The Nambu–Goto and Polyakov Actions

A relativistic particle action measures the invariant length of a worldline. A relativistic string action should measure the invariant area of a worldsheet. This gives the Nambu—Goto action, the most geometric form of the classical string action.

For quantization, however, the area action is not the most convenient starting point. Just as the point-particle square root was replaced by an einbein action, the string square root is replaced by an action with an independent worldsheet metric. This is the Polyakov action. Classically it is equivalent to the Nambu—Goto action; quantum mechanically it is the gateway to two-dimensional conformal field theory.

Conventions

We use a mostly-plus target-space metric,

$$\eta_{\mu\nu}=\operatorname{diag}(-,+,\ldots,+),$$

and worldsheet coordinates

$$\sigma^\alpha=(\tau,\sigma), \qquad \alpha=0,1.$$

We write

$$\dot X^\mu\equiv \partial_\tau X^\mu, \qquad X^{\prime\mu}\equiv \partial_\sigma X^\mu, \qquad A\cdot B\equiv \eta_{\mu\nu}A^\mu B^\nu.$$

The string tension is

$$T={1\over2\pi\alpha'},$$

so $\alpha'$ has dimensions of length squared. We keep $\alpha'$ explicit.

Open and closed worldsheets

A string moving in $D$-dimensional spacetime is described by embedding functions

$$X^\mu(\tau,\sigma), \qquad \mu=0,1,\ldots,D-1.$$

For an open string, $\sigma$ runs over an interval. A standard convention is

$$0\leq \sigma\leq \pi.$$

The worldsheet is topologically a strip; its two boundaries are the histories of the two endpoints.

For a closed string, $\sigma$ is periodic. We often use

$$X^\mu(\tau,\sigma+2\pi)=X^\mu(\tau,\sigma).$$

The worldsheet is topologically a cylinder.

An open string sweeps out a strip with boundaries. A closed string sweeps out a cylinder with periodic coordinate $\sigma$.

The coordinate range of $\sigma$ is partly conventional. The invariant object is not the coordinate rectangle, strip, or cylinder drawn on paper, but the surface embedded in spacetime.

The induced metric

The target-space metric induces a metric on the worldsheet. In a general background $G_{\mu\nu}(X)$,

$$\gamma_{\alpha\beta} =\partial_\alpha X^\mu\partial_\beta X^\nu G_{\mu\nu}(X).$$

In flat spacetime,

$$\gamma_{\alpha\beta}=\partial_\alpha X\cdot\partial_\beta X.$$

In coordinates $(\tau,\sigma)$,

$$\gamma_{\alpha\beta} = \begin{pmatrix} \dot X^2 & \dot X\cdot X' \\ \dot X\cdot X' & X'^2 \end{pmatrix}.$$

The determinant is

$$\gamma\equiv \det\gamma_{\alpha\beta} =\dot X^2X'^2-(\dot X\cdot X')^2.$$

For a physical Lorentzian worldsheet, one tangent direction is timelike and one is spacelike, so $\gamma<0$. The invariant area element is

$$dA=d\tau d\sigma\sqrt{-\gamma} =d\tau d\sigma\sqrt{(\dot X\cdot X')^2-\dot X^2X'^2}.$$

The induced metric records the inner products of the tangent vectors $\partial_\tau X^\mu$ and $\partial_\sigma X^\mu$. Its determinant gives the invariant area of the infinitesimal parallelogram.

The analogy with the particle is exact at the geometric level:

$$\text{particle:}\qquad ds=d\lambda\sqrt{-\dot X^2},$$ $$\text{string:}\qquad dA=d^2\sigma\sqrt{-\det(\partial_\alpha X\cdot\partial_\beta X)}.$$

The Nambu—Goto action

The simplest reparametrization-invariant action for a classical string is area times tension:

$$S_{\rm NG} =-T\int d^2\sigma\sqrt{-\det\gamma_{\alpha\beta}}.$$

Using $T=1/(2\pi\alpha')$,

$$S_{\rm NG} =-{1\over2\pi\alpha'} \int d\tau d\sigma\sqrt{(\dot X\cdot X')^2-\dot X^2X'^2}.$$

The sign is the Lorentzian analogue of $S=-m\int ds$ for a particle. For a static string of length $L$,

$$S=-T\int dt\,L,$$

so its energy is

$$E=TL.$$

This is why $T$ is called the string tension: it is energy per unit length.

