The Kasner Solution

Abstract

The Schwarzschild solution describes a space-time geometry whose three-dimensional spatial sections are rotational invariant and thus isotropic, without preferred directions. In this chapter we will present another exact solution of the Einstein equations with a spatial geometry which is homogeneous, i.e. position-independent, but anisotropic, and thus characterized by a different curvature along different spatial directions.

Appendices

Exercises Chap. 11

11.1

The Milne Space-Time Show that the Milne line-element (11.25) can be obtained from the Minkowski line-element through the global transformation

$$\begin{aligned} ct = ct' \cosh \left( x'\over \lambda \right) , ~~~~~~~ x= ct' \sinh \left( x'\over \lambda \right) , \end{aligned}$$

(11.26)

where $\lambda$ is a constant parameter, and (ct, x) are the coordinates of the Minkowski plane. Compute the Riemann tensor for the Milne metric, and show that the Milne coordinates $(ct', x')$ only cover the region inside the light cone of the Minkowski plane.

11.2

Anisotropic Einstein Equations from a Variational Principle Deduce Eqs. (11.8), (11.9), in the absence of sources, starting from the effective Einstein action for a Bianchi-I type metric and applying the variational principle.

Solutions

11.1

Solution By differentiating Eq. (11.26) we obtain

$$\begin{aligned}&c dt= c dt' \cosh \left( x'\over \lambda \right) + dx' {c t'\over \lambda } \sinh \left( x'\over \lambda \right) , \nonumber \\&dx= c dt' \sinh \left( x'\over \lambda \right) + dx' {c t'\over \lambda } \cosh \left( x'\over \lambda \right) . \end{aligned}$$

(11.27)

Inserting this result into the Minkowski line-element we obtain the Milne line-element,

$$\begin{aligned} ds^2= c^2 dt^2 - dx^2= c^2 dt^{{\prime }2}- \left( ct' \over \lambda \right) ^2 dx^{{\prime }2}, \end{aligned}$$

(11.28)

with the same Milne metric as that of Eq. (11.25),

$$\begin{aligned} g_{00}=1, ~~~~~~~~ g_{11}= - \left( t'\over t_0\right) ^2, \end{aligned}$$

(11.29)

where $t_0= \lambda /c$.

The Riemann tensor for this metric is identically vanishing. Using for the connection components the results (11.4) we have, in fact,

$$\begin{aligned} \varGamma _{01}{}^1= {1\over c t'}, ~~~~~~~ ~~~ \varGamma _{11}{}^0= {t'\over c t_0^2}, \end{aligned}$$

(11.30)

so that

$$\begin{aligned}&R_{101}\,^0= -{1\over c^2 t_0^2}+{1\over c^2 t^2_0} \equiv 0, \nonumber \\&R_{100}\,^1={1\over c^2 t^{{\prime }2}}-{1\over c^2 t^{{\prime }2}} \equiv 0. \end{aligned}$$

(11.31)

Let us finally notice that, from the transformation (11.26), we have:

$$\begin{aligned} {x\over ct}= \tanh \left( x'\over \lambda \right) , ~~~~~~~~~~ c^2 t^2- x^2 = c^2 t^{{\prime }2}. \end{aligned}$$

(11.32)

The first equation, at fixed $x'$, represents a straight line crossing the origin of the Minkowski plane, and forming with the ct axis an angle between $-\pi /4$ and $\pi /4$. The second equation, at fixed $t'$, represents a hyperbola centered on the origin, with asymptotes given by the lines $x=\pm ct$, which intersects the ct axis for $t= \pm t'$. By varying $x'$ and $t'$ the two curves span the portion of Minkowski plane internal to the light cone, defined by

$$\begin{aligned} ct>|x|, ~~~~~~~~~~ ct<-|x|, \end{aligned}$$

(11.33)

namely the region also called “Milne space-time”. This region is just the complement of the so-called Rindler space-time, covering the region of Minkowski plane exterior to the light cone (see Exercise 6.1 ) .

11.2

$\mathbf{. Solution }$ In order to obtain all equations of motion, and in particular the (0, 0) component of the Einstein equations, the effective action must contain all relevant metric components. So, let us start with anisotropic metric (11.2) without imposing the synchronous gauge $g_{00}=1$, and define

$$\begin{aligned} g_{00}= N^2(t), ~~~~~~~~~~~ g_{ij}= - a^2_i(t) \delta _{ij}. \end{aligned}$$

(11.34)

The nonvanishing components of the connection are now given by

$$\begin{aligned} \varGamma _{0i}{}^j= H_i \delta _i^j, ~~~~~~~~~~ \varGamma _{ij}{}^0= {a_i \dot{a}_i\over N^2} \delta _{ij}, ~~~~~~~~~~ \varGamma _{00}{}^0 = F, \end{aligned}$$

(11.35)

where $F= \dot{N}/N$, and the scalar curvature becomes

$$\begin{aligned} R= {1\over N^2} \left[ 2 F \sum _i H_i- \sum _i \left( 2 \dot{H}_i + H_i^2 \right) - \left( \sum _i H_i \right) ^2 \right] . \end{aligned}$$

