New interpretation of the extended geometric deformation in isotropic coordinates

Abstract

We study the particular case in which extended geometric deformation does consists of consecutive deformations of temporal and spatial components of the metric, in Schwarzschild-like and isotropic coordinates. In the latter, we present two inequivalent ways to perform this two-step GD. This was done in such a way that the method may be applied to different seed solutions. As an example, we use Tolman IV as seed solution, in order to obtain two inequivalent physical solutions with anisotropy in the pressures in Schwarzschild-like coordinates. In the isotropic sector, we obtained four different solutions with anisotropy in the pressures that satisfies physical acceptability conditions, using Gold III as seed solution.

Appendices

Appendix A: Physical acceptability conditions

Solving Einstein’s equations does not ensure that the solution will describe any physical system. Indeed, among all the known solutions of Einstein’s equations, only a part of them fulfill the physically acceptable conditions (see for example [5]).

Then, in order to ensure that the solutions of Einstein’s equations are physically acceptable, we must verify if the following conditions are satisfied

• $P_r$, $P_t$ and $\rho$ are positive and finite inside the distribution.

• $\frac{dP_r}{dr}$, $\frac{dP_t}{dr}$ and $\frac{d\rho }{dr}$ are monotonically decreasing.

• Dominant energy condition: $\frac{P_r}{\rho }\le 1$  ,   $\frac{P_t}{\rho }$ $\le 1$.

• Causality condition: $0<\frac{dP_r}{d\rho }<1$ ,  $0<\frac{dP_t}{d\rho }<1$.

• The local anisotropy of the distribution should be zero at the center and increasing towards the surface.

Appendix B: The Ricci invariants

Let us verify that solutions obtained by EGD, and therefore, by any of its limits, are in general inequivalent to the seed solution considered. In order to see this, let us notice that the Ricci invariants can be written in terms of the trace-free Ricci tensor

$$\begin{aligned} S^\mu _\nu = R^\mu _\nu -\delta ^\mu _\nu \frac{R}{4}, \end{aligned}$$

(232)

where $R^\mu _\nu$ and R are the Ricci tensor and the scalar curvature , respectively. Besides the scalar curvature, Ricci invariants are defined as

$$\begin{aligned} r_1= & {} \frac{1}{4}S^\mu _\nu S^\nu _\mu , \end{aligned}$$

(233)

$$\begin{aligned} r_2= & {} -\frac{1}{8}S^\mu _\nu S^\rho _\mu S^\nu _\rho , \end{aligned}$$

(234)

$$\begin{aligned} r_3= & {} S^\mu _\nu S^\rho _\mu S^\lambda _\rho S^\nu _\lambda . \end{aligned}$$

(235)

Let us assume that $\{{\tilde{\nu }},{\tilde{\mu }}\}$ and $\{\nu ,\mu \}$ represent two solutions of the of Einstein’s equations for an spherically symmetric fluid with a line element of the form (1). Defining ${\tilde{\nu }}$ and ${\tilde{\mu }}$ as

$$\begin{aligned} {\tilde{\mu }} = \mu + \alpha f(r), \quad {\tilde{\nu }} = \nu + \beta h(r), \end{aligned}$$

(236)

then it can be shown that the scalar curvature and the the trace-free Ricci tensor satisfies the following relations

$$\begin{aligned} {\tilde{R}} = R + \alpha \zeta _1 (r) + \beta \zeta _2 (r) +\alpha \beta \zeta _3 (r), \end{aligned}$$

(237)

and

$$\begin{aligned} {\tilde{S}}^\mu _\nu= & {} S^\mu _\nu + \frac{\delta ^\mu _0\delta ^0_\nu }{2}\left[ \alpha \left( \frac{\zeta _1}{2}-\frac{2}{r}\left( \frac{f}{r}+f'\right) \right) +\frac{\beta }{2}(\zeta _2+\alpha \zeta _3)\right] \nonumber \\&\quad +\frac{\delta ^\mu _1\delta ^1_\nu }{2}\left[ \alpha \left( \frac{\zeta _1}{2}-\frac{2f}{r}\left( \nu '+\frac{1}{r}\right) \right) + \beta \left( \frac{\zeta _2}{2}-\frac{2\mu h'}{r}\right) \right. \nonumber \\&+ \left. \alpha \beta \left( \frac{\zeta _3}{2}-\frac{2fh'}{r}\right) \right] + \frac{\delta ^\mu _2\delta ^2_\nu }{2}\Bigg [\alpha \left( \frac{f}{r}\left( \nu '+\frac{2}{r}\right) + \frac{f'}{r}-\frac{\zeta _1}{2}\right) \nonumber \\&\quad +\beta \left( \frac{\mu h'}{r}-\frac{\zeta _2}{2}\right) + \alpha \beta \left( \frac{fh'}{r}-\frac{\zeta _3}{2}\right) \Bigg ] \nonumber \\&+ \frac{\delta ^\mu _3\delta ^3_\nu }{2} \Bigg [\alpha \left( \frac{f}{r}\left( \nu '+\frac{2}{r}\right) + \frac{f'}{r}-\frac{\zeta _1}{2}\right) \nonumber \\&\quad +\beta \left( \frac{\mu h'}{r}-\frac{\zeta _2}{2}\right) + \alpha \beta \left( \frac{fh'}{r}-\frac{\zeta _3}{2}\right) \Bigg ] \end{aligned}$$

