On page 116 of this book it is said, that reparameterization invariance of the string action is analogous to the gauge invariance in electrodynamices.

Whereas Maxwell's equations are symmetric under a gauge transformation which allows to describe the same $\vec{E}$ and $\vec{B}$ fields by different potentials $A_{\mu}$, reparameterization invariance makes it possible to use different grids on the world sheet to describe the same physical motion of a string.

So is reparameterization invariance some kind of gauge invariance too? What would take the role of the gauge potential $A_{\mu}$ and what would the corresponding "gauge invariant" action look like?

  • $\begingroup$ Please tell me in a comment if I am again reading way too much into a trivial issue here, then I will remove the question ... $\endgroup$ – Dilaton Aug 18 '13 at 19:52
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    $\begingroup$ Or possibly it's more closely analogous to coordinate-independence (diffeomorphism invariance) in GR, which is the gauge symmetry of GR...? $\endgroup$ – Ben Crowell Aug 18 '13 at 19:57
  • $\begingroup$ @BenCrowell that sounds plausible somehow, I dont know which is why I am asking so stupidly ;-) $\endgroup$ – Dilaton Aug 18 '13 at 20:01
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    $\begingroup$ Related: physics.stackexchange.com/q/4359/2451 , physics.stackexchange.com/q/12461/2451 , physics.stackexchange.com/q/46324/2451 and links therein. $\endgroup$ – Qmechanic Aug 18 '13 at 20:03
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    $\begingroup$ @Dilaton I don't see why you would need to delete the question. Future users might be similarly terminologically confused, and I think these comments could be helpful. $\endgroup$ – joshphysics Aug 18 '13 at 20:13

Reparametrization invariance here means worldsheet diffeomorphism invariance, which is a gauge symmetry in the sum over string histories formulation of perturbative string theory. It acts non-trivially on the worldsheet metric and on the sigma model fields, and tells you that most things you can build using them aren't valid observables. The relevant 'gauge theory' is 2d gravity, which is why QMechanic gave you those links.

  • $\begingroup$ Thanks for this nice clarification :-). I like the links Qmechanic has given to me, but as he first called them "possible duplicates" I felt very dumb because I did not even see why ... $\endgroup$ – Dilaton Aug 19 '13 at 9:17

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