Geodesic family of spherical instantons and cosmic quantum creation

Abstract

The Einstein field equations for any spherically symmetric metric and a geodesic perfect fluid source are cast in a canonical simple form, both for Lorentzian metrics and for instantons. Both kinds of metrics are explicitly written for the Lemaître–Tolman–Bondi family and for a general $\varLambda$-Friedmann–Lemaître–Robertson–Walker universe. In the latter case (including of course the instanton version) we study whether the probability of quantum creation of our Universe vanishes or not. It is found, in accordance with previous results, that only the closed model can have a nonzero probability for quantum creation. To obtain this result, we resort to general assumptions, which are satisfied in the particular creation case considered by Vilenkin. On the other hand, Fomin and Tryon suggested that the energy of a quantically creatable universe should vanish. This is in accordance with the above result in which only the closed $\varLambda$FLRW model is quantically creatable while the open and flat models are not. That is so since it can be seen that this closed model has vanishing energy while the open model and the limiting flat case (suitably perturbed) have both infinite energy.

Appendices

Appendix 1: Isotropic instanton and “conservation equations”, $\nabla \cdot {T} = 0$

Let $V_4$ be a four-dimensional manifold and g a metric on $V_4$ whose signature is $(\epsilon , +, +, +)$, with $\epsilon = -1$ (Lorentzian case) or $\epsilon = 1$ (Euclidean case) and let us consider an isotropic stress-energy tensor T (see [14])

$$\begin{aligned} T = \mu \, n \otimes n + p \, \gamma , \end{aligned}$$

(63)

where $\gamma \equiv g - \epsilon \, n \otimes n$,   $n^2 \equiv g(n, n) = \epsilon$, and

$$\begin{aligned} \lambda _s \equiv \epsilon \, \mu , \qquad \lambda _t \equiv p, \end{aligned}$$

(64)

with $\lambda _s$ ($\lambda _t$) the simple (triple) eigenvalue of T,

$$\begin{aligned} T(n) = \epsilon \, \mu \, n, \quad T(v) = p \, v \end{aligned}$$

(65)

for any vector v ortogonal to n, $g(n,v) = 0$. Then

$$\begin{aligned} T = \epsilon \, (\lambda _s - \lambda _t) \, n \otimes n + \lambda _t \, g = (\mu - \epsilon \, p) \, n \otimes n + p \, g, \end{aligned}$$

(66)

that, in the Lorentzian case, is the expression for the stress-energy tensor of a perfect fluid,

$$\begin{aligned} T_{L} = (\mu + p) \, n \otimes n + p \, g, \end{aligned}$$

(67)

and, in the Euclidean case, it becomes

$$\begin{aligned} T_{E} = (\mu - p) \, n \otimes n + p \, g, \end{aligned}$$

(68)

describing the “stresses” associated to an instanton field.

For a conserved T, $\nabla \cdot T =0$, the divergence of (63) gives

$$\begin{aligned} \nabla _\alpha T^{\alpha \beta }= & {} u^{\alpha } \partial _\alpha (\mu - \epsilon p) \, n^{\beta } \nonumber \\&+ (\mu - \epsilon p) [\theta \, n^{\beta } + a^{\beta }] + (\partial _\rho p) \, g^{\rho \beta } = 0, \end{aligned}$$

(69)

with $\theta \equiv \nabla _\alpha n^{\alpha }$, $a^{\beta } \equiv n^{\alpha } \nabla _\alpha n^{\beta }$, whose decomposition in parts parallel and orthogonal to $n^{\alpha }$ leads to

$$\begin{aligned} \dot{\mu } = - (\mu - \epsilon p) \, \theta , \end{aligned}$$

(70)

and

$$\begin{aligned} (\mu - \epsilon p) \, a_\beta = - \partial _\beta p + \epsilon \, \dot{p} \, n_\beta , \end{aligned}$$

(71)

respectively, where the dot denotes derivation along $n^{\alpha }$.

