Editing Polyakov action (section)

== Relation with Nambu–Goto action ==

Writing the [[Euler–Lagrange equation]] for the [[metric tensor]] <math> h^{ab} </math> one obtains that
: <math> \frac{\delta S}{\delta h^{ab}} = T_{ab} = 0. </math>
Knowing also that:
: <math> \delta \sqrt{-h} = -\frac12 \sqrt{-h} h_{ab} \delta h^{ab}. </math>
One can write the variational derivative of the action:
: <math> \frac{\delta S}{\delta h^{ab}} = \frac{T}{2} \sqrt{-h} \left( G_{ab} - \frac12 h_{ab} h^{cd} G_{cd} \right), </math>
where <math> G_{ab} = g_{\mu \nu} \partial_a X^\mu \partial_b X^\nu </math>, which leads to
: <math>\begin{align}
  T_{ab} &= T \left( G_{ab} - \frac12 h_{ab} h^{cd} G_{cd} \right) = 0, \\
  G_{ab} &= \frac12 h_{ab} h^{cd} G_{cd}, \\
       G &= \operatorname{det} \left( G_{ab} \right) = \frac14 h \left( h^{cd} G_{cd} \right)^2.
\end{align}</math>

If the auxiliary [[worldsheet]] [[metric tensor]] <math>\sqrt{-h}</math> is calculated from the equations of motion:
: <math> \sqrt{-h} = \frac{2 \sqrt{-G}}{h^{cd} G_{cd}} </math>
and substituted back to the action, it becomes the [[Nambu–Goto action]]:
: <math> S = {T \over 2}\int \mathrm{d}^2 \sigma  \sqrt{-h} h^{ab} G_{ab} = {T \over 2}\int \mathrm{d}^2 \sigma \frac{2 \sqrt{-G}}{h^{cd} G_{cd}} h^{ab} G_{ab} = T \int \mathrm{d}^2 \sigma \sqrt{-G}.</math>

However, the Polyakov action is more easily [[quantization (physics)|quantized]] because it is [[linear]].