Polyakov action

Local symmetries

The action is invariant under worldsheet diffeomorphisms (or coordinates transformations) and Weyl transformations.

Diffeomorphisms

Assume the following transformation:

${\displaystyle {\sigma }^{\alpha }\to {\tilde{\sigma }}^{\alpha }\left(\sigma ,\tau \right).}$

It transforms the metric tensor in the following way:

${\displaystyle h^{ab}(\sigma )\to {\tilde{h}}^{ab}=h^{cd}(\tilde{\sigma })\frac{\mathrm{\partial }{\sigma }^a}{\mathrm{\partial }{\tilde{\sigma }}^c}\frac{\mathrm{\partial }{\sigma }^b}{\mathrm{\partial }{\tilde{\sigma }}^d}.}$

One can see that:

${\displaystyle {\tilde{h}}^{ab}\frac{\mathrm{\partial }}{\mathrm{\partial }{\sigma }^a}{X}^{\mu }(\tilde{\sigma })\frac{\mathrm{\partial }}{\mathrm{\partial }{\sigma }^b}{X}^{\nu }(\tilde{\sigma })=h^{cd}\left(\tilde{\sigma }\right)\frac{\mathrm{\partial }{\sigma }^a}{\mathrm{\partial }{\tilde{\sigma }}^c}\frac{\mathrm{\partial }{\sigma }^b}{\mathrm{\partial }{\tilde{\sigma }}^d}\frac{\mathrm{\partial }}{\mathrm{\partial }{\sigma }^a}{X}^{\mu }(\tilde{\sigma })\frac{\mathrm{\partial }}{\mathrm{\partial }{\sigma }^b}{X}^{\nu }(\tilde{\sigma })=h^{ab}\left(\tilde{\sigma }\right)\frac{\mathrm{\partial }}{\mathrm{\partial }{\tilde{\sigma }}^a}{X}^{\mu }(\tilde{\sigma })\frac{\mathrm{\partial }}{\mathrm{\partial }{\tilde{\sigma }}^b}{X}^{\nu }(\tilde{\sigma }).}$

One knows that the Jacobian of this transformation is given by

${\displaystyle \mathrm{J}=\det \left(\frac{\mathrm{\partial }{\tilde{\sigma }}^{\alpha }}{\mathrm{\partial }{\sigma }^{\beta }}\right),}$

which leads to

${\displaystyle \begin{array}{cc}{\mathrm{d}}^2\tilde{\sigma }& =\mathrm{J}{\mathrm{d}}^2\sigma \\ h& =\det \left(h_{ab}\right)\\ \Rightarrow \tilde{h}& ={\mathrm{J}}^{2}h,\end{array}}$

and one sees that

${\displaystyle \sqrt{-\tilde{h}}{\mathrm{d}}^2\sigma =\sqrt{-h\left(\tilde{\sigma }\right)}{\mathrm{d}}^2\tilde{\sigma }.}$

Summing up this transformation and relabeling ${\displaystyle \tilde{\sigma }=\sigma }$, we see that the action is invariant.

Weyl transformation

Assume the Weyl transformation:

${\displaystyle h_{ab}\to {\tilde{h}}_{ab}=\mathrm{\Lambda }(\sigma )h_{ab},}$

then

${\displaystyle \begin{array}{cc}{\tilde{h}}^{ab}& ={\mathrm{\Lambda }}^{-1}(\sigma )h^{ab},\\ \det \left({\tilde{h}}_{ab}\right)& ={\mathrm{\Lambda }}^2(\sigma )\det (h_{ab}).\end{array}}$

And finally:

And one can see that the action is invariant under Weyl transformation. If we consider n-dimensional (spatially) extended objects whose action is proportional to their worldsheet area/hyperarea, unless n = 1, the corresponding Polyakov action would contain another term breaking Weyl symmetry.

One can define the stress–energy tensor:

${\displaystyle T^{ab}=\frac{-2}{\sqrt{-h}}\frac{\delta S}{\delta h_{ab}}.}$

Let's define:

${\displaystyle {\hat{h}}_{ab}=\exp \left(\varphi (\sigma )\right)h_{ab}.}$

Because of Weyl symmetry, the action does not depend on ${\displaystyle \varphi }$:

${\displaystyle \frac{\delta S}{\delta \varphi }=\frac{\delta S}{\delta {\hat{h}}_{ab}}\frac{\delta {\hat{h}}_{ab}}{\delta \varphi }=-\frac{1}{2}\sqrt{-h}{T}_{ab}{e}^{\varphi }{h}^{ab}=-\frac{1}{2}\sqrt{-h}{T}_a^a{e}^{\varphi }=0\Rightarrow T_a^a=0,}$

where we've used the functional derivative chain rule.

