This is an elementary question in string theory. In Polyakov action, it is often explained that worldsheet metric is independent dynamic variable and target manifold often (though it does not have to be) is a Minkowski one.

Intuitively, we think of strings in worldsheet sweeping through the submanifold of the target spacetime - and that means that once target manifold is set, then what we actually describe should follow the metric tensor of the target manifold.

So what does metric tensor of worldsheet actually do to change our descriptions of physics? Is it that it contributes additional metric tensor term to the background manifold metric tensor?

Or does this mean that metric tensor in worldsheet "fluctuates" relative to target manifold's metric tensor, or what?


1 Answer 1


The metric on the worldsheet is an auxiliary variable, it has no direct physical meaning and it has nothing to do with the metric on the target space.

You can see this by observing that the starting point for the classical string, the Nambu-Goto action $$ S_\text{NG}[X] = -T \int_\Sigma \sqrt{-\det\left(\frac{\partial X^\mu}{\partial \sigma^a}\frac{\partial X_\mu}{\partial \sigma^b}\right)\mathrm{d}^2\sigma},$$ does not involve the worldsheet metric, only the target space metric (hidden in the contraction of the Greek indices). It is only when we pass to the Polyakov action $$ S_\text{P}[X,h] = -\frac{T}{2}\int_\Sigma \sqrt{-\det{h}} h^{ab} \frac{\partial X^\mu}{\partial \sigma^a}\frac{\partial X_\mu}{\partial \sigma^b}\mathrm{d}^2\sigma$$ by artificially enlarging the space of dynamical variables that a worldsheet metric $h$ appears. $h$ is a gauge degree of freedom, it can be eliminated again - by solving the equations of motion for $h$ and plugging in - to return to the Nambu-Goto action that knows nothing of it. The only reason we introduce it is that the Nambu-Goto action is difficult to deal with due to the square root, and quantization of the Polyakov action, in constract, is simpler and well-understood.

  • $\begingroup$ I can't remember where I read it, but I recall that it was said there is a deeper, more technical reason that has nothing to do with the square root for preferring the Polyakov action - do you have any idea? $\endgroup$
    – JamalS
    Oct 3, 2017 at 11:49

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