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\def\sign{\mathop{\rm sign}\nolimits}
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\def\arcsec{\mathop{\rm arcsec}\nolimits}
\def\arccsc{\mathop{\rm arccsc}\nolimits}
\def\sech{\mathop{\rm sech}\nolimits}
\def\csch{\mathop{\rm csch}\nolimits}
\def\arcsinh{\mathop{\rm arcsinh}\nolimits}
\def\arccosh{\mathop{\rm arccosh}\nolimits}
\def\arctanh{\mathop{\rm arctanh}\nolimits}
\def\arccoth{\mathop{\rm arccoth}\nolimits}
\def\arcsech{\mathop{\rm arcsech}\nolimits}
\def\arccsch{\mathop{\rm arccsch}\nolimits}
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\begin{document}
\thesaurus{
           02        % A&A Section 
          (12.03.4;  % cosmology: theory
           03.13.4   % methods: numerical
           12.04.3;  % cosmology: distance scale
           12.03.2;  % cosmology: miscellaneous
           12.07.1)  % gravitational lensing
}
\title{
A general and practical method for calculating cosmological distances
} 
\author{
Rainer Kayser\thanks{\tt rkayser@hs.uni-hamburg.de}\inst{1}, 
Phillip Helbig\thanks{\tt phelbig@hs.uni-hamburg.de}\inst{1} \and  
Thomas Schramm\thanks{\tt schramm@tu-harburg.d400.de}\inst{1,2}
}
\offprints{
P.~Helbig
}
\institute{
           Hamburger Sternwarte,
           Gojenbergsweg 112,
           D-21029 Hamburg-Bergedorf, Germany \and
           Rechenzentrum, Technische Universit\"{a}t Hamburg-Harburg,
           Denickestra\ss e 17, D-21071  Hamburg-Harburg, Germany
}
\date{
received ; accepted
}
\maketitle
\markboth{Kayser et~al.: Cosmological distances}{}
%***********************************************************************


\appendix



%***********************************************************************
\section{Symmetry: The relation between $D_{xy}$ and $D_{yx}$}
\label{symmetry}


