When you write the Dirac equation in a curved spacetime, in the context of General Relativity (which allows curvature, but not torsion) , you have a spin connection :

$$\nabla_\mu\psi=\left(\partial_{\mu}-\frac i4\omega_{\mu}^{IJ}\sigma_{IJ}\right)\psi$$

Now, the Einstein-Cartan theory is not General Relativity, because it allows curvature, but also torsion, which is proportionnal to your term $\kappa$.

So, it turns, that, in presence of torsion ($\kappa$), everything happens as if there was a Lagrangian with a quadratic term, so the equation of movement has a cubic term (see for instance, this Ref, formula $16$ for the torsion, and formula $31$ for the Dirac equation.

[EDIT]

Due to OP comments, some precisions :

(Always working with the same Ref):

$\nabla_\mu \psi$, is the covariant derivative for a spinor in general relativity, that is without torsion. This means that the Dirac equation in general relativity is $(i\gamma^{\mu}\nabla_{\mu}-m)\psi=0$. But it is no more true with a space-time with torsion, so it is better to use an other notation $D_\mu$ if you want to write a Dirac equation like $(i\gamma^{\mu}D_{\mu}-m)\psi=0$. The quartic term (in $\psi$) for $R_{ab}$ (formula $30$), or the quadratic term in $T_a$ (formula $29$), which lead to the cubic term for the Dirac equation (formula $31$), come from the coupled Euler-Lagrange equations $15,16$. For instance, you see, in formula $4$, that the connection has a supplementary term $K^a_b$ (contorsion), which is linked to the torsion (formula $6$), which is itself quadratic in the $\psi$ (formula $29$). So, in some way, you may simply consider this supplementary connection, if you remember where connection appears in a "derivative". Yes $\kappa$ is the gravitational constant, see formula $9$. Now, you may look, in this ref, that the torsion (defined in formula $2$), in the Einstein-Cartan model (formulae $8,9$), is proportionnal to $\kappa$ (formula $16$)

This post imported from StackExchange Physics at 2014-03-07 16:43 (UCT), posted by SE-user Trimok