Cosmography

In the following problems we use an approach to the description of the evolution of the Universe, which is called "cosmography"$^*$. It is based entirely on the cosmological principle and on some consequences of the equivalence principle. The term "cosmography" is a synonym for "cosmo-kinematics". Let us recall that kinematics represents the part of mechanics which describes motion of bodies regardless of the forces responsible for it. In this sense cosmography represents nothing else than the kinematics of cosmological expansion.

In order to construct the key cosmological quantity $a(t)$ one needs the equations of motion (the Einstein's equation) and some assumptions on the material composition of the Universe, which enable one to obtain the energy-momentum tensor. The efficiency of cosmography lies in the ability to test cosmological models of any kind, that are compatible with the cosmological principle. Modifications of General Relativity or introduction of new components (such as dark matter or dark energy) certainly change the dependence $a(t)$, but they absolutely do not affect the kinematics of the expanding Universe.

The rate of Universe's expansion, determined by Hubble parameter $H(t)$, depends on time. The deceleration parameter $q(t)$ is used to quantify this dependence. Let us define it through the expansion of the scale factor $a(t)$ in a Taylor series in the vicinity of current time ${{t}_{0}}$: \[a(t)=a\left( {{t}_{0}} \right) +\dot{a}\left( {{t}_{0}} \right)\left[ t-{{t}_{0}} \right] +\frac{1}{2}\ddot{a}({t}_{0}) {{\left[ t-{{t}_{0}} \right]}^{2}}+\cdots\] Let us present this in the form \[\frac{a(t)}{a\left( {{t}_{0}} \right)} =1+{{H}_{0}}\left[ t-{{t}_{0}} \right] -\frac{{{q}_{0}}}{2}H_{0}^{2} {{\left[ t-{{t}_{0}} \right]}^{2}}+\cdots\] where the deceleration parameter is \[q(t)\equiv -\frac{\ddot{a}(t)a(t)}{{{{\dot{a}}}^{2}}(t)} =-\frac{\ddot{a}(t)}{a(t)}\frac{1}{{{H}^{2}}(t)}.\]

Note that the accelerated growth of scale factor takes place for $q<0$. When the sign of the deceleration parameter was originally defined, it seemed evident that gravity is the only force that governs the dynamics of Universe and it should slow down its expansion. The choice of the sign was determined then by natural wish to deal with positive quantities. This choice turned out to contradict the observable dynamics and became an example of historical curiosity.

In order to describe the kinematics of the cosmological expansion in more detail it is useful to consider the extended set of the parameters: \begin{align} H(t)\equiv &\frac{1}{a}\frac{da}{dt}\\ q(t)\equiv& -\frac{1}{a}\frac{{{d}^{2}}a}{d{{t}^{2}}}{{\left[ \frac{1}{a}\frac{da}{dt} \right]}^{-2}}\\ j(t)\equiv &\frac{1}{a}\frac{{{d}^{3}}a}{d{{t}^{3}}}{{\left[ \frac{1}{a}\frac{da}{dt} \right]}^{-3}}\\ s(t)\equiv &\frac{1}{a}\frac{{{d}^{4}}a}{d{{t}^{4}}}{{\left[ \frac{1}{a}\frac{da}{dt} \right]}^{-4}}\\ l(t)\equiv&\frac{1}{a}\frac{{{d}^{5}}a}{d{{t}^{5}}}{{\left[ \frac{1}{a}\frac{da}{dt} \right]}^{-5}} \end{align}

$^*$See Weinberg, Gravitation and Cosmology, chapter 14.