(238)

where

$$\begin{aligned} \zeta _1 (r)= & {} \frac{1}{2}\left\{ f\left[ (\nu ')^2+2\nu ''+\frac{4}{r}\left( \nu '+\frac{1}{r}\right) \right] + f'\left( \nu '+\frac{4}{r}\right) \right\} , \\ \zeta _2 (r)= & {} \frac{1}{2}\left\{ \mu \left[ 2h''+2h'\nu '+\beta (h')^2+\frac{4}{r}h'\right] +\mu 'h' \right\} , \\ \zeta _3 (r)= & {} \frac{1}{2}\left\{ f \left[ 2h''+2h'\nu '+\beta (h')^2+\frac{4}{r}h'\right] +f'h' \right\} . \end{aligned}$$

Now, using (1), the Ricci invariants can be written as

$$\begin{aligned} {\tilde{r}}_1= & {} \frac{1}{4} \left[ ({\tilde{S}}^0_0)^2+({\tilde{S}}^1_1)^2+2({\tilde{S}}^2_2)^2\right] , \end{aligned}$$

(239)

$$\begin{aligned} {\tilde{r}}_2= & {} -\frac{1}{8} \left[ ({\tilde{S}}^0_0)^3+({\tilde{S}}^1_1)^3+2({\tilde{S}}^2_2)^3\right] , \end{aligned}$$

(240)

$$\begin{aligned} {\tilde{r}}_3= & {} ({\tilde{S}}^0_0)^4+({\tilde{S}}^1_1)^4+2({\tilde{S}}^2_2)^4. \end{aligned}$$

(241)

Thus, from (238), it is evident that in general

$$\begin{aligned} {\tilde{r}}_1 \not = r_1, \quad {\tilde{r}}_2 \not = r_2, \quad {\tilde{r}}_3 \not = r_3, \end{aligned}$$

(242)

and therefore the two solutions of Einstein’s equations, represented by $\{{\tilde{\nu }},{\tilde{\mu }}\}$ and $\{\nu ,\mu \}$, are inequivalent. It is easy to see, from equations (237) and (238), that the solutions are different even when we take $\alpha =0$ or $\beta =0$, which corresponds to the to two limits of the EGD method mentioned in Section (2). Since the right and left paths of the two-step GD produce different results for the deformations functions f and g, then the Ricci invariant of the final solutions obtained with each path will be in general different. This indicates that the right path and left paths leads, in general, to inequivalent solutions. For the case of the isotropic coordinates the analysis is analogous.

Appendix C: Two-step GD and the BVW theorems

In this appendix we will present the unnoticed relation that exists between the two-step GD (and therefore, EGD) with the first four theorems presented by the authors in [87]. Let us consider as seed solution a perfect fluid sphere represented by the set of functions $\{\nu ,\mu ,\rho ,P\}$.

1.1 Left path

Starting with the pure spatial deformation (MGD) and imposing the condition $(\theta ^f_L)^1_1=(\theta ^f_L)^2_2$, implies that the system (36)-(38) leads to the following differential equation

$$\begin{aligned} 2f_L \left[ r^2 B''-(rB)'\right] + f'_Lr(rB)'=0, \end{aligned}$$

(243)

where $B(r)^2=\exp {\nu }$. The solution to (243) can be written as

$$\begin{aligned} f_L = \frac{r^2}{[(rB)']^2}\exp {\int \frac{4B'}{(rB)'}dr}. \end{aligned}$$

(244)

The new solution is given by the set of functions $\{\nu ,{\bar{\mu }},{\bar{\rho }},{\bar{P}}\}$, where

$$\begin{aligned} {\bar{\mu }}= & {} \mu +\alpha f_L, \end{aligned}$$

(245)

$$\begin{aligned} {\bar{\rho }}= & {} \rho +\alpha (\theta ^f_L)^0_0, \end{aligned}$$

(246)

$$\begin{aligned} {\bar{P}}= & {} P-\alpha (\theta ^f_L)^1_1. \end{aligned}$$

(247)

