But where the heck is the spinor?

  1. But where the heck is the spinor??

    Hi lads,

    I've been reading the other thread "...intrinsic vs orbital
    angular momentum..." which talks about spinors, r x p, etc, etc.
    It mentioned ch41 of Misner, Thorne & Wheeler's "Gravitation"
    book. I've studied all the math there quite thoroughly, no problems
    with that. But seems like something's missing...

    In Fig 41.6 on p1149 (the one with two concentric spheres, - the inner
    sphere connected to the outer by threads, which you're then supposed
    to
    twist through 2pi or 4pi, and contort the inner sphere around to show
    whether you can/can't untwist the threads using only translations of
    the inner sphere). OK,... yeah, I get it. I've done the related "Dirac
    belt" thing, and I get that 2pi rotation ain't necessarily the same as
    4pi. But where the heck is the actual spinor in MTW's diagram??

    Later in the chapter MTW rave on about poles and flags, but isn't
    that just a combination of a 4-vector and a bivector? Where is the
    spinor in "flag+pole"??. If I rotate the flag about its axis through
    2pi, the flag returns to its original appearance and I don't see
    anything spinor-like until I start trying to move the pole around (as
    if the pole was elastic).

    So I still have no clue where the spinor is in MTW's diagram. Should
    I be thinking instead of an army of tiny flags all the way along an
    elastic
    flagpole? (That way, rotating only one end of the pole through 2pi/4pi
    leaves obviously different orientations of the flags all the way along
    the pole, and it's more obvious that all this 2pi-4pi monkey business
    has something to do with rotation "here" without a matching rotation
    at
    "infinity".)

    LOL,

    Neuropulp.
     
  2. jcsd
  3. Re: But where the heck is the spinor??

    neurop...@yahoo.com.au wrote:

    > Hi lads,
    >
    > I've been reading the other thread "...intrinsic vs orbital
    > angular momentum..." which talks about spinors, r x p, etc, etc.
    > It mentioned ch41 of Misner, Thorne & Wheeler's "Gravitation"
    > book. I've studied all the math there quite thoroughly, no problems
    > with that. But seems like something's missing...
    >
    > In Fig 41.6 on p1149 (the one with two concentric spheres, - the inner
    > sphere connected to the outer by threads, which you're then supposed
    > to
    > twist through 2pi or 4pi, and contort the inner sphere around to show
    > whether you can/can't untwist the threads using only translations of
    > the inner sphere). OK,... yeah, I get it. I've done the related "Dirac
    > belt" thing, and I get that 2pi rotation ain't necessarily the same as
    > 4pi. But where the heck is the actual spinor in MTW's diagram??
    >
    > Later in the chapter MTW rave on about poles and flags, but isn't
    > that just a combination of a 4-vector and a bivector? Where is the
    > spinor in "flag+pole"??. If I rotate the flag about its axis through
    > 2pi, the flag returns to its original appearance and I don't see
    > anything spinor-like until I start trying to move the pole around (as
    > if the pole was elastic).
    >
    > So I still have no clue where the spinor is in MTW's diagram. Should
    > I be thinking instead of an army of tiny flags all the way along an
    > elastic
    > flagpole? (That way, rotating only one end of the pole through 2pi/4pi
    > leaves obviously different orientations of the flags all the way along
    > the pole, and it's more obvious that all this 2pi-4pi monkey business
    > has something to do with rotation "here" without a matching rotation
    > at
    > "infinity".)
    >
    > LOL,
    >
    > Neuropulp.


    Dear Neuropulp,
    I agree with you, there does not appear to be any connection between a
    spinor and Dirac's spanner illustrated by figure 41.6 on page 1148 of
    MTW. Dirac's spanner illustrates the fact that the rotation group
    SO(3) is not simply connected. The rotations in MTW's figure are
    examples of the defining representation of SO(3) which is the spin 1
    rep. A spinor rep such as spin 1/2 is not carried by ordinary 3-d
    Euclidean space (for example).