Problem 1: scale factor series

Using the cosmographic parameters introduced above, expand the scale factor into a Taylor series in time.

solution

\begin{align} a(t) &= {a_0}[1 + {H_0}\left( {t - {t_0}} \right) - \frac{1} {2}{q_0}H_0^2{\left( {t - {t_0}} \right)^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} + \;{\rm O}\left( {\left| {t - {t_0}} \right|} \right)] . \end{align}

Problem 2: redshift series

Using these cosmographic parameters, expand the redshift into a Taylor series in time.

solution

\begin{align*} a(t) &= a(t_0) + \dot a(t_0)(t - t_0) +\frac{1}{2}\ddot a(t_0)(t - {t_0})^2 + \cdots =\\ &= a(t_0)\left[ 1 + H_0(t - t_0) -\frac{1}{2}q_0H_0^2(t - t_0)^2+ \cdots \right]. \end{align*} Use $1 + z = a(t_0)/a(t)$ to obtain \begin{align*} 1 + z &= \left[ 1 + H_0(t - t_0) - \frac{1}{2}q_0H_0^2(t - t_0)^2 + \cdots\right]^{ - 1};\\ z& = H_0(t_0 - t) + \left( 1 + \frac{q_0}{2}\right) H_0^2(t - t_0)^2 + \cdots \end{align*} It is worth to invert the latter relation as it is the redshift $z$ which is the measured quantity. This is easy to do for $z\ll 1$: \[t_0 - t = \frac{1}{H_0} \left[ z - \left(1 + \frac{q_0}{2}\right)z^2 + \cdots\right]\]

Problem 3: $q$, $z$ and $H$

Obtain the following relations between the deceleration parameter and Hubble's parameter \[q(t)=\frac{d}{dt}\left( \frac{1}{H} \right)-1;\quad q(z)=\frac{1+z}{H}\frac{dH}{dz}-1;\quad q(z)=\frac{d\ln H}{dz}(1+z)-1.\]

Problem 4: $q(a)$

Show that for the deceleration parameter the following relation holds: \[q\left( a \right)=-\left( 1+ \frac{\frac{dH}{dt}}{{{H}^{2}}} \right) -\left( 1+\frac{a\frac{dH}{da}}{H} \right).\]

Problem 5: derivatives

Show that derivatives of lower cosmographic parameters can expressed through the higher ones.

Problem 6: $q(z)$

Prove that \[\frac{dq}{d\ln (1+z)}=j-q(2q+1).\]

Problem 7: $d^{n}H/dz^{n}$

Show that the derivatives $\frac{dH}{dz}$ and $\frac{{{d}^{2}}H}{d{{z}^{2}}}$ can be expressed through the parameters $q$ and $j$.

solution

\[\frac{dH}{dz}=\frac{1+q}{1+z}H; \quad \frac{{{d}^{2}}H}{d{{z}^{2}}} =\frac{j-1+2(1+q)-(1+q)^{2}} {( 1+z)^{2}}\]

Problem 8: $d^{n}H/dt^{n}$

Show that the time derivatives of the Hubble's parameter can be expressed through the cosmographic parameters as follows: \begin{eqnarray} \dot{H} &=& -{{H}^{2}}(1+q); \\ \ddot{H} &=& {{H}^{3}}\left( j+3q+2 \right); \\ \dddot{H}&=& {{H}^{4}}\left[ s-4j-3q(q+4)-6 \right]; \\ \ddddot{H}&=& {{H}^{5}}\left[ l-5s+10\left( q+2 \right)j+30(q+2)q+24\right]. \end{eqnarray}

Problem 9: Ricci scalar

Consider the case of spatially flat Universe and express the scalar (Ricci) curvature and its time derivatives in terms of the cosmographic parameters $q,j,s,l$.

solution

The scalar curvature as function of the Hubble's parameter is \[R = - 6({H^2} + 2{\dot H^2}).\] Consequent differentiation of the latter with respect to time gives: \[\begin{array}{l} \dot R = - 6\left( {\ddot H + 4H\dot H} \right);\\ \ddot R = - 6\left( {\dddot H + 4H\ddot H +4{{\dot H}^2}} \right);\\ \dddot R = - 6\left( {\ddddot H + 4H\dddot H + 12\dot H\ddot H} \right). \end{array}\] Using the expressions for the time derivatives of the Hubble's parameter through the cosmographic parameters, derived in the previous problem \[\begin{array}{l} \dot H = - {H^2}(1 + q);\\ \ddot H = {H^3}\left( {j + 3q + 2} \right);\\ \dddot H = {H^4}\left[ {s - 4j - 3q(q + 4) - 6} \right];\\ \ddddot H = {H^5}\left[ {l - 5s + 10\left( {q + 2} \right)j + 30(q + 2)q + 24} \right], \end{array}\] one finds \[\begin{array}{l} R = - 6{H^2}(1 - q);\\ \dot R = - 6{H^3}\left( {j - q - 2} \right);\\ \ddot R = - 6{H^4}\left( {{q^2} + 8q + s + 6} \right);\\ \dddot R = - 6{H^5}\left[ { - 6\left( {3q + 8} \right)q + 2\left( {q + 4} \right)j - s + l - 24} \right] \end{array}\]

