Inflation in energy-momentum squared gravity in light of Planck2018

Abstract

We study cosmological dynamics of the energy-momentum squared gravity. By adding the squared of the matter field’s energy-momentum tensor ($\zeta \, \mathbf{T} ^{2}$) to the Einstein–Hilbert action, we obtain the Einstein’s field equations and study the conservation law. We show that, the presence of $\zeta \, \mathbf{T} ^{2}$ term, breaks the conservation of the energy-momentum tensor of the matter fields. However, an effective energy-momentum tensor in this model is conserved in time. By considering the FRW metric as the background, we find the Friedmann equations and by which we explore the cosmological inflation in $\zeta \,\mathbf{T} ^{2}$ model. We perform a numerical analysis on the perturbation parameters and compare the results with Planck2018 different data sets at $68\%$ and $95\%$ CL, to obtain some constraints on the coupling parameter $\zeta$. We show that, for $0< \zeta \le 2.1\times 10^{-5}$, the $\zeta \, \mathbf{T} ^{2}$ gravity is an observationally viable model of inflation.

Appendix

We have the following Lienard differential equation

$$\begin{aligned} {\ddot{H}} + A H \dot{H} + B H^3 = 0\,, \end{aligned}$$

where

$$\begin{aligned} A = \frac{4 \, \zeta }{\kappa } \end{aligned}$$

and

$$\begin{aligned} B = \frac{- 27 \, \zeta }{\kappa }\pm \frac{2 \, \Lambda }{\kappa ^2}\, \end{aligned}$$

By assuming $H \equiv y$, we can rewrite the above Lienard equation as follows

$$\begin{aligned} \ddot{y} + f(y) \dot{y} + g(y) =0\,, \end{aligned}$$

where

$$\begin{aligned} f(y) = A \, y \,, \end{aligned}$$

and

$$\begin{aligned} g(y) = B \, y^3 \,. \end{aligned}$$

Now, we define $w(y) = \dot{y}$ and by which we convert our Lienard equation to the Abel differential equation of the second kind as

$$\begin{aligned} w w' + f(y) w + g(y) = 0\,, \end{aligned}$$

with $w w' = \ddot{y}$ and $w' \equiv \frac{\hbox {d}w}{\hbox {d}y}$. By introducing $z=\int F(y) \hbox {d}y$, with $F(y) = -A y$ and $G(y) = - B y^3$, the Abel equation takes the following canonical form

$$\begin{aligned} w w_{,z} = w + \phi (z)\,, \end{aligned}$$

where “, z” demonstrates derivative with respect to z, and

$$\begin{aligned} \phi (z) = \frac{G(y)}{F(y)}\,. \end{aligned}$$

Considering that

$$\begin{aligned} z = - \frac{A y^2}{2}\,, \end{aligned}$$

we find the following expression for $\phi (z)$

$$\begin{aligned} \phi (z) = \frac{B}{A}\, y^2 \,, \end{aligned}$$

leading to

$$\begin{aligned} w w_{,z}=w + \frac{B}{A} \, y^2 =w+D\,z\,. \end{aligned}$$

By defining $k(z)=Dz$, we get

$$\begin{aligned} w w_{,z}=w+k(z)\,, \end{aligned}$$

which has the following solution

$$\begin{aligned} z={{\mathcal {C}}} \exp \bigg ( - \int {\frac{\sigma \hbox {d}\sigma }{ \sigma ^2 - \sigma - D } } \bigg )\,,\quad w={{\mathcal {C}}}\sigma \exp \bigg ( - \int {\frac{\sigma \hbox {d}\sigma }{ \sigma ^2 - \sigma - D } } \bigg ). \end{aligned}$$

Now, we can obtain the Hubble parameter and its derivatives. From $y\equiv H$ and $z=-\frac{Ay^2}{2}$, we find

$$\begin{aligned} H = \left( \frac{-2z}{A}\right) ^{\frac{1}{2}}. \end{aligned}$$

Using ${\dot{H}}={\dot{y}}=w$, we obtain

$$\begin{aligned} \dot{H} = \sigma \,z\,. \end{aligned}$$

Also, from ${\ddot{H}} = w w'$ and considering that $\frac{\hbox {d}w}{\hbox {d}y}\equiv \frac{\hbox {d}w}{\hbox {d}z}\frac{\hbox {d}z}{\hbox {d}y}$, we get

$$\begin{aligned} {\ddot{H}} = -A \left( \frac{\sigma + D}{\sigma }\right) H \dot{H}\,. \end{aligned}$$