Non singular M theory universe in loop quantum cosmology—inspired models

Abstract

We study an M theory universe in the loop quantum cosmology—inspired models which involve a function, the choice of which leads to a variety of evolutions. The M theory universe is dominated by four stacks of intersecting brane–antibranes and, in general relativity, it becomes effectively four dimensional in future while its seven dimensional internal space reaches a constant size. We analyse the conditions required for non singular evolutions and obtain explicit solutions in the simplified case of a bi-anisotropic universe and a piece-wise linear function for which the evolutions are non singular. One may now ask whether the physics in the Planckian regime can enhance the internal volume to phenomenologically interesting values. In the simplified case considered here, there is no non trivial enhancement. We make some comments on it.

Appendices

Appendix A: Anisotropic solutions in general relativity

Consider the general relativity Eqs. (3)–(5) for the anisotropic case. When the equations of state are linear, it is straightforward to solve these equations and obtain analytic solutions [23]. It follows from Eqs. (14) that, upon replacing $\kappa ^2$ by $c^2 \kappa ^2$, these solutions are applicable to the LQC—inspired models when $f(x) = c x + c_0$.

We now present these solutions. First, define a new variable $\tau$ by

$$\begin{aligned} d t = e^\Lambda \; d \tau \; \; \; \longleftrightarrow \; \; \; t - t_0 = \int _{\tau _0}^\tau d \tau \; e^\Lambda \end{aligned}$$

(53)

where $t_0$ and $\tau _0$ are initial times. Then, for any function $\psi (t(\tau ))$, we have

$$\begin{aligned} \psi _\tau \; = \; e^\Lambda \; \psi _t \; \; , \; \; \; \psi _{\tau \tau } \; = \; e^{2 \Lambda } \; (\psi _{t t} + \Lambda _t \psi _t) . \end{aligned}$$

Defining $(\hat{*}) = e^{2 \Lambda } \; (*)$ for $(*) = (\rho , \; p_i, \;r^i)$, Eqs. (3)–(5) become

$$\begin{aligned} \sum _{i j} G_{i j} \; \lambda ^i_\tau \; \lambda ^j_\tau= & {} 2 \kappa ^2 \; \hat{\rho } \end{aligned}$$

(54)

$$\begin{aligned} \lambda ^i_{\tau \tau }= & {} \kappa ^2 \; \hat{r}^i \end{aligned}$$

(55)

$$\begin{aligned} (\hat{\rho })_\tau= & {} \sum _i (\hat{\rho } - \hat{p}_i) \; \lambda ^i_\tau . \end{aligned}$$

(56)

Let the equations of state be linear and be given by

$$\begin{aligned} p_i = (1 - u_i) \; \rho \end{aligned}$$

(57)

where $u_i$ are constants. Define $l, \; v^i$, and ${{\mathcal {G}}}$ by

$$\begin{aligned} l = \sum _i u_i \; \lambda ^i, \; \; \; v^i = \sum _j G^{i j} \; u_j, \; \; \; {{\mathcal {G}}} = \sum _i v^i \; u_i = \sum _{i j} G^{i j} \; u_i \; u_j \end{aligned}$$

(58)

and let the initial values of various quantities at $t = t_0$ be given by

$$\begin{aligned}&\left( \lambda ^i, \; \lambda ^i_t, \; \rho \; ; \; \Lambda , \; l, \; l_t \; ; \; \tau , \; \lambda ^i_\tau , \; l_\tau , \; \hat{\rho } \right) _{t = t_0} \nonumber \\&\quad = \left( \lambda ^i_0, \; k^i, \; \rho _0\; ; \; \Lambda _0, \; l_0, \; l_{t 0} \; ; \; \tau _0, \; \lambda ^i_{\tau 0}, \; l_{\tau 0}, \; \hat{\rho }_0 \right) \end{aligned}$$

