Light and distances

Problem 1: comoving distance in a flat Universe

Find the comoving distance to a galaxy as function of redshift in a spatially flat Universe

solution

The equation of the photon's worldline is $ds^{2}=0$. Let us consider the a trajectory with observer placed at the origin. For spatially flat metric in terms of conformal-comoving variables the equation is \[ds^{2}=a^{2}(t)(d\eta^{2}-d\chi^{2})=0.\] Using the relation between differentials \begin{equation}\label{Deta(Dz)} d\eta=\frac{d\eta}{dt}\frac{dt}{da}\frac{da}{dz}dz=-\frac{\dot{a}}{a}dz=-\frac{dz}{H(z)}, \end{equation} we get \[\chi(z)=\int\limits_{0}^{z}\frac{dz'}{H(z')}.\]

Problem 2: the proper distance

Determine the "physical" distance -- the proper distance measured along the hypersurface of constant cosmological time -- to an object that is observed with redshift $z$

solution

If $\chi(t)$ is the comoving distance to the source, then the proper distance is \[R(t)=a(t)\chi(t),\] or in terms of redshift \[R(z)=\frac{1}{1+z}\chi(z),\] where the scale factor is normalized by the value at the moment of observation. Using the result of the previous problem, then we obtain \[R(z)=\frac{1}{1+z}\int\limits_{0}^{z}\frac{dz'}{H(z')}.\]

Problem 3: comoving distance in Einstein-de Sitter

Solve the previous problem for a flat Universe with domination of non-relativistic matter (the Einstein-de Sitter model)

solution

In this case \[H(z)=H_{0}(1+z)^{3/2}\] and evaluation of the integral gives \[\chi(z)=\frac{2}{H_0}\big[1-(1+z)^{-1/2}\big].\]

Problem 4: recession velocity

Determine the recession velocity caused by the cosmological expansion for an object with redshift $z$ in a flat Universe

solution

\[V\equiv \frac{dR}{dt}=\frac{da}{dt}\chi=H(z)R(z)=\frac{H(z)}{1+z}\int\limits_{0}^{z}\frac{dz'}{H(z')}.\]

In cosmology the are other types of distances used, besides the proper and comoving one. One of the most frequently used is the photometric distance. Let $E$, $[E]=J/s$, be the internal absolute luminosity of some source. The observer on Earth detects energy flux $F$, $[F]=J/s\cdot m^2$. The luminosity distance to the source $d_{L}$ is then defined through \[F=\frac{E}{4\pi d_{L}^{2}}.\] Thus this would be the distance to the observed object, given its absolute and observed luminosities, in a flat and stationary Universe. Non-stationarity and curvature imply that $d_{L}$ in general does not coincide with the proper distance.

Problem 5: luminosity distance in a flat Universe

Express the luminosity distance in terms of observed redshift for a spatially flat Universe

solution

Energy emitted by the source per time $dt_{e}$ is $Edt_{e}$. Due to redshift, which reduces the energy of each photon emitted, during the time the scale factor changes from $a_{emission}=a_{e}$ to $a_{observation}=a_{o}$, this energy becomes \[Edt_{e}\cdot \frac{1}{1+z}=Edt_{e}\frac{a_e}{a_o}.\] The observed luminosity is the energy detected per given interval of local time $dt_{o}$ by observers that cover the sphere of the proper radius $R$, to which the photons travel by the time of observation. This local time of observation is subject to cosmological dilation: the worldline of one photon is given by $d\chi^{2}-d\eta^{2}=0$, and thus its equation is $\chi=\eta-\eta_{0}$. As the comoving distance between the source and detector is constant, the conformal time the photons travel is also the same, so for two photons emitted and detected one after the other \[d\eta=\frac{dt_{o}}{a_{o}}=\frac{dt_e}{a_e}.\] Then the observed luminosity is \[L=\frac{Edt_{e}\frac{a_e}{a_o}}{dt_{o}}=E\cdot \Big(\frac{a_e}{a_o}\Big)^{2},\] and, as this energy flux is distributed over a sphere of proper radius $R$, the detected energy flux is \[F=\frac{L}{4\pi R^2}.\] On the other hand, the luminosity distance is defined though \[F=\frac{E}{4\pi d_{L}^{2}},\] so we obtain \begin{equation}\label{dL} d_{L}=\frac{a_o}{a_e}R=(1+z)R \end{equation} In a spatially flat Universe $R=a_{o}\chi$, where $\chi$ is the comoving distance to the source, so \[d_{L}=(1+z)a_{o}\chi.\]

