Observational cosmology
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Observational cosmology problems
Problem 1
problem id: 150_o1-problem-1
Find ratio between the illuminance of the Earth surface by the Sun and other light sources in the Universe (it is a quantitative resolution of the Olbers paradox).
As we mentioned above (in the Warm up chapter), finiteness of the Universe's lifetime enables to qualitatively resolve the Olbers paradox: the night sky is dark because light from the distant galaxies do not have time to reach the Earth during the lifetime of the Universe equal to $t_0\sim H_0^{-1}$. Let us now make some qualitative estimates. The observations show that the luminosity density of the galaxies in the observed Universe equals to \[nL\approx2\times10^8L_\odot Mpc^{-3}.\] Compared to the terrestrial standards, the Universe is not a well illuminated place: the above cited luminosity density is equivalent to one $40W$ lamp per a sphere of $1$ à.u. radius. Illuminance created by all the galaxies situated inside the horizon $R_{hor}\sim cH_0$ is \[E_g\approx nL\int\limits_0^{c/H_0}dr=nL(c/H_0)\sim2\times10^{-11}L_\odot AU^{-2}.\] Comparing this quantity with the illuminance on the Earth due to the Sun $E_{sun}=\frac{L_\odot}{4\pi AU^2}\approx0.08L_\odot AU^{-2},$ one finds that \[E_g/E_{sun}\sim3\times10^{-10}.\] Due to the cosmological principle the Universe is weakly illuminated as the Earth. That is why darkness of the night sky should not amaze us.
Problem 2
problem id: 150_o2
Consider light emitted by surface of the Sun and observed on the Earth. Determine the shift of the observed frequency compared to the analogues light frequency emitted by atoms on the Earth.
For a given atom transition ratio of the light frequencies (observed in the point $1$) between that emitted from the points $2$ and $1$, equals to \[\frac{\nu_2}{\nu_1}=\left(\frac{g_{00}(x_2)}{g_{00}(x_1)}\right)^{1/2}.\] In the weak field limit \[g_{00}\approx -1-2\Phi,\] $\Phi\ll1$, so that \[\frac{\Delta\nu}\nu=\frac{\nu_2}{\nu_1}-1=\Phi(x_2)-\Phi(x_1).\] Let us apply this relation to the case when light is emitted by surface of the Sun and is observed on the Earth. Gravity potential of the Sun is \[\Phi_\odot=-G\frac{M_\odot}{R_\odot},\] where solar mass and radius respectively are \begin{align} \nonumber M_\odot&=1.97\times10^{33}g;\\ \nonumber R_\odot&=0.695\times10^{6}km. \end{align} Finally one obtains \[\Phi_\odot=-2.12\times10^{-6}.\] The terrestrial gravity potential can be neglected compared to the solar one. Therefore one finally obtains that the relative redshift of the light arrived from the Sun approximately equals to $-2.12\times10^{-6}$ (compared to the light frequency emitted by atoms on the Earth).
Problem 3
problem id: 150_o3
Find the distance to a distant galaxy using recessional velocity as measured by the Doppler redshift.
The Hubble distance is given by \[R=\frac{V}{H_0}=\left[\frac{(z+1)^2-1}{(z+1)^2+1}\right]\frac{c}{H_0}\] and can be calculated from the wavelength shift of any spectral line.
Problem 4
problem id: 150_o4
Show that galaxies situated outside the Hubble sphere have imaginary redshift.
\[z=\sqrt{c+V}{c-V}-1=\sqrt{c+HR}{c-HR}-1.\] The Hubble radius is \[R_H=\frac c H,\] so for $R>R_H$ we obtain imaginary values of the redshift.
Problem 5
problem id: 150_o5
Find the exact relativistic Doppler velocity-redshift relation.
\[(1+z)_{Dop}=\sqrt{c+V}{c-V}.\] The exact relativistic Doppler velocity-redshift relation is \[V_{Dop}(z)=c\frac{2z+z^2}{2+2z+z^2}.\] For $z\to\infty$, the velocity $V_{Dop}(z)\to c$ corresponding to the limit $V_{Dop}\le c$.
