Difference between revisions of "Light and distances"

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

\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}

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 \[d_{L}=a(t_0)(1+z)\chi=\frac{1}{H_0}\big[z+\frac{1}{2}(1-q_0)z^{2}+\ldots\big]\]

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 $q_0=1/2$.

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 \[d_{A}\equiv\frac{\delta l_{\bot}}{\delta\theta}=\frac{\Sigma(\chi)}{1+z}.\]

\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}

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 $q(z)$.

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

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 \[\frac{dd_{A}}{dz}=0\quad \Leftrightarrow\quad 2\sqrt{1+z}-3=0,\quad\Leftrightarrow\quad z_{max}=1.25.\]

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