for a string in Minkowski space-time. Use the static gauge, which ¬xes

the longitudinal directions X 0 = „ , X 1 = σ, while leaving the transverse

directions X i free. Show that the kinetic energy contains only the transverse

velocity. Determine the mass per unit length of the string.

28 The bosonic string

SOLUTION

In the static gauge

‚„ X µ ‚„ Xµ ‚„ X µ ‚σ Xµ

det G±β = det

‚σ X µ ‚„ Xµ ‚σ X µ ‚σ Xµ

’1 + ‚„ X i ‚„ Xi ‚„ X i ‚σ Xi

= det .

‚σ X i ‚„ Xi 1 + ‚ σ X i ‚σ Xi

Then,

det G±β ≈ ’1 + ‚„ X i ‚„ Xi ’ ‚σ X i ‚σ Xi + . . .

Here the dots indicate higher-order terms that can be dropped in the non-

relativistic limit for which the velocities are small. In this limit the action

becomes (after a Taylor expansion)

| ’ 1 + ‚ „ X i ‚„ Xi ’ ‚ σ X i ‚σ Xi |

SNG = ’T d„ dσ

1 1

d„ dσ ’1 + ‚„ X i ‚„ Xi ’ ‚σ X i ‚σ Xi .

≈T

2 2

The ¬rst term in the parentheses gives ’m d„ , if L is the length of the σ

interval and m = LT . This is the rest-mass contribution to the potential

energy. Note that L is a distance in space, because of the choice of static

gauge. Thus the tension T can be interpreted as the mass per unit length, or

mass density, of the string. The last two terms of the above formula are the

kinetic energy and the negative of the potential energy of a nonrelativistic

string of tension T . 2

EXERCISE 2.8

Show that if a cosmological constant term is added to the string sigma-model

action, so that

√ √

T

d2 σ ’hh±β ‚± X µ ‚β Xµ + Λ d2 σ ’h,

Sσ = ’

2

it leads to inconsistent classical equations of motion.

SOLUTION

The equation of motion for the world-sheet metric is

2 δSσ 1

= ’T [‚γ X µ ‚δ Xµ ’ hγδ (h±β ‚± X µ ‚β Xµ )] ’ Λhγδ = 0,

√

’h δhγδ 2

2.2 The string action 29

where we have used Eq. (2.15). Contracting with hγδ gives

1

hγδ hγδ Λ = T ( hγδ hγδ ’ 1)h±β ‚± X µ ‚β Xµ .

2

Since hγδ hγδ = 2, the right-hand side vanishes. Thus, assuming h = 0,

consistency requires Λ = 0. In other words, adding a cosmological constant

term gives inconsistent classical equations of motion. 2

EXERCISE 2.9

Show that the sigma-model form of the action of a p-brane, for p = 1,

requires a cosmological constant term.

SOLUTION

Consider a p-brane action of the form

√ √

Tp

d σ ’hh±β ‚± X · ‚β X + Λp

p+1 p+1

Sσ = ’ σ ’h.

d (2.16)

2

The equation of motion for the world-volume metric is obtained exactly as

in the previous exercise, with the result

1

Tp [‚γ X · ‚δ X ’ hγδ (h±β ‚± X · ‚β X)] + Λp hγδ = 0.

2

This equation is not so easy to solve directly, so let us instead investigate

whether it is solved by equating the world-volume metric to the induced

metric

h±β = ‚± X · ‚β X. (2.17)

Substituting this ansatz in the previous equation and dropping common

factors gives

1

Tp (1 ’ h±β h±β ) + Λp = 0.

2

Substituting h±β h±β = p + 1, one learns that

1

(p ’ 1)Tp .

Λp = (2.18)

2

Thus, consistency requires this choice of Λp .2 This con¬rms the previous

result that Λ1 = 0 and shows that Λp = 0 for p = 1. Substituting the

value of the metric in Eq. (2.17) and the value of Λp in Eq. (2.18), one ¬nds

that Eq. (2.16) is equivalent classically to Eq. (2.6). For the special case of

2 A di¬erent value is actually equivalent, if one makes a corresponding rescaling of h±β . However,

this results in a multiplicative factor in the relation (2.17).

30 The bosonic string

p = 0, this reproduces the result in Eq. (2.5) if one makes the identi¬cations

T0 = m and h00 = ’m2 e2 . 2

2.3 String sigma-model action: the classical theory

In this section we discuss the symmetries of the string sigma-model action in

Eq. (2.14). This is helpful for writing the string action in a gauge in which

quantization is particularly simple.

Symmetries

The string sigma-model action for the bosonic string in Minkowski space-

time has a number of symmetries:

• Poincar´ transformations. These are global symmetries under which the

e

world-sheet ¬elds transform as

δX µ = aµ ν X ν + bµ δh±β = 0.

and (2.19)

Here the constants aµ ν (with aµν = ’aνµ ) describe in¬nitesimal Lorentz

transformations and bµ describe space-time translations.

• Reparametrizations. The string world sheet is parametrized by two coor-

dinates „ and σ, but a change in the parametrization does not change the

action. Indeed, the transformations

‚f γ ‚f δ

± ± ±

σ ’ f (σ) = σ and h±β (σ) = hγδ (σ ) (2.20)

‚σ ± ‚σ β

leave the action invariant. These local symmetries are also called di¬eo-

morphisms. Strictly speaking, this implies that the transformations and

their inverses are in¬nitely di¬erentiable.

• Weyl transformations. The action is invariant under the rescaling

h±β ’ eφ(σ,„ ) h±β δX µ = 0,

and (2.21)

√ √

since ’h ’ eφ ’h and h±β ’ e’φ h±β give cancelling factors. This

local symmetry is the reason that the energy“momentum tensor is trace-

less.

Poincar´ transformations are global symmetries, whereas reparametriza-

e

tions and Weyl transformations are local symmetries. The local symmetries

can be used to choose a gauge, such as the static gauge discussed earlier, or

else one in which some of the components of the world-sheet metric h±β are

of a particular form.

2.3 String sigma-model action: the classical theory 31