In this expression the pion and nucleon covariant derivatives are given by [@Wei68]

$$\begin{aligned}
D^\mu {\mbox{\boldmath $\phi $}} &=&\partial _\mu {\mbox{\boldmath $\phi $}}-
\frac 1{f+f_\pi }\partial ^\mu f{\mbox{\boldmath $\phi $}}, \\
D^\mu \psi &=&\left[ \partial ^\mu +i\frac 1{f_\pi \left( f+f_\pi \right) }
\frac{{\mbox{\boldmath $\tau $}}}2 \cdot \left( {\mbox{\boldmath $\phi $}}\times
\partial ^\mu {\mbox{\boldmath $\phi $}}\right) \right] \psi .\end{aligned}$$

This Lagrangian has been used recently in the study of $\pi N$ form factors in constituent quarks models [@Mae+97] and it is worth noting that its last term describes a chiral invariant coupling between the scalar and the nucleon.

In the case of $PS$ coupling, one has

$${\cal L}_N^{PS}=\bar Ni{\not \partial }N -g\bar N\left( f+i{\mbox{\boldmath
$\tau $}}\cdot {\mbox{\boldmath $\phi $}}\gamma _5\right) N 
-\frac{g_s}{f_\pi }S\bar N\left( f+i{\mbox{\boldmath $\tau $}}\cdot {\mbox{\boldmath
$\phi $}}\gamma _5\right) N,$$

\[33\]

where $N$ is the nucleon field that transforms linearly. The last term of this expression has the same meaning as the corresponding one in eq. (\[Lpv\]).

On general grounds one knows that, in the framework of chiral symmetry, results should not depend on the choice of ${\cal L}_N$ [@ref11; @ref12]. The equivalence between the PS and PV Lagrangians was verified explicitly in the case of TPEP [@Rob+] and 