# Polyakov From Nambu-Goto Directly, for Strings?

The following derivation, for a classical relativistic point particle, of the 'Polyakov' form of the action from the 'Nambu-Goto' form of the action, without any tricks - no equations of motion or Lagrange multipliers just a direct set of equalities, is as follows:

$$S = - m \int ds = - m \int \sqrt{-g_{\mu \nu} \dot{X}^{\mu} \dot{X}^{\nu}} d \tau = - m \int \sqrt{- \dot{X}^2}d \tau = \frac{-m}{2} \int \frac{-2\dot{X}^2}{\sqrt{-\dot{X}^2}}d \tau \\ = \frac{1}{2} \int \frac{\dot{X}^2 + \dot{X}^2}{\sqrt{-\dot{X}^2/m^2}}d \tau = \frac{1}{2} \int \frac{\dot{X}^2 - m^2(-\dot{X}^2/m^2)}{\sqrt{-\dot{X}^2/m^2}}d \tau = \frac{1}{2} \int (e^{-1} \dot{X}^2 - e m^2)d \tau$$

Apart from randomly adding $\frac{m^2}{m^2}$ to only one of the $\dot{X}^2$ terms in the second last equality (can anybody explain this without referring to the EOM or LM's?), this derivation is completely straightforward.

Can a similarly straightforward derivation of the Polyakov string action from the Nambu-Goto string action be given, without knowing the Polyakov action in advance?

The best hope comes from reversing the last line of this wikipedia calculation: but it's so random, unmotivated and unexplained I cannot see it as obvious to do such a calculation. I can loosely motivate adding $\frac{h^{ab}G_{ab}}{h^{cd}G_{cd}}$ by noting $\sqrt{-G}$ is like the general relativity volume element telling us to add in $1 =$ stuff built from what's under the square root over itself, but that's it, the $2$'s are quite random too...

[This is nice but (maybe I'm wrong) I see it as too distinct from what I'm asking].

I) OP is asking for a direct/forward derivation from the Nambu-Goto (NG) action to the Polyakov (P) action (as opposed to the opposite derivation). This is non-trivial since the Polyakov action contains the world-sheet (WS) metric $$h_{\alpha\beta}$$ with 3 more variables as compared to the Nambu-Goto action.

Although we currently do not have a natural forward derivation of all 3 new variables, we have for 2 of the 3 variables, see section IV below.

II) Let us first say a few words about the derivation of the relativistic point particle,

$$L~:=~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2}\tag{1}$$

from the square root Lagrangian

$$L_0~:=~-m\sqrt{-\dot{x}^2}.\tag{2}$$

Note that OP's derivation does not explain/illuminate the fact that the einbein/Lagrange multiplier

$$e~>~0\tag{3}$$

can be taken as an independent variable, and not just a trivial renaming of the quantity $$\frac{1}{m}\sqrt{-\dot{x}^2}>0$$. It is an important property of the Lagrangian (1) that we can vary the einbein/Lagrange multiplier (3) independently. OP's request to not use Lagrange multipliers seems misguided, and we will not follow this instruction.

III) It is possible to directly/forwardly/naturally derive the Lagrangian (1) with its Lagrange multiplier $$e$$ from the square root Lagrangian (2) as follows:

1. Derive the Hamiltonian version of the square root Lagrangian (2) via a (singular) Legendre transformation. This is a straightforward application of the unique Dirac-Bergmann recipe. This leads to momentum variables $$p_{\mu}$$ and one constraint with corresponding Lagrange multiplier $$e$$. The constraint reflects world-line reparametrization invariance of the square root action (1). The Hamiltonian $$H$$ becomes of the form 'Lagrange multiplier times constraint': $$H~=~\frac{e}{2}(p^2+m^2).\tag{4}$$ See also e.g. this & this Phys.SE posts.

2. The corresponding Hamiltonian Lagrangian reads $$L_H~=~p \cdot \dot{x} - H ~=~p \cdot \dot{x} - \frac{e}{2}(p^2+m^2). \tag{5}$$

3. If we integrate out the momentum $$p_{\mu}$$ again (but keep the Lagrange multiplier $$e$$), the Hamiltonian Lagrangian density (5) becomes the sought-for Lagrangian (1). $$\Box$$

IV) The argument for the string is similar.

1. Start with the NG Lagrangian density $${\cal L}_{NG}~:=~-T_0\sqrt{{\cal L}_{(1)}}, \tag{6}$$ $${\cal L}_{(1)}~:=~-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} X\right)_{\alpha\beta} ~=~(\dot{X}\cdot X^{\prime})^2-\dot{X}^2(X^{\prime})^2~\geq~ 0, \tag{7}$$

2. Derive the Hamiltonian version of the NG string via a (singular) Legendre transformation. This leads to momentum variables $$P_{\mu}$$ and two constraints with corresponding two Lagrange multipliers, $$\lambda^0$$ and $$\lambda^1$$, cf. my Phys.SE answer here. The two constraints reflect WS reparametrization invariance of the NG action (6).