The action is invariant under worldsheet reparametrizations

$$\sigma^\alpha\mapsto \widetilde\sigma^\alpha(\sigma),$$

because $\gamma_{\alpha\beta}$ transforms as a two-dimensional tensor and

$$\sqrt{-\widetilde\gamma}\,d^2\widetilde\sigma =\sqrt{-\gamma}\,d^2\sigma.$$

Thus the Nambu—Goto action depends only on the geometric surface, not on the coordinates used to describe it.

Particle-to-string analogy

Particle	String
Worldline $X^\mu(\lambda)$	Worldsheet $X^\mu(\tau,\sigma)$
Induced metric $\dot X^2$	Induced metric $\gamma_{\alpha\beta}=\partial_\alpha X\cdot\partial_\beta X$
Proper length $\int d\lambda\sqrt{-\dot X^2}$	Area $\int d^2\sigma\sqrt{-\det\gamma}$
Mass $m$	Tension $T=1/(2\pi\alpha')$

Static gauge and transverse oscillations

The Nambu—Goto square root is geometrically clear but difficult to quantize directly. To see the physical degrees of freedom, consider a long string stretched along the $X^1$ direction and choose the static gauge

$$X^0=\tau, \qquad X^1=\sigma, \qquad X^i=Y^i(\tau,\sigma), \qquad i=2,\ldots,D-1.$$

The $Y^i$ are transverse displacements. In this gauge,

$$\gamma_{\tau\tau}=-1+\dot Y^i\dot Y^i,$$ $$\gamma_{\sigma\sigma}=1+Y^{\prime i}Y^{\prime i},$$

and

$$\gamma_{\tau\sigma}=\dot Y^iY^{\prime i}.$$

Therefore

$$-\gamma =1+Y^{\prime i}Y^{\prime i}-\dot Y^i\dot Y^i -\left(\dot Y^i\dot Y^i\right)\left(Y^{\prime j}Y^{\prime j}\right) +\left(\dot Y^iY^{\prime i}\right)^2.$$

The last two terms are quartic in the transverse fluctuations. To quadratic order,

$$\sqrt{-\gamma} =1+{1\over2}Y^{\prime i}Y^{\prime i}-{1\over2}\dot Y^i\dot Y^i+O(Y^4).$$

Substituting into the Nambu—Goto action gives

$$S_{\rm NG} =-T\int d\tau d\sigma +{T\over2}\int d\tau d\sigma \left(\dot Y^i\dot Y^i-Y^{\prime i}Y^{\prime i}\right) +O(Y^4).$$

So the small oscillations of a long string are $D-2$ massless scalar fields on the worldsheet.

In static gauge the longitudinal directions are identified with worldsheet coordinates. The physical small oscillations are the transverse fields $Y^i(\tau,\sigma)$.

This result previews a central fact: a relativistic string does not have independent longitudinal oscillations. In covariant quantization, the Virasoro constraints remove the unphysical timelike and longitudinal excitations.

The Lüscher term

The quadratic transverse action already contains a beautiful universal quantum correction. Take an open string with endpoints fixed a distance $L$ apart. The transverse fields obey Dirichlet boundary conditions,

$$Y^i(\tau,0)=Y^i(\tau,L)=0.$$

The normal modes are

$$Y^i(\tau,\sigma) \sim \sin\left({n\pi\sigma\over L}\right)e^{-i\omega_n\tau}, \qquad \omega_n={n\pi\over L}, \qquad n=1,2,\ldots.$$

Each oscillator contributes zero-point energy $\omega_n/2$. With $D-2$ transverse fields,

$$E_0={D-2\over2}\sum_{n=1}^\infty \omega_n ={D-2\over2}{\pi\over L}\sum_{n=1}^\infty n.$$

Using zeta-function regularization,

$$\sum_{n=1}^\infty n=\zeta(-1)=-{1\over12},$$

we find

$$E(L)=TL-{\pi(D-2)\over24L}+\cdots.$$

The correction

$$-{\pi(D-2)\over24L}$$

is the Lüscher term. It is universal for a long open string with fixed endpoints because it depends only on the number of massless transverse fields.

Dirichlet transverse modes have frequencies $\omega_n=n\pi/L$. Their regulated zero-point energy gives the universal $-1/L$ correction to the long-string energy.

