The Dirac Equation in Curved spacetime makes a difference between Lorentzian indicies and Covariant indicies. In the equation we find a $\partial_\mu$. Is this actually $e^a_\mu\partial_a$ where $e$ is the tetrad (or vielbein)? I.e. does this derivative look different than the regular derivative operator in flat space?

(To be clear I am not asking about the spin connection and covariant derivative, just if the partial will have addition factors from the tetrad.)


FWIW, the notation $\partial_a$ with a so-called "flat" index $a\in\{0,1,2,3\}$ does not make sense as a partial derivative wrt. some "flat" variable. It can only be understood as $$E^{\mu}{}_{a} \partial_{\mu}, \qquad \partial_{\mu}~\equiv~\frac{\partial}{\partial x^{\mu}},$$ where $\mu\in\{0,1,2,3\}$ is a so-called "curved" index; where $x^{\mu}$ is local coordinate of spacetime; and where $E^{\mu}{}_{a}$ is an (inverse) vielbein.

  • $\begingroup$ And just to be clear, this $\partial_\mu = (\frac{1}{c}\partial_t, \nabla)$ even though we're working in curved space? $\endgroup$ – Craig Apr 4 '19 at 18:25
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Apr 4 '19 at 19:25

$D_\mu$ and thus $\partial_\mu$ is being explicitly contracted on $\mu$ with the tetrad $e^\mu_a$ in the first equation. There is no implicit contraction of $\partial_\mu$ with the tetrad in the second equation. There is, however, an implicit $4\times4$ identity matrix multiplying it.

  • $\begingroup$ I assumed $e_a^\mu$ was being contracted on the $\gamma$ matrices to transform then from Lorentz to curvilinear cooridnates. Would this choice of contraction ever matter? $\endgroup$ – Craig Apr 4 '19 at 16:23
  • $\begingroup$ The tetrad is being doubly contracted, on $\mu$ with the covariant derivative and on $a$ with the gamma matrix. So the first term, if expanded out, would be 16 terms. $\endgroup$ – G. Smith Apr 4 '19 at 16:25

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