# Is the partial derivative in the Dirac equation in curved space contracted with a tetrad?

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.
• And just to be clear, this $\partial_\mu = (\frac{1}{c}\partial_t, \nabla)$ even though we're working in curved space? – Craig Apr 4 '19 at 18:25
$$D_\mu$$ 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.
• 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? – Craig Apr 4 '19 at 16:23
• 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. – G. Smith Apr 4 '19 at 16:25