Problem 10: acceleration and deceleration

Show that the accelerated growth of expansion rate $\dot{H}>0$ takes place on the condition $q<-1$.

Problem 11: $H[z]$

Obtain the relation \[H(z)=-\frac{1}{1+z}\frac{dz}{dt}.\]

Problem 12: inflection point

Let $t_a$ be the moment in the history of the Universe when the decelerated expansion turned to the accelerated one, i.e. $q(t_a)=0$, and let $t_{1}<t_{a}$ and $t_{2}>t_{a}$ be two moments in the vicinity of $t_{a}$. Show that \[\Delta t\equiv t_1-t_2 =\frac{1}{H_1}-\frac{1}{H_2}.\]

solution

The average value of the deceleration parameter reads: \[\bar q = \frac{1}{t_1 - t_2 } \int_{t_2 }^{t_1 } {q(t)dt}.\] As \[q(t) = \frac{d}{dt}\left( {\frac{1}{H}} \right) - 1,\] we have \[1 + \bar q = \frac{1}{\Delta t}\left( {\frac{1}{H_1 } - \frac{1}{H_2 }} \right),\] where \[\Delta t = t_1 - t_2 = [t_0 - t(z_2 )] - \left[ {t_0 - t(z_1)} \right]\] and \[t_0 - t(z) = \int_0^z {\frac{dz'}{(1 + z')H(z')}}.\] In the vicinity of the inflection point $\bar q(t_a ) \simeq 0$ and \[\Delta t \simeq \frac{1}{H_1 } - \frac{1}{H_2}.\]

Problem 13: $H[q]$

Obtain the following integral relation between the Hubble's parameter and the deceleration parameter \[H=H_0\exp\left[ \int\limits_0^z [q(z^\prime)+1] d\ln(1+z^\prime)\right].\]

Problem 14: Hubble law for close galaxies

Reformulate the Hubble's law in terms of redshift for close galaxies $(z\ll 1)$.

solution

For nearby galaxies $(z \ll 1)$ \[z \approx H_0(t_0-t_e).\] In units $c=1$ \[z = H_0d,\] where $d$ is the proper distance to the object.
It was Vesto Melvin Slipher who for the first time observed redshift of extragalactic objects in 1912, in the Lowell observatory in Flagstaff, Arizona. In 1922 he published the redshifts for $41$ spiral galaxies, 36 from which were positive (greater 0.006) and 5 -- negative. The Andromeda Nebula is the fastest to approach us, its redshift is $z=-0.001$.

Problem 15: $\tfrac{d}{dt}[\tfrac{d}{dz}]$

Show that \[\frac{d}{dt}=-(1+z)H\frac{d}{dz}.\]

solution

Substitute the definition of the deceleration parameter \[q(t)=-\frac{\ddot{a}a}{{{{\dot{a}}}^{2}}}=\frac{d}{dt}\left( \frac{1}{H} \right)-1\] to obtain that \[\bar{q}({{t}_{0}})=-1+\frac{1}{{{t}_{0}}{{H}_{0}}}\]

Problem 16: $\tfrac{d^{n}}{dt^{n}}[\tfrac{d}{dz}]$

Obtain the transformation law from the higher time derivatives to the derivatives with respect to redshift: \[\frac{d^{(i)}}{dt}\to \frac{d^{(i)}}{dz}.\]

Problem 17: $\tfrac{d^{n}H^2}{dz^n}$

Calculate the derivatives of Hubble's parameter squared with respect to redshift \[\frac{d^{(i)}H^2}{dz^{(i)}},\quad i=1,2,3,4\] and express them in terms of the cosmographic parameters.

