which on comparison with equation 1.38 of section 1.2.8, in the limit ϵ → 0, allows us to extract the decay constant F$_{ρ}$ : Our analysis of the axial vector mesons closely follows that of the vector mesons. However things are complicated by the mixing of the pion field π and the longitudinal axial gauge field φ . We start with the action, expanded to quadratic order: where we have defined v ( z ) = m$_{q}$z + σz $^{2}$, A = ( A$_{L}$ − A$_{R}$ ) / 2 and X = X$_{0}$ exp( i 2 π $^{a}$t$^{a}$). We make the gauge choice A$_{z}$ = 0 and A$_{µ}$ = $^{A$_{µ}$}$⊥ + ∂$_{µ}$φ . Then the equations of motion become The a$_{1}$ meson is a spin-1 particle, and so is the solution to equation (5.18) with the same boundary conditions we imposed for the vector mesons, namely 78 ψ$_{a}$$_{1}$ (0) = 0 ∂$_{z}$ψ$_{a}$$_{1}$ ( z$_{m}$ ) = 0 (5.21) − q $^{2}$∂$_{z}$ φ a + g 2 $_{5}$v 2 z 2 ∂$_{z}$π a = 0 (5.20) ∂$_{z}$ ( 1 z ∂$_{z}$φ a ) + g 2 $_{5}$v 2 z 3 ( π a − φ $^{a}$) = 0 (5.19) [ ∂$_{z}$ ( 1 z ∂$_{z}$A a µ ) + q 2 z A a µ − g 2 $_{5}$v 2 z 3 A a µ ]$_{⊥}$ = 0 (5.18) S = ∫ d $^{5}$x ( − 1 4 g 2 $_{5}$z F a $_{A}$F a A + v ( z ) 2 2 z 3 ( ∂π a − A $^{a}$) 2 ) (5.17) F 2 ρ = 1 g 2 5 ( $_{ψ}$′ $_{ρ}$( ϵ ) /ϵ ) 2 (5.16) 5.3.2 Axial Vector Mesons