The Einstein Equations for the Gravitational Field

Abstract

With the inclusion of the Riemann tensor, introduced in the previous chapter, we have completed the list of the geometric ingredients needed to formulate a relativistic theory of gravity: the metric, the affine connection, and the curvature.

Appendices

Exercises Chap. 7

7.1

Variational Contribution of the Ricci Tensor Show that the variational contribution of the Ricci tensor, appearing in Eq. (7.8), can be written in explicitly covariant form as

$$\begin{aligned} g^{\mu \nu } \delta R_{\mu \nu }= \nabla _\mu \left( g_{\alpha \beta } \nabla ^\mu \delta g^{\alpha \beta } - \nabla _\nu \delta g^{\mu \nu }\right) . \end{aligned}$$

(7.84)

Check that this expression immediately leads to the boundary contribution (7.16).

7.2

Energy-Momentum Conservation for a Scalar Field Show that the covariant divergence of the dynamical energy-momentum tensor (7.40) is vanishing, provided the equations of motion of the scalar field are satisfied.

7.3

Second Covariant Derivative of a Killing Vector Derive the result of Eq. (7.76) by exploiting the properties of the Killing vectors and of the Riemann curvature tensor.

Solutions

7.1

Solution In order to obtain Eq. (7.84) we may conveniently adopt a local inertial system (see Sect. 2.2), where $g=$ cost, $\varGamma =0$, $\partial \varGamma \not =0$, and where we can set $\delta g =0$, keeping, however, $\partial \delta g \not =0$. Starting with the definition (6.21) of the Ricci tensor we have, in this system,

$$\begin{aligned} \delta R_{\mu \nu }\big |_{\varGamma =0}&= \partial _\alpha \left( \delta \varGamma _{\mu \nu }\,^\alpha \right) -\partial _\mu \left( \delta \varGamma _{\alpha \nu }\,^\alpha \right) \nonumber \\&={1\over 2}g^{\alpha \beta } \partial _\alpha \big ( \partial _\mu \delta g_{\nu \beta } +\partial _\nu \delta g_{\mu \beta }-\partial _\beta \delta g_{\mu \nu }\big )-{1\over 2} g^{\alpha \beta }\partial _\mu \partial _\nu \delta g_{\alpha \beta }. \nonumber \\ \end{aligned}$$

(7.85)

By tracing we obtain

$$\begin{aligned} \Big (g^{\mu \nu } \delta R_{\mu \nu }\Big )_{\varGamma =0}= \partial ^\beta \partial ^\nu \delta g_{\nu \beta }-g^{\alpha \beta } \partial _\mu \partial ^\mu \delta g_{\alpha \beta }. \end{aligned}$$

(7.86)

By switching from the inertial frame to a more general coordinate frame our partial derivatives become covariant derivatives, and we are led to

$$\begin{aligned} g^{\mu \nu } \delta R_{\mu \nu }= \nabla ^\mu \nabla ^\nu \delta g_{\mu \nu }-g^{\alpha \beta } \nabla _\mu \nabla ^\mu \delta g_{\alpha \beta }. \end{aligned}$$

(7.87)

By recalling that $g^{\alpha \beta } \delta g_{\alpha \beta }=- g_{\alpha \beta } \delta g^{\alpha \beta }$, and using the metric property $\nabla g=0$, we finally arrive at the result (7.84):

$$\begin{aligned} g^{\mu \nu } \delta R_{\mu \nu }&=g_{\alpha \beta } \nabla _\mu \nabla ^\mu \delta g^{\alpha \beta }-\nabla _\mu \nabla _\nu \delta g^{\mu \nu } \nonumber \\&\equiv \nabla _\mu \left( g_{\alpha \beta } \nabla ^\mu \delta g^{\alpha \beta }- \nabla _\nu \delta g^{\mu \nu }\right) . \end{aligned}$$

(7.88)

The boundary contribution (7.16) can be easily obtained starting from the covariant result (7.87). The integration of Eq. (7.87) over the space-time domain $\varOmega$, and the application of the Gauss theorem, gives, in fact,

$$\begin{aligned} -{1\over 2 \chi }\int _{\partial \varOmega } d^3 \xi \sqrt{|h|}\,n^\mu \left( g^{\nu \alpha } \nabla _\alpha \delta g_{\mu \nu } - g^{\alpha \beta } \nabla _\mu \delta g_{\alpha \beta } \right) . \end{aligned}$$

(7.89)

On the boundary, however, $\delta g=0$, so that the only non-zero contributions to the above integral are provided by the partial-derivative terms inside the round brackets. We thus find a boundary contribution proportional to

$$\begin{aligned} n^\mu g^{\nu \alpha } \partial _\alpha \delta g_{\mu \nu }- g^{\alpha \beta } n^\mu \partial _\mu \delta g_{\alpha \beta }, \end{aligned}$$

(7.90)

which is exactly the same as the result given in Eq. (7.16).

