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I have seen the Dirac equation in curved space-time written as $$[i\bar{\gamma}^{\mu}\frac{\partial}{\partial x^{\mu}}-i\bar{\gamma}^{\mu}\Gamma_{\mu}-m]\psi=0 $$

This $-i\bar{\gamma}^{\mu}\Gamma_{\mu} $ is the spin connection:

a) Why is it needed?

b) Why is it minus could it be a plus (is it due to metric signature)?

c) Is the a suitable method to calculate it for a given metric e.g FRW space-time?

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2 Answers 2

This is standard theory. Try

  • Birrell, N. D., & Davies, P. C. W. (1982). Quantum Fields in Curved Space. Cambridge: Cambridge University Press. Bog standard Curved space QFT text. Don't remember how much is said specifically about spinors though.

  • Brill, D., & Wheeler, J. (1957). Interaction of Neutrinos and Gravitational Fields. Reviews of Modern Physics, 29(3), 465–479. doi:10.1103/RevModPhys.29.465 <-- This paper was particularly clear from memory.

  • Yepez, J. (2011). Einstein’s vierbein field theory of curved space. General Relativity and Quantum Cosmology; History of Physics. Retrieved from Great discussion. Thanks twistor59.

  • Boulanger, N., Spindel, P., & Buisseret, F. (2006). Bound states of Dirac particles in gravitational fields. Physical Review D, 74(12). doi:10.1103/PhysRevD.74.125014 Technical examples worked out in painful detail

  • Lasenby, A., Doran, C., & Gull, S. (1998). Gravity, gauge theories and geometric algebra. Philosophical Transactions of the Royal Society A: Mathematical, Physical and Engineering Sciences, 356(1737), 487–582. General Relativity and Quantum Cosmology; Astrophysics. doi:10.1098/rsta.1998.0178 Geometric algebra technique - a powerful and elegant modern formalism that I'm hardly an expert on. See Muphrid's answer for more details. :)

These are less specific to the question but still with material pertaining to it:

There are other references. I'll put them in as I think of them or others point them out (thanks guys!).

The reason you need the spin connection is because fundamentally you need the tetrad or orthonormal frame fields. These fields give a set of "laboratory frames" at every point in spacetime:

$$ e^\mu_a(x),\ \text{with}\ e^\mu_a(x) e_{\mu b}(x) = \eta_{ab}, $$

where $a$ labels the field $a=\hat{t},\hat{x},\cdots$ and $\mu$ is the spacetime vector index. The intuitive meaning of this is that $e^\mu_{\hat{t}}$ represents the 4-velocity of the lab, $e^\mu_{\hat{x}}$ is bolted down on the lab bench oriented along the $x$-axis etc. You can prove the relationship

$$ g_{\mu\nu}(x) = \eta_{ab} e^a_\mu(x) e^b_\nu(x). $$

For this reason the tetrad is commonly known as the "square root of the metric," which is not an entirely satisfactory notion. Anyway, you can see that the tetrad is not uniquely defined. Any tetrad related to another by $e'^a_\mu(x) = \Lambda^a_b(x) e^b_\mu(x)$ where $\Lambda^a_b(x)$ is a local Lorentz transformation is just as good. This corresponds to the freedom of different labs to rotate their axes and boost themselves independently.

This means the theory has a huge built in redundancy - local Lorentz invariance - which plays a similar role for spinors as coordinate tranformation invariance does in GR. The tetrads are necessary because spinor representations are defined in relation to the double cover of the Lorentz group SL(2,C), which cannot be represented in terms of tensors under the diffeomorphism group. You can however define spinors relative to a locally Minkowski frame:

$$ \psi(x) \rightarrow \left( 1 - \frac{i}{2} \omega_{ab}(x) \sigma^{ab} \right) \psi(x), $$

where $\omega_{ab}(x)$ is a local Lorentz transformation and $\sigma^{ab}\propto [\gamma^a,\gamma^b]$ are the generators of spinor transformations. The spinors basically live in an internal space. The next key idea is that you have to then be able to "solder" SL(2,C) representations in these frames together consistently to cover the spacetime. The consistency conditions you desire are that:

  • $\bar{\psi} \psi$ is a scalar field
  • The product rule and linearity work for covariant derivatives
  • The tetrad postulate, i.e. compatibility of the spinor covariant derivative with the ordinary covariant derivative

These conditions form the relationship between the internal spin and the spacetime, and they give the formula for the spin connection:

$$ \omega_{\mu b}^{a}=e_{\lambda}^{a}\Gamma_{\mu\nu}^{\lambda}e_{b}^{\nu}-\left(\partial_{\mu}e_{\nu}^{a}\right)e_{b}^{\nu}. $$

In older literature you may see curved space gamma matrices defined by contraction with the tetrad:

$$ \gamma^\mu(x) = \gamma^a e^\mu_a(x). $$

This is fine but I find it less confusing to keep the tetrads explicit. Note that the $\gamma^a$ are constant numerical matrices, whereas $\gamma^\mu(x)$ are spacetime functions.

