Difference between revisions of "Cosmography"

\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}

\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 $z\ll 1$: \[t_0 - t = \frac{1}{H_0} \left[ z - \left(1 + \frac{q_0}{2}\right)z^2 + \cdots\right]\]

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

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 \[\begin{array}{l} R = - 6{H^2}(1 - q);\\ \dot R = - 6{H^3}\left( {j - q - 2} \right);\\ \ddot R = - 6{H^4}\left( {{q^2} + 8q + s + 6} \right);\\ \dddot R = - 6{H^5}\left[ { - 6\left( {3q + 8} \right)q + 2\left( {q + 4} \right)j - s + l - 24} \right] \end{array}\]

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 \[\Delta t \simeq \frac{1}{H_1 } - \frac{1}{H_2}.\]

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 $z=-0.001$.

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 \[\bar{q}({{t}_{0}})=-1+\frac{1}{{{t}_{0}}{{H}_{0}}}\]

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.

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}

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.)

\[\frac1{1+z}=\frac{a(t)}{a_0}=1+H_0(t-t_0)-\frac{q_0 H_0^2}{2!}(t-t_0)^2+\frac{j_0 H_0^3}{3!}(t-t_0)^3+O(|t-t_0|^4).\] 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)

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

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 \[z>1+2z_{pivot}.\]

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: \[d_L(y)=d_H y\left\{1-\frac12(-3+q_0)y+O(y^2)\right\}.\]

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)

\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}

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

\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}

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

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

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))-->.

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!

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 \[\frac{\Delta z}{\Delta t_{obs}}=(1+z)H_{obs}-H(z).\]

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

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

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

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

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.

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}

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

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

\[8\pi G_N=\frac{d\rho}{dt}=-6c^2H\left[(1+q)H^2+\frac{kc^2}{a^2}\right],\] \[8\pi G_N=\frac{dp}{dt}=2c^2H\left[(1-j)H^2+\frac{kc^2}{a^2}\right].\] 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)

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 \[\left.\frac{d^2p}{d\rho^2}\right|_0 = -\frac{s_0(1+q_0)+j_0(1+j_0+4q_0+q_0^2)+q_0(1+2q_0)}{6\rho_0(1+q_0)^3}.\]