Appendix 2: Spherically symmetric Einstein equations for both instantons and Lorentzian solutions with geodesic perfect flows

The material in this appendix can be found in most relativity textbooks that mainly deal with the Lorentzian case (see for example [18]). The case of a general diagonal metric is considered in [45]. This material is included here for sake of completeness and in order to be precise about the sign conventions we use.

To begin with, we take for the Riemann tensor definition of $(V_4, g)$

$$\begin{aligned} R^{\alpha }_{\,\, \beta \mu \nu } = \partial _\mu \varGamma ^{\alpha }_{\nu \beta } - \partial _\nu \varGamma ^{\alpha }_{\mu \beta } + \varGamma ^{\alpha }_{\mu \lambda } \varGamma ^\lambda _{\nu \beta } - \varGamma ^{\alpha }_{\nu \lambda } \varGamma ^\lambda _{\mu \beta }, \end{aligned}$$

(72)

and for the Ricci tensor

$$\begin{aligned} R_{\mu \nu } = R^{\alpha }_{\,\,\mu \alpha \nu }. \end{aligned}$$

(73)

If the metric $V_4$ is spherically symmetric, in a Gauss coordinate system, $\{u, \rho , \theta , \phi \}$, adapted to the symmetry the line element is written as:

$$\begin{aligned} \hbox {d}s^2 = \epsilon \,\hbox {d}u^2 + B(u, \rho )\,\hbox {d}\rho ^2 + C(u, \rho )\, \hbox {d}\sigma ^2, \end{aligned}$$

(74)

where $\hbox {d}\sigma ^2 = \hbox {d}\theta ^2 + \sin ^2 \theta \, \hbox {d}\phi ^2$ stands for the metric on the unit two-sphere. According to the sign conventions (72) and (73), the essential components of the Ricci tensor in these coordinates are given by:

$$\begin{aligned} R_{uu}= & {} - \frac{\ddot{B}}{2B} - \frac{\ddot{C}}{C} + \frac{\dot{B}^2}{4B^2} + \frac{\dot{C}^2}{2C^2}, \end{aligned}$$

(75)

$$\begin{aligned} R_{\rho \rho }= & {} - \frac{\epsilon }{2} \left( \ddot{B} - \frac{\dot{B}^2}{2B} + \frac{\dot{B} \dot{C}}{C}\right) + \frac{1}{C} \left( - C'' + \frac{C'^2}{2C} + \frac{B' C'}{2 B} \right) , \end{aligned}$$

(76)

$$\begin{aligned} R_{\theta \theta }= & {} - \frac{\epsilon }{2} \left( \ddot{C} + \frac{\dot{B} \dot{C}}{2B}\right) - \frac{1}{2B}\left( C'' - \frac{B' C'}{2B} \right) +1,\end{aligned}$$

(77)

$$\begin{aligned} R_{\phi \phi }= & {} R_{\theta \theta } \sin ^2 \theta , \end{aligned}$$

(78)

$$\begin{aligned} R_{u\rho }= & {} \frac{1}{C} \left( - \dot{C}' + \frac{\dot{C} C'}{2C} + \frac{\dot{B} C'}{2B}\right) , \end{aligned}$$

(79)

the remaining components being identically zero by virtue of the assumed symmetry.

The Einstein field equations

$$\begin{aligned} Ric(g) - \frac{R}{2} g + \varLambda g = \kappa T, \quad \kappa \equiv 8 \pi G, \end{aligned}$$

(80)

are written in equivalent form as

$$\begin{aligned} Ric(g) = \kappa T + (\varLambda - \frac{\kappa }{2} \mathrm{tr} T) \, g. \end{aligned}$$

(81)

Then for an isotropic source T given by (63), Eq. (81) becomes

$$\begin{aligned} Ric(g) = \epsilon [\varLambda + \frac{\kappa }{2} (\lambda _s - 3 \lambda _t)] \, n \otimes n + [\varLambda - \frac{\kappa }{2} (\lambda _s - \lambda _t)] \, \gamma . \end{aligned}$$

(82)