This semiclassical calculation should not be confused with the full quantum spectrum of the free bosonic string. But it foreshadows the normal-ordering constants that appear in exact quantization.

The Polyakov action

To remove the square root, introduce an independent worldsheet metric

$$h_{\alpha\beta}(\tau,\sigma).$$

The Polyakov action is

$$S_P[X,h] =-{T\over2}\int d^2\sigma\sqrt{-h}\, h^{\alpha\beta}\partial_\alpha X^\mu\partial_\beta X_\mu.$$

Using $T=1/(2\pi\alpha')$,

$$S_P[X,h] =-{1\over4\pi\alpha'} \int d^2\sigma\sqrt{-h}\, h^{\alpha\beta}\partial_\alpha X^\mu\partial_\beta X_\mu.$$

Here

$$h\equiv \det h_{\alpha\beta}.$$

The metric $h_{\alpha\beta}$ is not the induced metric. It is an independent field on the worldsheet. The induced metric is still

$$\gamma_{\alpha\beta}=\partial_\alpha X\cdot\partial_\beta X.$$

Thus

$$S_P[X,h] =-{T\over2}\int d^2\sigma\sqrt{-h}\, h^{\alpha\beta}\gamma_{\alpha\beta}.$$

For fixed $h_{\alpha\beta}$, the Polyakov action is quadratic in $X^\mu$. This is the key technical advantage. Once we gauge fix $h_{\alpha\beta}$, the matter theory becomes a two-dimensional field theory of free scalar fields, supplemented by constraints and eventually ghosts.

Equations of motion and the stress tensor

Varying $X^\mu$ gives the bulk equation of motion

$$\nabla_\alpha\nabla^\alpha X^\mu=0,$$

or equivalently

$${1\over\sqrt{-h}}\partial_\alpha \left(\sqrt{-h}\,h^{\alpha\beta}\partial_\beta X^\mu\right)=0.$$

For open strings, the variation also produces boundary terms. The systematic discussion of Neumann and Dirichlet boundary conditions is postponed to the next page, where conformal gauge makes them especially transparent.

The variation with respect to the metric gives the worldsheet stress tensor. Using

$$\delta\sqrt{-h} =-{1\over2}\sqrt{-h}\,h_{\alpha\beta}\delta h^{\alpha\beta},$$

we obtain

$$\delta_h S_P =-{T\over2}\int d^2\sigma\sqrt{-h} \left( \partial_\alpha X\cdot\partial_\beta X -{1\over2}h_{\alpha\beta}h^{\rho\lambda} \partial_\rho X\cdot\partial_\lambda X \right) \delta h^{\alpha\beta}.$$

Define the dimensionless matter stress tensor by

$$T^{\rm m}_{\alpha\beta} =-{2\over T\sqrt{-h}}{\delta S_P\over\delta h^{\alpha\beta}}.$$

Then

$$T^{\rm m}_{\alpha\beta} =\partial_\alpha X\cdot\partial_\beta X -{1\over2}h_{\alpha\beta}h^{\rho\lambda} \partial_\rho X\cdot\partial_\lambda X.$$

Because $h_{\alpha\beta}$ has no kinetic term in the classical Polyakov action, its equation of motion is a constraint:

$$T^{\rm m}_{\alpha\beta}=0.$$

These are the classical Virasoro constraints.

The trace vanishes identically in two dimensions:

$$h^{\alpha\beta}T^{\rm m}_{\alpha\beta}=0,$$

because $h^{\alpha\beta}h_{\alpha\beta}=2$. This classical tracelessness is the local expression of Weyl invariance.

Classical equivalence to Nambu—Goto

The Polyakov action appears to contain more fields than the Nambu—Goto action, but $h_{\alpha\beta}$ is auxiliary. The equation $T^{\rm m}_{\alpha\beta}=0$ can be written as

$$\gamma_{\alpha\beta} ={1\over2}h_{\alpha\beta}h^{\rho\lambda}\gamma_{\rho\lambda}.$$

