Derivation of measure for summation over surfaces, including the polyakov action

In his 1981 paper "Quantum geometry of bosonic strings" Polyakov defines a measure for the summation over continuous surfaces. This measure must count all surfaces of a given area with the same weight. This means that if we have a transformation $$\Omega$$ which maps a surface $$S$$ to a surface $$\widetilde{S}$$ such that $$A(S) = A(\widetilde{S})$$ we must have for any functional \$\phi[S],

$$\int d\mu(S) \phi(S) = \int d\mu(S) \phi(\widetilde{S}).$$

He claims that this leads to the following measure (for which he has no derivation):

$$\int d\mu(S) \phi(S)$$ $$=\int[Dg_{ab}(\xi)]\exp(-\lambda\int\sqrt{g}d^2\xi)$$ $$\times \int Dx(\xi)[\exp(-\frac{1}{2}\int_D\sqrt{g}g^{ab}\partial_a x_\mu\partial_b x_\mu d^2\xi)]$$ $$\times \phi[x(\xi)],$$ where $$D$$ is a unit disk in the $$\xi$$-plane.

My question: how to derive this measure from the general considerations at the start?

Possibly related: Integration measure for Polyakov's path integral

• Are you aware of the Nambu-Goto action? Commented May 30 at 5:21
• Yeah, what about it? Commented May 30 at 7:01
• The Nambu-Goto action gives the area of the worldsheet. The Nambu-Goto path integral is exactly the measure you are looking for. But the Nambu-Goto path integral is equivalent to the Polyakov path integral. Commented May 30 at 10:58
• @ɪdɪətstrəʊlə, the Nambu-Goto action is classically(!) equivalent to the Polyakov action. But when we would write down a path integral they would be different, for Nambu-Goto we have to intgrate over all possible embeddings in the target space, while for the Polyakov action we would have to integrate over all possible embeddings in the target space as well as all possible metrics on the worldsheet. So to me it seems that as a path integral they are rather different. Could you clarify how they would be equivalent at the path integral level? Commented May 30 at 11:25