When the field n is geodesic, that is, when $\partial _\rho p =0$, adapting Gauss coordinates $(u, \rho , \theta , \phi )$ so that $n= \epsilon \, \hbox {d}u = (\epsilon , 0, 0, 0)$, the Einstein equations are:

$$\begin{aligned}&- \frac{\ddot{B}}{2B} - \frac{\ddot{C}}{C} + \frac{\dot{B}^2}{4B^2} + \frac{\dot{C}^2}{2C^2} = \epsilon \, [ \varLambda + \frac{\kappa }{2} (\lambda _s - 3 \lambda _t)],\end{aligned}$$

(83)

$$\begin{aligned}&- \frac{\epsilon }{2} \left( \ddot{B} - \frac{\dot{B}^2}{2B} + \frac{\dot{B} \dot{C}}{C}\right) + \frac{1}{C} \left( - C'' + \frac{C'^2}{2C} + \frac{B'C'}{2 B} \right) = [ \varLambda - \frac{\kappa }{2} (\lambda _s + \lambda _t)] B,\qquad \quad \end{aligned}$$

(84)

$$\begin{aligned}&- \frac{\epsilon }{2} \left( \ddot{C} + \frac{\dot{B} \dot{C}}{2B}\right) - \frac{1}{2B}\left( C'' - \frac{B' C'}{2B} \right) +1 = [ \varLambda - \frac{\kappa }{2} (\lambda _s + \lambda _t)] C,\end{aligned}$$

(85)

$$\begin{aligned}&- \dot{C}' + \frac{\dot{C} C'}{2C} + \frac{\dot{B} C'}{2B} = 0. \end{aligned}$$

(86)

The integration of these equations involves two separated cases. The case $C'=0$ leads to a generalized family of Datt metrics with its associated instanton family, and will be analyzed elsewhere. Here, let us consider the generic case $C' \ne 0$. Then (86) is written as:

$$\begin{aligned} \frac{\partial }{\partial u}\left( \ln \frac{C'^{2}}{CB}\right) = 0, \end{aligned}$$

(87)

and then $\chi \equiv C'^{2}/ CB$ does not depend on the u coordinate, i.e.,

$$\begin{aligned} B(u, \rho ) = \frac{C'^{2}}{C \chi (\rho )} = \frac{4 A'^{2}}{\chi (\rho )}, \end{aligned}$$

(88)

where we have put $C \equiv A^2$. Defining $\chi (\rho ) \equiv 4 (1 + \epsilon K(\rho ))$, the metric (74) becomes:

$$\begin{aligned} \hbox {d}s^2 = \epsilon \hbox {d}u^2 + \frac{A'^{2}}{1 + \epsilon K(\rho )} \hbox {d}\rho ^2 + A^2 \hbox {d}\sigma ^2. \end{aligned}$$

(89)

The following linear combination

$$\begin{aligned} \displaystyle {\mathrm{Eq. \, (83)} - \frac{\epsilon }{B} \, \mathrm{Eq. \, (84)} + 2 \, \frac{\epsilon }{C} \, \mathrm{Eq. \, (85)}}, \end{aligned}$$

(90)

leads to

$$\begin{aligned} -2 \frac{\ddot{C}}{C} + \frac{\dot{C}^2}{2 C^2} + \frac{\epsilon }{C}\left( 2 - \frac{C'^2}{2 B C}\right) = 2 \epsilon \, (\varLambda - \kappa \, \lambda _t), \end{aligned}$$

(91)

and taking into account (88) with $C=A^2$, we obtain the following equation for A:

$$\begin{aligned} \dot{A}^2+2A\ddot{A} + K + \epsilon \, \varLambda \, A^2 = \epsilon \, \kappa \, \lambda _t \, A^2. \end{aligned}$$

(92)

By substituting (92) in (83), we arrive at

$$\begin{aligned} 2 \frac{\ddot{A}}{A}+\frac{\ddot{A'}}{A'} + \, \epsilon \, \varLambda =- \frac{\kappa }{2} \, \epsilon \, (\lambda _s -3 \lambda _t). \end{aligned}$$

(93)

Then, the remaining Einstein equations (84) and (85) are identically satisfied by substitution of (92) and (93) in them.