Problem 6
problem id: 150_o7
Find $V_{exp}(z)$ for three cosmological models: Einstein-de Sitter, Milne and de Sitter.
For $\Omega_m=1$, $\Omega_\Lambda=0$ \[V_{exp}=H_0R(z)=H_0\int\limits_0^z\frac{cdz'}{H(z')}=\frac{2\left(\sqrt{1+z}-1\right)}{\sqrt{1+z}}c.\] For $\Omega_m=0$, $\Omega_\Lambda=0$ \[V_{exp}=\frac{z\left(1+z/2\right)}{1+z}c.\] For $\Omega_m=0$, $\Omega_\Lambda=1$ \[V_{exp}=zc.\]
Problem 7
problem id: 150_o8
Some luminous object has apparent stellar magnitude $m=20$, and the absolute one is $M=-15$. Determine distance to it.
Use the relation \[m-M=5\log_{10}\left(\frac{d_L}{Mpc}\right)+25.\]
Problem 8
problem id: 150_o9
Due to a random coincidence the Balmer series of singly ionized helium atom in a distant star overlap with the Balmer series of hydrogen in the Sun. How fast this star recedes from us?
Energy levels of any system, composed of two particles with opposite charges ($-e,Ze$) coupled solely by the electrostatic attractive forces, are \[E_n\sim Z^2.\] Thus the photon wavelength generated by the electron transitions is \[\lambda_n\sim\frac1{Z^2}.\] Therefore for stationary atoms \[\lambda_{He}=\frac14\lambda_H.\] For a distant star moving with velocity $V$, \[\lambda_{He}'=\lambda_{He}\sqrt{\frac{1+V/c}{1-V/c}}.\] The condition of the spectral lines overlapping reads \[\lambda_{He}'=\lambda_H.\] Then \[V\approx0.88c.\]
Problem 9
problem id: 150_o10
A gas cloud rotates around a super-massive black hole with mass equal to $M=3.6\times10^6M_\odot$ (it is a possible interpretation of recent observations). Assuming that distance between these objects is of order of $60$ lightyears, determine the expected Doppler shift.
\[G\frac{m_{cl}M}{R^2}=\frac{m_{cl}V^2}{R}.\] \[V\approx 2.9\times10^4m/s.\] In assumption of small Doppler shift one obtains \[\frac{\Delta\lambda}{\lambda}\approx\frac V c\approx 9.6\times10^{-5}.\]
Problem 10
problem id: 150_o11
Explain why transition from the cosmological time to the redshift can be considered as a quantitative characteristic describing evolution of the Universe.
During evolution of the Universe the wavelength $\lambda$ of freely propagating light increases as \[\lambda(t_0)=\lambda(t)\frac{a(t_0)}{a(t)}.\] The redshift is \[z=\frac{\lambda(t_0)-\lambda(t)}{\lambda(t_0)}.\] Using the first equation, one finds \[a=\frac a{1+z}.\] Measuring redshift of the observed object we can know how large the Universe was in the moment of the emission of the light. relative size of the Universe provides us with a measurable characteristic of the time in cosmology. The cosmologists use the redshift as a measure of time: $z=0$ means today, $z=\infty$ at the moment of the Big Bang. The advantage of the redshift before time is that the former is a directly measurable quantity.
Problem 11
problem id: 150_o12
Show, that for small separations the "Hubble law" $dz=Hdr$ holds. In other words, we have replaced the velocity by the redshift.
