Parameter estimation of a nonlinear magnetic universe from observations

Abstract

The cosmological model consisting of a nonlinear magnetic field obeying the Lagrangian $\mathcal {L}= \gamma F^{\alpha },\, F$ being the electromagnetic invariant, coupled to a Robertson-Walker geometry is tested with observational data of Type Ia Supernovae, Long Gamma-Ray Bursts and Hubble parameter measurements. The statistical analysis show that the inclusion of nonlinear electromagnetic matter is enough to produce the observed accelerated expansion, with not need of including a dark energy component. The electromagnetic matter with abundance $\varOmega _B$, gives as best fit from the combination of all observational data sets $\varOmega _B=0.562^{+0.037}_{-0.038}$ for the scenario in which $\alpha =-1, \varOmega _B=0.654^{+0.040}_{-0.040}$ for the scenario with $\alpha =-1/4$ and $\varOmega _B=0.683^{+0.039}_{-0.043}$ for the one with $\alpha =-1/8$. These results indicate that nonlinear electromagnetic matter could play the role of dark energy, with the theoretical advantage of being a mensurable field.

Appendices

Appendix 1: Scaling between the scale factor $a$ and the electromagnetic invariant $F$

The energy conservation $T_{;\mu }^{\mu \nu } = 0$, leads to the equation

$$\begin{aligned} \dot{\rho } + 3 H (\rho +p)=0, \end{aligned}$$

(22)

which also can be derived from Eq. (6). So, using the expressions of $\rho$ and $p$, Eq. (2), in terms of the electromagnetic Lagrangian, the scaling between the scale factor $a$ and the electromagnetic invariant $F$ can be determined for a Lagrangian with arbitrary dependence on the two electromagnetic invariants $\mathcal {L}(F,G)$ as

$$\begin{aligned} -\!\dot{F}\mathcal {L}_F + 3 \left( {\frac{\dot{a}}{a}}\right) \left( {- \frac{4}{3} (2E^2+2B^2)\mathcal {L}_F}\right) =0. \end{aligned}$$

(23)

Now, if one restricts to the case $G=0$ (i.e. no electric field $E=0$), then $F=2B^2$ and

$$\begin{aligned} -\!\mathcal {L}_F \left\{ \dot{F} + 3 \left( {\frac{\dot{a}}{a}}\right) \left( { \frac{4}{3}F}\right) \right\} =0, \end{aligned}$$

(24)

whose solution, given by $Fa^4=\mathrm{const}$, is independent of the particular form of $\mathcal {L}(F)$.

Appendix 2: The scale factor as a function of time

The expressions of Friedmann equations for the nonlinear magnetic terms are

$$\begin{aligned} \begin{aligned} \left( {\frac{\dot{a}}{a}}\right) ^2&= - \frac{\mathcal {L}}{3}, \\ \frac{\ddot{a}}{a}&= - \frac{1}{3} (\mathcal {L}-2F\mathcal {L}_F). \end{aligned} \end{aligned}$$

(25)

Knowing that $\mathcal {L}a^{4 \alpha }=\mathrm{const}$, and using the Friedmann equations, the expression for $a(t)$ can be determined. Let us consider the following derivative,

$$\begin{aligned} \frac{d}{dt} \left( {a^{(4 \alpha -1)} \dot{a}}\right) = a^{4 \alpha } \left\{ { \frac{\ddot{a}}{a} + (4 \alpha -1) \frac{\dot{a}^2}{a^2}}\right\} , \end{aligned}$$

(26)

and substituting Friedmann’s equation, Eq. (25), we realize that the right hand term is constant,

$$\begin{aligned} \frac{d}{dt} \left( {a^{(4 \alpha -1)} \dot{a}}\right) = {- \frac{2 \alpha \mathcal {L}a^{4 \alpha }}{3}} = \text {const}. \end{aligned}$$

(27)

Finally, integrating for $a(t)$, it is obtained that $a(t)= \mathrm{const}(t-t_0)^{1/2 \alpha }$.