(59)

where

$$\begin{aligned} \rho _0 \;> & {} \; 0 ,\quad \sum _{i j} G_{i j} \; k^i \; k^j \; = \; 2 \kappa ^2 \; \rho _0 \nonumber \\ \Lambda _0= & {} \sum _i \lambda ^i_0 ,\quad l_0 = \sum _i u_i \; \lambda ^i_0 \; \; , \; \; \; l_{t 0} = \sum _i u_i \; k^i \nonumber \\ \lambda ^i_{\tau 0}= & {} e^{\Lambda _0} \; k^i , \quad l_{\tau 0} = e^{\Lambda _0} \; l_{t 0} \; \; , \; \; \; \hat{\rho }_0 = e^{2 \Lambda _0} \; \rho _0 . \end{aligned}$$

(60)

Then Eqs. (55) and (56) give

$$\begin{aligned} \lambda ^i_{\tau \tau }= & {} \kappa ^2 \; v^i \; \hat{\rho } \end{aligned}$$

(61)

$$\begin{aligned} l_{\tau \tau }= & {} \kappa ^2 \; {{\mathcal {G}}} \; \hat{\rho } \end{aligned}$$

(62)

$$\begin{aligned} \hat{\rho }= & {} \hat{\rho }_0 \; e^{l - l_0} \end{aligned}$$

(63)

and it follows from Eqs. (61) and (62) that

$$\begin{aligned} \lambda ^i - \lambda ^i_0 \; = \; \frac{v^i}{{\mathcal {G}}} \; (l - l_0) + L^i \; (\tau - \tau _0) . \end{aligned}$$

(64)

Since $l = \sum _i u_i \lambda ^i$, it follows that the integration constants $L^i$ must satisfy the constraint $\sum _i u_i L^i = 0$. This constraint is identically satisfied if $L^i = e^{\Lambda _0} \left( k^i - \frac{v^i}{{\mathcal {G}}} \; \; l_{t 0} \right)$ where $l_{t 0} = \sum _i u_i k^i$, see Eqs. (60). Thus, the set of d number of initial values $\{ k^i \}$ is equivalent to the set of $(1 + d)$ number of initial values $\{ l_{t 0}, \; L^i \}$ together with one constraints on $L^i$. Upon using $\sum _i u_i L^i = 0$, Eq. (54) gives

$$\begin{aligned} (l_\tau )^2 \; = \; 2 \; {{\mathcal {G}}} \; \left( E + \kappa ^2 \; \hat{\rho } \right) , \; \; \; 2 \; E = - \; \sum _{i j} G_{i j} \; L^i \; L^j . \end{aligned}$$

(65)

Now, in principle, Eqs. (62), (63), and (65) give $l(\tau )$ and Eqs. (64) and (53) give $\lambda ^i (\tau )$ and $t(\tau )$ from which $\tau (t), \; l (t)$, and $\lambda ^i (t)$ follow. Also, it can be shown that if $\sum _i u_i L^i = 0$ and ${{\mathcal {G}}} = \sum _{i j} G^{i j} \; u_i u_j > 0$ then $E \ge 0$ and E will vanish if and only if all $L^i$ vanish [23]. Henceforth, we assume that ${{\mathcal {G}}} > 0$ and $E > 0$.

For the case of bi-anisotropic universe considered in this paper, see Eqs. (26) and (27), it follows straightforwardly that

$$\begin{aligned}&\tilde{p} = (1 - \tilde{u}) \; \rho ,\quad p = (1 - u) \; \rho \nonumber \\&\tilde{v} = \frac{n u - (n - 1) \tilde{u}}{\tilde{n} + n - 1} ,\quad v = \frac{\tilde{n} \tilde{u} - (\tilde{n} - 1) u}{\tilde{n} + n - 1} \nonumber \\&l = \tilde{n} \tilde{u} \tilde{\lambda } + n u \lambda ,\quad {{\mathcal {G}}} = \tilde{n} \tilde{u} \tilde{v} + n u v \nonumber \\&\tilde{n} \tilde{u} \tilde{L} + n u L = 0 , \quad 2 E = \frac{n (n + \tilde{n} - 1) {{\mathcal {G}}} L^2}{\tilde{n} \tilde{u}^2} \end{aligned}$$