The photon trajectory is given by $ds^2=0$. Introducing the conformal time $\eta$, one obtains \[ds^2=0\rightarrow d\eta^2-dr^2=0.\] The solution for the radial comoving distance between source and observer is \[r=\pm\eta+const.\] Having made the transformation \[d\eta=\frac{d\eta}{dt}\frac{dt}{da}\frac{da}{dz}dz=-\frac a{\dot a}dz=-\frac{dz}{H(z)},\] we find the "Hubble law" for the solution $d\eta=-dr$ \[dz=Hdr.\]
Problem 12
problem id: 150_o13
Solve the previous problem using the Taylor series of the scale factor in terms of time.
Small redshifts correspond to small time difference between the moments of light emission $t_e$ and its observation $t_0$. It allows to limit the decomposition of $a(t_e)$ by the first terms \[a(t_e)\approx a(t_0)-\dot a(t_0-t_e)\approx a_0(1-H_0 r).\] Using the relation between the red shift and the scale factor \[1+z=\frac{a(t_0)}{a(t_e)}\approx\frac1{1-H_0 r}\approx1+H_0r.\] Therefore for $z\ll1$ the Hubble law can be represented in the form \[z=H_0r.\]
Problem 13
problem id: 2501_02o
Why the Linear Distance-Redshift Law in Near Space?
Consider a nearby galaxy and the change of the scale factor during the small time interval $dt$ when the light travels from this galaxy to the observer: \[a(t_{obs})=a(t_{em})+\left.\frac{da}{dt}\right|_{t=t_{em}}dt.\] When the distance is short and the expansion speed much less than the speed of light, then the distance $R$ measured by the astronomer using any of the available ways is practically equal to the metric distance $a(t)r$ and the time interval can be approximated as $dt\approx a(t)r/c=R/c$. From this it follows that \[a(t_{obs})/a(t_{em})=1+z=1+\frac{\dot a}a \frac R c.\] and finally \[cz=HR.\]
Problem 14
problem id: 150_o1-problem-2
Determine physical distance to an object emitted light with redshift $z$ in the flat expanding Universe.
The physical distance reads \[R(t)=a(t)r(t),\] where $r(t)$ is the comoving distance \[r(t)=\int\limits_t^{t_0}\frac{dt'}{a(t')}.\] Therefore \[R(t)=\frac1{1+z}\int\limits_t^{t_0}\frac{dt'}{a(t')}=\frac1{1+z}\int\limits_t^{t_0}d\eta.\] The differentials of the conformal time $d\eta$ and redshift $dz$ are related by \[d\eta=\frac{d\eta}{dt}\frac{dt}{da}\frac{da}{dz}dz=-\frac a{\dot a}dz=-\frac{dz}{H(z)}.\] Thus the physical distance to an object is related to the redshift $z$ of the emitted light by \[R(t)=\frac1{1+z}\int\limits_0^z\frac{dz'}{H(z')}.\]
Problem 15
problem id: 150_o16
Find comoving distance to a presently observed galaxy as a function of the redshift.
The photon equation of motion reads $ds^2=0$. Consider a radial trajectory with the observer in the origin. In this case one obtains for the spatially flat metrics \[ds^2=a^2(t)(d\eta^2-dr^2)=0,\] where $r$ is the radial comoving coordinate. Taking into account that \[d\eta=-\frac{dz}{H(z)},\] one obtains \[r(z)=\int\limits_0^z\frac{dz'}{H(z')}.\] The same result can be obviously obtained using result of the previous problem \[r(z)=\frac{R(z)}{a(z)}=\int\limits_0^z\frac{dz'}{H(z')}.\]
Problem 16
problem id: 150_o17
Solve the previous problem for the flat Universe dominated by non-relativistic matter.
In this case \[H(z)=H_0(1+z)^{3/2}\] and \[r(z)=\frac2{H_0}\left(1-\frac1{\sqrt{1+z}}\right).\]
Problem 17
problem id: 150_o18
Determine the recession velocity due to the cosmological expansion for an object emitted light with the redshift $z$ in the flat expanding Universe.
\[V=\dot R(t)=\dot a(t)=H(z)R(z)=\frac{H(z)}{1+z}\int\limits_0^z\frac{dz'}{H(z')}.\]
Problem 18
problem id: 150_o19
Obtain relations between velocity of cosmological expansion and redshift.
