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  Non-linear Dirac equation in Einstein Cartan theory

+ 2 like - 0 dislike

Can someone explain this Wikipedia article, specifically the section on Einstein-Cartan theory? I have no idea how the equation \begin{equation}(i\gamma^{\mu}D_{\mu}-m)\psi=i\gamma^{\mu}\nabla_{\mu}\psi+\frac{3\kappa}{8}(\bar\psi\gamma_{\mu}\gamma^{5}\psi)\gamma^{\mu}\gamma^{5}\psi -m\psi\end{equation} is justified. They state that $D_{\mu}$ is a covariant derivative acting on spinors, and then later state that $\nabla_{\mu}$ is a general-relativistic covariant derivative acting on spinors. I don't know what the difference between the two would be. Lastly, where does the cubic term come from? That's what confuses me the most.

This post imported from StackExchange Physics at 2014-03-07 16:43 (UCT), posted by SE-user TeeJay
asked Dec 6, 2013 in Theoretical Physics by TeeJay (20 points) [ no revision ]

1 Answer

+ 4 like - 0 dislike

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.


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
answered Dec 6, 2013 by Trimok (955 points) [ no revision ]
Two questions, if $\nabla_{\mu}\psi$ is what you've described, then what is $D_{\mu}\psi$? I'm beginning to understand the origin of the cubic term, but I'm not quite there yet. I'm pretty sure $\kappa$ is just the gravitational constant here, so I'm unsure why you've referenced it with the inclusion of torsion. Could you perhaps elaborate a bit more on these?

This post imported from StackExchange Physics at 2014-03-07 16:43 (UCT), posted by SE-user TeeJay
@TeeJay : I edited the answer.

This post imported from StackExchange Physics at 2014-03-07 16:43 (UCT), posted by SE-user Trimok
thank you very much. This helped tremendously.

This post imported from StackExchange Physics at 2014-03-07 16:43 (UCT), posted by SE-user TeeJay
@Adobe : Well, guys, all this was my fault, I should not use a copy of an image...

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

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