(66)

where the expression for E follows after some algebra. If $\tilde{v} = 0$, which is necessary for the $(\tilde{n} + n)$ dimensional space to become effectively n dimensional in the limit $e^\Lambda \rightarrow \infty$, then one has

$$\begin{aligned}&\tilde{u} = \frac{n u}{n - 1} ,\quad v = \frac{u}{n - 1} \nonumber \\&l = \tilde{u} \; \left( \tilde{n} \tilde{\lambda } + (n - 1) \lambda \right) ,\quad {{\mathcal {G}}} = \frac{n u^2}{n - 1} \nonumber \\&\tilde{n} \tilde{L} + (n - 1) L = 0 ,\quad 2 E = \frac{(n - 1) (n + \tilde{n} - 1) L^2}{\tilde{n}} . \end{aligned}$$

(67)

Consider now the solution $l(\tau )$ for Eqs. (62), (63), and (65). As can be verified easily, it is given by

$$\begin{aligned} \kappa ^2 \; \hat{\rho } \; = \; \kappa ^2 \; \hat{\rho }_0 \; e^{l - l_0} \; = \; \frac{E}{Sinh^2 \; \sigma (\tau _\infty - \tau )} \end{aligned}$$

(68)

where $2 \sigma ^2 = {{\mathcal {G}}} E$. Note that the sign of $\sigma$ is immaterial; that

$$\begin{aligned} l_\tau \; = \; 2 \; \sigma \; Coth \; \sigma (\tau _\infty - \tau ); \end{aligned}$$

(69)

and that setting $l = l_0$ and $\tau = \tau _0$ in Eq. (68) gives $\tau _\infty$ in terms of E and $\hat{\rho }_0$. Eqs. (64) and (53) will now give $\lambda ^i (\tau )$ and $t(\tau )$ from which $\tau (t), \; l (t)$, and $\lambda ^i (t)$ follow. Taking $\sigma > 0$ and $l_{\tau 0} > 0$ for the sake of definiteness, we now mention some features of these solutions.

• Since $l_{\tau 0} > 0$, it follows from Eq. (69) that $\tau _\infty > \tau _0$. It follows from Eq. (68) that $l(\tau )$ varies monotonically between $- \infty$ and $+ \infty$, that $l \rightarrow - \infty$ as $\tau \; \rightarrow \; - \infty$, and that $l \rightarrow \infty$ as $\tau \; \rightarrow \; \tau _\infty$.

• In the limit $\tau \; \rightarrow \; \tau _\infty$ from below, one has $l(\tau ) \; \sim \; - 2 \; ln \; (\tau _\infty - \tau ) \; \rightarrow \; \infty$. Eqs. (64) and (53) then give, upto unimportant constants,

  $$\begin{aligned} t\sim & {} (\tau _\infty - \tau )^{- \frac{2 B - {{\mathcal {G}}}}{{\mathcal {G}}}}, \; \; \; B = \sum _j v^j \end{aligned}$$

  (70)
  $$\begin{aligned} e^{\lambda ^i}\sim & {} (\tau _\infty - \tau )^{- \frac{2 v^i}{{\mathcal {G}}}} \; \sim \; t^{\frac{2 v^i}{2 B - {{\mathcal {G}}}}} \end{aligned}$$

  (71)
  $$\begin{aligned} e^\Lambda\sim & {} (\tau _\infty - \tau )^{- \frac{2 B}{{\mathcal {G}}}} \; \sim \; t^{\frac{2 B}{2 B - {{\mathcal {G}}}}} . \end{aligned}$$

  (72)

  For the bi-anisotropic universe, it follows from Eqs. (66) that

  $$\begin{aligned} 2 B - {{\mathcal {G}}} = \tilde{n} \tilde{v} \; (2 - \tilde{u}) + n v \; (2 - u) . \end{aligned}$$

  (73)

  Hence, for $\tilde{v} = 0$, one has $e^{ \tilde{\lambda }} \sim const$ and $e^{\lambda } \sim t^{\frac{2}{n (2 - u)}}$ which is the standard n dimensional result.

