# Explicit form of $\gamma^\mu \partial_\mu$ in the Dirac equation

I'm in an introductory particle physics class, and in performing manipulations on the Dirac equation, my instructor expands the $\gamma^\mu \partial_\mu$ term as:

$$\gamma^\mu \partial_\mu = \gamma^0 \frac{\partial}{\partial t} + \vec{\gamma}\cdot \vec{\nabla}$$

where $\vec{\gamma} = (\gamma^1,\gamma^2,\gamma^3)$, but to my knowledge,

$$\gamma^\mu \partial_\mu = \gamma^\mu \eta_{\mu \nu} \partial^\nu = \gamma^0\frac{\partial}{\partial t} - \vec{\gamma} \cdot \vec{\nabla}$$

using the convention $\eta_{\mu \nu}=\operatorname{diag}(+,-,-,-)$. Am I missing something here?

• you a forgeting an extra minus sign on the $\partial^\nu = (\frac{\partial}{\partial t}, - \nabla)$ – Hydro Guy May 24 '13 at 21:15

Yes. You are missing the fact that he is using the convention $$\nabla = (\partial_1, \partial_2, \partial_3)$$ as opposed to $$\nabla = (\partial^1, \partial^2, \partial^3)$$ The first convention is by far the most common in my experience.
• I see. Now that I think about it, we didn't actually discuss the difference between $\partial_\mu$ and $\partial^\mu$. Is the only difference a minus sign in the spatial part? – Gino May 24 '13 at 21:22
• @PeterKravchuk Well, sure; it's a well-motivated convention that is consistent with other notational conventions used for vectors and covectors, but in this context the distinction is largely artificial anyway because of the tangent-cotangent isomorphism on semi-Riemannian manifolds. Also, besides all of this highbrow talk, at some point you have to choose a definition for the symbol $\nabla$, and you might think one definition is more natural than another, but you certainly could have picked the other one if you really wanted to. – joshphysics May 24 '13 at 21:43