phoenix95 said:
Polyakov action in what he calls a first-order
Let us set the string tension T = 1 and write the following matrix form of the Plyakov action
S[X , \gamma] = -\frac{1}{2} \int d^{2}\sigma \ \sqrt{- \gamma} \left( \dot{X} , X^{\prime}\right)^{\mu} \begin{pmatrix} \gamma^{00} & \gamma^{01} \\ \gamma^{10} & \gamma^{11} \end{pmatrix} \begin{pmatrix} \dot{X} \\ X^{\prime}\end{pmatrix}_{\mu} . \ \ \ (1) Next, we define the following symmetric 2 \times 2 matrix M^{\alpha \beta} = \sqrt{- \gamma} \gamma^{\alpha \beta}. Since the determinant of the inverse world-sheet metric \mbox{det}(\gamma^{\alpha \beta}) = \frac{1}{\mbox{det}(\gamma_{\alpha \beta})} \equiv \frac{1}{\gamma}, it follows that \mbox{det}(M) = (- \gamma) \frac{1}{\gamma} = -1. And, since M^{\alpha \beta} = M^{\beta \alpha}, we can parametrize M by two non-zero numbers (\lambda_{1} , \lambda_{2}) and write \sqrt{-\gamma} \begin{pmatrix} \gamma^{00} & \gamma^{01} \\ \gamma^{10} & \gamma^{11} \end{pmatrix} = \frac{1}{\lambda_{1}} \begin{pmatrix} -1 & \lambda_{2} \\ \lambda_{2} & \lambda_{1}^{2} - \lambda_{2}^{2} \end{pmatrix} . \ \ \ \ (2) From (2) you read off \lambda_{1} = - \frac{1}{\sqrt{-\gamma} \gamma^{00}}, \ \ \ \lambda_{2} = - \frac{\gamma^{01}}{\gamma^{00}} . \ \ \ \ \ \ \ (3) Now, we substitute (2) in the Polyakov action (1) and do the simple algebra to obtain [In the followings, I will supress the world indices on the derivatives of X. This should not cause any confusion because they are contracted]:
S[X;\lambda_{1},\lambda_{2}] = \frac{1}{2} \int d^{2}\sigma \left\{ \lambda_{1}^{-1} \dot{X} \cdot \left( \dot{X} - \lambda_{2} X^{\prime}\right) - \lambda_{2} \lambda_{1}^{-1} X^{\prime} \cdot \left( \dot{X} - \lambda_{2} X^{\prime}\right) - \lambda_{1} X^{\prime} \cdot X^{\prime} \right\} .
Now, we define the following new variable P_{\mu} = \lambda_{1}^{-1} \left( \dot{X}_{\mu} - \lambda_{2}X_{\mu}^{\prime}\right) . Substituting P in S[X;\lambda_{1},\lambda_{2}], we find S[X,P; \lambda_{1},\lambda_{2}] = \int d^{2}\sigma \left\{ \frac{1}{2} \left( \dot{X} - \lambda_{2} X^{\prime}\right) \cdot P - \frac{1}{2} \lambda_{1} (X^{\prime} \cdot X^{\prime}) \right\} . This can be rewritten as
S = \int d^{2}\sigma \left\{ \lambda_{1} P \cdot P - \frac{1}{2}\lambda_{1} ( P \cdot P - X^{\prime} \cdot X^{\prime})\right\} . Finally, for one of the P’s in the first term, we substitute \lambda_{1}P = \dot{X} - \lambda_{2}X^{\prime} to obtain
S[X,P;\lambda_{1},\lambda_{2}] = \int d^{2}\sigma \left\{ P \cdot \dot{X} - \frac{\lambda_{1}}{2}(P \cdot P - X^{\prime} \cdot X^{\prime}) - \lambda_{2} (P \cdot X^{\prime})\right\} . \ \ (4) We recognise this as the phase-space NG-action incorporating the first-class constraints (P)^{2} - (X^{\prime})^{2} \approx 0, P \cdot X^{\prime} \approx 0 through the Lagrange multipliers (\lambda_{1},\lambda_{2}), and the fact that the Hamiltonian vanishes on the constraint surface, H \approx 0. The phase-space Polyakov-action is obtained by substituting (3) in (4): S[X,P; \gamma ] = \int d^{2}\sigma \left\{ P \cdot \dot{X} + \frac{P \cdot P - X^{\prime} \cdot X^{\prime}}{2 \sqrt{- \gamma} \gamma^{00}} + \frac{\gamma^{01}(P \cdot X^{\prime})}{\gamma^{00}} \right\} .
It is important to know that the integrand in a phase-space action S[q,p] is not a Lagrangian. Lagrangians are functions of the tangent bundle coordinates (q , \dot{q}) \in T(M), whereas (q,p) are local coordinates on the cotangent bundle T^{*}(M).