(11.36)

This is more general than Eq. (11.6), because of the contributions of $g_{00}=N^2$. We have, also,

$$\begin{aligned} \sqrt{-g}= N \prod _i a_i, \end{aligned}$$

(11.37)

and the effective Einstein action takes the form

$$\begin{aligned} S= & {} -{1\over 2\chi } \int d^{d+1}x \sqrt{-g} \, R \nonumber \\= & {} -{1\over 2\chi } \int d^{d}x {dt\over N} \prod _i a_i \left[ 2 F \sum _i H_i- \sum _i \left( 2 \dot{H}_i + H_i^2 \right) - \left( \sum _i H_i \right) ^2 \right] . \nonumber \\&\end{aligned}$$

(11.38)

We can note, at this point, that

$$\begin{aligned}&{d \over dt} \left[ {2\over N} \prod _i a_i\sum _i H_i\right] = \nonumber \\&~~~ ={1\over N}\prod _i a_i \left[ 2 \sum _i \dot{H}_i- 2 F \sum _i H_i +2\left( \sum _iH_i\right) ^2 \right] . \end{aligned}$$

(11.39)

By eliminating through the above relation the terms linear in F and $\dot{H}$ of Eq. (11.38), we can rewrite the action (modulo a total time derivative) in the following standard quadratic form:

$$\begin{aligned} S=-{1\over 2\chi } \int {dt\over N} \prod _i a_i \left[ \left( \sum _i H_i\right) ^2 - \sum _i H_i^2\right] . \end{aligned}$$

(11.40)

Notice that the variable N has no kinetic term, hence it plays the role of auxiliary field (or Lagrange multiplier): it is not a dynamical field, and can be set to a constant—after the variation—with an appropriate gauge choice.

We are now in the position of deriving the field equations by varying with respect to the variables $N, a_i$, and imposing the condition of stationary action, $\delta S=0$. The variation with respect to N gives the constraint

$$\begin{aligned} \left( \sum _i H_i\right) ^2- \sum _i H_i^2=0, \end{aligned}$$

(11.41)

corresponding to Eq. (11.8) with $\rho =0$.

In order to vary with respect to the spatial metric components we can conveniently set $a_i= \exp \alpha _i$, so that $H_i= \dot{\alpha }_i$, and the effective action becomes

$$\begin{aligned} S=-{1\over 2 \chi } \int dt \, L(\alpha _i, \dot{\alpha }_i), \end{aligned}$$

(11.42)

where

$$\begin{aligned} L= {\exp (\sum _i \alpha _i)\over N} \left[ \left( \sum _i \dot{\alpha }_i \right) ^2- \sum _i \dot{\alpha }_i^2 \right] . \end{aligned}$$

(11.43)

The variation with respect to $\alpha _i$ leads to the Lagrange equations of motion for this new variable. Computing the derivatives, and then imposing the synchronous gauge $N=1$, we obtain

$$\begin{aligned} { \partial L\over \partial \alpha _i}= & {} \exp \left( \sum _k \alpha _k \right) \left[ \left( \sum _k \dot{\alpha }_k\right) ^2- \sum _k \dot{\alpha }_k^2 \right] , \nonumber \\ {\partial L\over \partial \dot{\alpha }_i}= & {} \exp \left( \sum _k \alpha _k \right) \left[ 2 \sum _k \dot{\alpha }_k- 2 \dot{\alpha }_i \right] , \nonumber \\ {d \over dt} {\partial L\over \partial \dot{\alpha }_i}= & {} \exp \left( \sum _k \alpha _k \right) \sum _k \dot{\alpha }_k \left[ 2 \sum _k \dot{\alpha }_k- 2 \dot{\alpha }_i \right] + \nonumber \\+ & {} \exp \left( \sum _k \alpha _k \right) \left[ 2 \sum _k \ddot{\alpha }_k- 2 \ddot{\alpha }_i \right] . \end{aligned}$$

(11.44)

The Lagrange equations for $\alpha _i$ thus imply:

$$\begin{aligned} \left( \sum _k \dot{\alpha }_k\right) ^2 - 2 \dot{\alpha }_i \sum _k \dot{\alpha }_k+ 2 \sum _k \ddot{\alpha }_k - 2 \ddot{\alpha }_i +\sum _k \dot{\alpha }_k^2=0. \end{aligned}$$

(11.45)

Multiplying by $-1/2$, and replacing $\dot{\alpha }_i$ with $H_i$, we can rewrite the above equation in the form

$$\begin{aligned} \dot{H}_i+ H_i \sum _k H_k -\sum _k \dot{H}_k -{1\over 2} \sum _k H_k ^2 -{1\over 2} \left( \sum _k H_k \right) ^2 =0, \end{aligned}$$

(11.46)

which exactly coincides with the component $i=j$ of Eq. (11.9), written in the absence of sources ($p_i=0$).