The exact velocity-proper distance relation (Hubble law) reads \[V_{exp}=HR.\] For the observer on the Earth living at the time $t=t_0$ the observed redshift serves as a measure of distance at this epoch. For photons moving along the comoving coordinate $r$ one obtains $cdt=\pm adr$. Consequently \[a_0 dr=\frac{cdz}{H(z)}.\] Integration over the redshift gives the proper distance-redshift relation \[R(t_0,z)\equiv R(z)=\int\limits_0^z\frac{cdz'}{H(z')}.\] Using the $R(z)$ relation one gets the exact velocity-redshift relation \[V_{exp}(z)=H_0R(z)=H_0\int\limits_0^z\frac{cdz'}{H(z')},\] where the expansion velocity is for an object with the redshift $z$ observed at the time $t=t_0$.
Problem 19
problem id: 150_o20
Find dependence of the Hubble parameter $H$ on the redshift $z$ for the case of one-component Universe composed of the non-relativistic matter.
For the non-relativistic matter dominated Universe one has \[\rho=\rho_0/a^3=\rho_0(1+z)^3.\] By the definition of the relative density \[\Omega\equiv\frac{\rho}{\rho_{cr}};\quad \rho_{cr}\equiv\frac{3H^2}{8\pi G},\] then \[\frac\Omega{\Omega_0}=\frac\rho{\rho_0}\frac{H_0^2}{H^2}=(1+z)^3\frac{H_0^2}{H^2};\] \[\Omega H^2=(1+z)^3\Omega_0H_0^2.\] Rewrite the first Friedmann equation in the form \begin{align} \nonumber a^2 H^2(1-\Omega) & =-k;\\ \nonumber a^2 H^2(1-\Omega) & =a_0^2 H_0^2(1-\Omega_0), \end{align} as $k=const$. Taking into account that $a_0/a=1+z$, one obtains \[H^2(1-\Omega) =(1+z)^2 H_0^2(1-\Omega_0).\] Substituting the above obtained expression for the $\Omega H^2$, one finally obtains \[H=H_0(1+z)(1+\Omega_0 z)^{1/2}.\]
Problem 20
problem id: 150_o21
Find dependence of relative density $\Omega$ on the redshift $z$ for the case of one-component Universe composed of the non-relativistic matter.
Substituting the expression for the relative density obtained in the previous problem \[\Omega = (1+z)^3\Omega_0\frac{H_0^2}{H^2}\] into the obtained $ibid$ expression for the Hubble parameter, one finds \[\Omega=\Omega_0\frac{1+z}{1+\Omega_0z}.\]
Problem 21
problem id: 150_o22
Construct a scheme to determine sign of acceleration of the scale factor, based on measurement of the supernovae bursts characteristics.
The Hubble law tells nothing about the magnitude, sign and the very possibility of non-uniform expansion of the Universe. It is valid under the approximation which is insensitive to the acceleration. In order to investigate the non-linear effects one needs data for high redshifts. If the observations detect deviations from the linear law, then it enables us to determine the acceleration sign basing on the magnitude and sign of the deviation. If the detected deviation lies towards increasing distance at fixed redshift then the acceleration is positive. The distance is estimated by luminosity of the source under assumption that the considered set of the sources represents the standard candles, i.e. has fixed luminosity with sufficiently high accuracy.
