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

  1. Jun 22, 2006 #1
    I wrote:

    >One way to see why the products of some rotations will be boosts is as
    >follows.
    >
    >Suppose that for a pair of linearly independent vectors u and v in R^3 we
    >can find an element of O(2,1) with the following properties:
    >
    > * it preserves both u and v
    > * its determinant is -1
    > * its square is the identity
    >
    >If we can find such an element, we'll call it ref(u,v).
    >
    >Now, if we pick three linearly independent vectors u, v and w so that
    >ref(u,v), ref(u,w) and ref(v,w) all exist, we can construct the following
    >elements of SO(2,1):
    >
    > g1 = ref(u,v) ref(v,w) which preserves v
    > g2 = ref(v,w) ref(u,w) which preserves w
    > g3 = ref(u,v) ref(u,w) which preserves u
    >
    >and we have g3 = g1.g2 because ref(v,w)^2 = I.
    >
    >So if we pick a spacelike u and a timelike v and w, we'll have a boost
    >that's equal to a product of rotations.
    >
    >The one question that remains is, when can we find ref(u,v)? This is
    >trivial in O(3), but O(2,1) is trickier. Generically we can construct a
    >normal n to the plane spanned by u and v:
    >
    > n^a = g^{ab} eps_{bcd} u^c v^d
    >
    >and if it's not a null vector we can construct ref(u,v) as the projector
    >into the plane minus the projector onto n. But if n is null, I don't
    >think ref(u,v) exists.


    I think the only way the normal to the plane can be null is if the plane
    is tangent to the light cone, in which case it will contain a single null
    ray (which is itself normal to the plane), and all the other vectors in
    it will be spacelike.

    So if (but not only if) u and v are timelike, ref(u,v) will exist.

    What's the significance of this construction failing if two of the
    vectors lie on a tangent plane to the light cone? Well, if u and v are
    *not* on such a plane, then you can choose w to be almost anywhere:
    anywhere such that (u,w) and (v,w) also don't lie on tangent planes to
    the light cone. In other words, given such generic u and v, you can find
    SO(2,1) elements g_u and g_v that preserve them, *and* such that the
    eigenvector of g_u g_v lies *almost* anywhere in R^3.

    However, if u and v lie on a tangent plane to the light cone, then I
    think the eigenvector of (g_u g_v) will always lie on that same tangent
    plane, for all g_u and g_v that preserve u and v respectively.

    The upshot of this for collisions in 2+1 gravity would be that certain
    tachyon-tachyon and tachyon-luxon collisions (which yield either tachyons
    or luxons as the outgoing particle, never tardyons) involve *coplanar*
    vector-valued momenta.

    What I'm claiming amounts to the statement that for each null ray n there
    is a subgroup H(n) of SO(2,1), consisting of those elements whose
    eigenvectors are normal to n.

    So the simplest way to check this would be to look for a subalgebra of
    so(2,1) consisting of elements whose null space lies in the plane normal
    to n.

    If we take the example of n being the null ray generated by (1,1,0), then
    so(2,1) elements of the form:

    | 0 a b |
    | a 0 b |
    | b -b 0 |

    have their null space generated by (b,b,-a), which is always normal to
    (1,1,0). And this set turns out to be closed under the Lie bracket, with
    [e_a,e_b]=e_b (where by e_a I mean the element above with a=1, b=0, and
    by e_b vice versa).

    So it looks like those subgroups of H(n) do exist. But they won't be
    Abelian! So those weird coplanar collisions in 2+1 gravity still won't
    obey the old vector addition law.
     
  2. jcsd
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