Revision as of 22:54, 18 June 2012

Newtonian cosmology

In next problems we use an approach to the description of the evolution of the Universe's, 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 the kinematics represent 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 an 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 details 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.

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

solution

\[\begin{gathered} a(t) = {a_0}[1 + {H_0}\left( {t - {t_0}} \right) - \frac{1} {2}{q_0}H_0^2{\left( {t - {t_0}} \right)^2} + \hfill \\ + \frac{1} [[:Template:3!]]{j_0}H_0^3{\left( {t - {t_0}} \right)^3} + \frac{1} [[:Template:4!]]{s_0}H_0^4{\left( {t - {t_0}} \right)^4} + \;{\rm O}\left( {\left| {t - {t_0}} \right|} \right)] \hfill . \end{gathered} \]

Problem 1.

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 1.

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

solution

Problem 1.

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

solution

Problem 1.

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

solution

Problem 1.

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

solution

Problem 1.

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 1.

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}

solution

Problem 1.

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 results of 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 1.

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

solution

Problem 1.

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

solution

Problem 1.

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: \[\bar q = \frac{1}[[:Template:T 1 - t 2]] \int_{t_2 }^{t_1 } {q(t)dt}.\] As \[q(t) = \frac{d}[[:Template:Dt]]\left( {\frac{1}{H}} \right) - 1,\] we have \[1 + \bar q = \frac{1}[[:Template:\Delta t]]\left( {\frac{1}[[:Template:H 1]] - \frac{1}[[:Template:H 2]]} \right),\] where \[\Delta t = t_1 - t_2 = [t_0 - t(z_2 )] - \left[ {t_0 - t(z_1)} \right]\] and \[t_0 - t(z) = \int_0^z {\frac[[:Template:Dz']][[:Template:(1 + z')H(z')]]}.\] 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 1.

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

solution

Problem 1.

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 1.

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 1.

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

solution

Problem 1.

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.

solution

Problem 1.

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

solution

Problem 1.

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

solution

Problem 1.

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

solution

Problem 1.

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.

solution

Problem 1.

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 1.

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

solution

Difference between revisions of "Cosmography"

Revision as of 22:54, 18 June 2012

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Newtonian cosmology

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@@ Line 43: / Line 43: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;"></p>
+     <p style="text-align: left;">\[\begin{gathered}
+  a(t) = {a_0}[1 + {H_0}\left( {t - {t_0}} \right) - \frac{1}
+{2}{q_0}H_0^2{\left( {t - {t_0}} \right)^2} +  \hfill \\
++  \frac{1} {{3!}}{j_0}H_0^3{\left( {t - {t_0}} \right)^3} +
+\frac{1} {{4!}}{s_0}H_0^4{\left( {t - {t_0}} \right)^4} + \;{\rm
+O}\left( {\left| {t - {t_0}} \right|} \right)] \hfill .
+\end{gathered} \]</p>
    </div>
 </div>
@@ Line 54: / Line 60: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;"></p>
+     <p style="text-align: left;">\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*}
++ 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  $z\ll 1$:
+\[t_0 - t = \frac{1}{H_0}
+	\left[ z - \left(1 + \frac{q_0}{2}\right)z^2
+	+  \cdots\right]\]</p>
    </div>
 </div>
@@ Line 116: / Line 139: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;"></p>
+     <p style="text-align: left;">\[\frac{dH}{dz}=\frac{1+q}{1+z}H;
+	\quad \frac{{{d}^{2}}H}{d{{z}^{2}}}
+	=\frac{j-1+2(1+q)-(1+q)^{2}}
+		{( 1+z)^{2}}\]</p>
    </div>
 </div>
@@ Line 144: / Line 170: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;"></p>
+     <p style="text-align: left;">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:
+\[\begin{array}{l}
+\dot R =  - 6\left( {\ddot H + 4H\dot H} \right);\\
+\ddot R =  - 6\left( {\dddot H + 4H\ddot H
+	+4{{\dot H}^2}} \right);\\
+\dddot R =  - 6\left( {\ddddot H +
+H\dddot H + 12\dot H\ddot H} \right).
+\end{array}\]
+Using the results of [[#equ61_6|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
+\[\begin{array}{l}
+R =  - 6{H^2}(1 - q);\\
+\dot R =  - 6{H^3}\left( {j - q - 2} \right);\\
+\ddot R =  - 6{H^4}\left( {{q^2} + 8q + s + 6} \right);\\
+\dddot R =  - 6{H^5}\left[ { - 6\left( {3q + 8} \right)q + 2\left( {q + 4} \right)j - s + l - 24} \right]
+\end{array}\]</p>
    </div>
 </div>
@@ Line 179: / Line 229: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;"></p>
+     <p style="text-align: left;">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
+\[\Delta t \simeq \frac{1}{H_1 } - \frac{1}{H_2}.\]</p>
    </div>
 </div>
@@ Line 203: / Line 269: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;"></p>
+     <p style="text-align: left;">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 $z=-0.001$.</p>
    </div>
 </div>
@@ Line 215: / Line 287: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;"></p>
+     <p style="text-align: left;">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
+\[\bar{q}({{t}_{0}})=-1+\frac{1}{{{t}_{0}}{{H}_{0}}}\]</p>
    </div>
 </div>
@@ Line 298: / Line 373: @@
    <div class="NavHead">solution</div>
    <div style="width:100%;" class="NavContent">
-     <p style="text-align: left;"></p>
+     <p style="text-align: left;">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.</p>
    </div>
 </div>