Problem 22
problem id: 150_o23
Show that the time derivative of the redshift for the light emitted at time $t$ and registered at time $t_0$ can be determined as \[\dot z\equiv\frac{dz}{dt_0}=H(t_0)(1+z)-H(t).\]
Make use of the definition \[1+z=\frac{a(t_0)}{a(t)},\] then \[\frac{dz}{dt_0}=\frac{\dot a(t_0)a(t)-a(t_0)\dot a(t)\frac{dt}{dt_0}}{a^2(t)}.\] Taking into account that \[\frac{dt}{dt_0}=\frac1{1+z},\] one obtains \begin{align} \nonumber \dot z & = \frac{\dot a(t_0)}{a(t)}-\frac{a(t_0)\dot a(t)}{a^2(t)}\frac1{1+z}=\\ \nonumber & = \frac{\dot a(t_0)}{a(t_0)}\frac{a(t_0)}{a(t)} - \frac{a(t_0)}{a(t)}\frac{\dot a(t)}{a(t)}\frac1{1+z}=\\ \nonumber & = H(t_0)(1+z)-H(t)=H_0(1+z)-H(z). \end{align}
Problem 23
problem id: 150_o24
In a flat one-component Universe with the state equation $p=w\rho$ at time $t_0$ one registers a light signal with redshift $z$. What values of the EoS parameter $w$ lead to \[\frac{dz}{dt_0}>0?\] Explain physical sense of the obtained result.
In the flat one-component Universe one has \[H(z)=H_0(1+z)^{\frac32(1+w)}.\] Using result of the previous problem one obtains \[\frac{dz}{dt_0}=H_0(1+z)-H_0(1+z)^{\frac32(1+w)}=H_0(1+z)\left[1-(1+z)^{\frac32(1+w)-1}\right]\] Thus $dz/dt_0>0$ if \[w<-1/3.\] In the flat one-component case this condition provides the accelerated expansion of Universe because in this case \[\frac{\ddot a}{a}=-\frac{4\pi G}{3}(\rho+3p).\]
Problem 24
problem id: 150_o26
Show that the luminosity distance can be generally presented as \[d_L=\frac{1+z}{H_0\sqrt{\Omega_{k0}}}\sinh\left(H_0\sqrt{\Omega_{k0}}\int\limits_0^z\frac{dz}{H(z)}\right),\] where $\Omega_{k0}$ is the relative contribution of the spatial curvature.
Using that \[\rho_{curv}=-\frac{3}{8\pi G}\frac{k}{R^2},\] where $R$ is the (comoving) radius of curvature of the open or closed Universe, one finds \[\Omega_{k0}=-\frac{k}{RH_0}.\] It is easy to see that the expression \[d_L=\frac{1+z}{H_0\sqrt{\Omega_{k0}}}\sinh\left(H_0\sqrt{\Omega_{k0}}\int\limits_0^z\frac{dz}{H(z)}\right)\] reproduces the well known one for the luminosity distance \[d_L=(1+z)\left\{ \begin{array}{ll} R\sinh\left[\frac{c}{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & open;\\ \frac{c}{H_0}\int\limits_0^z\frac{dz'}{E(z')}, & open;\\ R\sin\left[\frac{c}{H_0R}\int\limits_0^z\frac{dz'}{E(z')}\right], & closed; \end{array}\right.\] where $E(z)\equiv H(z)/H_0$.
Problem 25
problem id: 150_o27
\it When galaxy formation started in the history of the Universe remains unclear. Studies of the cosmic microwave background indicate that the Universe, after initial cooling (following the Big Bang), was reheated and reionized by hot stars in newborn galaxies at a redshift in the range $6<z<14$. Here we report a spectroscopic redshift of $z=6.96$. [M. Iue et al., A galaxy at a redshift z=6.96 , arXiv:0609393]
\bf Estimate age of the Universe when the first galaxies were created.
According to the SCM, age of the Universe corresponding to the redshift $z$ can be determined by the formula \[t(z)=\frac1{H_0}\int_0^{1/(1+z)}\frac{dx}{x\sqrt{\Omega_{0\Lambda}+\Omega_{0M}x^{-3}}}.\] Redshift $z\simeq7$ corresponds to just $750$ Myr after the Big Bang or only $6\%$ of its present age.