Problem 18: $q[H(1+z)]$

Show that the deceleration parameter $q$ can be presented in the form \[q(x) = \frac{H'(x)}{H(x)}x - 1;\; x = 1 + z.\]

Problem 19: $q[H(z)]$

Show that \[q(z)=\frac{1}{2}\frac{d\ln {H^2}}{d\ln (1+z)}.\]

Problem 20: $dH/dz$

Express the derivatives $dH/dz$ and $d^2H/dz^2$ throgh the parameters $q$ and \[r \equiv \frac{\dddot a}{aH^3}.\]

Problem 21: averaged $q$

Consider the time average of the deceleration parameter \[\bar{q}\left( {{t}_{0}} \right) =\frac{1}{{{t}_{0}}}\int_{0}^{{{t}_{0}}}{q(t)dt}\] and show that it can be evaluated without integration of equation of motion for the scale factor.

Problem 22: age of the Universe via $q$

Show that the current age of the Universe is proportional to $H_{0}^{-1}$ and the proportionality coefficient is determined by the average value of the deceleration parameter.

solution

Use the solution of the previous problem to find the following \[{{t}_{0}}=\frac{H_{0}^{-1}}{1+\bar{q}}\] It is worth noting that this purely kinematic result depends neither on the curvature of the Universe, nor on the number of components composing the Universe, nor on the particular kind of the theory of gravity used.

Problem 23: proper distance through $q_0$

Show that the proper distance to an object with redshift $z$ is related to the current deceleration parameter $q_0$ as \[R=\frac{c}{H_{0}q_{0}^{2}}\,\frac{1}{1+z}\, \Big[q_{0}z +(q_{0}-1)\big(\sqrt{1+2q_0}-1\big)\Big].\]

Problem 24: current value of deceleration parameter

Express the current values of deceleration parameter $q_0 \equiv \left. { - \frac{1}{a}\frac{d^2 a}{dt^2}\left[ {\frac{1}{a}\frac{da}{dt}} \right]^{ - 2} } \right|_{t = t_9 }$ and the jerk parameter $j \equiv \left. {\frac{1}{a}\frac{d^3 a}{dt^3}\left[ {\frac{1}{a}\frac{da}{dt}} \right]^{ - 3} } \right|_{t = t_0 }$ in terms of $N \equiv - \ln \left( {1 + z} \right)$

solution

For the current time $t = t_0 $ we have $N = 0$. \begin{align} q_0 & = \left. { - \frac{1}{H^2}\left\{ {\frac{1}{2}\frac{d\left( {H^2 } \right)}{dN} + H^2 } \right\}} \right|_{N = 0}.\\ j_0 &= \left. {\frac{1}{2H^2}\frac{d^2 H^2 }{dN^2} + \frac{3}{2H^2}\frac{dH^2}{dN} + 1} \right|_{N = 0} . \\ \end{align}

Problem 25

problem id:

Using $d_l (z)=a_0 (1+z)f(\chi)$, where \[\chi=\frac1{a_0}\int\limits_0^z\frac{du}{H(u)},\quad f(\chi)=\left\{\begin{array}{rcl} \chi & - & flat\ case\\ \sinh(\chi) & - & open\ case\\ \sin(\chi) & - & closed\ case. \end{array}\right.\] find the standard luminosity distance-versus-redshift relation up to the second order in $z$: \[d_L=\frac z{H_0}\left[1+\left(\frac{1-q_0}2\right)z+O(z^2)\right].\]

solution

Simply break up the Hubble parameter under the integral into a series, integrate, and use the numerous formulas already in the Cosmography section. (see Expanding Universe: slowdown or speedup?, or the Visser convergence article.)

