- #1
buddychimp
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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)
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)