q \vec x}\
{-N_c\over 6I}
 \int d^3 y \ \Bigg\lbrace\ \sum_{n\in VAL\atop m\in ALL}
{\psi^{\dagger}_n(\vec x)\tau^A\psi_m(\vec x)\
\psi^{\dagger}_m(\vec y)\tau^A
\psi_n(\vec y) \over \epsilon_n - \epsilon_m}\ + \cr
&+ \ {1\over 4\sqrt\pi}
\sum_{m,n\in ALL}
\int_{1\over\Lambda^2}^{\infty} {d\tau\over\sqrt\tau} \,
\Bigg\lbrack {\epsilon_n e^{-\tau\epsilon_n^2} + \epsilon_m
e^{-\tau\epsilon_m^2}\over \epsilon_n + \epsilon_m } +
{1\over\tau} {e^{-\tau\epsilon_n^2} - e^{-\tau\epsilon_m^2}
\over \epsilon_n^2 - \epsilon_m^2}\ \Bigg\rbrack \ \times \cr
&\times \ \psi_n^{\dagger}(\vec x)\tau^A \psi_m(\vec x)\
\psi_m^{\dagger}(\vec y)\tau^A\psi_n(\vec y)\ \Bigg\rbrace \ , \cr
} \eqno(\ELVEC)$$

The formula () for $q^2=0$ is equivalent to the quotient of the moment of inertia by itself and: $G^{T=1}_E(q^2=0) = 1$, as it should be.

We proceed to compute the magnetic form factors. We start with the isoscalar one. In this case, the contribution comes from the imaginary part of the action and the regularization is not necessary. The quark current matrix $\hat Q$ contributes [*via*]{} the isoscalar term, $\sim {1\over 6}\,\,\hat 1$, diagonal in the isospin space. Hence, the rotation matrices $R(t)$ cancel each other and, like for the electric scalar form factor, there is no time ordering problem for the ${\cal D}R$ integration in (). However, there is still a nonzero contribution from terms linear in $\Omega$. Starting from the reversed Hermicity of $A_{\mu}$ 