    Stephen Blake
    http://www.stebla.pwp.blueyonder.co.uk
     
  4. Re: But where the heck is the spinor??

    On Apr 19, 3:52 am, neurop...@yahoo.com.au wrote:
    > Hi lads,
    >
    > I've been reading the other thread "...intrinsic vs orbital
    > angular momentum..." which talks about spinors, r x p, etc, etc.
    > It mentioned ch41 of Misner, Thorne & Wheeler's "Gravitation"
    > book. I've studied all the math there quite thoroughly, no problems
    > with that. But seems like something's missing...
    >
    > In Fig 41.6 on p1149 (the one with two concentric spheres, - the inner
    > sphere connected to the outer by threads, which you're then supposed
    > to
    > twist through 2pi or 4pi, and contort the inner sphere around to show
    > whether you can/can't untwist the threads using only translations of
    > the inner sphere). OK,... yeah, I get it. I've done the related "Dirac
    > belt" thing, and I get that 2pi rotation ain't necessarily the same as
    > 4pi. But where the heck is the actual spinor in MTW's diagram??


    I haven't got MTW here, but I do have a bit of a take on looking at
    spinors (for spin 1/2 systems) - which might not be perfectly rigorous
    (be warned!). First though, it's important to keep in mind that
    spinors are what are acted on by SO(3) (rotations), they are not the
    rotations themselves.

    You can think of spinors as a pair of objects: a unit direction in
    space, and a +1 or -1 sign. So, the spinor can be represented as a
    pair (n, +/-), where n.n=1.

    What then is important is to ask is how the rotation group acts on a
    spinor. For that, you need a picture of the rotation group, and how
    it is doubly connected. Intuitively, one would think *at first* that
    a good way of representing rotations is as a ball in 3-d space,
    centred at the origin, and of radius pi. A point a in the ball then
    represents a right-handed rotation about the a-direction, (i.e., in
    the direction a/|a|), by an amount equal to the distance of a from the
    origin (i.e., by an angle equal to |a|). Since a rotation by pi about
    some direction a is equivalent to a rotation by -pi about the
    direction -a, one has to identify opposite points on the surface of
    this ball. The effect of a rotation R_a, represented by point a of
    the ball, would then change a spinor (n,+/-) to the spinor (R_a n,
    +/-).

    However, this picture obviously doesn't quite work, as nothing happens
    to the +/- sign of the spinor. The doubly connected nature of SO(3)
    has not shown up. To fix this, one needs a *second* 3d ball of radius
    pi. One can sit this ball in a different space to the first, but it
    is mentally convenient to think of the second ball as sitting close
    by. However, the second ball is inverted with respect to the first.
    In particular, a point b in the second ball corresponds to a *left-
    handed* rotation about the direction b/|b|, by an angle |b|. Now,
    instead of identifying antipodal points of the first ball with each
    other, one instead simply identifies the surface of the first ball
    with the surface of the second ball (for essentially the same reason:
    a right-handed rotation of pi about a given direction is equivalent to
    a left-handed rotation of pi about the negative direction). This
    gives us what we need.

    A rotation is now characterised by a point a in *one* of the two
    balls. If it is a point in the first ball, the effect on the spinor
    (n,+/-) is to map it to the spinor (R_a n, +/-) as before. However,
    if it is in the second ball, then the effect on the spinor is to map
    it to the spinor (R'_a n, -/+), i.e., the second component of the
    spinor is reversed in sign. Here, R'_a denotes a left-handed rotation
    by angle |a|.

    To see the 4 pi effect, consider the action on a spinor as one moves
    through rotation space, starting at the centre of the first ball
    (i.e., no rotation). First, move up along the z-axis to the north
    pole. This is the same as the north pole is of the second ball, so
    move over to the second ball's north pole (you can imagine a "virtual
    trajectory" through the intervening 3d space if you like). Second,
    move down the z-axis of the second ball to its centre. What has
    happened to a spinor (n,+) ? Well, you ended up in the centre of the
    second ball, so R'_0 is the identity matrix, and n is mapped to n - no
    change there. But, you are in the second ball, so the second
    component of n changes sign. Hence, (n,+) is mapped to (n,-).