The proof in this appendix follows closely the proof presented in Kayser
(\cite{RKayser85a}).  For completeness we note that after the original
derivation by Kayser (\cite{RKayser85a}) an equivalent equation was
derived by Linder (\cite{ELinder88a}). We rewrite the differential
equation, Eq.~(\ref{RKeqG}), for the angular size distance in the normal
form: 
\begin{equation}\label{RKAeqA}
a_{2}D''(z) + a_{1}(z)\,D'(z) + a_{0}(z)\,D(z) = 0
\end{equation}
with the coefficient functions
\begin{equation}\label{RKAeqB}
a_{2}(z) = Q(z)
\end{equation}
\begin{equation}\label{RKAeqBB}
a_1(z) = {2Q(z)\over 1+z} + {1\over 2}\,Q'(z)
\end{equation}
\begin{equation}\label{RKAeqBBB}
a_{0}(z) = {3\over 2}\,\eta\,\Omega_0(1+z)\quad.
\end{equation}
Now let $D^{(1)}$ and $D^{(2)}$ be two solutions of Eq.~(\ref{RKAeqA})
which build a fundamental system, i.e.~the Wronskian for these two
solutions does not vanish: 
\begin{equation}
W(z) = \left|
    \begin{array}{cc}
     D^{(1)} & D^{(2)} \\
     {\D D^{(1)}\over \D z} & 
     {\D D^{(2)}\over \D z}
    \end{array}
    \right| \neq 0 \quad \forall z \quad.
\end{equation}
Every solution $D_i$ of Eq.~(\ref{RKAeqA}) can then be written as a
linear combination of $D^{(1)}$ and $D^{(2)}$: 
\begin{equation}\label{RKAeqD3}
D_i = \alpha_{i}\,D^{(1)} + \beta_{i}\,D^{(2)};
\quad{\rm with}\,\,\alpha_{i},\,\beta_{i}={\rm const}\quad.
\end{equation}
The angular size distances are special solutions $D_x$ of
Eq.~(\ref{RKAeqA}) fulfilling the following boundary conditions: 
\begin{equation}
D_x = 0 \quad {\rm for}\quad z=z_x
\end{equation}
and
\begin{equation}
{\D D_x\over\D z} = b(z_x)\quad{\rm for}\quad z=z_x
\end{equation}
with
\begin{equation}\label{RKAeqG}
b(z_x) = {c\over H_0}\,\left((1+z_x)\sqrt{Q(z_x)}\right)^{-1}
          \sign(t_y-t_x) \quad,
\end{equation}
compare Eq.~(\ref{RKeqH}). From Eq.~(\ref{RKAeqD3}) we obtain 
\begin{equation}
0 = \alpha_i\,D^{(1)}(z_x) + \beta_i\,D^{(2)}(z_x)
\end{equation}
and
\begin{equation}
b(z_x) = \alpha_i\,\left.{\D D^{(1)}\over\D z}\right|_{z=z_x}
      \, +\, \beta_i\, \left.{\D D^{(2)}\over\D z}\right|_{z=z_x}
\quad.
\end{equation}
These equations can easily be solved for $\alpha_i$ and $\beta_i$:
\begin{equation}
\alpha_i=\beta_i\,{D^{(2)}(z_x)\over D^{(1)}(z_x)}
\end{equation}
\begin{equation}
\beta_i={b(z_x)\,D^{(1)}(z_x)\over W(z_x)}
\end{equation}
and inserting $\alpha_i$ and $\beta_i$ back  into Eq.~(\ref{RKAeqD3}) we
obtain for the special solutions $D_x$: 
\begin{eqnarray}
D_x(z) &=& {b(z_x)\over W(z_x)}\,
  \left( D^{(1)}(z_x)\,D^{(2)}(z)\right.\nonumber\\ 
    & &\left. -\,  D^{(1)}(z)\,D^{(2)}(z_x)\right)
\quad.
\end{eqnarray}
If we now consider a second special solution $D_y$ we find the relation 
\begin{equation}
{D_x(z_y)\over D_y(z_x)} = -\,{b(z_x)\over b(z_y)}\,
    {W(z_y)\over W(z_x)}\quad.
\end{equation}
The Wronskians can be calculated using Liouville's formula:
\begin{equation}
W(z) = W(z_0)\,\exp\int\limits_z^{z_0} a_{2}(z)\,\D z
\quad,
\end{equation}
where $z_0$ is arbitrary. Thus
\begin{equation}
{D_x(z_y)\over D_y(z_x)} = -\,{b(z_x)\over b(z_y)}\,
\exp\int\limits_{z_y}^{z_x} \frac{a_1(z)}{a_{2}(z)}\,\D z
\quad
\end{equation}
and after inserting $a_{0}$, $a_1$ and $a_{2}$ from Eqs.~(\ref{RKAeqB}),
(\ref{RKAeqBB}) and (\ref{RKAeqBBB}) as well as $b(z_x)$ and $b(z_y)$
from Eq.~(\ref{RKAeqG}) and integration we finally obtain for the
angular size distances (cf.~Eq.~(\ref{RKeqI})) the relation 
\begin{equation}
{D_{xy}\over D_{yx}} = {1+z_x\over 1+z_y}\quad. 
\end{equation}



%***********************************************************************


\section{Special cases}\label{special}


For certain special cases the differential equation can be simplified
and sometimes analytically solved. 



\subsection{$\Omega_{0} = 0$}\label{omeganull}


A glance at Eq.~(\ref{RKeqG}) shows that for $\Omega_{0}=0$ the third
term on the left hand side of Eq.~(\ref{RKeqG}) vanishes; one thus has a
first order differential equation for $D'$. (Of course $\eta$ has no
meaning for $\Omega_{0}=0$.)  Due to the fact that a vanishing
$\Omega_{0}$ also simplifies $Q(z)$, it is possible to calculate the
angular size distance analytically.  Since in this case the angular size
distance is determined exclusively by global effects, one can use an
approach based on global geometry.\footnote{See the discussion in
Sect.~\ref{eta1}.}  Depending on the value of $\lambda_{0}$, one can use the 
following expression to calculate 
$\chi_{xy} = \chi(y) - \chi(x)$ (Feige \cite{BFeige92a}) 
\begin{equation}\label{omeganull-g}
\chi(z) = 
\left\{ \begin{array}{rcrr}
\arccosh(\psi) & & \rm for &     \lambda_{0} < 0 \\& & \\
\ln(1+z)       & & \rm for &     \lambda_{0} = 0 \\& & \\
\arcsinh(\psi) & & \rm for & 0 < \lambda_{0} < 1 \\& & \\
z              & & \rm for &     \lambda_{0} = 1 \\& & \\
\arcsin(\psi)  & & \rm for &     \lambda_{0} > 1 
\end{array}\right.,
\end{equation} 
where $\psi:=(1+z)\sqrt{\frac{|1-\lambda_{0}|}{|\lambda_{0}|}}$.  The
relationship between $\chi$ and the angular size distance~$D$ is 
\begin{equation}
D_{xy} = \frac{R_{0}}{(1 + z_{y})}\left\{ \begin{array}{rcrrrr}
\sinh\chi & & \rm for & k & = & -1 \\
        \chi & & \rm for & k & = &  0 \\
\sin \chi & & \rm for & k & = & +1 
\end{array}\right.,
\end{equation}
as discussed below in Sect~\ref{eta1}. 