Problem 26

problem id:

Many supernovae give data in the $z>1$ redshift range. Why is this a problem for the above formula for the $z$-redshift? (Problems 2) - 10) are all from the Visser convergence article)

solution

\[\frac1{1+z}=\frac{a(t)}{a_0}=1+H_0(t-t_0)-\frac{q_0 H_0^2}{2!}(t-t_0)^2+\frac{j_0 H_0^3}{3!}(t-t_0)^3+O(|t-t_0|^4).\] This is the power series for $a(t)$, and it serves as the basis of all of our power series. However, it has a pole at $z=-1$. By complex variable theory, this means that the radius of convergence is at most $1$, so data with $z>1$ is outside of the convergence radius. (NOTE: I'm a bit confused about the fine points of this - I don't fully understand why we look at the convergence radius of this particular relation, to be honest. This is something you should discuss)

Problem 27

problem id:

Give physical reasons for the above divergence at $z=-1$.

solution

$z=-1$ corresponds to $a=\infty$ - we can't "look past" infinity.

Problem 28

problem id:

Sometimes a "pivot" is used: \[z=z_{pivot}+\Delta z,\] \[\frac1{1+z_{pivot}+\Delta z}=\frac{a(t)}{a_0}=1+H_0(t-t_0)-\frac1{2!}{q_0 H_0^2}(t-t_0)^2+\frac1{3!}{j_0 H_0^3}(t-t_0)^3+O(|t-t_0|^4).\] What is the convergence radius now?

solution

The pole is now located at \[\Delta z=-(1+z_{pivot}),\] which again physically corresponds to a Universe that had undergone infinite expansion, $a=\infty$. The radius of convergence is now \[|\Delta z|\le(1+z_{pivot}),\] and we expect the pivoted version of the Hubble law to fail for \[z>1+2z_{pivot}.\]

Problem 29

problem id:

The most commonly used definition of redshift is \[z=\frac{\lambda_0-\lambda_e}\lambda_e=\frac{\Delta\lambda}\lambda_e.\] Let's introduce a new redshift: \[z=\frac{\lambda_0-\lambda_e}\lambda_0=\frac{\Delta\lambda}\lambda_0.\]

solution

a) What is the "new $y$-redshift" in terms of the old $z$? \[y=\frac z{1+z},\quad z=\frac y{1-y}.\] b) The past and the future correspond, in terms of $z$-redshift to $z\in(0,\infty)$ and $z\in(-1,0)$ respectively. What are these intervals in terms of $y$-redshift? Past: $y\in(0,1)$, future: $y\in(-\infty,0)$.
c) get the correlation between luminosity distance and $y$-redshift up to the second power: \[d_L(y)=d_H y\left\{1-\frac12(-3+q_0)y+O(y^2)\right\}.\]

Problem 30

problem id:

Argue why, on physical grounds, we can't extrapolate beyond $y=1$.

solution

We can't extrapolate through the big bang. Thus, we assume that the convergence radius of $y$-redshift-based formulas should be $1$ (see the picture)

Problem 31

problem id: cg_7

The most commonly used distance is the luminosity distance, and it is related to the distance modulus in the following way: \[\mu_D=5\log_10[d_L/(10\ pc)]=5\log_10[d_L/(1\ Mpc)]+25.\] However, alternative distances are also used (for a variety of mathematical purposes):
1) The "photon flux distance": \[d_F=\frac{d_L}{(1+z)^{1/2}}.\]
2) The "photon count distance": \[d_P=\frac{d_L}{(1+z)}.\]
3) The "deceleration distance": \[d_Q=\frac{d_L}{(1+z)^{3/2}}.\]
4) The "angular diameter distance": \[d_A=\frac{d_L}{(1+z)^{2}}.\]

Obtain the Hubble law for these distances (in terms of $z$-redshift, up to the second power by z)

solution

\begin{align} \nonumber d_F(z) & =d_H z\left\{1-\frac12q_0z+O(z^2)\right\}.\\ \nonumber d_P(z) & =d_H z\left\{1-\frac12(1+q_0)z+O(z^2)\right\}.\\ \nonumber d_Q(z) & =d_H z\left\{1-\frac12(2+q_0)z+O(z^2)\right\}.\\ \nonumber d_A(z) & =d_H z\left\{1-\frac12(3+q_0)z+O(z^2)\right\}. \end{align}

Problem 32

problem id:

Do the same for the distance modulus directly.

solution

\[\mu_D(z)=25+\frac5{\ln(10)}\left\{\ln(d_H/Mpc)+\ln z+\frac12(1-q_0)z+O(z^2)\right\}.\]

