Constraints on Janus Cosmological model from recent observations of supernovae type Ia

Abstract

From our exact solution of the Janus Cosmological equation we derive the relation of the predicted magnitude of distant sources versus their red shift. The comparison, through this one free parameter model, to the available data from 740 distant supernovae shows an excellent fit.

Appendices

Appendix A: Bolometric magnitude

Starting from the cosmological equations corresponding to positive species and neglectible pressure (dust universe) establish in Petit and D’Agostini (2014a):

$$\begin{aligned} a^{(+) 2} \ddot{a}^{(+)} + \frac{8 \pi G}{3}E = 0 \end{aligned}$$

(9)

with $E \equiv a^{(+) 3} \rho ^{(+)} + a^{(-) 3} \rho ^{(-)} = \mathit{constant} < 0$. For the sake of simplicity we will write ${a \equiv a^{(+)}}$ in the following. A parametric solution of Eq. (9) can be written as:

$$\begin{aligned} a(u) = \alpha ^{2} \mathit{ch}^{2}(u) \qquad t(u) = \frac{\alpha ^{2}}{c} \biggl( 1+\frac{\mathit{sh}(2 u)}{2} + u \biggr) \end{aligned}$$

(10)

with

$$\begin{aligned} \alpha ^{2} = -\frac{8 \pi G}{3 c^{2}} E \end{aligned}$$

(11)

This solution imposes $k = -1$. Writing the usual definitions:

$$\begin{aligned} q \equiv -\frac{a \ddot{a}}{\dot{a}^{2}} \quad \text{and} \quad H \equiv \frac{\dot{a}}{a} \end{aligned}$$

(12)

we can write:

$$\begin{aligned} q = -\frac{1}{2 \mathit{sh}^{2}(u)} = -\frac{4 \pi G}{3} \frac{|E|}{a ^{3} H^{2}} \end{aligned}$$

(13)

and also

$$\begin{aligned} ( 1-2 q ) = \frac{c^{2}}{a^{2} H^{2}} \end{aligned}$$

(14)

In terms of the time $t$ used in the FRLW metric, the light emitted by $G_{e}$ at time $t_{e}$ is observed on $G_{0}$ at a time $t_{0}$ ($t _{e} > t_{0}$) and the distance $l$ traveled by photons ($ds^{2}=0$) is related to the time difference $t$ and then to the $u$ parameter through the relation:

$$\begin{aligned} l= \int _{t_{e}}^{t_{0}}\frac{c dt}{a(t)} = \int _{u_{e}} ^{u_{0}}\frac{ ( 1+\mathit{ch}(2 u ) }{ \mathit{ch}^{2}(u)} du =2 u_{0} - 2 u_{e} \end{aligned}$$

(15)

We can also relate the distance $l$ to the distance marker $r$ by (using Friedman’s metric with $k=-1$):

$$\begin{aligned} l= \int _{t_{e}}^{t_{0}}\frac{c dt}{a(t)} = \int _{0}^{r}\frac{dr'}{ \sqrt{1+r^{\prime 2}}} =\mathit{argsh}(r) \end{aligned}$$

(16)

So we can write:

$$\begin{aligned} r = \mathit{sh}(2 u_{0}-2 u_{e}) = 2 \mathit{sh}(u_{0}-u_{e}) \mathit{ch}(u_{0}-u_{e}) \end{aligned}$$

(17)

We need now to link $u_{e}$ and $u_{0}$ to observable quantities $q_{0}$, $H_{0}$, and $z$. From Eq. (10) we get:

$$\begin{aligned} u = \mathit{argch} \biggl( \sqrt{\frac{a}{\alpha ^{2}}} \biggr) \end{aligned}$$

(18)

Equation (15) gives the usual redshift expression:

$$\begin{aligned} a_{e}=\frac{a_{0}}{1+z} \end{aligned}$$

(19)

From Eqs. (13) and (18) we get:

$$\begin{aligned} u_{0} = \mathit{argch}\sqrt{\frac{2q_{0}-1}{2q_{0}}} = \mathit{argsh}\sqrt{- \frac{1}{2q _{0}}} \end{aligned}$$

(20)

From Eqs. (13), (18) and (19) we get:

$$\begin{aligned} u_{e} = \mathit{argch}\sqrt{\frac{2q_{0}-1}{2q_{0}(1+z)}} = \mathit{argsh}\sqrt{- \frac{1+2q _{0}z}{2q_{0}(1+z)}} \end{aligned}$$

(21)

Inserting Eqs. (20) and (21) into Eq. (17), after a ‘few’ technical manipulations, using at the end Eq. (14) and considering the constraint that $1+2q_{0}z>0$, we get:

$$\begin{aligned} r = \frac{c}{a_{0} H_{0}} \frac{q_{0} z+(1-q_{0}) ( 1-\sqrt{1+2q _{0}z} ) }{q_{0}^{2}(1+z)} \end{aligned}$$

(22)

Which is similar to Mattig’s work (Mattig 1959) with usual Friedmann solutions where $q_{0}>0$, here we have always $q_{0}<0$.

The total energy received per unit area and unit time interval measured by bolometers is related to the luminosity:

$$\begin{aligned} E_{\mathit{bol}} = \frac{L}{4\pi a^{2}_{0} r^{2} (1+z)^{2}} \end{aligned}$$

(23)

Using Eq. (22), the bolometric magnitude can therefore be written as:

$$\begin{aligned} m_{\mathit{bol}}=5 \mathit{log}_{10} \biggl[ \frac{q_{0} z+(1-q_{0}) ( 1-\sqrt{1+2q _{0}z} ) }{q_{0}^{2}} \biggr] + \mathit{cte} \end{aligned}$$

(24)

This relation rewrites as (Terell 1977):

$$\begin{aligned} m_{\mathit{bol}}=5 \mathit{log}_{10} \biggl[ z+ \frac{z^{2}(1-q_{0})}{1+q_{0}z+\sqrt{1+2q _{0}z}} \biggr] + \mathit{cst} \end{aligned}$$

(25)

which is valid for $q_{0}=0$.

Appendix B: Age of the universe

Below we will establish the relation between the age of the universe $T_{0}$ with $q_{0}$ and $H_{0}$. This age is defined by:

$$\begin{aligned} T_{0} = \frac{\alpha ^{2}}{c} \biggl( \frac{\mathit{sh}(2u_{0})}{2}+u_{0} \biggr) \end{aligned}$$

(26)

From Eqs. (11), (13), (14) we get:

$$\begin{aligned} \frac{\alpha ^{2}}{c} = -\frac{2q}{H} ( 1-2q ) ^{- \frac{3}{2}} = \frac{2q_{0}}{H_{0}} ( 1-2q_{0} ) ^{- \frac{3}{2}} \end{aligned}$$

(27)

and so:

$$\begin{aligned} T_{0}.=-2q_{0} ( 1-2q_{0} ) ^{-\frac{3}{2}} \biggl( \frac{\mathit{sh}(2u _{0})}{2}+u_{0} \biggr) \frac{1}{H_{0}} \end{aligned}$$

(28)

Inserting Eq. (20) in Eq. (28) we finally get:

$$\begin{aligned} T_{0}.H_{0}=2q_{0} ( 1-2q_{0} ) ^{-\frac{3}{2}} \biggl( \mathit{argsh}\sqrt{\frac{-1}{2q _{0}}}-\frac{\sqrt{1-2q_{0}}}{2q_{0}} \biggr) \end{aligned}$$

(29)

This relation is shown in Fig. 8.

Fig. 8

Age of the universe time Hubble’s constant versus $q_{0}$