Problem 26
problem id: 150_o30
Express probability to find two galaxies in the infinitesimally small volumes $dV_1$ and $dV_2$ in terms of the correlation function $\xi(\vec{r_1},\vec{r_2})$, if average density of the galaxies in the considered volume equals $\bar n$ and total number of the galaxies is $N$. Point out main properties of the correlation function.
For absolutely random homogeneous distribution of galaxies the probability $dP_1$ to find a galaxy in a infinitesimally small volume $dV_1$ is proportional to the volume $dV_1$ and to the average galaxy number $\bar n$: \[dP_1 = \frac{\bar n}{N}dV_1,\] where $N$ is total number of galaxies in the sample. If there were no galaxy clusterization effect, then the probability to find galaxies in the volumes $dV_1$ and $dV_2$ would be equal to the product $dP_1dP_2$. Any clusterization will lead to deviation from this simple formula. It allows to define the two-point correlation function $\xi(\vec{r_1}-\vec{r_2})$: \[dP_{12}=\frac{\bar n^2}{N^2}[1+\xi(\vec{r_1}-\vec{r_2})]dV_1dV_2.\] The homogeneity implies that $\xi$ depends only on $|\vec{r_1}-\vec{r_2}|$. Assumption that the distribution is random on sufficiently large scales leads to the condition: \[\xi(\vec{r_1}-\vec{r_2})\to0\] for sufficiently large $r=|\vec{r_1}-\vec{r_2}|$. The function $xi$ must have values in the range $-1<\xi<\infty$. Positive value of $xi$ means that the galaxies tend to the cluster, and negative value --- that they repel. If $xi$ is positive on short distances then it must become negative with increasing distance in order to keep constant total density.
Problem 27
problem id: 150_o31
\bf Find relation between the spatial $\xi(r)$ and angular $w(\theta)$ correlation functions.
In order to obtain the spatial correlation function $\xi(r)$ one needs three-dimensional information on the galaxy distribution. Such information requires intensive observation of the redshifts and therefore long time of the telescope processing time. Until the recent time all large catalogues of galaxies recorded their position on the sky regardless their depth (redshift or distance). Therefore we can immediately obtain only the angular correlation function $w(\theta)$. The spatial correlation function $\xi(r)$ can be reconstructed from $w(\theta)$ under assumption of homogeneity and isotropy. The angular correlation function $w(\theta)$ is defined as the probability to observe a pair of galaxies in the solid angles $d\Omega_1$ and $d\Omega_2$ separated by angle $\theta$ on the celestial sphere. If average surface density in the unit solid angle equals to $\bar\sigma$, then \begin{equation}\label{150_o31_e1} dP_{12}\propto \bar\sigma^2[1+w(\theta)]d\Omega_1d\Omega_2. \end{equation} Let us assume for simplicity that all galaxies have the same absolute magnitude (luminosity). It means that all the galaxy have the same limiting distance up to which they are still detectable. Let this limiting distance equals to $D$. Then the angular distribution is obtained from the three-dimensional spatial distribution by integration up to the depth $D$: \begin{equation}\label{150_o31_e2} dP_{12}\propto \left\{\int\limits_0^Dr_1^2dr_1\int\limits_0^Dr_2^2dr_2[1+\xi(r)]\right\}d\Omega_1d\Omega_2. \end{equation} For the angular distance $\theta$ the corresponding spatial separation $r$ is \[r^2=r_1^2+r_2^2-2r_1r_2\cos\theta.\] Comparing (\ref{150_o31_e1}) and (\ref{150_o31_e2}) one finds \[w(\theta)\propto\int\limits_0^Ddr_1\int\limits_0^Ddr_2\,r_1^2r_2^2\ \xi\left[(r_1^2+r_2^2-2r_1r_2\cos\theta)^{1/2}\right].\]
Problem 28
problem id: 150_o32
Show that power decay law of the spatial correlations leads to power decay law for the angular correlations.