Problem 6: generalization to arbitrary curvature

Generalize the result of the previous problem to the case of arbitrary curvature

solution

The above derivation holds mostly, with the exception that now the observed luminosity is distributed over the sphere of proper surface \[S=4\pi a_{o}^{2}\Sigma^{2}(\chi),\] where \[\Sigma(x)=\left\{\begin{array}{ll}\sin x,&\quad \Omega_{0}>1,\\ x,&\quad \Omega_{0}=1\\ \sinh x,&\quad \Omega_{0}<1.\end{array}\right.\] and $\chi$ is the proper distance to the source. Expressing it through the observed quantities \[\chi=\int\frac{dt}{a}=\int\frac{da}{a^2 H}=\frac{1}{a_o}\int\frac{dz}{H}\] and using that $a=H^{-1}(\Omega-1)^{-1/2}$, we finally obtain \begin{align} d_{L}(z)&=(1+z)a_{o}\;\Sigma(\chi)\\ &=(1+z)(1-\Omega_0)^{-1/2}\cdot H_{0}^{-1}\Sigma\Big[(1-\Omega_0)^{1/2}H_{0} \int\limits_{0}^{z}\frac{dz'}{H(z')}\Big], \end{align}

Problem 7: multi-component flat Universe

Find the expression for the luminosity distance for the multi-component flat Universe

solution

\[d_{L}(z)=(1+z)H_{0}^{-1}\int\limits_{0}^{z}\frac{dz'}{\sqrt{\sum\limits_{i}\Omega_{i0}(1+z')^{3(1+w_{i})}}}\]

Problem 8: luminocity distance in terms of deceleration parameter

Express the luminosity distance in a flat Universe in terms of the redshift dependence of deceleration parameter $q(z)$

solution

\[d_{L}(z)=(1+z)\int\limits_{0}^{z}\frac{dz'}{H(z')}=(1+z)H_{0}^{-1}\int\limits_{0}^{1}du \exp\Big\{-\int\limits_{0}^{u}\big[1+q(z')\big]d\ln (1+z')\Big\}\]

Problem 9: Einstein-de Sitter

Express the luminosity distance in terms of redshift for the Einstein-de Sitter model

solution

\begin{align} \chi(z)&=\frac{2}{H_0}\Big[1-\frac{1}{\sqrt{1+z}}\Big];\\ d_{L}(z)&=(1+z)\chi(z)=\frac{2}{H_0}\big[1+z-\sqrt{1+z}\big]. \end{align}

Problem 10: small redshifts limit

Show that in the first order by $z\ll 1$ luminosity distance is $d_{L}\approx z /H_{0}$ and find the second order correction

solution

Let us use the series, valid for $z\ll 1$, \[1+z=\frac{a(t_0)}{a(t)}=1+H_{0}\Delta t +\frac{1}{2}(q_{0}+2)H_{0}^{2}\Delta t ^{2}+\ldots.\] The relation can be inverted to give \[H_{0}\Delta t =z-\frac{1}{2}(q_{0}+2)z^{2}.\] For a photon tnat is emitted at a source with comoving coordinate $r$ at time $t$ and detected at some later time $t_{0}$ then \[\chi=\int\limits_{t}^{t_0}\frac{dt}{a(t)}\approx \frac{\Delta t}{a(t_0)}+\frac{H_{0}\Delta t^{2}}{2a(t_0)}\] and therefore \[d_{L}=a(t_0)(1+z)\chi=\frac{1}{H_0}\big[z+\frac{1}{2}(1-q_0)z^{2}+\ldots\big]\]

Problem 11: the closed and open dusty Universes

Derive the luminosity distance as function of redshift for the closed and open models of the Universe, dominated by non-relativistic matter (dust)

solution

For the closed model \begin{align} \chi(z)&=\frac{\sqrt{2q_0 -1}}{q_0^2 (1+z)}\Big[q_0 z +(q_0 -1)\big(\sqrt{1+2q_0 z}\;-1\big)\Big],\\ a_{0}&=\frac{H_0^{-1}}{\sqrt{2q_0 -1}},\\ d_{L}(z)&=(1+z) a_{0}\chi \\ &=\frac{1}{q_0^2 H_0}\Big[q_0 z +(q_0 -1)\big(\sqrt{1+2q_0 z}\;-1\big)\Big]. \end{align} For the open model \begin{align} \chi(z)&=\frac{\sqrt{1-2q_0}}{q_0^2 (1+z)}\Big[q_0 z +(q_0 -1)\big(\sqrt{1+2q_0 z}\;-1\big)\Big],\\ a_{0}&=\frac{H_0^{-1}}{\sqrt{1-2q_0 }},\\ d_{L}(z)&=\frac{1}{q_0^2 H_0}\Big[q_0 z +(q_0 -1)\big(\sqrt{1+2q_0 z}\;-1\big)\Big]. \end{align} The final expressions for $d_{L}$ are the same for all models ($k=0,\pm 1$). Note, that for Einstein-de Sitter $q_0=1/2$.

Another distance used is the angular diameter distance. It is defined through the angular dimension of the object $\delta \theta$ and its proper transverse size $\delta l$ as \[d_{A}=\frac{\delta l}{\delta \theta}.\] Again, in a stationary and flat Universe this is reduced to the ordinary distance, while in general they differ.