Problem 33

problem id:

Do problem #cg_7 but for $y$-redshift.

solution

\begin{align} \nonumber d_F(y) & =d_H y\left\{1-\frac12(-2+q_0)y+O(y^2)\right\}.\\ \nonumber d_P(y) & =d_H y\left\{1-\frac12(-1+q_0)y+O(y^2)\right\}.\\ \nonumber d_Q(y) & =d_H y\left\{1-\frac{q_0}2y+O(y^2)\right\}.\\ \nonumber d_A(y) & =d_H y\left\{1-\frac12(1+q_0)y+O(y^2)\right\}. \end{align}

Problem 34

problem id: cg_10

Obtain the THIRD-order luminosity distance expansion in terms of $z$-redshift

solution

\[d_L(z) =d_H z\left\{1-\frac12(-1+q_0)z +\frac16[q_0+3q_0^2-(j_0+\Omega_0)]z^2+O(z^3)\right\},\] \[\Omega_0=1+\frac{kc^2}{H_0^2a_0^2}=1+\frac{kd_H^2}{a_0^2}.\]

Problem 35

problem id:

Alternative (if less physically evident) redshifts are also viable. One promising redshift is the $y_4$ redshift: $y_4 = \arctan(y)$. Obtain the second order redshift formula for $d_L$ in terms of $y_4$.

solution

\[d_L=\frac1{H_0}\cdot\left[y_4+y_4^2\cdot\left(\frac12-\frac{q_0}2\right)\right].\]

Problem 36

problem id:

$H_0a_0/c\gg1$ is a generic prediction of inflationary cosmology. Why is this an obstacle in proving/measuring the curvature of pace based on cosmographic methods?

solution

Recalling problem #cg_10, we see that the term that includes $K$ is proportional to $(H_0a_0/c\gg1)^{-2}$, so for all practical (cosmographic) purposes, our Universe is indistinguishable from a flat Universe. If the Universe is really flat, then we will never be able to strictly prove its flatness from cosmographic methods - we will only be able to put ever-stricter bounds on $H_0a_0/c\gg1$. If the Universe is not flat, then with good enough data, we will be able to determine the non-flatness <!-(see Visser3(Simple good))-->.

Problem 37

problem id:

When looking at the various series formulas, you might be tempted to just take the highest-power formulas you can get and work with them. Why is this not a good idea?

solution

The more parameters you have, the larger the region in parameter space, the less strict the limitations on the parameters and the higher the possibility of degeneracies. So the complexity of the formula should be in correspondence with the amount and accuracy of available experimental data!

Problem 38

problem id:

Using the definition of the redshift, find the ratio \[\frac{\Delta z}{\Delta t_{obs}},\] called "redshift drift" through the Hubble Parameter for a fixed/commoving observer and emitter: \[\int\limits_{t_s}^{t_o}\frac{dt}{a(t)}=\int\limits_{t_s+\Delta t_s}^{t_o+\Delta t_o}\frac{dt}{a(t)}.\]

solution

From the formula in the formulation of the problem, take $\Delta t/t\ll1$. This gives $\Delta t_s=[a(t_s)/a(t_o)]\Delta t_o$. Then, \[\Delta z\equiv\frac{a(t_o+\Delta t_o)}{a(t_s+\Delta t_s)}-\frac{a(t_o)}{a(t_s)}\approx\left[\frac{\dot a(t_o)-\dot a(t_s)}{a(t_s)}\right]\Delta t_o,\] which automatically gives \[\frac{\Delta z}{\Delta t_{obs}}=(1+z)H_{obs}-H(z).\]

Problem 39

problem id:

A photon's physical distance traveled is \[D=c\int dt=c(t_0-t_s).\] Using the definition of the redshift, construct a power series for $z$-redshift based on this physical distance.

solution

\[z(D)=\frac{H_0D}c+\frac{1+q_0}2\frac{H_0^2D^2}{c^2}+\frac{6(1+q_0)+j_0}6\frac{H_0^3D^3}{c^3} +O\left(\left[\frac{H_0^4D^4}{c^4}\right]\right).\]

Problem 40

problem id:

Invert result of the previous problem (up to $z^3$).