Thus the induced metric is proportional to $h_{\alpha\beta}$:

$$\gamma_{\alpha\beta}=\Lambda(\tau,\sigma)h_{\alpha\beta}.$$

Equivalently,

$$h_{\alpha\beta}=e^{2\omega(\tau,\sigma)}\gamma_{\alpha\beta}.$$

The function $\omega$ is not fixed because the Polyakov action is classically invariant under Weyl rescalings,

$$h_{\alpha\beta}\mapsto e^{2\omega}h_{\alpha\beta}.$$

Choose the Weyl representative $h_{\alpha\beta}=\gamma_{\alpha\beta}$. Then

$$S_P[X,\gamma] =-{T\over2}\int d^2\sigma\sqrt{-\gamma}\, \gamma^{\alpha\beta}\gamma_{\alpha\beta}.$$

In two dimensions,

$$\gamma^{\alpha\beta}\gamma_{\alpha\beta}=2,$$

so

$$S_P[X,\gamma] =-T\int d^2\sigma\sqrt{-\gamma} =S_{\rm NG}[X].$$

The Polyakov metric equation sets $h_{\alpha\beta}$ equal to the induced metric up to a Weyl factor. Substituting back gives the Nambu—Goto area action.

This equivalence is classical. Quantum mechanically, the Polyakov path integral includes a sum over worldsheet metrics modulo gauge equivalences. The survival of Weyl invariance becomes a real dynamical condition, and it is what will eventually select the critical dimension.

Summary

The two classical string actions are

$$S_{\rm NG} =-{1\over2\pi\alpha'} \int d^2\sigma\sqrt{-\det(\partial_\alpha X\cdot\partial_\beta X)},$$

and

$$S_P =-{1\over4\pi\alpha'} \int d^2\sigma\sqrt{-h}\,h^{\alpha\beta} \partial_\alpha X\cdot\partial_\beta X.$$

The Nambu—Goto action is geometrically direct: it is tension times area. The Polyakov action is technically superior: it is quadratic in $X^\mu$ and has local worldsheet diffeomorphism and Weyl gauge symmetries. Varying the auxiliary metric gives $T^{\rm m}_{\alpha\beta}=0$, the classical ancestor of the Virasoro constraints.

Exercises

Exercise 1: Determinant of the induced metric

Starting from

$$\gamma_{\alpha\beta} = \begin{pmatrix} \dot X^2 & \dot X\cdot X' \\ \dot X\cdot X' & X'^2 \end{pmatrix},$$

show that the Nambu—Goto Lagrangian density in flat spacetime is

$$\mathcal L_{\rm NG} =-T\sqrt{(\dot X\cdot X')^2-\dot X^2X'^2}.$$

Solution

The determinant is

$$\gamma =\det\gamma_{\alpha\beta} =\dot X^2X'^2-(\dot X\cdot X')^2.$$

For a Lorentzian worldsheet, $\gamma<0$, so

$$-\gamma=(\dot X\cdot X')^2-\dot X^2X'^2.$$

The Nambu—Goto action is

$$S_{\rm NG}=-T\int d^2\sigma\sqrt{-\gamma},$$

hence

$$\mathcal L_{\rm NG} =-T\sqrt{(\dot X\cdot X')^2-\dot X^2X'^2}.$$

Exercise 2: Static-gauge expansion

Use

$$X^0=\tau, \qquad X^1=\sigma, \qquad X^i=Y^i(\tau,\sigma)$$

and show that, to quadratic order in $Y^i$,

$$S_{\rm NG} =-T\int d\tau d\sigma +{T\over2}\int d\tau d\sigma \left(\dot Y^i\dot Y^i-Y^{\prime i}Y^{\prime i}\right)+O(Y^4).$$

Solution

In static gauge,

$$\partial_\tau X^\mu=(1,0,\dot Y^i), \qquad \partial_\sigma X^\mu=(0,1,Y^{\prime i}).$$

Using $\eta_{\mu\nu}=(-,+,\ldots,+)$,

$$\gamma_{\tau\tau}=-1+\dot Y^i\dot Y^i,$$ $$\gamma_{\sigma\sigma}=1+Y^{\prime i}Y^{\prime i},$$

and

$$\gamma_{\tau\sigma}=\dot Y^iY^{\prime i}.$$

Therefore

$$\begin{aligned} -\gamma &=(\gamma_{\tau\sigma})^2-\gamma_{\tau\tau}\gamma_{\sigma\sigma} \\ &=(\dot Y^iY^{\prime i})^2 -(-1+\dot Y^i\dot Y^i)(1+Y^{\prime j}Y^{\prime j}) \\ &=1+Y^{\prime i}Y^{\prime i}-\dot Y^i\dot Y^i+O(Y^4). \end{aligned}$$

Thus

$$\sqrt{-\gamma} =1+{1\over2}Y^{\prime i}Y^{\prime i} -{1\over2}\dot Y^i\dot Y^i+O(Y^4).$$

Substituting into $S_{\rm NG}=-T\int d^2\sigma\sqrt{-\gamma}$ gives the stated result.