3. If we integrate out the momenta $$P_{\mu}$$ again (but keep the two Lagrange multipliers, $$\lambda^0$$ and $$\lambda^1$$), the Hamiltonian Lagrangian density for the NG string becomes $${\cal L}~=~T_0\frac{\left(\dot{X}-\lambda^0 X^{\prime}\right)^2}{2\lambda^1} -\frac{T_0\lambda^1}{2}(X^{\prime})^2,\tag{8}$$ cf. my Phys.SE answer here.

4. [As a check, if we integrate out the two Lagrange multipliers, $$\lambda^0$$ and $$\lambda^1$$, with the additional assumption that $$\lambda^1~>~0\tag{9}$$ to avoid a negative square root branch, we unsurprisingly get back the original NG Lagrangian density (6).]

5. Eq. (8) is as far as our forward derivation goes. It can be viewed as the analogue of our derivation for the relativistic point particle in section III.

6. Now we will cheat and work backwards from the Polyakov Lagrangian density

$${\cal L}_P~=~-\frac{T_0}{2} \sqrt{-h} h^{\alpha\beta} \partial_{\alpha}X \cdot\partial_{\beta}X ~=~\frac{T_0}{2} \left\{\frac{\left(h_{\sigma\sigma}\dot{X}- h_{\tau\sigma}X^{\prime}\right)^2}{\sqrt{-h}h_{\sigma\sigma}} - \frac{ \sqrt{-h}}{h_{\sigma\sigma}}(X^{\prime})^2 \right\} . \tag{10}$$

1. By classical Weyl symmetry, only 2 out of the 3 degrees of freedom in the WS metric $$h_{\alpha\beta}$$ enter the Polyakov Lagrangian density (10). If we identify $$\lambda^0~=~\frac{h_{\tau\sigma}}{h_{\sigma\sigma}}\quad\text{and} \quad\lambda^1~=~\frac{\sqrt{-h}}{h_{\sigma\sigma}}~>~0, \tag{11}$$ then the Lagrangian (8) becomes the Polyakov Lagrangian density (10). $$\Box$$
• While I like your answer and think the Dirac-Bergmann approach is really the correct one, I'm afraid this won't satisfy OP since they demand a derivation "without any tricks - no equations of motion or Lagrange multipliers just a direct set of equalities" [emphasis mine]. – ACuriousMind Feb 26 '16 at 16:58
• The Dirac-Bergmann recipe is not a trick. It is possible to artificially introduce the two Lagrange multipliers via their on-shell value, but it would not infuse any additional clarification or further naturalness into the derivation. In fact, it would just make the derivation look more contrived. It is more natural to rely on the Lagrange multipliers, which are directly dictated by the Dirac-Bergmann recipe. – Qmechanic Feb 26 '16 at 17:15
• I updated the answer. – Qmechanic Feb 26 '16 at 19:33

Given $$S = - T \int d \tau d \sigma \sqrt{- h}$$ we have \begin{align} \delta S &= - T \delta \int d \tau d \sigma \sqrt{-h} = - T \int d \tau d \sigma \delta \sqrt{-h} \\ &= - \frac{T}{2} \int d \tau d \sigma \sqrt{-h}h^{ab} \delta h_{ab} = - \frac{T}{2} \int d \tau d \sigma \sqrt{-h}h^{ab} \delta (\partial_a X^{\mu} \partial_b X_{\mu}) \\ &= - T \int d \tau d \sigma \sqrt{-h}h^{ab}( \partial_a \delta X^{\mu} )\partial_b X_{\mu} = - T \int d \tau d \sigma \partial_a ( \sqrt{-h}h^{ab} \partial_b X_{\mu})\delta X^{\mu} + 0 \\ &= - T \int d \tau d \sigma \partial_a ( \sqrt{-h}h^{ab} \partial_b X_{\mu})\delta X^{\mu} \\ &= 0 \end{align} so that, for a metric $h$ not depending on $x$, we have $$\dfrac{\partial S}{\partial x^{\mu}} = - T \partial_a( \sqrt{-h}h^{ab} \partial_b X_{\mu}) = 0 \rightarrow S = - \frac{T}{2}\int d \tau d \sigma ( \sqrt{-h}h^{ab} \partial_a X^{\mu} \partial_b X_{\mu})$$

• I often see equations of motion for the NG action that contain a matter term come with a $1/\sqrt{-h}$ term in front of the outermost $\partial_a$. (e.g. the NG equation of motion is $\frac{1}{\sqrt{-h}}\partial_a(\sqrt{-h}h^{ab}\partial_b X_\mu) + \text{matter terms} =0$). Without a matter term, it seems like we can just factor the $1/\sqrt{-h}$ out, but I would like to understand better where this comes from. Thanks for any insight. – Bob Feb 8 '17 at 21:47