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Residual conformal symmetry in GSW

  1. Mar 29, 2004 #1
    Hello,

    I hope you don't mind this elementary question, but I got stuck with
    it this evening:

    GSW state in their equ. (2.1.44) that a combined reparametrisation and
    a Weyl scaling obeying

    d^aA^b + d^bA^a = Gamma n^ab

    don't change the choosen gauge h^ab=n^ab. Now I tried to proof this.
    The metric varies according to

    delta n^ab = Gamma n^ab

    and this should be zero. As n^ab is definitly not zero, Gamma = 0.
    Then d^aA^b = - d^bA^a, which automatically implies eq. (2.1.19) to
    vanish too. But then, in the coordinates sigma^+ and sigma^-,

    d/d^sigma^+ A^+ = 0 = d/d^sigma^- A^-

    Thus the statement that A^+=A^+(sigma^+) and similarily for A^- is
    wrong, though I know is it right from other sources. So what is wrong
    with my line of reasoning?

    Hope someone can find some time and help me.

    René.

    --
    René Meyer
    Student of Physics & Mathematics
    Zhejiang University, Hangzhou, China
     
  2. jcsd
  3. Mar 29, 2004 #2
    On Mon, 29 Mar 2004, Rene Meyer wrote:

    > GSW state in their equ. (2.1.44) that a combined reparametrisation and
    > a Weyl scaling obeying
    >
    > d^aA^b + d^bA^a = Gamma n^ab
    >
    > don't change the choosen gauge h^ab=n^ab. Now I tried to proof this...


    Dear Rene,

    I am afraid that there are too many errors in your reasoning, and I won't
    comment on your argument because already your starting point is
    incorrect, and therefore all other steps are incorrect, too.

    The first fact that you don't seem to appreciate (or know) is that under a
    small diffeomorphism given by an infinitesimal vector field A^a, the
    metric changes by

    delta h^{ab} = d^a A^b + d^b A^a (###)

    where "d" is the covariant derivative (one can also lower the indices if
    one changes the sign). You seem to say that it (or everything) changes
    either by zero or by a multiple of itself, which suggests that you are not
    aware of (###). It is not hard to prove (###), for example by
    Taylor-expanding the general formula for the transformation of a tensor.

    h^{ab} (x) = h^{a'b'} (x(x')) . M^a_a' . M^b_b.

    wbere M^a_a' is the matrix of derivatives of x^a with respect to x'^a' -
    in this case you want to choose x'^a = x^a + A^a (where A^a is
    infinitesimal).

    Let me just motivate (###). It has the right structure of indices; it is
    dimensionless as the metric should be (A^a cancels the derivative) and it
    uses covariant derivatives. And it is ab-symmetric. And if "A^2" depends
    linearly on "x^2", for example, you know that you are "stretching" the
    coordinate "x^2", and therefore "h^{22}" will change by a constant, which
    is exactly what (###) says.

    The tensor field d^a A^b + d^b A^a is a general tensor field, but if you
    choose the vector field A^a appropriately, it may equal a multiple of the
    metric that you started with (at each point of your manifold). In this
    case you can "undo" the change of the metric by making a Weyl
    transformation. Under a Weyl transformation, the metric changes by a
    multiple of itself, i.e.

    delta h^{ab} = epsilon . h^{ab} ($$$)

    The total variation of the metric is the sum of the (###) piece and the
    ($$$) piece, and for a special choice of A^a and epsilon, this total
    variation can vanish.

    Best wishes
    Lubos
    ______________________________________________________________________________
    E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
    eFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call)
    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
     
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