solution

\[D(z)=\frac{cz}{H_0}\left[1-\left(1+\frac{q_0}2\right)z +\left(1+q_0+\frac{q_0^2}2-\frac{q_0}6\right)z^2 O(z^3)\right].\]

Problem 41

problem id:

Obtain a power series (in $z$) breakdown of redshift drift up to $z^3$ (hint: use $(dH)/(dz)$ formulas)

solution

\[\frac{\Delta z}{\Delta t_0} = -H_0q_0z+\frac12H_0(q_0^2-j_0)z^2+ \frac12H_0\left[\frac13(s_0+4q_0j_0)+j_0-q_0^2-q_0^3\right]z^3 +O(z^4).\]

Problem 42

problem id:

Using the Friedmann equations, continuity equations, and the standard definitions for heat capacity(that is, \[C_V=\frac{\partial U}{\partial T},\quad C_P=\frac{\partial h}{\partial T},\] where $U=V_0\rho_ta^3$, $h=V_0(\rho_t+P)a^3$, show that \[C_P=\frac{V_0}{4\pi G}\frac{H^2}{T'}\frac{j-1}{(1+z)^4},\] \[C_V=\frac{V_0}{8\pi G}\frac{H^2}{T'}\frac{2q-1}{(1+z)^4}.\]

Problem 43

problem id:

What signs of $C_p$ and $C_v$ are predicted by the $\Lambda-CDM$ model? ($q_{0\Lambda CDM}=-1+\frac32\Omega_m$, $j_{0\Lambda CDM}=1$ and experimentally, $\Omega_m=0.274\pm0.015$)

solution

\[C_P=0,\quad C_V<0.\]

Problem 44

problem id: cg_20

In cosmology, the scale factor is sometimes presented as a power series \[a(t)=c_0|t-t_\odot|^{\eta_0}+c_1|t-t_\odot|^{\eta_1}+c_2|t-t_\odot|^{\eta_2}+c_3|t-t_\odot|^{\eta_3}+\ldots\]. Get analogous series for $H$ and $q$.

solution

A partial solution reads: \[\dot a(t)=c_0\eta_0(t-t_\odot)^{\eta_0-1}+c_1\eta_1(t-t_\odot)^{\eta_1-1}+\ldots\] Also note: the most general lesson to be learned is this: If in the vicinity of any cosmological milestone, the input scale factor $a(t)$ is a generalized power series, then all physical observables ($H$, $q$, the Riemann tensor, etc.) will likewise be generalized power series, with related indicial exponents that can be calculated from the indicial exponents of the scale factor. Whether or not the particular physical observable then diverges at the cosmological milestone is "simply" a matter of calculating its dominant indicial exponent in terms of those occurring in the scale factor.

Problem 45

problem id:

What values of the powers in the power series are required for the following singularities?
a) Big bang/crunch (scale factor $a=0$)
b) Big Rip (scale factor is infinite)
c) Sudden singularity ($n$th derivative of the scale factor is infinite)
d) Extremality event (derivative of scale factor $= 1$) \end{description}

solution

a) ($0<\eta_0<\eta_1\ldots$).
b) ($\eta_0<\eta_1\ldots$), $\eta_0<0$ and $c_0\ne0$.
c) $\eta_0=0$ and $\eta_1>0$ with $c_0>0$ and $\eta_1$ non-integer.
d) $\eta_0=0$ and $\eta_i\in Z^+$. \end{description}

Problem 46

problem id:

Analyze the possible behavior of the Hubble parameter around the cosmological milestones (hint: use your solution of problem #cg_20).