    Physically, what has happened? Well, first you rotated to the right
    by pi, corresponding to moving up to the north pole of the first
    ball. Then you hopped over to the second north pole, as you had
    equivalently rotated to the left by pi. But then you undid the
    rotation to the left by pi, by moving down to the centre of the second
    ball, which is the same as rotating to the right by pi again. Hence,
    overall, you rotated by 2 pi, but the spinor flipped its sign.

    To get back to the original spinor, (n,+), one can imagine keeping
    moving down to the south pole of the 2nd ball, then over to the south
    pole of the 1st ball (draw another virtual trajectory between the two
    balls), and up to the origin of the first ball. This corresponds
    rotating by another 2 pi angle, i.e., 4 pi altogether, and the spinor
    flips back from (n,-) to (n,+) again.

    Hope this picture helps. The important thing is remember that the
    balls aren't spinors, they are ways of visualising the action of the
    rotation group on spinors.

    The above picture also helps seeing how the rotation group SO(3) is
    connected. Drawing a path from one ball over to the other, as in the
    first 2 pi rotation above (with "virtual" portions), one clearly can't
    shrink this path to a point. However,extending such a path to go back
    to the original ball allows a loop to be formed, which can clearly be
    continuously shrunk to a point.
     
  5. Re: But where the heck is the spinor??

    a student said:

    > I haven't got MTW here, but I do have a bit of a take


    Hi salacious student! Thanks for your "take". I like
    being taken.

    > on looking at spinors (for spin 1/2 systems) - which might
    > not be perfectly rigorous (be warned!).


    I'm fairly warned, but it's ok. Rigorous or non-rigorous,
    I like it both ways.

    > First though, it's important to keep in mind that spinors
    > are what are acted on by SO(3) (rotations), they are not
    > the rotations themselves.


    Yep, but did you mean SO(3) or SU(2)? Oh, you're
    representing the spinor as (n, +/-), so I guess you really
    do mean SO(3).

    > .... as a ball in 3-d space, centred at the origin,
    > and of radius pi. A point a in the ball then represents
    > a right-handed rotation about the a-direction, ....


    OK... this is like one of Penrose's other diagrams (the one
    that looks a bit like a mostly bald head with some hair-like
    comb-overs from one side to the other).

    > However, this picture obviously doesn't quite work, as
    > nothing happens to the +/- sign of the spinor. The doubly
    > connected nature of SO(3) has not shown up. To fix this, one
    > needs a *second* 3d ball of radius pi. ....


    Ah, and this is a bit different from Penrose. He's only got
    one ball with antipodal points identified, which can be
    confusing. I like your picture better - two balls plus a
    free a tardis trip between corresponding points on their
    surfaces. Yep, two balls are definitely better than one.

    > However, the second ball is inverted with respect to the
    > first.


    OK, by "invert" you mean a 69'er with every point swapped
    with its antipodal point? Not a conformal-type inversion
    where the shell gets exchanged with the center, right? (BTW,
    how do your spinors behave under those conformal inversions?
    Still a factor of -1? But no, spinors in that case would be
    twistors, wouldn't they?)

    > The above picture also helps seeing how the rotation group
    > SO(3) is connected. Drawing a path from one ball over to the
    > other, as in the first 2 pi rotation above (with "virtual"
    > portions), one clearly can't shrink this path to a point.
    > However, extending such a path to go back to the original
    > ball allows a loop to be formed, which can clearly be
    > continuously shrunk to a point.


    Yes, I've heard shrinkage can be tricky to manage. Homotopy
    groups, right? By "shrinking" I guess you mean "shrink while
    keeping endpoints fixed", which is only possible if the
    endpoints coincide.

    LOL.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook