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Physics
Beyond the Standard Models
First order formalism of Polyakov action
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[QUOTE="samalkhaiat, post: 6342936, member: 35381"] Let us set the string tension [itex]T = 1[/itex] and write the following matrix form of the Plyakov action [tex]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)[/tex] Next, we define the following symmetric [itex]2 \times 2[/itex] matrix [tex]M^{\alpha \beta} = \sqrt{- \gamma} \gamma^{\alpha \beta}.[/tex] Since the determinant of the inverse world-sheet metric [tex]\mbox{det}(\gamma^{\alpha \beta}) = \frac{1}{\mbox{det}(\gamma_{\alpha \beta})} \equiv \frac{1}{\gamma},[/tex] it follows that [tex]\mbox{det}(M) = (- \gamma) \frac{1}{\gamma} = -1.[/tex] And, since [itex]M^{\alpha \beta} = M^{\beta \alpha}[/itex], we can parametrize [itex]M[/itex] by two non-zero numbers [itex](\lambda_{1} , \lambda_{2})[/itex] and write [tex]\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)[/tex] From (2) you read off [tex]\lambda_{1} = - \frac{1}{\sqrt{-\gamma} \gamma^{00}}, \ \ \ \lambda_{2} = - \frac{\gamma^{01}}{\gamma^{00}} . \ \ \ \ \ \ \ (3)[/tex] 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 [itex]X[/itex]. This should not cause any confusion because they are contracted]: [tex]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\} .[/tex] Now, we define the following new variable [tex]P_{\mu} = \lambda_{1}^{-1} \left( \dot{X}_{\mu} - \lambda_{2}X_{\mu}^{\prime}\right) .[/tex] Substituting [itex]P[/itex] in [itex]S[X;\lambda_{1},\lambda_{2}][/itex], we find [tex]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\} .[/tex] This can be rewritten as [tex]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\} .[/tex] Finally, for one of the [itex]P[/itex]’s in the first term, we substitute [itex]\lambda_{1}P = \dot{X} - \lambda_{2}X^{\prime}[/itex] to obtain [tex]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)[/tex] We recognise this as the phase-space NG-action incorporating the first-class constraints [itex](P)^{2} - (X^{\prime})^{2} \approx 0[/itex], [itex]P \cdot X^{\prime} \approx 0[/itex] through the Lagrange multipliers [itex](\lambda_{1},\lambda_{2})[/itex], and the fact that the Hamiltonian vanishes on the constraint surface, [itex]H \approx 0[/itex]. The phase-space Polyakov-action is obtained by substituting (3) in (4): [tex]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\} .[/tex] It is important to know that the integrand in a phase-space action [itex]S[q,p][/itex] is not a Lagrangian. Lagrangians are functions of the tangent bundle coordinates [itex](q , \dot{q}) \in T(M)[/itex], whereas [itex](q,p)[/itex] are local coordinates on the cotangent bundle [itex]T^{*}(M)[/itex]. [/QUOTE]
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Beyond the Standard Models
First order formalism of Polyakov action
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