The Dirac Equation in a Gravitational Field

Abstract

The aim of this chapter is to introduce a topic which is often omitted in books adopting a traditional approach to general relativity (with the due exceptions, see e.g. [24]): the gravitational interactions of spinors.

Appendices

Exercises Chap. 13

13.1

Dirac Equation in a Conformally Flat Space-Time Write down the explicit form of the Dirac equation for a massive spinor in a conformally flat space-time geometry, described by the metric

$$\begin{aligned} g_{\mu \nu }(x)= f^2 (x) \eta _{\mu \nu }. \end{aligned}$$

(13.58)

13.2

Trace of the Lorentz Connection Derive Eq. (13.55) for the trace of the Lorentz connection.

13.3

Energy-Momentum Tensor of a Dirac Spinor Compute the dynamical energy-momentum tensor (7.27) for a free massive Dirac field embedded in a curved Riemannian geometry.

Solutions

13.1

Solution The vierbein field associated to the metric (13.58), and defined in such a way as to satisfy Eq. (12.5), is given by

$$\begin{aligned} V_\mu ^a= f \delta _\mu ^a, \quad V^\mu _a = f^{-1} \delta ^\mu _a. \end{aligned}$$

(13.59)

The computation of the corresponding Ricci rotation coefficients, from Eq. (12.43), leads to

$$\begin{aligned} C_{ab}{}^c = {1\over 2 f^2}\left( \delta ^c_b \delta ^\mu _a- \delta ^c_a \delta ^\mu _b\right) \partial _\mu f. \end{aligned}$$

(13.60)

The trace of the Lorentz connection, according to Eq. (13.38), is then:

$$\begin{aligned} \omega _b{}^b{}_a=2 C_{ab}{}^b ={3\over f^2} \delta ^\mu _a\partial _\mu f. \end{aligned}$$

(13.61)

Let us now compute the antisymmetric part of the connection. From Eq. (13.60) we have

$$\begin{aligned} C_{abc} = {1\over 2 f^2}\left( \eta _{cb}\delta ^\mu _a- \eta _{ca} \delta ^\mu _b\right) \partial _\mu f. \end{aligned}$$

(13.62)

Hence, according to Eq. (13.38):

$$\begin{aligned} \omega _{[abc]}= C_{[abc]} \equiv 0. \end{aligned}$$

(13.63)

It follows that the Dirac equation (13.39) (or (13.57)) reduces to

$$\begin{aligned} \left( i f^{-1}\gamma ^a \delta ^\mu _a \partial _\mu - m +{3i\over 2 f^2} \gamma ^a \delta ^\mu _a \partial _\mu f \right) \psi =0. \end{aligned}$$

(13.64)

Multiplying by f we finally obtain

$$\begin{aligned} \left( i \gamma ^a \delta ^\mu _a \partial _\mu - mf +i{3\over 2} \gamma ^a \delta ^\mu _a \partial _\mu \ln f \right) \psi =0. \end{aligned}$$

(13.65)

We may thus conclude that the coupling to a conformally flat geometry generates an effective, position-dependent mass term $\widetilde{m} =mf$, and an “effective potential” described by the last term of the above equation.

13.2

Solution Let us start with the metric condition for the vierbein field, Eq. (12.39), which we can rewrite as

$$\begin{aligned} \omega _\mu {}^a{}_\nu = \varGamma _{\mu \nu }{}^a - \partial _\mu V_\nu ^a. \end{aligned}$$

(13.66)

Let us compute its trace, by applying $V^\mu _a$:

$$\begin{aligned} \omega _a\,^a\,_\nu&= \varGamma _{\mu \nu }\,^\mu - V^\mu _a \partial _\mu V_\nu ^a \nonumber \\&= {1\over \sqrt{-g}} \partial _\nu \sqrt{-g} + V_\nu ^a \partial _\mu V^\mu _a. \end{aligned}$$

(13.67)

Note that in the second line we have used the trace of the Christoffel connection and the relation

$$\begin{aligned} \partial _\mu \left( V_\nu ^a V^\mu _a\right) = \partial _\mu \left( \delta ^\mu _\nu \right) =0. \end{aligned}$$

(13.68)

By multiplying Eq. (13.67) by $V^\nu _b$ we finally obtain

$$\begin{aligned} \omega _a{}^a{}_b= {1\over \sqrt{-g} } \partial _b \sqrt{-g} + \partial _\mu V^\mu _b= {1\over \sqrt{-g} } \partial _\mu \left( \sqrt{-g} V^\mu _b\right) , \end{aligned}$$

(13.69)

which exactly reproduces Eq. (13.55) used in Sect. 13.4.

13.3

Solution Let us consider the covariant action (13.41), symmetrized with respect to $\psi$ and $\overline{\psi }$. Using our previous results for the Lagrangian (13.48) we can rewrite the symmetrized action in compact form as follows:

$$\begin{aligned} S= \int d^4 x \sqrt{-g} \left[ {i\over 2} g^{\mu \nu } \left( \overline{\psi }\gamma _\mu D_\nu \psi - D_\nu \overline{\psi }\gamma _\mu \psi \right) - m \overline{\psi }\psi \right] , \end{aligned}$$

(13.70)

where:

$$\begin{aligned}&D_\nu \psi = \partial _\nu \psi +{1\over 4} \omega _{\nu a b} \gamma ^{[a} \gamma ^{b]} \psi , \nonumber \\&D_\nu \overline{\psi }= \partial _\nu \overline{\psi }-{1\over 4} \omega _{\nu a b}\overline{\psi }\gamma ^{[a} \gamma ^{b]}. \end{aligned}$$

(13.71)

In order to obtain the dynamical energy-momentum tensor we have now to vary the action with respect to the metric, by imposing that the equations of motion of the Dirac field are satisfied (see e.g. Sect. 7.2). By applying the standard definition (7.27), in particular, we obtain

$$\begin{aligned} \delta S= {1\over 2} \int d^4x \sqrt{-g} \, T_{\mu \nu } \delta g^{\mu \nu }, \end{aligned}$$

(13.72)

where

$$\begin{aligned} T_{\mu \nu } = i \overline{\psi }\gamma _{(\mu } D_{\nu )} \psi -i D_{(\nu }\overline{\psi }\gamma _{\mu )} \psi \end{aligned}$$

(13.73)

is the sought energy-momentum tensor.

It should be noted that the variation of $\sqrt{-g}$ (which is present in the action) does not contribute to $T_{\mu \nu }$ as a consequence of the equations of motion (13.28), which impose on the Dirac field the conditions:

$$\begin{aligned} i\gamma ^\mu D_\mu \psi = m\psi , \quad i D_\mu \overline{\psi }\gamma ^\mu = -m\overline{\psi }. \end{aligned}$$

(13.74)