• In the limit $\tau \; \rightarrow \; - \infty$, one has $l(\tau ) \; \sim \; 2 \; \sigma \tau \; \rightarrow \; - \infty$. Eqs. (64) then imply that $\lambda ^i (\tau )$ are all linear in $\tau$. Let $\tau \rightarrow - \infty$ and $e^\Lambda \rightarrow 0$ in this limit and, upto unimportant constants, let

  $$\begin{aligned} \lambda ^i \; \sim \; q^i \; \tau , \; \; \; \Lambda \; \sim \; q \; \tau , \; \; \; q = \sum _i q^i > 0 . \end{aligned}$$

  Then, after some algebra, it follows from Eq. (53) that

  $$\begin{aligned} e^\Lambda \; \sim \; e^{q \; \tau } \; \sim \; q \; t \; \; \rightarrow \; 0 , \; \; \; \; \; e^{\lambda ^i} \; \sim \; e^{q^i \; \tau } \; \sim \; \left( q \; t \right) ^{\frac{q^i}{q}} \end{aligned}$$

  (74)

  which are the Kasner-type solutions.

Appendix B: Isotropic solutions in LQC—inspired models

Consider the fully isotropic case where

$$\begin{aligned} (m^i, \; f^i, \; g_i, \; X_i, \; \lambda ^i, \; p_i, \; r^i ) \; = \; (m, \; f, \; g, \; X, \; \lambda , \; p, \; r ) \end{aligned}$$

for $i = 1, 2, \ldots , d$. Then

$$\begin{aligned} g \; = \; \frac{d \; f}{d m}, \; \; \; X \; = \; (d - 1) \; g f, \; \; \; r \; = \; \frac{\rho - p}{d - 1} \end{aligned}$$

and Eqs. (10)–(12) give

$$\begin{aligned} f^2= & {} \frac{2 \; \gamma ^2 \lambda _{qm}^2 \kappa ^2 \; \rho }{d \; (d - 1)} \end{aligned}$$

(75)

$$\begin{aligned} m_t= & {} - \; \frac{\gamma \lambda _{qm} \kappa ^2}{d - 1} \; (\rho + p) \end{aligned}$$

(76)

$$\begin{aligned} \lambda _t= & {} \frac{g \; f}{\gamma \lambda _{qm}} \; \; \; \Longrightarrow \; \; \; (\lambda _t)^2 \; = \; \frac{2 \kappa ^2 \; (\rho \; g^2)}{d (d - 1)} . \end{aligned}$$

(77)

Let the equation of state be linear and be given by $p = (1 - u) \rho$ where $u < 2$ is a constant. Then Eqs. (5) and (75)–(77) may be solved explicitly if certain integrations and functional inversions can be performed. Eqs. (5) and (75) give

$$\begin{aligned} \frac{\rho }{\rho _0} \; = \; \frac{f^2}{f^2_0}\; = \; e^{ - (2 - u) \; d \; (\lambda - \lambda _0)} \end{aligned}$$

(78)

which leads to $\lambda (m)$. Eqs. (75) and (76) then lead to t(m) given by

$$\begin{aligned} c_{qm} \; (t - t_0) \; = \; - \; \int ^m_{m_0} \frac{d m}{f^2} \end{aligned}$$

(79)

where $c_{qm} = \frac{(2 - u) \; d}{2 \; \gamma \lambda _{qm}}$. Inverting t(m) then gives m(t) and $\lambda (t)$. The integrations and functional inversions required here can be performed explicitly for $f(x) = c x + c_0$ and also for $f(x) = sin \; x$ but not for a generic f(x) . The resulting solutions are given in [45, 46].

