# First order formalism of Polyakov action

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• phoenix95
In summary: The cotangent bundle is the phase-space where Hamiltonian evolution takes place. The former is the space of histories, the latter is space of initial conditions and conserved quantities. The objects on which the two different functions (Lagrangian and Hamiltonian) act are also different. The Hamiltonian acts on functions (real-valued differentiable functions) on the cotangent bundle. It is exactly the cotangent vector field that you read off from the Hamiltonian 1-form on the cotangent bundle. The Lagrangian, on the other hand, is a function on the tangent bundle. It is the vertical lift of the Lagrangian function on the configuration space to a function on T(M). The two functions are related by a Legend
phoenix95
Gold Member
In the notes of Arutyunov, he writes down the equation of Polyakov action in what he calls a first-order formalism(equation 3.19). But here I did not understand how he got this equation. Can someone help?

Moreover, can someone explain how he got the constraints in equation 3.25? And why they are not individually equal to zero?

In the thesis

https://www.rug.nl/research/portal/...ed(fb063f36-42dc-4529-a070-9c801238689a).html

the Nambu-goto action for the string is treated at page 54 and beyond. Worldsheet indices are indicated by bars, and the constraints in your (3.25) are also treated there. These constraints are, as far as I understand, indeed equal to zero.

My Hamiltonian formalism is a bit rusty now, but I'd suggest you take another look at this topic. I found it rather confusing in the beginning.

phoenix95
phoenix95 said:
In the notes of Arutyunov, he writes down the equation of Polyakov action in what he calls a first-order formalism(equation 3.19). But here I did not understand how he got this equation. Can someone help?
He probably guessed it by using experience with simpler systems. The ultimate proof that the guess is correct is deriving the equations of motion and showing that they are equivalent to the standard ones.

phoenix95
The Hamiltonian is introduced in the calculus of variations as part of breaking up a 2nd order ode into a system of first order ode's, and of course one knows the Hamiltonian is related to the Lagrangian by a Legendre transform, ##L = p \dot{x} - H##, where now the derivatives are at most of first order hence the first order formalism, but since the Hamiltonian of the NG action is zero we have ##L = p \dot{x}##, however we also have two constraints [equations (3.8) and (3.9) of Arutyunov], so one can factor the constraints into the problem with Lagrange multipliers, which is just (3.19), where you can see (3.8) and (3.9) are just scaled by strangely written Lagrange multipliers. Compare also equation (2.1.32) of Kaku's Intro to Superstrings book (see chapter 1 for the point particle analogue of this procedure) and (3.14) of Townsend

http://www.damtp.cam.ac.uk/user/examples/3P6.pdf

Interestingly, eliminating the momentum in the first order action gives you the Polyakov action (see sec 3.3.1 of the last set of notes).

(3.25) is just explicitly stating that (3.8) and (3.9) are constraints.

phoenix95, haushofer and Demystifier
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 $T^{*}(M)$.

phoenix95

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