Therefore, the Einstein equations for the, let us say, generalized $\varLambda$LTB family of metrics (89) (now the pressure p is homogeneous, $p = p(u)$ and can be different from zero) reduce to (92) and (93), or equivalently, to the following equations:

$$\begin{aligned} 3 \, \frac{\dot{A}^2}{A^2} + 2 \left( \frac{\ddot{A}}{A} - \frac{\ddot{A'}}{A'}\right) + 3 \, \frac{K}{A^2} + \epsilon \, \varLambda = \kappa \, \epsilon \, \lambda _s, \end{aligned}$$

(94)

and

$$\begin{aligned} 2 \, \frac{\ddot{A}}{A} + \frac{\dot{A}^2}{A^2} + \frac{K}{A^2} + \epsilon \, \varLambda = \epsilon \, \kappa \, \lambda _t. \end{aligned}$$

(95)

In the case $p= \varLambda = 0$, we recover the LTB-family and the associated instanton, which are both governed by the following equations:

$$\begin{aligned}&\dot{A}^2+2A\ddot{A} + K=0,\end{aligned}$$

(96)

$$\begin{aligned}&2 \frac{\ddot{A}}{A}+\frac{\ddot{A'}}{A'}=-4\pi \mu , \end{aligned}$$

(97)

that do not depend on $\epsilon$. Nevertheless, despite the above pair of equations, (96) and (97), are signature independent, the corresponding metric components, $g_{\rho \rho } = A'^{2}/(1 + \epsilon K(\rho ))$ depend on the $\epsilon$ sign.

From (96), it results that $A^2 \ddot{A}$ does not depend on u. Defining $M(\rho ) \equiv - A^2 \ddot{A}$, Eqs. (96) and (97) are respectively written,

$$\begin{aligned} \dot{A}^2= & {} \frac{2}{A} M(\rho ) - K(\rho ),\end{aligned}$$

(98)

$$\begin{aligned} 4\pi \mu= & {} \frac{M'(\rho )}{A^2 A'}. \end{aligned}$$

(99)

Integration of (98) leads to (9)–(11).

Appendix 3: “Perfect fluid” instanton Lagrangian

In this Appendix, we detail the steps allowing to determine the instanton Lagrangian $L_E = \mu _E$ whose stress-energy tensor is the one considered in “Appendix 1”, that is (68). We follow closely the method developed in references [29, 33] for action functionals of matter fields, starting from the expression

$$\begin{aligned} T^{\mu \nu } = \frac{2}{\sqrt{|\mathrm {g}|}} \, \frac{\partial (L \sqrt{|\mathrm {g}|})}{\partial g_{\mu \nu }}, \end{aligned}$$

(100)

where $\mathrm{g}$ stands for the metric determinant, $\mathrm{g} \equiv \mathrm{det}g_{\alpha \beta }$. Let us consider the Lagrangian $L=f(\omega , \xi )$ and its associated action $S = \int f \sqrt{\epsilon \mathrm {g}} \, d^4x$. Substituting this L in (100) we will determine the function $f(\omega , \xi )$, which leads to the general expression of an isotropic tensor given by (63). We assume that $\omega$ is alike a thermodynamic variable that has associated a “conserved flux of particles”, say N:

$$\begin{aligned} N = \omega \, n, \qquad n^2 = \epsilon . \end{aligned}$$

(101)

In addition, we also assume that the $\xi$ variable is constant on each integral curve of the vector field N, that is, $n^{\alpha } \partial _\alpha \xi = \dot{\xi }= 0$ (“isoentropic flux”) and then, that f only depends on $\omega$, $f = f(\omega )$ (“barotropic instanton” for $\epsilon = +1$). Then, developing (100) we obtain

$$\begin{aligned} T^{\mu \nu } = 2 \, \frac{\partial f}{\partial g_{\mu \nu }} + \frac{f}{\mathrm{g}} \, \frac{\partial {\mathrm{g}}}{\partial g_{\mu \nu }} = 2 \, f' \, \frac{\partial \omega }{\partial g_{\mu \nu }} + f g^{\mu \nu }, \end{aligned}$$

(102)

where $f' \equiv \hbox {d}f/\hbox {d}\omega$, and we have considered the relation $g^{\mu \nu } = \frac{1}{\mathrm{g}} \frac{\partial \mathrm{g}}{\partial g_{\mu \nu }}$.