Consider now the isotropic solutions for the simplified, piece-wise linear function f(x) given in Eq. (36). Equation (78) gives the density $\rho (m)$ and the scale factor $e^{\lambda (m)}$. Let the initial value $m_0$ at time $t_0$ lie in the range $0< m_0 < A$. It then follows that as m increases from 0 to $m_0$ to A to $A + 2 \Delta$ to $2 m_*$, the function f increases from 0 to $m_0$ to A, remaining at A, and then decreasing to 0. Hence, correspondingly, the scale factor $e^{\lambda (m)}$ decreases from $\infty$ to $e^{\lambda _0}$, decreases further, then remains constant, and then increases again to $\infty$.

The time t(m) follows straightforwardly upon performing the integration in Eq. (79), and is given by

$$\begin{aligned} c_{qm} \; (t - t_0)= & {} - \frac{1}{m_0} + \frac{1}{m} \; \; \; for \; \; 0 \le m \le A \nonumber \\= & {} \frac{2}{A} - \frac{1}{m_0} - \frac{m}{A^2} \; \; \; for \; \; A \le m \le A + 2 \Delta \nonumber \\= & {} \frac{2}{A} - \frac{1}{m_0} - \frac{2 \Delta }{A^2} + \frac{1}{m - 2 m_*} \; \; \; for \; \; A + 2 \Delta \le m \le 2 m_* . \end{aligned}$$

(80)

Hence, as m increases from 0 to $m_0$ to A to $A + 2 \Delta$ to $2 m_*$, the time t decreases monotonically from $\infty$ to $t_0$ to $- \infty$, first as $\frac{1}{m}$, then linearly as $- \frac{m}{A^2}$, and then as $\frac{1}{m - 2 m_*}$.

Appendix C: Bi-anisotropic solutions when only $\in ({\mathbf {A, \; A + 2}}\Delta )$

During $t_b> t > t_e$, let m(t) lie in the interval $(A, \; A + 2 \Delta )$ and let $\tilde{m} (t)$ lie in $(0, \; A)$ or in $(A + 2 \Delta , \; 2 m_*)$. Then $f = A$, $\; g = X = 0$, $\; \tilde{f} = c \tilde{m} + c_0$ where $(c, \; c_0) = (1, \; 0)$ or $(- 1, \; 2 m_*)$, $\; \tilde{g} = c$, and $\tilde{X} = c \left( (\tilde{n} - 1) \tilde{f} + n A \right)$. The times $t_b$ and $t_e$ are defined by the equalities in the following expressions for the values of $\tilde{m}$ and m at $t_b$ and $t_e$ :

$$\begin{aligned}&\tilde{m}_b \;< \; A , \quad m_b \; = \; A \nonumber \\&\tilde{m}_e \; = \; A , \quad A \;< \; m_e \;< \; A + 2 \Delta \nonumber \\&or \; \; \; \; \; \; \tilde{m}_b \; = \; A + 2 \Delta ,\,\, A \;< \; m_b \; < \; A + 2 \Delta \nonumber \\&\tilde{m}_e \; > \; A + 2 \Delta , \quad m_e \; = \; A + 2 \Delta . \end{aligned}$$

(81)

Several equations are different if $\tilde{n} > 1$ or $\tilde{n} = 1$. Hence we consider these two cases seperately.

