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Re: This Week's Finds in Mathematical Physics (Week 232)

  1. Jun 9, 2006 #1
    I wrote:

    >In article <4486EF88.1010902@aic.nrl.navy.mil>, Ralph Hartley
    ><hartley@aic.nrl.navy.mil> wrote:


    [snip]

    >>[T]here is a limit to how many wedges you can cut out of a plane, and
    >>still have the topology of a plane. If the deficit angle is 2Pi the
    >>plane closes up into a sphere.
    >>
    >>If the deficit angle is more than 2Pi then it will become disconnected.


    >Good point. I did wonder about this, but I clearly haven't given it
    >enough thought. I'll have to check the literature more carefully; I
    >expect someone has analysed this issue.


    One tricky way around this would be to allow some particles of "negative
    mass", i.e. with negative deficit angles. That way the total deficit
    angle in a spacelike slice could be limited to 2 pi, but you could still
    analyse a group of particles whose collective deficit angle would exceed
    2 pi.

    For example:

    to infinity
    ^ ^
    A | | B
    | |
    | |
    | |
    . 3 4 .
    / \
    / \
    / \
    1 . . 2
    \ /
    \ /
    \ /
    . 3 4 .
    | |
    | |
    | |
    A | | B
    v v
    to infinity

    Take the interior of this diagram as flat space, and identify the pairs
    of lines that run between 1-3, 3-A, 2-4, and 4-B. The points marked
    1,2,3 and 4 are singularities, while A and B are just marked to clarify
    the identification of the edges.

    The angular deficit around the points 1 and 2 individually both exceed
    pi, and as a group their total deficit exceeds 2pi. The negative
    deficits around 3 and 4 mean that the total angular deficit of this
    connected spacelike slice does not exceed 2pi.

    Of course, there might be good reasons to rule out these negative mass
    particles as unphysical.
     
  2. jcsd
  3. Jun 9, 2006 #2
    Greg Egan wrote:
    > <hartley@aic.nrl.navy.mil> wrote:
    >> [T]here is a limit to how many wedges you can cut out of a plane, and
    >> still have the topology of a plane. If the deficit angle is 2Pi the
    >> plane closes up into a sphere.

    >
    >> If the deficit angle is more than 2Pi then it will become disconnected.


    Oops, I was off by a factor of 2. A sphere actually has a total deficit
    of 4Pi. I think that gives "Big Bang" solutions.

    An example of a surface with deficit 2Pi is an infinite prism truncated
    on one end. I still think 2Pi is where you start seeing unavoidable
    topology changes.

    If the total deficit is 2Pi < M < 4Pi then there have to be additional
    particles, bringing the total to 4Pi (excluding negative mass as noted
    below). Proof: Consider curve surrounding particles with total deficit
    M>2Pi. The curve is *concave*, it gets shorter as it gets farther away.
    If the region outside the curve were flat, it could be shrunk to a point
    without changing its holonomy, which is not zero unless M is a multiple
    of 2Pi.

    This is based on a static picture. If the velocities are large there may
    be relativistic corrections.

    > One tricky way around this would be to allow some particles of "negative
    > mass", i.e. with negative deficit angles. That way the total deficit
    > angle in a spacelike slice could be limited to 2 pi, but you could still
    > analyse a group of particles whose collective deficit angle would exceed
    > 2 pi.


    Very tricky! With more points you can get any total deficit. Also, you
    can have any total deficit if your space has a boundary (which will be
    concave if M>2Pi).

    I'm not sure this helps you with collisions though. As the positive
    curvature particles approach each other the other required particles
    have to move with them (at least I think they do). You end up with a
    total deficit involved in the collision of less that 2Pi (or exactly 4Pi).

    Ralph Hartley
     
  4. Jun 12, 2006 #3
    Ralph Hartley wrote:
    > [T]here is a limit to how many wedges you can cut out of a plane, and
    > still have the topology of a plane. If the deficit angle is 4Pi [corrected]
    > the plane closes up into a sphere.
    > If the deficit angle is more than 2Pi then it will become disconnected.


    After a little more thought, I don't think this is as bad as it seems.

    It just means that in 2+1 (as in 3+1) GR, there may not be any globally
    defined reference frames, relative to which you can define surfaces of
    constant time. If you follow a path with holonomy that has a boost
    component, then your concept of what constitutes "constant time" will
    inevitably change.

    An *arbitrary* spacelike surface can have curvature anywhere, not just
    at the particles.

    It is true that an isolated particle has a (local) rest frame, and that
    the surfaces of constant time in that frame are (locally) cones, but for
    a collection of particles there is not in general a well defined "center
    of mass" frame, nor is there always a spacelike slice that is flat
    except at the particles.

    There is no problem defining the group valued momentum for a particle,
    as long as you specify both a loop, a base point and a coordinate frame
    at the base point. You also need to be careful defining the "ordinary"
    velocity. A reference frame is not enough. You need a reference frame at
    a some chosen point, *and* a path from that point to the particle.

    There is still the question of how the distribution of particles
    interacts with the topology of the whole 3D space. I don't know about
    that, the classification of manifolds is a *bit* harder in 3D than in 2D.

    Ralph Hartley
     
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