3
$\begingroup$

I have been studying the spin-connection for the Dirac equation. The covariant derivative is defined as:

$$D_\mu \psi = (\partial_\mu - {i \over 4} {\omega_\mu}^{ab} \sigma_{ab}) \psi$$

where $\sigma_{ab} = i[\gamma^a,\gamma^b]$ and ${\omega_\mu}^{ab}$ is the spin-connection.

When working out the consequence I find that this $i$ factor before the spin-connection leads to pseudo-vector terms which treat left and right differently. (Perhaps this is good as maybe left and right handed fermions move differently under gravity but this seems to go against the idea that all things moves the same under gravity)

So my very technical question is, is this $i$ really necessary. Or would the theory still be self consistent if we removed the $i$? What is the justification for this $i$? It would be a different theory but would it be self-consistent? Could we even add a factor of $\gamma^5$ before the spin-connection and keep it self-consistent? (Since this would be the same except having a factor of -1 for right-handed fermions?) Basically I'm thinking of any ways that this equation can be changed and remain consistent with general covariance.

$\endgroup$
1
$\begingroup$

The $i$ is necessary.

In order to derive the Dirac equations in curved spacetime, the approach is the following: We know how Lorentz vectors can be transported in a generic curved spacetime using covariant derivatives. We use the fact that bilinears of $\psi$(a spinor) should transform like a vector. This then gives rise to the above formula.

You may wanna look at this paper for derivation.

$\endgroup$
  • $\begingroup$ What do you make of that spinning fermions will move differently near a spinning gravitational force. Is this compatible with the equivalence principle? Hmm... I guess this is how gravity-probe-b works? $\endgroup$ – zooby Mar 26 at 5:30
  • $\begingroup$ It is compatible. You should look at Soft Theorems for gravity and gauge theories. For gravity these predicts that at first order gravity couples to mass but in next order it couples to the spin. $\endgroup$ – Ari Mar 26 at 6:01
  • $\begingroup$ Thanks @Ari I will check this out. I can't intuitively see why this means that things made of different stuff fall differently under gravity. It worries me. But I'll have a look at the Soft Theorems. $\endgroup$ – zooby Mar 26 at 14:08
  • $\begingroup$ In fact Soft theorems are a good way to study the equivalence principle in a QFT setting. You'll see as energy-> 0, the gravitons couple to everything just my their momentum. The corrections to these are the spin ones that I'm talking about. $\endgroup$ – Ari Mar 26 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.