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Conformal invariance / reparametrization invariance

  1. Aug 13, 2009 #1
    Hi! I have little questions about symmetries. I begin in the field, so...

    First about conformal symmetry. As I studied, in 2-d, a transformation [tex](\tau, \sigma) \to (\tau', \sigma')[/tex] changing the metric by a multiplicative factor implies that the transformation [tex](\tau, \sigma) \to (\tau', \sigma')[/tex] satisfies Cauchy-Riemann equations : [tex]\partial_\tau \tau' (\tau, \sigma) = \partial_\sigma \sigma'(\tau, \sigma)[/tex] and [tex]\partial_\sigma \tau' (\tau, \sigma) = - \partial_\tau \sigma'(\tau, \sigma)[/tex]. Under such a transformation [tex](\tau, \sigma) \to (\tau', \sigma')[/tex], one can verify that the Polyakov action remains unchanged and we say the action is conformally invariant. (Correct?)

    What is not clear to me is the following. We also have reparametrization invariance. But I would be tempted to say that reparametrization implies conformal symmetry since it seems to be more general: we still start from a transformation [tex](\tau, \sigma) \to (\tau', \sigma')[/tex], but without the constraints [tex]\partial_\tau \tau' (\tau, \sigma) = \partial_\sigma \sigma'(\tau, \sigma)[/tex]. I'm wrong somewhere, but I can't figure out where.

    Thanks for your help.

    (I have created a new topic for clarity)
     
  2. jcsd
  3. Aug 14, 2009 #2
    A mapping is said to be conformal if it locally preserves angles. So the derivatives of the new parameters must be orthogonal and of the same length. It must be orthogonal since the derivatives of the old parameters are naturally orthogonal ( (1,0), (0,1) ), and of the same length because every vector need to preserves angles. This is only ensured when the Cauchy-Riemann equations are satisfied.

    If you have an invariance, but the Cauchy-Riemann equations for your transformation are not satisfied, it's not a conformal symmetry.
     
  4. Aug 14, 2009 #3
    You are right, but I think the appearance of conformal symmetries comes because of one subtlety.
    The gauge fixing of the worldsheet metric to an euclidean metric using reparametrization and weyl symmetries leaves behind some more residual symmetries (which are roughly transformation of one being canceled by the other) which actually turns out to be the conformal symmetries of the worldsheet metric.
    So roughly, conformal symmetry "=" reparametrization "-" weyl symmetry
     
  5. Aug 15, 2009 #4
    Thanks for the answers, this is more clear. But...

    I run into problem checking conformal invariance. I consider the Polyakov action with gauge fixing [tex]h=\text{diag}(1,-1)[/tex]. We obtain (up to a constant factor)

    [tex]S_P=\int d\tau d\sigma \left( \frac{\partial X^\mu}{\partial \tau}
    \frac{\partial X_\mu}{\partial \tau} - \frac{\partial X^\mu}{\partial \sigma}
    \frac{\partial X_\mu}{\partial \sigma} \right) [/tex]

    where [tex]X^\mu[/tex] are the coordinate functions of the worldsheet.

    Now, if we consider a conformal transformation, i.e. a map [tex](\tau, \sigma) \mapsto (\tau', \sigma')[/tex] satisfying [tex]\partial_\tau \tau' = \partial_\sigma \sigma'[/tex] and [tex]\partial_\sigma \tau' = - \partial_\tau \sigma'[/tex],

    we obtain

    [tex]d\tau' d\sigma' = \left( \frac{\partial \tau'}{\partial \tau}
    \frac{\partial \sigma'}{\partial \sigma} - \frac{\partial \tau'}{\partial \sigma}
    \frac{\partial \sigma'}{\partial \tau} \right) d\tau d\sigma = \Big\lbrack \left(\frac{\partial \tau'}{\partial\tau}\right)^2 - \left(\frac{\partial \tau'}{\partial\sigma}\right)^2 \Big\rbrack d\tau d\sigma [/tex]

    Now the action becomes

    [tex]S_P = \int d\tau d\sigma \Big\lbrack \left( \frac{\partial X^\mu }{\partial \tau'}\frac{\partial \tau' }{\partial \tau} + \frac{\partial X^\mu }{\partial \sigma'}\frac{\partial \sigma' }{\partial \tau} \right) \left( \frac{\partial X_\mu }{\partial \tau'}\frac{\partial \tau' }{\partial \tau} + \frac{\partial X_\mu }{\partial \sigma'}\frac{\partial \sigma' }{\partial \tau} \right) - \left( \frac{\partial X^\mu }{\partial \tau'}\frac{\partial \tau' }{\partial \sigma} + \frac{\partial X^\mu }{\partial \sigma'}\frac{\partial \sigma' }{\partial \sigma} \right) \left( \frac{\partial X_\mu }{\partial \tau'}\frac{\partial \tau' }{\partial \sigma} + \frac{\partial X_\mu }{\partial \sigma'}\frac{\partial \sigma' }{\partial \sigma} \right) \Big\rbrack[/tex]

    and after some computations, I almost get it right. I obtain :

    [tex]S_P = A + B[/tex]

    with

    [tex]A = \int d\tau d\sigma \left( \frac{\partial X^\mu}{\partial \tau'}
    \frac{\partial X_\mu}{\partial \tau'} - \frac{\partial X^\mu}{\partial \sigma'}
    \frac{\partial X_\mu}{\partial \sigma'} \right) \Big\lbrack \left(\frac{\partial \tau'}{\partial\tau}\right)^2 - \left(\frac{\partial \tau'}{\partial\sigma}\right)^2 \Big\rbrack = S'_P[/tex]

    and unfortunatly a extra non-vanishing term

    [tex]B= 4 \int d\tau d\sigma \frac{\partial \tau'}{\partial \tau} \frac{\partial \sigma'}{\partial \tau} \frac{\partial X^\mu}{\partial \tau'} \frac{\partial X_\mu}{\partial \sigma'}[/tex]

    Any idea to get rid off this B term? I have double checked, triple checked my computations, it's still there. I would really appreciate some help, I thought it would be an easy check... and I'm stuck on this for a while now.
     
  6. Aug 15, 2009 #5
    I think I get it. I forgot that we here consider (orientation preserving) conformal transformations with respect to a Minkowsky metric, which leads to the equations [tex]\partial_\tau \tau' = \partial_\sigma \sigma'[/tex] and [tex]\partial_\sigma \tau' = \partial_\tau \sigma'[/tex] for a metric h=diag(1,-1). Cauchy-Riemann equations are obtained for the euclidean metric...
     
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