Firstly, take into account that the correlation is essential only for small angular distances. It allows to set \[\cos\theta\simeq1-\frac12\theta^2.\] Secondly, $\xi$ essentially differs from zero only on small $r$. Therefore $r_1-r_2\ll r_1+r_2.$ Then \[4r_1r_2=(r_1+r_2)^2-(r_1-r_2)^2\approx(r_1+r_2)^2.\] Due to fast decay of the correlation functions we can extend the upper integration bound to infinity without much loss of accuracy. Let us introduce new variables \begin{align} \nonumber u & = \frac1{2D}(r_1+r_2);\\ \nonumber v & = \frac1{D\theta}(r_1-r_2). \end{align} Then \[w(\theta)\propto \int\limits_0^\infty du\int\limits_0^\infty dv\ u^4\xi\left[(u^2+v^2)^{1/2}D\theta\right]\theta.\] Let \[\xi(r)=Ar^{-\alpha}.\] Then \[w(\theta)\propto\left\{\int\int\ du\ dv\ u^4(u^2+v^2)^{-\alpha/2}D^{-\alpha}\right\}\theta^{-\alpha+1}.\] The conclusion is that power law decay of the spatial correlation function results in the power law decay of the angular correlation.
Problem 29
problem id: 150_o33
Find two-point correlation function for $N$ galaxies distributed on a straight line with average density $\bar n$ in non-overlapping clusters of length $a$. Density of galaxies inside the cluster is constant and equals to $n_c$. The clusters are randomly distributed.
In this case we can construct the correlation function $\xi(r)$ in the following way: Take a random galaxy. If $r>a$ then the second galaxy will be situated in another cluster randomly distributed with respect to the first, so that $\xi=0$. If $r<a$ then the second galaxy will be in the same cluster so that the probability to find it in the infinitesimal volume $dV_2$ equals to $n_cdV_2/N$. Thus \[dP_{12}=\frac{\bar nn_c}{N^2}=\frac{\bar n^2}{N^2}\left(1+\frac{n_c-\bar n}{\bar n}\right)\] and therefore $\xi(r)=(n_c-\bar n)/\bar n$, so \[\xi(r)=\left\{\begin{array}{ll}(n_c-\bar n)/\bar n, & r<a;\\ 0, & r>a.\end{array}\right.\] If the clusterization takes place on scales of order $a$ then on the same scales we must observe a sharp bend in behavior of the correlation function. And vise versa, presence of such a sharp bend in the observational data should point on the clusterization effect present.
Problem 30
problem id: 150_o34
In a flat matter-dominated Universe of age $t_0$ light from a certain galaxy exhibits a redshift $z=0.95$. How long has it taken the light signal to reach us from this galaxy?
Let the light signal starts off at $t=t_1$ to reach us at $t=t_0$. Consider the time interval $dt'$ where $t_1<t'<t_0$. In this time interval the light signal covers a distance $cdt'$, but by the time $t=t_0$, this will have expanded to $cdt'\times a(t=t_0)/a(t')=cdt'a(0)/a(t')$. In a matter-dominated Universe $a(0)/a(t')=(t_0/t')^{2/3}$. Hence, the total distance travelled by the light signal will be \[D=a(0)\int\limits_{t_1}^{t_0}\frac{cdt'}{a(t')}=ct_0^{2/3} \int\limits_{t_1}^{t_0}\frac{dt'}{t'^{2/3}}=3ct_0\left[1-\left(\frac{t_1}{t_0}\right)^{1/3}\right].\] The redshift is given by $1+z=a(0)/a(t_1)=(t_0/t_1)^{2/3}$. Hence, the time elapsed \[t_{el}=3t_0\left[1-\frac1{(1+z)^{1/2}}\right]\approx0.85t_0.\]