Difference between revisions of "New Cosmography"

We can write the scale factor in terms of the present time cosmographic parameters: \[a(t)\sim 1+H_{0} \Delta t-\frac{1}{2} q_{0} H_{0}^{2} \Delta t^{2} +\frac{1}{6} j_{0} H_{0}^{3} \Delta t^{3} +\frac{1}{24} s_{0} H_{0}^{4} \Delta t^{4} +120l_{0} H_{0}^{5} \Delta t^{5} \] This decomposition describes evolution of the Universe on the time interval $\Delta t$ directly through the measurable cosmographic parameters. Each of them describes certain characteristic of the evolution. In particular, the sign of deceleration parameter $q$ indicates whether the dynamics is accelerated or decelerated. In other words, a positive\textbf{ }acceleration parameter indicates that standard gravity predominates over the other species, whereas a negative sign\textbf{ }provides a repulsive e\textbf{ff}ect which overcomes the standard attraction due to gravity. Evolution of the deceleration parameter is described by the jerk parameter $j$. In particular, a positive jerk parameter would\textbf{ }indicate that there exists a transition time when the Universe modifies its expansion. In the vicinity of this transition the modulus of deceleration parameters tends to zero and then changes its sign\textbf{. }The two terms, i.e., $q$ and $j$ fix the local dynamics, but they may be not sufficient to remove the degeneration between different cosmological models and one will need higher terms of the decomposition.

\[\begin{array}{l} {1+z=\left[\begin{array}{cc} {} & {1+H_{0} (t-t_{0} )-\frac{1}{2} q_{0} H_{0}^{2} (t-t_{0} )^{2} +\frac{1}{3!} j_{0} H_{0}^{3} \left(t-t_{0} \right)^{3} +\frac{1}{4!} s_{0} H_{0}^{4} \left(t-t_{0} \right)^{4} } \\ {} & {+\frac{1}{5!} l_{0} H_{0}^{5} \left(t-t_{0} \right)^{5} \; +{\rm O}\left(\left(t-t_{0} \right)^{6} \right)} \\ {} & {} \end{array}\right]^{-1} ;} \\ {z=H_{0} (t_{0} -t)+\left(1+\frac{q_{0} }{2} \right)H_{0}^{2} (t-t_{0} )^{2} +\cdots .} \end{array}\]

The cosmographic parameters are model-independent quantities for the simple reason: these parameters are not functions of the EoS parameters $w$ or $w_{i} $ of the cosmic fluid filling the Universe in a concrete model.

\[q=-\frac{d\dot{a}}{Hda} =-\frac{\ddot{a}dt}{Hda} =-\frac{\ddot{a}}{aH^{2} } .\] It corresponds to the standard definition of the deceleration parameter \[q-\frac{\ddot{a}}{aH^{2} } .\]

When the rate of expansion never changes, and $\dot{a}$ is constant, the scaling factor is proportional to time $t$ , and the deceleration term is zero. When the Hubble term is constant, the deceleration term $q$ is also constant and equal to $\mathrm{-}$1, as in the de Sitter and steady-state Universes. In most models of Universes the deceleration term changes in time. One can classify models of Universe on the basis of time dependence of the two parameters. All models can be characterized by whether they expand or contract, and accelerate or decelerate:

• (a) $H>0,\; q>0$: expanding and decelerating
• (b) $H>0,\; q<0$: expanding and accelerating
• (c) $H<0,\; q>0$: contracting and decelerating
• (d) $H<0,\; q<0$: contracting and accelerating
• (e) $H>0,\; q=0$: expanding, zero deceleration
• (f) $H<0,\; q=0$: contracting, zero deceleration
• (g) $H=0,\; q=0$ : static.

Of course, generally speaking, both the Hubble parameter and deceleration parameter can change their sign during the evolution. Therefore the evolving Universe can transit from one type to another. It is one of the basic tasks of cosmology to follow this evolution and clarify its causes. There is little doubt that we live in an expanding Universe, and hence only (a), (b), and (e) are possible candidates. Evidences in favor of the fact that the expansion is presently accelerating continuously grows in number and therefore the current dynamics belongs to type (b).