Problem 12: angular diameter distance in terms of redshift

Express the angular diameter distance in terms of the observed redshift

solution

First, let us choose the comoving coordinate system with the observer at the origin. Let the comoving coordinates of the end points of the source be $(\chi,\theta,\phi)$ and $\chi+\delta\chi,\theta+\delta\theta,\phi$. Take the FLRW metric in the form \[ds^{2}=dt^{2}-a^{2}(t)\big[d\chi^{2}-\Sigma^{2}(\chi)(d\theta^{2}+\sin^{2}\theta d\phi^{2})\big].\] Then the transverse size $\delta l_{\bot}$ of the source at the time of emission is \[\delta l_\bot=a(t_e)\Sigma(\chi)\delta \theta\] and \[d_{A}\equiv\frac{\delta l_{\bot}}{\delta\theta} =a_{e}\Sigma(\chi)=\frac{a_{o}}{1+z}\Sigma(\chi)=\frac{d_{L}}{(1+z)^{2}}.\]

Problem 13: in terms of $q(z)$

Express the angular diameter distance in terms of $q(z)$

solution

\begin{align} d_{A}&=\frac{H_0^{-1}}{1+z}\int\limits_{0}^{z}\frac{dz'}{H(z')}\\ &=\frac{H_0^{-1}}{1+z}\int\limits_{0}^{z}du\exp\Big\{-\int\limits_{0}^{z}\big[1+q(z')\big]d\ln(1+z')\Big\}. \end{align}

Problem 14: a two-parametric expansion

Find $d_{A}(z)$ in a flat Friedman Universe for the linear two-parametric expansion$^*$ \[q(z)=q_{0}+q_{1}(z).\] $^*$ J. Lima et al. arXiv:0905.2628

solution

For the expansion $q(z)=q_{0}+zq_{1}$ the integral obtained in the previous problem can be evaluated analytically \[d_{A}(z)=\frac{H_0^{-1}}{1+z}e^{q_1}q_{1}^{q_{0}-q_{1}} \big[\gamma(q_{1}-q_{0},(z+1)q_{1})-\gamma(q_{1}-q_{0},q_{1})\big],\] where $\gamma(\alpha,x)$ is the incomplete gamma-function \[\gamma(\alpha,z)=\int\limits_{0}^{z}dt\,e^{-t}t^{\alpha-1},\qquad (\mathrm{Re}\alpha>0).\] Using this expression, we can obtain information on $q_0$ and $q_1$, and consequently, on the global evolution of the deceleration parameter $q(z)$.

Problem 15: another expansion

Find $d_{A}(z)$ in a flat Friedman Universe for the linear two-parametric expansion \[q(z)=q_{0}+q_{1}\frac{z}{1+z}.\]

solution

\[d_{A}(z)=\frac{H_{0}^{-1}}{1+z}e^{q_{1}q_{1}^{-(q_{0}+q_{1})}} \big[\gamma(q_0 +q_1, q_1)-\gamma(q_0 +q_1 , (1+z)^{-1}q_{1})\big].\]

Problem 16: maximum of angular diameter distance in Einstein-de Sitter

Find the redshift for which the angular diameter distance of an object in the Einstein-de Sitter Universe reaches its maximum

solution

The angular diameter distance for Einstein-de Sitter is \[d_{A}=\frac{2}{H_0}\Big(\frac{1}{1+z}-\frac{1}{(1+z)^{3/2}}\Big),\] so \[\frac{dd_{A}}{dz}=0\quad \Leftrightarrow\quad 2\sqrt{1+z}-3=0,\quad\Leftrightarrow\quad z_{max}=1.25.\]

Problem 17: the maximum in Friedman models

Find the redshift for which the angular diameter distance reaches the maximum in the closed and open Friedman models$^*$

$^*$ Juri Shtanov, Lecture notes on theoretical cosmology, 2010

solution

For the closed model \[\Sigma(\chi)=\sin\chi\] and \[a(\eta)=\frac{\alpha}{2}(1-\cos\eta),\] where $\alpha$ is some constant. Then \[d_{A}(\eta)=a_{e}\Sigma(\chi)\sim (1-\cos\eta_{e})\sin(\eta_{o}-\eta_{e}),\] where subscripts "e" and "o" denote emission and observation respectively, and condition \[\frac{dd_{A}}{d\eta_{e}}=0\] is transformed, after renaming $\eta_{e}\equiv \eta$, to \[0=\sin\eta\; \sin(\eta_o -\eta)-(1-\cos\eta)\cos(\eta_o -\eta)\sim \cos(2\eta-\eta_o)-\cos(\eta-\eta_o) \sim \sin \frac{3\eta-2\eta_o}{2}\sin\frac{\eta}{2}.\] So the maximum $d_A$ is realized for $\eta_{e}=\frac{2}{3}\eta_{o}$. The corresponding redshift is found from \[1+z_{\max}=\frac{a_o}{a_e}=\frac{1-\cos \eta_o}{1-\cos\eta_e}=\frac{1-\cos \eta_o}{1-\cos\frac{2}{3}\eta_o}.\] In the open model, following the same procedure, one obtains the same result with trigonometric functions replaced by hyperbolic ones.