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I have a question concerning fermions in curved space-time. Please read it to the end before suggesting the spin-connection and vierbein-based approach.

I heard that there is a special way of thinking about spin-1/2 particles (Dirac fermions) in flat space-time: the spinor field $\psi(x)$ is considered a (Grassmanian) scalar multiplet (under the Lorentz transformations), but the matrix-valued 4-vector $\gamma^{\mu}$ transforms as an actual 4-vector.

The value of the $\psi$ field here is in correspondence with the value of the usual spinor-transforming field, but taken at some fixed frame of reference (in which $\gamma^{\mu}$ take the usual fixed values). Quantities like $\bar{\psi} \gamma^{\mu} \psi$ transform like vectors, which is basically why this formalism is equivalent to the standard (with $\psi$ transforming as spinor and constant $\gamma^{\mu}$).

The Dirac action is then just $$ S[\psi] = \int d^4 x \: \bar{\psi} \left( i \gamma^{\mu} \partial_{\mu} - m \right) \psi, $$ which is manifestly Lorentz-invariant in this strange formalism.

My question is about curved space-time of GR. The idea is to write something like $$ S[\psi] = \int d^4 x \: \sqrt{-g} \: \bar{\psi} \left( i \gamma^{\mu} \partial_{\mu} - m \right) \psi, $$ where $\gamma^{\mu}$ transforms as matrix-valued vector under GCTs, $\nabla_{\mu} \psi$ and $\partial_{\mu} \psi$ are equivalent since $\psi$ is basically a (Grassmanian) scalar multiplet. So this new action is manifestly diffeomorphism-invariant, and agrees with the Dirac field in flat space limit. Also (since $\left\{ \gamma^{\mu}, \gamma^{\nu} \right\} = 2 g^{\mu \nu} \cdot 1_{4 \times 4}$) the metric field can be constructed out of (more fundamental?) matrix-valued vector field $\gamma^{\mu}$.

My teacher says it is incorrect, and I am pretty sure it is, but he can't explain why (and that's what really bothers me). One guess is that the interaction between fermions and gravity probably not correct since there is no spin-connection term (like in the standard vierbein-based approach).

So the question then becomes: what should I add in this action to make the fermion-gravity interaction term correct, given that I don't want to abandon this strange formalism and consider the spinor transformation of $\psi$.

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  • $\begingroup$ This formalism is under-specified, and it's not obvious to me that it makes sense. By GCT, did you mean general coordinate transform, i.e., an element of $Diff(\mathbb{R}^4)$? Can you say more precisely how $\gamma^\mu$ is supposed to transform under these? $\endgroup$
    – user1504
    Jul 6, 2016 at 19:35

2 Answers 2

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The problem with the argument (even in Minkowski spacetime) is that spinors are not scalar multiplets. The usual way of defining spinors is by specifying their transformation rule under Lorentz transformations (see section 4.1.1 of http://www.damtp.cam.ac.uk/user/tong/qft/four.pdf), and this is not how a collection of scalar fields would transform.

However, you could push on and try to define some field in the way proposed above. The issue is that then the objects $\gamma^\mu$ that you have defined are matrix valued vectors (i.e. there is no additional transformation rule for the matrix indices). The combination $$\overline{\psi}(i\gamma^\mu\partial_\mu - m)\psi$$ is then just a sum of terms involving scalar fields $$\overline{\psi}_\alpha(i\gamma^\mu_{\alpha\beta}\partial_\mu - m)\psi_\beta.$$ While this is Lorentz invariant, it depends on the arbitrary choice of vectors $\gamma^\mu_{\alpha\beta}$. This is the same problem that comes up in trying to define a first order differential operator that is Lorentz invariant and does not depend on some initial choice of preferred vector. It is the motivation for introducing spinors in the Dirac equation.

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The formalism you suggest is perfectly correct, and not "strange", since it generalizes to curved spacetimes while the usual transformation law of spinors do not. The only problem is your integrand, which does not have a coordinate invariant meaning. Just replace partial derivatives with covariant ones and it's ok.

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  • $\begingroup$ 1. When acting on scalars, covariant derivatives are just partial derivatives. 2. I disagree with “everything is ok”, because what I get back is not the usual Dirac theory in curved space time. $\endgroup$ Jan 23, 2019 at 17:16

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