\[\begin{array}{l} {\frac{dC_{n} }{d\ln (1+z)} =-a\frac{dC_{n} }{da} =-\frac{1}{H} \frac{dC_{n} }{dt} ;} \\ {\frac{dC_{n} }{d\ln (1+z)} =-\gamma _{n} \left(\frac{a^{\left(n+1\right)} }{aH^{n+1} } -\frac{a^{\left(n\right)} }{aH^{n} } -n\frac{a^{(n)} \dot{H}}{aH^{n+2} } \right)=} \\ {=-\gamma _{n} \left(\frac{1}{\gamma _{n+1} } C_{n+1} -\frac{1}{\gamma _{n} } C_{n} -\frac{1}{\gamma _{n} } nC_{n} \frac{\dot{H}}{H^{2} } \right);} \\ {\frac{dC_{n} }{d\ln (1+z)} =-\frac{\gamma _{n} }{\gamma _{n+1} } C_{n+1} +C_{n} -nC_{n} (1+q)} \end{array}\]

\[\frac{dC_{n} }{dt} =-H\frac{dC_{n} }{d\ln (1+z)} =H\left[\frac{\gamma _{n} }{\gamma _{n+1} } C_{n+1} -C_{n} +nC_{n} (1+q)\right]\]

\[\begin{array}{l} {\dot{q}=-H\left(j-2q^{2} -q\right),} \\ {\dot{j}=H\left[s+j(2+3q)\right],} \\ {\dot{s}=H\left[l+s(3+4q)\right],} \\ {\dot{l}=H\left[m+l(4+5q)\right]} \end{array}\]

Using results of the previous problem, one can transit from the time derivatives to the derivative w.r.t. the red shift according to the relation $\frac{d}{dz} =-\frac{1}{H(1+z)} \frac{d}{dt} $: \[\begin{array}{l} {\frac{dH}{dz} =H\frac{1+q}{1+z} ,} \\ {\frac{dq}{dz} =\frac{j-2q^{2} -q}{1+z} ,} \\ {\frac{dj}{dz} =-\frac{s+j(2+3q)}{1+z} ,} \\ {\frac{ds}{dz} =-\frac{l+s(3+4q)}{1+z} ,} \\ {\frac{dl}{dz} =-\frac{m+l\left(4+5q\right)}{1+z} } \end{array}\]

