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

CHAPTER 8. COMPUTATIONS 102The second one needs some more care, as the factor preceding the partial derivatives <strong>of</strong>f is α-dependent. We see that(α∂t 2 2 − C 2) ((∂ ρ α τ ) = ∂2kα 2 ρ f ) ( − α2 α 2 + 1 ) − C 2 ( 3α 2 − 1 )kα 3 (α 2 − 1) 1/2 (∂ ρ f)+ (∂ ρ f) √ ( α 2α 2 − C 2 )− 1∂ αkα 2Using∂ α(α2 ( α 2 + 1 ) − C 2 ( 3α 2 − 1 )kα 3 (α 2 − 1) 1/2 )− f √ α 2 − 1∂ α(α2 ( α 2 + 1 ) − C 2 ( 3α 2 − 1 )kα 3 (α 2 − 1) 1/2 )∂ α( α 2 − C 2kα 2 )= − α4 + α 2 − 3C 2 α 2 + C 2[(α 2 − 1) α] 2 ,. (8.29)= α4 ( −3 + 6C 2) + α 2 ( 1 − 7C 2) + 3C 2kα 4 (α 2 − 1) 3/2 , (8.30)this amounts to∂ 2 t (∂ ρ α τ ) =(α 2 − C 2) (∂2kα 2 ρ f ) (α 4 + α 2 ( 1 − 5C 2) + 3C 2)−kα 3 (α 2 − 1) 1/2 (∂ ρ f)(3C 2 + α 2 ( 1 − 7C 2) + α 4 ( −3 + 6C 2))−kα 4 (α 2 f. (8.31)− 1)8.7 Differential equationSubtracting Eq. 8.28 from Eq. 8.31, we find that(∂2t f ) (α 2 − C 2) (= ∂2kα 2 ρ f ) (α 4 + α 2 ( 1 − 5C 2) + 3C 2)−kα 3 (α 2 − 1) 1/2 (∂ ρ f)(3C 2 + α 2 ( 1 − 7C 2) + α 4 ( −3 + 6C 2))−kα 4 (α 2 f− 1)(α 2 − C 2) (+ ∂2k (α 2 − 1) τ f ) . (8.32)We still have ρ derivatives working in on f. We replace these by α derivatives by using∂∂ρ = ∂α ∂∂ρ ∂α = (shρ)∂ α = ( α 2 − 1 ) 1/2∂α , (8.33)∂ 2∂ρ 2 = ∂α ( )∂ ∂α ∂= α∂ α + ( α 2 − 1 ) ∂ 2∂ρ ∂α ∂ρ ∂αα, (8.34)