This, combined with the references and some googling should hopefully get you started. If you are still really stuck after trying to work some of this out (Try to do it for yourself! There are so many different conventions in the literature it's hard to trust copy'n'pasting different people's results!) then I have an example calculation from go to woe in my honours thesis.

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Thanks, I have read that $$\Gamma_{\mu}=\frac{1}{2}\sum^{\alpha\beta}e_{\alpha}^{\mbox{ }\nu}(\frac{\partial}{\partial x^{\mu}}e_{\beta \nu}) $$ Is it correct that $\sum^{\alpha\beta}=\frac{1}{4}(\gamma^{\alpha}\gamma^{\beta} -\gamma^{\beta} \gamma^{\alpha})$ but how does this not equal zero, surely it would? Also since $e_{\alpha}^{\mbox{ }\nu}=(1,+a^2,+a^2,+a^2)$, what does $e_{\beta \nu}$ equal? Essentially I am trying to prove that $$\bar{\gamma}^{\mu}\Gamma_{\mu}=\frac{3}{2}\frac{\dot{a}}{a}$$ –  user21119 Mar 21 '13 at 11:24
Section VIII of is also useful background –  twistor59 Mar 21 '13 at 12:04
@user21119 I think the expression you are looking for is eq. 194 in Yepez (the article linked by twistor59). $\Sigma^{\alpha\beta}$ is not zero because the gamma matrices don't commute! In the case of an FRW metric the tetrads would have an $a$ instead of $a^2$ - remember the tetrad is the "square root" of the metric (ignoring issues if the metric is non-diagonal). Note that this notation is a bit different from my answer. That Yepez article will probably help. :) –  Michael Brown Mar 21 '13 at 22:07
@twistor59 Thanks. That is a good reference. –  Michael Brown Mar 21 '13 at 22:08
Thanks, what is the S term and how can one calculate it? –  user21119 Mar 21 '13 at 22:10

If you're like me, you might find the massive proliferation of indices, trying to distinguish between the coordinate basis and the tetrad basis, very soupy. Lasenby, Doran, and Gull have a paper using geometric algebra that enables abandoning index notation and differential forms to make the underlying mathematics clearer.

In that paper, the properties of gravity are derived using a "displacement gauge principle," saying that the laws of physics should be invariant under arbitrary differentiable remappings of positions in spacetime, and a "rotation gauge principle," saying that the laws should be covariant under local Lorentz rotations.

The displacement gauge principle generates the tetrad field, which they call $\underline h$. In their picture, there exist three spaces for objects to reside in: a tangent space and a cotangent space, which are standard, and between which objects can be moved through the metric, and a third space halfway in-between, a "gauge covariant" space. Objects are moved into and out of the gauge covariant space using $\underline h$, its inverse, or adjoint, and you can do all that you need in terms of gravitation largely in the gauge covariant space, rather than constantly moving things back and forth between the tangent and cotangent spaces.

Overall, I think GTG is simpler than standard tetrad formalism in terms of practical use because it doesn't rely on components and indices. You can use the same basis (the basis vectors $\gamma_0, \gamma_1, \gamma_2, \gamma_3$) for the (co)tangent space as for the gauge covariant space, and this is perfectly legal to talk about objects abstractly before trying to extract physical components later on. I think it's this point that makes it much easier to use.

GTG also gives a geometric interpretation for the spin connection $\underline \omega$ as a bivector-valued linear operator on a vector.

Regardless of whose work you use as a foundation (or if you use GTG vs. the standard formalism), I think you will find tetrad methods to be very powerful. I prefer them over metric methods for doing all problems involving general relativity.

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I am a bit confused still, since $\alpha$ must equal $\nu$ which must equal $\beta$ otherwise one of the e's would be zero, but isnt $\sum ^{ii}=0$ (for i equal to 1..3)? –  user21119 Mar 21 '13 at 23:28

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