/3 it always comes back to its original state, so
there is an element of SU(2) that's not the identity but gets mapped
to the identity by this rep."
And now I feel deeply troubled. ;-( -
Could it be that Baez has been mislead by a wrong
interpretation of the belt trick for spin 3/2?
Or is the spin-3/2 representation really not faithful?
A short explanation why I feel it should be faithful is as follows:
The spin-k/2 representations of SU(2) on C^(k+1)
induce projective representations of the group
PSU(2) := SU(2) / {+Id,-Id} = SO(3)
on the complex projective space CP^k = (C^(k+1) \ {0}) / (C \ {0}).
These representations are just k-fold symmetric products
of the projective representation of SO(3) on CP^1
that are induced by the spin-1/2 representation of SU(2) on C^2.
Now the representation of SO(3) on CP^1 = S^2
is (conjugate to) the usual representation of SO(3)
as rotations on S^2, and this representation is faithful.
As a consequence, _all_ projective representations of SO(3)
on CP^k that are induced by the spin-k/2 representations of SU(2)
are also faithful. (There is a canonical CP^1
in the symmetric product CP^k, given by the k-tuples of points in CP^1,
whose k compontents are all equal;
and SO(3) acts on this CP^1 in CP^k by rotations, i.e. faithfully.)
So, the induced projective SO(3) representations are all faithful.
Hence, the only elements of SU(2) that can act
under a spin-k/2 representation as the identity element
are those which map to the identity in SO(3),
i.e. (+Id) and (-Id) in SU(2).
But for _odd k_, (-Id) acts as multiplication by -1 on C^(k+1),
i.e. non-trivial; and thus the action of SU(2) must be faithful:
The only element which acts as identity on C^(k+1)
is the identity (+Id) in SU(2).
Should be correct, should't it?
Greg Egan
frank_k...@yahoo.co.uk wrote:
> To put it even nore specifically: is it possible to have
> an applet like the beautiful example at
> http://gregegan.customer.netspace.net.au/APPLETS/21/21.html
> but for spin 3/2 insteand of 1/2?
>
> That one shows the spinor state, the path through SO(3), and more.
>
> Frank
My view is that the only thing I could change in the applet to make it
apply to spin 3/2 would be to show the effect of the initial rotation of
the buckle on an element of C^4 (rather than C^2). So during that
rotation phase, you'd see the effect of the spin-3/2 representation on
four complex numbers. But the overall effect of the total, 2pi rotation
would just be to restore that C^4 elment to its original value (just as
the C^2 spinor I currently show is restored to its original value), and
everything that happened with the belt itself -- both during the 2pi
rotation of the buckle, and the part where the buckle is held fixed and
the belt is made flat -- and everything that happened with the path
through SO(3), would be identical.
The one thing that I think is (a bit) spin-1/2-specific about this is the
way you can find a very simple mathematical connection between the spinor
components and two orthogonal unit vectors in R^3 associated with the
buckle; see the technical notes on the page linked to the applet for more
on this:
http://gregegan.customer.netspace.net.au/APPLETS/21/DiracNotes.html
For higher spins, this becomes more complicated and abstract.
Greg Egan
>But the overall effect of the total, 2pi rotation would just be
^^^
Sorry, I meant 4pi
Cl.Massé
> >in the other direction, concealed behind a 360° rotation. Indeed, if
> >the loose end is bend 90° to the right, passed behind, and unbended,
> >the twist is obvious.
"Bossavit" <boss...@lgep.supelec.removethis.fr> a écrit dans le message de
news: 4553C2DF...@lgep.supelec.removethis.fr
> Which "twist" do you mean? As the loose end is moved
> around, its own frame doesn't rotate. As you say
The plane containing the loose end make a 360° rotation during the
operation. Everybody can see it by making the experiment. By bending it,
it's obvious that this rotation is actually a twist of the belt.
> >during the operation, the bended end is kept
> >parallel to itself
> and the belt flatten out.
No, as the experiment shows.
> >The belt trick is intended to illustrate SU2,
> >not SO3
> It "demonstrates" (with still the same caveat about what the
> word means) that SO3 is doubly connected.
No, because the twisted belt isn't equivalent to the flat belt, since during
the operation a stealthy untwist is performed. It is simple, but that
doesn't make it incorrect.
> >in SU2 [], a rotation of 360° isn't equivalent
> >to the identity
> But what justifies calling an element of SU2 a "rotation"?
That SU2 is a compact group. Non compact generators (well, generators
generating non compact subgroup) are usually called translation generator.
A physical object, like a Dirac spinor, is transformed through a space
rotation. The transformation represents an element of SU2. That is called
a spinor representation. If space is 360° rotated, the spinor doesn't
transform into itself, but do it if the rotation is of 720°.
> The group SU2 is acting on a different space than good
> old 3D space (where belts, plates, etc., live).
No, SU2 is acting on nothing except itself. A *representation* of SU2 is
acting in a representation space. A representation of SO3 acts in the same
space as a representation of SU2, if it is a spinor representation.
