$$\begin{aligned}
 L_{Dirac} = \sqrt{\tilde{G}} [e^{-\frac{2}{\sqrt{3}}\kappa\Phi} \bar{\Psi}i\tilde{\Gamma}^{a}\tilde{D}_{a}\Psi 
 - m_5e^{-\frac{5}{\sqrt{3}}\kappa\Phi} \bar{\Psi}\Psi]
 \label{1S}\end{aligned}$$

Recall that $\tilde{G}_{MN}$ (the quantities in tilde are reserved for ST theory) is the spacetime metric in ST theory and given by (see eqn.(\[grav.sol1.ST\_final\])): $$\begin{aligned}
 d\tilde{s}^2 = e^{- 2 A(\phi)} \eta_{\mu\nu} dx^{\mu} dx^{\nu} - dy^2
 \nonumber\end{aligned}$$

where $A(y)$ is obtained in eqn.(\[sol\_warp\_final\]). $\tilde{\Gamma}^{a} = (e^{A(y)}\tilde{\gamma}^{\mu}, -i\tilde{\gamma}^{5})$ is the five dimensional gamma matrices where $\tilde{\gamma}^{\mu}$ and $\tilde{\gamma}^{5}$ represent 4D gamma matrices in ST theory. By using the form of $\tilde{G}_{MN}$, the covariant derivative $D_{a}$ can be calculated and is given by, $$\begin{aligned}
   &\tilde{D}_{\mu} = \partial_{\mu} - \frac{1}{2}\tilde{\Gamma}_{\mu}\tilde{\Gamma}^{5}A'(y)e^{-A(y)}\nonumber\\
   &\tilde{D}_{5} = \partial_{y}
   \nonumber
  \end{aligned}$$

Further from eqn.(\[1S\]), it is evident that $\Psi(x^{\mu},y)$ is not canonical and thus the canonical Dirac field ($\tilde{\Psi}$) in ST theory is defined as, $$\begin{aligned}
   \tilde{\Psi}(x^{\mu},y) = e^{-\frac{1}{\sqrt{3}}\kappa\Phi} \Psi(x^{\mu,y})
   \nonumber
  \end{aligned}$$

In terms of such canonical field, the Dirac lagrangian turns out to be, $$\begin{aligned}
   L_{Dirac} = \sqrt{\tilde{G}} \bigg[\bar{\tilde{\Psi}}i\tilde{\Gamma}^{a}\tilde{D}_{a}\tilde{\Psi} 
 - m_5e^{-\frac{2}{\sqrt{3}}\kappa\Phi} \bar{\tilde{\Psi}}\tilde{\Psi} + \frac{1}{\sqrt{3}}\kappa\Phi'(y)\big(\bar{\tilde{\Psi}}\tilde{\gamma}^5\tilde{\Psi}\big)\bigg]
 \label{2S}
  \end{aligned}$$

As earlier, we consider $\xi(y)$ as the fluctuation of the scalar field ($\Phi(y)$) over its vev ($<\Phi>$) i.e $\Phi(y) = <\Phi> + \xi(y)$. Therefore $L_{Dirac}$ can be expressed as, $$\begin{aligned}
   L_{Dirac}&=&\sqrt{\tilde{G}} \bigg[\bar{\tilde{\Psi}}i\tilde{\Gamma}^{a}\tilde{D}_{a}\tilde{\Psi} 
 - m_5e^{-\frac{2}{\sqrt{3}}\kappa<\Phi>} \bar{\tilde{\Psi}}\tilde{\Psi}\nonumber\\
 &+&\frac{1}{2\sqrt{3}}\kappa m_5 \bar{\tilde{\Psi}}\tilde{\Psi} \xi(y) - \frac{1}{24}\kappa^2 m_5 \bar{\tilde{\Psi}}\tilde{\Psi} 