solution

Taking the exact form of the Hubble parameter as a ratio of series, and keeping only the most dominant terms, we have (assuming that the first power coefficient isn't $0$) \[H=\frac{\dot a}a\sim\frac{c_0\eta_0(t-t_\odot)^{\eta_0-1}}{c_0(t-t_\odot)^{\eta_0}}=\frac{\eta_0}{t-t_\odot};\quad (\eta_0\ne0).\] When the first power coefficient is zero (which corresponds to either a sudden singularity or extremality event), since the next power coefficient must be greater (by definition), we have the following: \[H\sim\frac{c_1\eta_1(t-t_\odot)^{\eta_1-1}}{c_0}=\eta_1\frac{c_1}{c_0}(t-t_\odot)^{\eta_1-1};\quad (\eta_0=0;\ \eta-1>0).\] Combining all of this, we get the following: \[\lim_{t\to t_\odot}H=\left\{ \begin{array}{lcl} +\infty & \eta_0>0; & {}\\ sign(c_1)\infty & \eta_0=0; & \eta_1\in(0,1);\\ c_1/c_0 & \eta_0=0; & \eta_1=1;\\ 0 & \eta_0=0; & \eta_1>1;\\ -\infty & \eta_0<0. & {}\\ \end{array} \right.\] Other things (cosmographic parameters, components of the Ricci tensor, validity of the Energy conditions) are analyzed analogously

Problem 47

problem id:

Going outside of the bounds of regular cosmography, let's assume the validity of the Friedmann equations. It is sometimes useful to expand the Equation of State as a series (like $p=p_0+\kappa_0(\rho-\rho_0)+O[(\rho-\rho_0)^2],$) and describe the EoS parameter ($w9t)=p/\rho$) at arbitrary times through the cosmographic parameters.

solution

\[w9t)=\frac p\rho=-\frac{H^2(1-2q)+kc^2/a^2}{3(H^2+kc^2/a^2)}= -\frac{(1-2q)+kc^2/(H^2a^2)}{3(1+kc^2/(H^2a^2))}.\]

Problem 48

problem id:

Analogously to the previous problem, analyze the slope parameter \[\kappa=\frac{dp}{d\rho}.\] Specifically, we are interested in the slope parameter at the present time, since that is the value that is seen in the series expansion.

solution

\[8\pi G_N=\frac{d\rho}{dt}=-6c^2H\left[(1+q)H^2+\frac{kc^2}{a^2}\right],\] \[8\pi G_N=\frac{dp}{dt}=2c^2H\left[(1-j)H^2+\frac{kc^2}{a^2}\right].\] Then \[\kappa_0=-\frac13\left[\frac{1-j_0+kc^2/(H_0^2a_0^2)}{1+q_0+kc^2/(H_0^2a_0^2)}\right]\] which approximates (using $H_0a_0/c\gg1$) to \[\kappa_0=-\frac13\left[\frac{1-j_0}{1+q_0}\right]\] We therefore see an important note: to get the FIRST term in the series expansion of the EOS contains the both the deceleration parameter and the jerk (third order observational term), the latter of which is very weakly bounded by observations. This is a significant problem in testing models (even with crude linear tests, as they require the slope parameter - see the previous problem)

Problem 49

problem id:

Continuing the previous problem, let's look at the third order term - $d^2 p/d\rho^2$.

solution

Starting off, it's easy to see that \[\frac{d^2p}{d\rho^2}=\frac{\ddot p-\kappa\ddot\rho}{(\dot\rho)^2}.\] Then \begin{align} \nonumber\left.\frac{d^2p}{d\rho^2}\right|_0 & = -\frac{(1+kc^2/[H_0^2a_0^2])}{6\rho_0(1+q_0+kc^2/[H_0^2a_0^2])^3} \left\{s_0(1+q_0)+j_0(1+j_0+4q_0+q_0^2)+q_0(1+2q_0)\right.\\ {} & \left.+(s_0+j_0+q_0+q_0j_0)\frac{kc^2}{H_0^2a_0^2}\right\}.\ \end{align} In the approximation $H_0a_0/c\gg1$ this reduces to \[\left.\frac{d^2p}{d\rho^2}\right|_0 = -\frac{s_0(1+q_0)+j_0(1+j_0+4q_0+q_0^2)+q_0(1+2q_0)}{6\rho_0(1+q_0)^3}.\]

Problem 50

problem id: 150_1

Show that \[\frac{d\dot a}{da}=-Hq.\]

solution

\[\frac{d\dot a}{da}=\frac{d\dot a}{dt}\frac{dt}{da}=\frac{\ddot a}{\dot a}=\frac{\ddot a}{aH}\equiv-Hq.\]