Explicitly, let a point of R3 be noted with the Pauli matrices:
A = a_i s^i, where a_i are the three coordinates.
Acting on both side of A by a two dimensional representation of SU2, we see
that the point is rotated like by a representation of SO3 acting in R3.
Now if the representation of SU2 acts only on one side, we have a
representation of SU2 witch has no SO3 counterpart, yet acting on a three
dimensional real space. Indeed, the left column of such a matrix defines it
completely. But in this case, when the equivalent of a 360° rotation is
performed, the result isn't A but -A. Intuitively, that is because there is
an action only on one side.
--
The Surprise Guest
Thomas Mautsch
schrieb Cl.MassÚ <ret...@contactprospect.com>:
>> "Thomas Mautsch" a ecrit:
>>>> Dirac's belt trick is _not_ a conjuring "trick", as the name might
>>>> suggest, but a proof of the fact that given a path of rotations around a
>>>> fixed axis with rotation angle going from 0 degrees to 360 degrees to
>>>> 720 degrees, this path can be homotopically deformed to become constant
>>>> without moving its endpoints.
>
>> Like the conjuring trick of the cord that seems to be winded two times, but
>> that is made loose because the second time was in the other direction.
>>
>> The first time, the buckle is passed beneath, the second time it is passed
>> above. All what have been done is reversing the direction of rotation with
>> respect to the belt, stealthily because it is supple. The axis isn't fixed,
>> it is stealthily changed by a small amount, that has a big topological
>> consequence because the belt is very thin. To see that, it suffices to make
>> a continuous deformation, such that the buckle remains in a plane
>> perpendicular to the axis of the belt, which is kept as straight as
>> possible.
^^^^^^^^
This nonsense should never have made it a second time
into a moderated newsgroup, but as it has appeared here,
let us take a look at it. - Let me consider just *one aspect*:
The parts I underlined indicate that Cl.Masse
wants to keep the middle line of the belt fixed,
even wants to keep it on a(n almost(?)) straight line.
This way the body of the belt can not move freely in space. -
Masse is basically only allowing the belt to describe
a path in SO(2). And since, unlike SO(3), there are
no elements of order two in the fundamental group pi_1(SO(2)),
he can not observe what the belt trick is about.
Well, you *can* twist the belt by an angle of 4*Pi as described
above - keeping the axis of the belt fixed, but
you can never untwist it if you keep the middle line of the belt fixed.
Masse does not seem to understand, that
the real belt trick is a demonstration that you can
untwist the belt from the "4Pi-twisted" position as above
when the ends are not allowed to be rotated
with respect to their respective positions, but are still
allowed to be moved around in space by parallel-translations.
I can only guess, that Masse wrongly thinks
the untwisting of the belt also has to be done
throuh a "path of rotations around a fixed axis".
Maybe he even confused the rotation axes
with the 'axis' (a.k.a. middle line) of the belt!
The next paragraph shows more reason why
Masse's message should not have made it into sci.physics.research:
>> That is a quite rigorous reasoning, based on attentive observation.
^^^^^^^^^^^^^^ ^^^^^^^^^^^
^^^^^ ??? Observation of what?
>> The "belt trick" has never been intended
>> to be wholly mathematically equivalent,
^^^^^^^^^^
"Equivalent"?! - Equivalent to what?
>> but to give a rough picture to lay persons.
^^^^^^^^^^^^^^
The audience that the belt trick is shown
to, usually consists of physics students. -
I wouldn't call them "lay persons". :-)
>> A true mathematical proof exists,
^^^^^ Proof of what?
>> and it's all what is needed by a physician.
I wish you good luck with this philosophy of yours! :-I
>> [Moderator's note: Should be "physicist". In French, a physicist is a
>> physicien, and a physician is a medicin.
[ ... ]
The apparent translation mistake "physician" - "physicist"
would not be my main concern with the content of Masse's message
if I was a moderator of this group.
frank_k...@yahoo.co.uk
He allowed me to post the answer here:
------------------
On Sat, Nov 18, 2006 at 10:09:24AM +0000, Frank Sheldon wrote:
> "Whoops, the spin-3/2 rep isn't faithful: when you
> rotate a spin-3/2
> particle by 4pi/3 it always comes back to its original
> state, so
> there is an element of SU(2) that's not the identity
> but gets mapped
> to the identity by this rep."
> And now I feel deeply troubled. ;-( -
> Could it be that Baez has been mislead by a wrong
> interpretation of the belt trick for spin 3/2?
I seem to have made a silly mistake. If a spin-3/2 system
is an eigenvector for angular momentum in some direction, with
eigenvalue 3/2, then rotating it 4pi/3 will return it to its
original state. But, it's not true that *every* state is mapped
to itself by this rotation. This is especially obvious for a
state that's an eigenvector with eigenvalue 1/2.
Since I wrote this erroneous article about 9 years ago,
perhaps I can claim that I was less smart back then.
Or, maybe I just didn't have anyone like you to correct me. :-)
If you wish, feel free to post what I'm saying to sci.physics.research.
Best,
jb
------------------