7.2

Solution Let first derive the covariant equation of motion for the scalar field $\phi$, coupled to the space-time geometry according to the action (7.37).

The variation of the action with respect to $\phi$ leads to the following Euler–Lagrange equations,

$$\begin{aligned} {\delta (\sqrt{-g}\mathcal{L}_m) \over \delta \phi } \equiv {\partial (\sqrt{-g}\mathcal{L}_m) \over \partial \phi }- \partial _\mu {\partial (\sqrt{-g}\mathcal{L}_m) \over \partial (\partial _\mu \phi )}=0, \end{aligned}$$

(7.91)

where, for the action density (7.38),

$$\begin{aligned}&{\partial (\sqrt{-g}\mathcal{L}_m) \over \partial \phi }=- \sqrt{-g} \,{\partial V\over \partial \phi }, \nonumber \\&{\partial (\sqrt{-g}\mathcal{L}_m) \over \partial (\partial _\mu \phi )}=\sqrt{-g} \,g^{\mu \nu } \partial _\nu \phi . \end{aligned}$$

(7.92)

The equation of motion thus takes the form

$$\begin{aligned} {1\over \sqrt{-g}} \partial _\mu \left( \sqrt{-g}g^{\mu \nu } \partial _\nu \right) \phi + {\partial V\over \partial \phi }=0, \end{aligned}$$

(7.93)

or, by recalling the definition (3.105) of the covariant D’Alembert operator:

$$\begin{aligned} \nabla _\mu \nabla ^\mu \phi + {\partial V\over \partial \phi }=0. \end{aligned}$$

(7.94)

Let us now compute the covariant divergence of the tensor (7.40):

$$\begin{aligned} \nabla _\nu T_\mu \,^\nu&= \nabla _\nu \big (\partial _\mu \phi \partial ^\nu \phi \big )-{1\over 2} \nabla _\mu \big (\partial _\alpha \phi \partial ^\alpha \phi \big )+ \nabla _\mu V \nonumber \\&= \big (\nabla _\nu \partial _\mu \phi \big ) \partial ^\nu \phi +\partial _\mu \phi \nabla ^2 \phi - \big (\nabla _\mu \partial _\alpha \phi \big ) \partial ^\alpha \phi + {\partial V\over \partial \phi } \partial _\mu \phi , \end{aligned}$$

(7.95)

where $\nabla ^2\equiv \nabla _\nu \nabla ^\nu$. In the second line, the second and fourth terms cancel out each other thanks to the equation of motion (7.94), while the first and third term cancel thanks to the index symmetry:

$$\begin{aligned} \nabla _\nu \partial _\mu \phi =\partial _\nu \partial _\mu \phi - \varGamma _{\nu \mu }{}^\alpha \partial _\alpha \phi = \nabla _\mu \partial _\nu \phi . \end{aligned}$$

(7.96)

Hence:

$$\begin{aligned} \nabla _\nu T_\mu \,^\nu =0. \end{aligned}$$

(7.97)

7.3

Solution Let us apply the result (6.19) for the commutator of two covariant derivatives acting on a vector,

$$\begin{aligned} \nabla _\mu \nabla _\nu \xi _\alpha - \nabla _\nu \nabla _\mu \xi _\alpha = -R_{\mu \nu \alpha }{}^\beta \xi _\beta , \end{aligned}$$

(7.98)

and consider the totally antisymmetrized part of this equation. From the Bianchi identity (6.19) we have $R_{[\mu \nu \alpha ]}\,^\beta =0$; hence we obtain

$$\begin{aligned} \nabla _\mu \nabla _\nu \xi _\alpha +\nabla _\nu \nabla _\alpha \xi _\mu +\nabla _\alpha \nabla _\mu \xi _\nu -\nabla _\nu \nabla _\mu \xi _\alpha -\nabla _\mu \nabla _\alpha \xi _\nu -\nabla _\alpha \nabla _\nu \xi _\mu =0. \end{aligned}$$

(7.99)

On the other hand, by using the property (3.107) of the Killing vectors,

$$\begin{aligned} \nabla _\nu \xi _\alpha =- \nabla _\alpha \xi _\nu , \end{aligned}$$

(7.100)

we can rewrite Eq. (7.99) as

$$\begin{aligned} \nabla _\mu \nabla _\nu \xi _\alpha -\nabla _\nu \nabla _\mu \xi _\alpha = \nabla _\alpha \nabla _\nu \xi _\mu . \end{aligned}$$

(7.101)

Combining the above equation with Eq. (7.98) we are finally led to

$$\begin{aligned} \nabla _\alpha \nabla _\nu \xi _\mu = - R_{\mu \nu \alpha }{}^\beta \xi _\beta , \end{aligned}$$

(7.102)

which reproduces the result of Eq. (7.76).