This correspond to the transformation in the first theorem in [87]. Now, let us consider $\{\nu ,{\bar{\mu }},{\bar{\rho }},{\bar{P}}\}$ as a seed solution and perform a pure temporal deformation with the constraint $(\theta ^h_L)^1_1=(\theta ^h_L)^2_2$, then the system (45)–(47) leads to the following differential equation

$$\begin{aligned} \frac{{\bar{\mu }}}{r}[r(Z_L''+Z_L'\nu )-Z_L']+\frac{{\bar{\mu }}'Z_L'}{2}=0, \end{aligned}$$

(248)

where $Z_L(r)^2= \exp {\beta h_L}$. The solution in this case is

$$\begin{aligned} Z_L(r) = A+B\int \frac{rdr}{e^{\nu }\sqrt{{\bar{\mu }}}}. \end{aligned}$$

(249)

The final solution is characterized by the set $\{{\tilde{\nu }},{\tilde{\mu }},{\tilde{\rho }},{\tilde{P}}\}$

$$\begin{aligned} {\tilde{\nu }}= & {} \nu +\beta h_L, \end{aligned}$$

(250)

$$\begin{aligned} {\tilde{\mu }}= & {} {\bar{\mu }} = \mu + \alpha f_L, \end{aligned}$$

(251)

$$\begin{aligned} {\tilde{\rho }}= & {} \rho + \alpha (\theta ^f_L)^0_0, \end{aligned}$$

(252)

$$\begin{aligned} {\tilde{P}}= & {} P - \alpha \left( (\theta ^f_L)^1_1+ \frac{\beta }{\alpha }(\theta ^h_L)^1_1\right) . \end{aligned}$$

(253)

The complete left path that goes from $\{\nu ,\mu ,\rho ,P\}$ to $\{{\tilde{\nu }},{\tilde{\mu }},{\tilde{\rho }},{\tilde{P}}\}$ corresponds to the transformation in the third theorem in [87].

1.2 Right path

As in the left path, let us begin by taking $\{\nu ,\mu ,\rho ,P\}$ as seed solution. Then the pure temporal deformation of the metric subject to the constraint $(\theta ^h_R)^1_1=(\theta ^h_R)^2_2$, leads to the following differential equation

$$\begin{aligned} \frac{\mu }{r}[r(Z_R''+Z_R'\nu )-Z_R']+\frac{\mu 'Z_R'}{2}=0, \end{aligned}$$

(254)

notice that it has the same form of Eq (248) but changing ${\tilde{\mu }}$ with $\mu$. Therefore, it can be check that

$$\begin{aligned} Z_R(r) = C+D\int \frac{rdr}{e^{\nu }\sqrt{\mu }}. \end{aligned}$$

(255)

The new solution is given by $\{{\bar{\nu }},\mu ,\rho ,{\bar{P}}\}$ where

$$\begin{aligned} {\bar{\nu }}= & {} \nu +\alpha h_L, \end{aligned}$$

(256)

$$\begin{aligned} {\bar{P}}= & {} P-\alpha (\theta ^h_L)^1_1. \end{aligned}$$

(257)

The transformation from $\{\nu ,\mu ,\rho ,P\}$ to $\{{\bar{\nu }},\mu ,\rho ,{\bar{P}}\}$, given by the pure temporal deformation, corresponds to the second theorem in [87]. Now we will take $\{{\bar{\nu }},\mu ,\rho ,{\bar{P}}\}$ as seed solution and perform a pure spatial deformation, subject to the constraint $(\theta ^f_R)^1_1=(\theta ^f_R)^2_2$. In this case the system (36)–(38) leads to the following differential equation

$$\begin{aligned} 2f_R \left[ r^2 {\bar{B}}''-(r{\bar{B}})'\right] + f'_Rr(r{\bar{B}})'=0, \end{aligned}$$

(258)

where ${\bar{B}}(r)^2=\exp {{\bar{\nu }}}$. Then

$$\begin{aligned} f_R = \frac{r^2}{[(r{\bar{B}})']^2}\exp {\int \frac{4{\bar{B}}'}{(r{\bar{B}})'}dr}. \end{aligned}$$

(259)

and the final solution of the right path can be written as

$$\begin{aligned} {\tilde{\nu }}= & {} {\bar{\nu }}=\nu +\alpha h_R, \end{aligned}$$

(260)

$$\begin{aligned} {\tilde{\mu }}= & {} \mu + \beta f_R, \end{aligned}$$

(261)

$$\begin{aligned} {\tilde{\rho }}= & {} \rho + \beta (\theta ^f_R)^0_0, \end{aligned}$$

(262)

$$\begin{aligned} {\tilde{P}}= & {} P - \beta \left( (\theta ^f_R)^1_1+ \frac{\alpha }{\beta }(\theta ^h_R)^1_1\right) . \end{aligned}$$

(263)

The complete right path that goes from $\{\nu ,\mu ,\rho ,P\}$ to $\{{\tilde{\nu }},{\tilde{\mu }},{\tilde{\rho }},{\tilde{P}}\}$ corresponds to the transformation in the fourth theorem in [87].