# Different way to demonstrate the equivalence of Nambu-Goto to Polyakov action?

The Nambu-Goto Action and Polyakov action are \begin{eqnarray} S_{NG} &=& -\frac{T}{c}\int d^2 \xi \sqrt{-h} \\ S_{poly} &=& -\frac{1}{2}\frac{T}{c} \int d^2\xi \sqrt{-h} h^{ab} g_{\mu\nu}\partial_a x^\mu \partial_b x^\nu \end{eqnarray} where $h_{ab}$ is the string metric and $h_{ab}=g_{\mu\nu}\partial_ax^\mu\partial_bx^\nu$ where $g_{\mu\nu}$ is the metric in the target space.

There are a number of posts about equivalence of the Nambu-Goto and Polyakov actions. I was wondering about this derivation:

\begin{eqnarray} S_{NG} &=& -\frac{T}{c}\int d^2 \xi \sqrt{-h} \\ &=& -\frac{T}{c}\int d^2 \xi \sqrt{-h} \frac{2}{2} \\ &=& -\frac{1}{2}\frac{T}{c}\int d^2 \xi \sqrt{-h} \delta^a\,_a \\ &=& -\frac{1}{2}\frac{T}{c}\int d^2 \xi \sqrt{-h} h^{ab}h_{ab} \\ &=& -\frac{1}{2}\frac{T}{c} \int d^2\xi \sqrt{-h} h^{ab} g_{\mu\nu}\partial_a x^\mu \partial_b x^\nu = S_{poly} \end{eqnarray}

Is there a reason that this derivation would not work? Does it have something to do with the fields in the actions (one field in NG, but two fields in Polyakov)?

(It seems simple, but I've never seen it done anywhere, so I suspect there is some reason that I can't do this.)