String Theory and M-Theory

2.2 The string action 29
where we have used Eq. (2.15). Contracting with h γδ gives
hγδh γδ Λ = T ( 1
2 hγδh γδ − 1)h αβ ∂αX µ ∂βXµ.
Since hγδh γδ = 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. ✷
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
Sσ = − Tp
d
2
p+1 σ √ −hh αβ
∂αX · ∂βX + Λp
d p+1 σ √ −h. (2.16)
The equation of motion for the world-volume metric is obtained exactly as
in the previous exercise, with the result
Tp[∂γX · ∂δX − 1
2 hγδ(h αβ ∂αX · ∂βX)] + Λphγδ = 0.
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
hαβ = ∂αX · ∂βX. (2.17)
Substituting this ansatz in the previous equation and dropping common
factors gives
Tp(1 − 1
2 hαβ hαβ) + Λp = 0.
Substituting h αβ hαβ = p + 1, one learns that
Λp = 1
2 (p − 1)Tp. (2.18)
Thus, consistency requires this choice of Λp. 2 This confirms 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 finds
that Eq. (2.16) is equivalent classically to Eq. (2.6). For the special case of
2 A different value is actually equivalent, if one makes a corresponding rescaling of hαβ. However,
this results in a multiplicative factor in the relation (2.17).