Exercise 3: The Lüscher term for fixed endpoints

For $D-2$ transverse massless fields on an interval of length $L$ with Dirichlet boundary conditions, use zeta-function regularization to show that

$$E_0=-{\pi(D-2)\over24L}.$$

Solution

Dirichlet boundary conditions give normal modes

$$Y_n(\sigma)\sim \sin\left({n\pi\sigma\over L}\right), \qquad n=1,2,\ldots.$$

The frequencies are

$$\omega_n={n\pi\over L}.$$

The zero-point energy of one scalar is

$${1\over2}\sum_{n=1}^\infty \omega_n ={\pi\over2L}\sum_{n=1}^\infty n.$$

Using $\zeta(-1)=-1/12$,

$${1\over2}\sum_{n=1}^\infty \omega_n =-{\pi\over24L}.$$

There are $D-2$ independent transverse fields, so

$$E_0=-{\pi(D-2)\over24L}.$$

Exercise 4: Stress tensor of the Polyakov action

Derive

$$T^{\rm m}_{\alpha\beta} =\partial_\alpha X\cdot\partial_\beta X -{1\over2}h_{\alpha\beta}h^{\rho\lambda} \partial_\rho X\cdot\partial_\lambda X$$

from the Polyakov action.

Solution

Write

$$S_P=-{T\over2}\int d^2\sigma\sqrt{-h}\,h^{\rho\lambda} \partial_\rho X\cdot\partial_\lambda X.$$

Varying with respect to $h^{\alpha\beta}$,

$$\delta\sqrt{-h}=-{1\over2}\sqrt{-h}\,h_{\alpha\beta}\delta h^{\alpha\beta}.$$

Therefore

$$\begin{aligned} \delta_h S_P &=-{T\over2}\int d^2\sigma \left[ \delta\sqrt{-h}\,h^{\rho\lambda}\partial_\rho X\cdot\partial_\lambda X +\sqrt{-h}\,\delta h^{\rho\lambda} \partial_\rho X\cdot\partial_\lambda X \right] \\ &=-{T\over2}\int d^2\sigma\sqrt{-h} \left[ \partial_\alpha X\cdot\partial_\beta X -{1\over2}h_{\alpha\beta}h^{\rho\lambda} \partial_\rho X\cdot\partial_\lambda X \right]\delta h^{\alpha\beta}. \end{aligned}$$

Using

$$T^{\rm m}_{\alpha\beta} =-{2\over T\sqrt{-h}}{\delta S_P\over\delta h^{\alpha\beta}},$$

we get the desired expression.

Exercise 5: Eliminating the auxiliary metric

Use $T^{\rm m}_{\alpha\beta}=0$ to show that the Polyakov action reduces to the Nambu—Goto action.

Solution

The metric equation of motion is

$$0=T^{\rm m}_{\alpha\beta} =\gamma_{\alpha\beta} -{1\over2}h_{\alpha\beta}h^{\rho\lambda}\gamma_{\rho\lambda}.$$

Thus

$$\gamma_{\alpha\beta}=\Lambda h_{\alpha\beta}, \qquad \Lambda={1\over2}h^{\rho\lambda}\gamma_{\rho\lambda}.$$

So $h_{\alpha\beta}$ is conformal to $\gamma_{\alpha\beta}$. Because the Polyakov action is Weyl invariant in two dimensions, we may choose

$$h_{\alpha\beta}=\gamma_{\alpha\beta}.$$

Then

$$\begin{aligned} S_P[X,\gamma] &=-{T\over2}\int d^2\sigma\sqrt{-\gamma}\, \gamma^{\alpha\beta}\gamma_{\alpha\beta} \\ &=-{T\over2}\int d^2\sigma\sqrt{-\gamma}\,(2) \\ &=-T\int d^2\sigma\sqrt{-\gamma} \\ &=S_{\rm NG}[X]. \end{aligned}$$