Relation with Nambu–Goto action

Writing the Euler–Lagrange equation for the metric tensor ${\displaystyle h^{ab}}$ one obtains that

${\displaystyle \frac{\delta S}{\delta h^{ab}}=T_{ab}=0.}$

Knowing also that:

${\displaystyle \delta \sqrt{-h}=-\frac{1}{2}\sqrt{-h}{h}_{ab}\delta h^{ab}.}$

One can write the variational derivative of the action:

${\displaystyle \frac{\delta S}{\delta h^{ab}}=\frac{T}{2}\sqrt{-h}\left(G_{ab}-\frac{1}{2}{h}_{ab}{h}^{cd}{G}_{cd}\right),}$

where ${\displaystyle G_{ab}=g_{\mu \nu }{\mathrm{\partial }}_a{X}^{\mu }{\mathrm{\partial }}_b{X}^{\nu }}$, which leads to

${\displaystyle \begin{array}{cc}{T}_{ab}& =T\left(G_{ab}-\frac{1}{2}{h}_{ab}{h}^{cd}{G}_{cd}\right)=0,\\ G_{ab}& =\frac{1}{2}{h}_{ab}{h}^{cd}{G}_{cd},\\ G& =\det \left(G_{ab}\right)=\frac{1}{4}h{\left(h^{cd}{G}_{cd}\right)}^2.\end{array}}$

If the auxiliary worldsheet metric tensor ${\displaystyle \sqrt{-h}}$ is calculated from the equations of motion:

${\displaystyle \sqrt{-h}=\frac{2\sqrt{-G}}{h^{cd}{G}_{cd}}}$

and substituted back to the action, it becomes the Nambu–Goto action:

${\displaystyle S=\frac{T}{2}\int {\mathrm{d}}^2\sigma \sqrt{-h}{h}^{ab}{G}_{ab}=\frac{T}{2}\int {\mathrm{d}}^2\sigma \frac{2\sqrt{-G}}{h^{cd}{G}_{cd}}{h}^{ab}{G}_{ab}=T\int {\mathrm{d}}^2\sigma \sqrt{-G}.}$

However, the Polyakov action is more easily quantized because it is linear.

Equations of motion

Using diffeomorphisms and Weyl transformation, with a Minkowskian target space, one can make the physically insignificant transformation ${\displaystyle \sqrt{-h}{h}^{ab}\to {\eta }^{ab}}$, thus writing the action in the conformal gauge:

${\displaystyle \mathcal{S}=\frac{T}{2}\int {\mathrm{d}}^2\sigma \sqrt{-\eta }{\eta }^{ab}{g}_{\mu \nu }(X){\mathrm{\partial }}_a{X}^{\mu }(\sigma ){\mathrm{\partial }}_b{X}^{\nu }(\sigma )=\frac{T}{2}\int {\mathrm{d}}^2\sigma \left({\dot{X}}^2-X^{\prime 2}\right),}$

where ${\displaystyle {\eta }_{ab}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)}$.

Keeping in mind that ${\displaystyle T_{ab}=0}$ one can derive the constraints:

${\displaystyle \begin{array}{cc}{T}_{01}& =T_{10}=\dot{X}{X}^{\prime }=0,\\ T_{00}& =T_{11}=\frac{1}{2}\left({\dot{X}}^2+X^{\prime 2}\right)=0.\end{array}}$

Substituting ${\displaystyle X^{\mu }\to X^{\mu }+\delta X^{\mu }}$, one obtains

${\displaystyle \begin{array}{cc}\delta \mathcal{S}& =T\int {\mathrm{d}}^2\sigma {\eta }^{ab}{\mathrm{\partial }}_a{X}^{\mu }{\mathrm{\partial }}_b\delta X_{\mu }\\ & =-T\int {\mathrm{d}}^2\sigma {\eta }^{ab}{\mathrm{\partial }}_a{\mathrm{\partial }}_b{X}^{\mu }\delta X_{\mu }+{\left(T\int d\tau X^{\prime }\delta X\right)}_{\sigma =\pi }-{\left(T\int d\tau X^{\prime }\delta X\right)}_{\sigma =0}\\ & =0.\end{array}}$

And consequently

${\displaystyle ◻X^{\mu }={\eta }^{ab}{\mathrm{\partial }}_a{\mathrm{\partial }}_b{X}^{\mu }=0.}$

The boundary conditions to satisfy the second part of the variation of the action are as follows.

• Closed strings:

  Periodic boundary conditions: ${\displaystyle X^{\mu }(\tau ,\sigma +\pi )=X^{\mu }(\tau ,\sigma ).}$

• Open strings:
  1. Neumann boundary conditions: ${\displaystyle {\mathrm{\partial }}_{\sigma }{X}^{\mu }(\tau ,0)=0,{\mathrm{\partial }}_{\sigma }{X}^{\mu }(\tau ,\pi )=0.}$
  2. Dirichlet boundary conditions: ${\displaystyle X^{\mu }(\tau ,0)=b^{\mu },X^{\mu }(\tau ,\pi )=b^{\prime \mu }.}$

Working in light-cone coordinates ${\displaystyle {\xi }^{\pm }=\tau \pm \sigma }$, we can rewrite the equations of motion as

${\displaystyle \begin{array}{cc}{\mathrm{\partial }}_{+}{\mathrm{\partial }}_{-}{X}^{\mu }& =0,\\ ({\mathrm{\partial }}_{+}X{)}^2=({\mathrm{\partial }}_{-}X{)}^2& =0.\end{array}}$

Thus, the solution can be written as ${\displaystyle X^{\mu }=X_{+}^{\mu }({\xi }^{+})+X_{-}^{\mu }({\xi }^{-})}$, and the stress-energy tensor is now diagonal. By Fourier-expanding the solution and imposing canonical commutation relations on the coefficients, applying the second equation of motion motivates the definition of the Virasoro operators and lead to the Virasoro constraints that vanish when acting on physical states.