It is easy to see that \textbf{$\dot{H}=-H'H/a$ }where\textbf{ $H'=dH/dx$. }Then \[q=-\frac{\dot{H}}{H^{2} } -1=\frac{H'}{H} x-1\] Calculating $j$, making use of $a'=-a^{2} $, we obtain \[j(x)=1-2\frac{H'}{H} x+\left(\frac{H'^{2} }{H^{2} } +\frac{H''}{H} \right)x^{2} \]</p> </div> </div></div> <div id="cs-15"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 15''' <p style="color: #999;font-size: 11px">problem id: cs-15</p> Express the derivatives $d^{2} H/dz^{2} $ , $d^{3} H/dz^{3} $ and $d^{4} H/dz^{4} $ in terms of the cosmographic parameters. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Using results of the two previous problems, one finds \[\begin{array}{l} {\frac{d^{2} H}{dz^{2} } =\frac{j-q^{2} }{(1+z)^{2} } H,} \\ {\frac{d^{3} H}{dz^{3} } =\frac{H}{(1+z)^{3} } \left(3q^{2} +3q^{3} -4qi-3j-s\right)} \\ {\frac{d^{4} H}{dz^{4} } =\frac{H}{(1+z)^{4} } \left(-12q^{2} -24q^{3} -15q^{4} =32qj+25q^{2} j+7qs+12j-4j^{2} +8s+1\right)} \end{array}\]</p> </div> </div></div> <div id="cs-16"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 16''' <p style="color: #999;font-size: 11px">problem id: cs-16</p> Find decomposition of the inverse Hubble parameter $1/H$ in powers of the red shift $z$. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\begin{array}{l} {\frac{d}{dz} \left(\frac{1}{H} \right)=-\frac{1}{H^{2} } \frac{dH}{dz} =-\frac{1+q}{1+z} \frac{1}{H} ;} \\ {\frac{d^{2} }{dz^{2} } \left(\frac{1}{H} \right)=2\left(\frac{1+q}{1+z} \right)^{2} \frac{1}{H} -\frac{j-q^{2} }{\left(1+z\right)^{2} } \frac{1}{H} =\frac{2+4q+3q^{2} -j}{(1+z)^{2} } \frac{1}{H} ;} \\ {\frac{1}{H(z)} =\frac{1}{H_{0} } \left[1-\left(1+q_{0} \right)z+\frac{2+4q_{0} +3q_{0}^{2} -j_{0} }{6} z^{2} +\ldots \right]} \end{array}\]</p> </div> </div></div> <div id="cs-17"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 17''' <p style="color: #999;font-size: 11px">problem id: cs-17</p> Obtain relations for transition from the time derivatives to that w.r.t. the red shift. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">\[\begin{array}{l} {\frac{d^{2} }{dt^{2} } =(1+z)H\left[H+(1+z)\frac{dH}{dz} \right]\frac{d}{dz} +(1+z)^{2} H^{2} \frac{d^{2} }{dz^{2} } ,} \\ {\frac{d^{3} }{dt^{3} } =-(1+z)H\left\{H^{2} +(1+z)^{2} \left(\frac{dH}{dz} \right)^{2} +(1+z)H\left[4\frac{dH}{dz} +(1+z)\frac{d^{2} H}{dz^{2} } \right]\right\}\frac{d}{dz} -3(1+z)^{2} H^{2} } \\ {\times \left[H+(1+z)\frac{dH}{dz} \right]\frac{d^{2} }{dz^{2} } -(1+z)^{3} H^{3} \frac{d^{3} }{dz^{3} } ,} \\ {\frac{d^{4} }{dt^{4} } =(1+z)H\left[H^{2} +11(1+z)H^{2} \frac{dH}{dz} +11(1+z)H\frac{dH}{dz} +(1+z)^{3} \left(\frac{dH}{dz} \right)^{3} +7(1+z)^{2} H\frac{d^{2} H}{dz^{2} } \right. } \\ {+\left. 4(1+z)^{3} H\frac{dH}{dz} \frac{d^{2} H}{d^{2} z} +(1+z)^{3} H^{2} \frac{d^{3} H}{d^{3} z} \right]\frac{d}{dz} +(1+z)^{2} H^{2} \left[7H^{2} +22H\frac{dH}{dz} +7(1+z)^{2} \left(\frac{dH}{dz} \right)^{2} \right. } \\ {+\left. 4H\frac{d^{2} H}{dz^{2} } \right]\frac{d^{2} }{dz^{2} } +6(1+z)^{3} H^{3} \left[H+(1+z)\frac{dH}{dz} \right]\frac{d^{3} }{dz^{3} } +(1+z)^{4} H^{4} \frac{d^{4} }{dz^{4} } +(1+z)^{4} H^{4} \frac{d^{4} }{dz^{4} } .} \end{array}\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 18''' <p style="color: #999;font-size: 11px">problem id: </p> Can $dH^{n} /dz^{n} $ generally be expressed in terms of the cosmographic parameters? </div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 19''' <p style="color: #999;font-size: 11px">problem id: </p> Show that UNIQ-MathJax5-QINU </div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 20''' <p style="color: #999;font-size: 11px">problem id: </p> Show that the time derivatives of the Hubble's parameter can be expressed through the cosmographic parameters as follows: UNIQ-MathJax6-QINU UNIQ-MathJax7-QINU UNIQ-MathJax8-QINU UNIQ-MathJax9-QINU generally be expressed in terms of the cosmographic parameters? </div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 21''' <p style="color: #999;font-size: 11px">problem id: </p> Show that \[j=\frac{\ddot{H}}{H^{3} } +3\frac{\dot{H}}{H^{2} } +1\] <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Using results of the previous problem, we find \[j=\frac{\ddot{H}}{H^{3} } -3q-2\] Substituting \[q=-\frac{\dot{H}}{H^{2} } -1\] one finally obtains \[j=\frac{\ddot{H}}{H^{3} } +3\frac{\dot{H}}{H^{2} } +1\]</p> </div> </div></div> <div id=""></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 22''' <p style="color: #999;font-size: 11px">problem id: </p> Express total pressure in flat Universe through the cosmographic parameters. <div class="NavFrame collapsed"> <div class="NavHead">solution</div> <div style="width:100%;" class="NavContent"> <p style="text-align: left;">Excluding the density $\rho $ from the Friedman equations \[\begin{array}{l} {H=\frac{1}{3} \rho ,} \\ {\frac{\ddot{a}}{a} =H^{2} +\dot{H}=-\frac{1}{6} \left(\rho +3p\right)} \end{array}\] one finds \[p=-\left(3H^{2} +2\dot{H}\right)\] Using the above obtained expression $\dot{H}=-H^{2} (1+q)$ , we obtain \[p=-H^{2} \left(1-2q\right)\]