1.1 $\tilde{\mathbf {n}} > \mathbf {1}$

Define $y, \; z$, and a by

$$\begin{aligned} y = (\tilde{n} - 1) \; \tilde{f} + n \; A, \; \; \; z = (m - \tilde{m}) , \; \; \; a = \sqrt{\frac{n \; (d - 1)}{\tilde{n}}} \; A . \end{aligned}$$

(82)

Note that $\tilde{X} = c y$ and that $n A< y < (d - 1) A$. After some algebra, it follows from Eqs. (28) and (29) that

$$\begin{aligned} \frac{\rho }{\rho _{qm}}= & {} \frac{\tilde{n}}{\tilde{n} - 1} \left( y^2 - a^2 \right) \end{aligned}$$

(83)

$$\begin{aligned} y_t= & {} - \; c_y \; (y^2 - a^2) \end{aligned}$$

(84)

$$\begin{aligned} z_t + b \; y \; z= & {} - \; c_z \; (y^2 - a^2) \end{aligned}$$

(85)

where

$$\begin{aligned} c_y \; = \; \frac{\tilde{n} \; c \; \left( \frac{2}{d - 1} - \tilde{v} \right) }{2 \; \gamma \lambda _{qm}}, \; \; \; c_z \; = \; \frac{\tilde{n} \; (\tilde{v} - v)}{2 \; (\tilde{n} - 1) \; \gamma \lambda _{qm}}, \; \; \; b \; = \; \frac{\tilde{n} \; c}{(d - 1) \; \gamma \lambda _{qm}} . \end{aligned}$$

The solutions y(t) and z(y) are given in Eqs. (45) and (46). Since $X = 0$, it follows that $\Lambda _t -\lambda _t = 0$ and hence, from Eqs. (32), (34), and (83), that

$$\begin{aligned} (\tilde{\lambda } - \tilde{\lambda }_0)= & {} - \; \left( \frac{n - 1}{\tilde{n}} \right) \; (\lambda - \lambda _0) \nonumber \\ e^{(2 \; - \; (d - 1) \; \tilde{v}) \; (\lambda _b - \lambda )}= & {} \frac{\rho }{\rho _b} \; = \; \frac{y^2 - a^2}{y^2_b - a^2} . \end{aligned}$$

(86)

1.2 $\tilde{\mathbf {n}} = \mathbf {1}$

Now $d = n + 1$. Define y and z by

$$\begin{aligned} y = 2 \; \tilde{f} + (n - 1) \; A, \; \; \; z = (m - \tilde{m}) . \end{aligned}$$

(87)

Note that $\; \tilde{X} = n c A$ and that $(n - 1) A< y < (n + 1) A$. After some algebra, it follows from Eqs. (28)–(29) that

$$\begin{aligned} \frac{\rho }{\rho _{qm}}= & {} n A \; y \end{aligned}$$

(88)

$$\begin{aligned} y_t= & {} 2 c \; \gamma \lambda _{qm} \kappa ^2 \; \left( \tilde{v} - \frac{2}{d - 1} \right) \; \rho \end{aligned}$$

(89)

$$\begin{aligned} z_t \; + \; \frac{n c \; A \; z}{(d - 1) \; \gamma \lambda _{qm}}= & {} \gamma \lambda _{qm} \kappa ^2 \; (v - \tilde{v}) \; \rho . \end{aligned}$$

(90)

Eqs. (88)–(90) lead to the solutions y(t) and z(y) given by

$$\begin{aligned} y \; = \; y_b \; e^{ - \frac{n c \; A }{ \gamma \lambda _{qm}} \; \left( \frac{2}{d - 1} - \tilde{v} \right) \; (t - t_b)} \end{aligned}$$

(91)

and

$$\begin{aligned} z \; = \; y^s \; \left( \frac{z_b}{y^s_b} \; + \; \sigma \; \; \int _{y_b}^y \; \frac{d y}{y^s} \right) . \end{aligned}$$

(92)

where $s = \frac{1}{2 - (d - 1) \; \tilde{v}}$ and $\sigma = \frac{(d - 1) \; (\tilde{v} - v)}{2 c \; \left( 2 - (d - 1) \; \tilde{v} \right) }$. Thus, if $\tilde{v} < \frac{2}{d - 1}$ then $y_t < 0$ and y increases monotonically from $y_b$ to $\infty$ as t decreases from $t_b$ to $- \infty$. Also, Eq. (86) give $\tilde{\lambda }$ and $\lambda$ in terms of y.