In all the Literature I have read, the covariant Dirac equation in curved spacetime is given as

\begin{equation} \left(i\hbar\gamma^{\mu}(x)\left[\frac{\partial}{{\partial}x^{\mu}}-{\Gamma}_{\mu}(x)\right]-mc\right)\psi(x)=0 \end{equation}

Where $\gamma^{\mu}(x)$ are the contravariant forms of curvature dependent Dirac matrices $\gamma_{\mu}(x)$ defined as

\begin{equation} \gamma_{\mu}(x)\gamma_{\nu}(x)+\gamma_{\nu}(x)\gamma_{\mu}(x)=2g_{\mu\nu}\end{equation}

None of the references I have gives any justification for using contravariant form of the curvature dependent Dirac matrices. My question is why do we use the contravariant form $\gamma^{\mu}(x)$ and not the covariant form $\gamma_{\mu}(x)$?

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    $\begingroup$ Why should we do that? $\endgroup$
    – Horus
    Sep 23, 2015 at 15:23
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    $\begingroup$ Uh...you have to couple a contravariant something to the covariant $\partial_\mu$ in there toget something diffeomorphism invariant, no? $\endgroup$
    – ACuriousMind
    Sep 23, 2015 at 15:27
  • $\begingroup$ Because then we can derive the Dirac equation in curved spacetime from the energy-momentum relation in curved spacetime, just like Dirace equation in flat spacetime is derived from energy-momentum relation in flat spacetime. See my other question link $\endgroup$ Sep 23, 2015 at 15:30
  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/208322/2451 $\endgroup$
    – Qmechanic
    Sep 23, 2015 at 15:34
  • $\begingroup$ @amateurRebel To whom is that comment meant for? $\endgroup$
    – Horus
    Sep 23, 2015 at 15:38

3 Answers 3


The equation you gave, as per my understanding, is only covariant under generalized Lorentz transformation in curved space-time. It is not generally covariant (though if I'm wrong on that someone please correct me).

You can use Clifford algebra to write Dirac's equation, and extend it to be generally covariant, but you get a slightly different equation than the one you posted.


While father Dirac already wrote down his Dirac equation with covariant derivatives, simply by formulating the Lagrangian and generalizing the $\gamma$-matrix algebra to the Clifford algebra for a Riemann manifold $$\gamma(x)^i \gamma(x)^k + \gamma(x)^k \gamma(x)^i = -2\ g(x)^{ik}$$ most of its content remained unclear, mostly because no theorist ever used position dependent $\gamma$ matrices.

The only leading principle for non-covariant, constant $\gamma$ formulations: the squared massive free Dirac operator has to yield the Klein-Gordon operator, coordinate independent, a form the arises by carefully introducing a curvilinear coordinate system in flat space with diagonal metrics.

$$g(x)_{ik} = \delta_{ik} h(x)_i h(x)_k$$

$$H^2 = (-i \vec \alpha \cdot \vec\nabla + \beta m)^2 = -\Delta_{Dirac} + m^2 = -\frac{1}{\sqrt{h_1 h_2 h_3}} \partial_{x^i}\left( \frac{\sqrt{h_1 h_2 h_3}}{h_i^2}\partial_{x_i}\right) + m^2$$

The destastrous presentation (from a pure mathematics view) in all physics text books comes from the fact, that the eigenvalue problem of the H-atom can be solved without ever using coordinate maps.

In spherical coordinates, the trick used is purely algebraic without any analysis

$$\vec \alpha \cdot\vec p = \left(\frac{\vec \sigma \cdot \vec r}{r}\right)^2 \ \vec \alpha \cdot\vec p= \frac{\vec \sigma \cdot \vec r}{r} \left( \frac{\vec \sigma \cdot r}{r} \ \ \vec \alpha \cdot\vec p\right) = \frac{\vec \alpha \cdot \vec r}{r} \ \hat r \cdot \vec p + \frac{i}{r} \vec \alpha \cdot \vec r \times \vec p = \alpha_r p_r + \frac{i}{r}\vec \alpha \cdot \vec L$$

$$\vec \alpha \cdot\vec p = \alpha_r \left( p_r -\frac{i}{r}\right) + \frac{i}{r}\left( \vec \alpha \cdot \vec L +\alpha_r \right)$$

The solutions can be deerived using pairs of the Schrödinger eigenfunctions.

Finally it became a dogma in QED, that any relevant calculation can be done coordinate free in the constant $\gamma$-basis.

The Dirac operator on smooth Riemannian manifolds was reinvented in topology in the 1960ties by Athiya, Singer as the most elementary elliptic operator.

They simply wrote

$$D = \sum e_i \nabla_i $$

where $e_i$ is the tangent coordinate basis and $\nabla_i$ the corresponding covariant derivative with the Christoffel basis transform: $$\nabla_i e_k - e_k \nabla_i = \sum_s e_s \Gamma^s_{ki}$$ and the tangent basis being the Clifford basis of $Cl(n)$ with $$e_i e_k+e_k e_i = -2 \ g_{ik}$$

This work was revolutionary, as it finally made possible to solve many old problems in differential geometry and eeven provided the path to find new solutions to Einsteins equations.

After development of the classes of smooth, complex, orientation or vector fields, used to classify compact manifolds, spinor bundles became the first step in the theory of general vector and fiber bundles.

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    $\begingroup$ The Dirac equation in curved space is due to Fock, and Ivanenko, "Geometrie quantique lineaire et deplacement parallel". , Compt. Rend. Acad. Sci. Paris 188, 1470-1472 (1929) and Fock alone " Geometrisierung der Diracschen Theorie des Elektrons" Zeits. Phys. 57, 261-277 (1929). $\endgroup$
    – mike stone
    Jan 26 at 13:23

It is much easier to explain the origin of contravariant $\gamma^{\mu}(x)$ using differential forms: In short, covariant $\gamma_{\mu}(x)$ corresponds to frame/tetrad/veirbein 1-form, whereas contravariant $\gamma^{\mu}(x)$ corresponds to frame/tetrad/veirbein 3-form. Why do we care about 3-form? Below is the explanation.

Let's define covariant 1-forms $$ \gamma(x) = \gamma_{\mu}(x)dx^{\mu} \\ D= (\frac{\partial}{{\partial}x^{\mu}}-{\Gamma}_{\mu}(x))dx^{\mu} $$ Dirac equation in curved spacetime in term of differential forms is: $$ \begin{equation} \left(i\hbar\gamma(x)^3\wedge D-mc\gamma(x)^4\right)\psi(x)=0 \end{equation} $$ where $$ \gamma(x)^3 = \gamma(x)\wedge \gamma(x)\wedge \gamma(x) \\ \gamma(x)^4 = \gamma(x)\wedge \gamma(x)\wedge \gamma(x)\wedge \gamma(x) $$ If we divide both sides of Dirac equation by $|\gamma(x)^4|$, you can see that the contravariant $\gamma^{\mu}(x)$ in equation $\begin{equation} \left(i\hbar\gamma^{\mu}(x)\left[\frac{\partial}{{\partial}x^{\mu}}-{\Gamma}_{\mu}(x)\right]-mc\right)\psi(x)=0 \end{equation}$ are nothing but components of the 3-form $\gamma(x)^3$ normalized by the volume 4-form $|\gamma(x)^4|$: $$ \frac{\gamma(x)^3}{|\gamma(x)^4|}. $$

In summary, whenever you couple a contravariant something to the covariant something to enforce diffeomorphism invariance, there is a 3-form (normalized by the volume 4-form) involved.


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