DBI Analysis of Open String Bound States on Non-compact D-branes

CHAPTER 8. COMPUTATIONS 100classical solution Eq. 8.12:O(1)[L <strong>DBI</strong> ] =4π 2 α 2 A√ F ρτ fk (α 2 − 1)√(α 2 − 1) + 4π 2 α 2 Fρτ2A 2πα 2 λ √ α= √ 2 − 1k (α 2 − 1) α √ f4π 2 α 2 A 2 − λ√(α 2( 2 − 1) + λ2 α 2 − 1 )4π 2 α 2 A 2 − λ 2= A √k2πλα√4π 2 α 2 A 2 − λ 2 √4π 2 α 2 A 2 − λ 2√(α 2 − 1)(4π 2 α 2 A 2 − λ 2 ) + λ 2 (α 2 − 1) f= A √k2πλα√(α 2 − 1)4π 2 α 2 A 2f=λf√k (α 2 − 1) . (8.20)We thus indeed find thatL <strong>DBI</strong> −λf√ = 0, (8.21)k (α 2 − 1)as we hoped. Further using the classical solution to simplify the factor in front <strong>of</strong> whatis left <strong>of</strong> our Lagrangian density, we see that√(α 2 − 1) + 4π 2 α 2 F 2 ρτk (α 2 − 1)=√1k + C 2k (α 2 − C 2 ) =α√k (α 2 − C 2 ) , (8.22)leaving us with{αL <strong>DBI</strong> = A √k (α 2 − C 2 ) − 2π2 α √ k (α 2 − C 2 )α 2 (∂ t α τ ) 2 − 2π2√ k (α 2 − C 2 )(∂ t α ρ ) 2− 1α[f 22π 2 α 2 [( α 2 − 1 ) + 4π 2 α 2 F 2 ] ]ρτ − 8π 4 α 4 Fρτ2 + √ [ ]k (α 2 − 1) + 4π 2 α 2 Fρτ2 3/2√ .α 2 − 1(8.23)If one further fills in the classical solution for F ρτ , one finds thatL <strong>DBI</strong> = 2π 2 A √ {k (α 2 − C 2 α)2π 2 [k (α 2 − C 2 )] − αα 2 − 1 (∂ tα τ ) 2 − (∂ tα ρ ) 2α(α 2 − C 2)+k (α 2 − 1)α[(∂ ρ α τ ) 2 − 2 (∂ ρ α τ ) (∂ τ α ρ ) + (∂ τ α ρ ) 2] } . (8.24)