On the other hand, from (101),

$$\begin{aligned} \omega ^2 = \epsilon \, g_{\alpha \beta } \, N^{\alpha } N^{\beta } = \frac{g_{\alpha \beta }}{\mathrm{g}} \left( \sqrt{\epsilon \mathrm{g}} N^{\alpha }\right) \left( \sqrt{\epsilon \mathrm{g}} N^{\beta }\right) , \end{aligned}$$

(103)

whose variation with respect to $g_{\mu \nu }$, the flow lines being given, leads to

$$\begin{aligned} 2 \omega \, \frac{\partial \omega }{\partial g_{\mu \nu }}= & {} \left[ \frac{\partial }{\partial g_{\mu \nu }}\, \left( \frac{g_{\alpha \beta }}{\mathrm{g}} \right) \right] \, \epsilon \, {\mathrm{g}} \, N^{\alpha } N^{\beta } \nonumber \\= & {} \epsilon \, {\mathrm{g}} \, \omega ^2 \, \left( \frac{1}{\mathrm{g}} \delta ^{\mu }_\alpha \delta ^{\nu }_\beta \, - \, \frac{1}{\mathrm{g}^2} \frac{\partial {\mathrm{g}}}{\partial g_{\mu \nu }} \, g_{\alpha \beta }\right) n^{\alpha } \, n^{\beta }\nonumber \\= & {} \omega ^{2} ( \epsilon \, n^{\mu } n^{\nu } - g^{\mu \nu }), \end{aligned}$$

(104)

because, given the integral curves, the “conserved current” $\sqrt{\epsilon {\mathrm{g}}} N^{\alpha }$ is uniquely determined on a flow line in terms of its initial value at some point on the same flow line. Then, this “conserved current” remains unchanged for variations of $g_{\mu \nu }$ that vanish on the boundary made out from the points where we fix these initial values.

Then, substituting in (102), we have:

$$\begin{aligned} T^{\mu \nu } = \omega \, f' ( \epsilon \, n^{\mu } n^{\nu } - g^{\mu \nu }) + f g^{\mu \nu } = \epsilon \, \omega \, f' n^{\mu } n^{\nu } + ( f - \omega \, f' ) g^{\mu \nu }, \end{aligned}$$

(105)

which is an isotropic 2-tensor T of the form (66), with

$$\begin{aligned} f = \lambda _s = \epsilon \, \mu , \qquad \lambda _t = p = f - \omega \, f'. \end{aligned}$$

(106)

In all, the functional action we are looking for an isotropic stress tensor is given by

$$\begin{aligned} S = \epsilon \int \mu \, \sqrt{\epsilon \, {\mathrm{g}}} \, d^4x, \end{aligned}$$

(107)

with $\epsilon = +1$ and $L_E = \mu _E$ for the instanton case.

Incidentally note that, in particular, the above expression for the Lagrangian density in (107), $L = \epsilon \mu$, allows us to deduce the $-\varLambda /\kappa$ term appearing in the instanton action $S_E$ [see Eq. (46)]. In the Einstein field equations, the $\varLambda g_{\alpha \beta }$ term at the left member corresponds to a source term $T = - (\varLambda /\kappa ) g$ at the right member. Thus, according to (66), this tensor has $\epsilon \mu = p = - \varLambda /\kappa$, and then, it follows from the variations of the Lagrangian density $L = \epsilon \mu = - \varLambda /\kappa$, with respect to the metric (in both the Lorentzian or the Euclidean cases).