\subsection{$\eta = 0$}


In the case $\eta=0$ the third term on the left hand side of
Eq.~(\ref{RKeqG}) vanishes; one thus has a first order differential
equation for $D'$.  Assuming $D'\neq0$,  Eq.~(\ref{RKeqG}) can be
written as 
\begin{equation}
\frac{D''}{D'}=-\frac{2}{1+z}-\frac{1}{2}\frac{Q'(z)}{Q(z)}\quad.
\end{equation}
This equation can be solved in two steps.  For $D'$ we obtain
\begin{equation}
D'={\frac {c_1}{\sqrt{Q(z)}\left (1+z\right )^2}}
\end{equation} 
and consequently for $D$
\begin{equation}
D = \int{\frac {c_1}{\sqrt{Q(z)}\left (1+z\right )^2}}+c_2\quad.
\end{equation}  
The constants $c_1, c_2$ are determined by the appropriate boundary
conditions (Eqs.~(\ref{boundary1-g}) and (\ref{boundary2-g})). We then
%find the solution (see also SEF for an equivalent discussion with
find the solution (see also Schneider et al.\ (\cite{PSchneiderEF92a}) -- 
hereafter SEF -- for an equivalent discussion with
$\lambda_{0}=0$) 
\begin{equation}\label{domega-g}
D_{xy} = \frac{\rm c}{{\rm H}_0}(1 + z_{x})(\omega(z_{y})-\omega(z_{x})), 
\end{equation}
where 
\begin{equation}\label{omega-g}
\omega(z) =              \int\limits_{0}^{z} \frac{{\rm d}z'}
                      {(1 + z')^{2}\sqrt{(1+z')^{2}(\Omega_{0}z'+1-
\lambda_{0})+\lambda_{0}}},
\end{equation}
or, perhaps more convenient, 
\begin{equation}
D_{xy}=\frac{c}{{H}_0}\left(1+z_x\right)\int_{z_x}^{z_y}\!
       \frac{{\rm d}z}{(1+z)^{2}\sqrt{Q(z)}}\quad.
\end{equation}

For $\lambda_{0}=0$ there is an analytic solution (see
Sect.~\ref{lambdanull}). 



\subsection{$\eta = 1$}\label{eta1}


The case $\eta=1$ has all matter distributed homogeneously. Due to
homogeneity, the matter {\em locally\/} affecting the propagation of
light is known when the {\em global\/} geometry is known, so that the
`classical' approach of relating global geometry to observable relations
is a better approach than using (the simplified form of)
Eq.~(\ref{RKeqG}).  This approach offers an analytic solution. Here, we
simply sketch the most important points; the interested reader can refer
to Feige (\cite{BFeige92a}) for a good description of this method. 

The angular size distance in this case is 
\begin{equation}\label{ang-g}
D_{xy} = R_{y}\sigma_{xy} = \frac{R_{0}\sigma_{xy}}{(1 + z_{y})},
\end{equation} 
where $\sigma$ is the radial coordinate in the Robertson-Walker metric
(cf.~Eq.~(\ref{rwm-g})) and thus
\begin{equation}
D_{yx} = \frac{R_{0}\sigma_{xy}}{1 + z_{x}} = 
D_{xy}\left(\frac{1 + z_{y}}{1 + z_{x}}\right),
\end{equation}
since this angle is inversely proportional to R for constant $\sigma$
and physical size.  (The value of $R$ at the time the light rays
defining the angle intersect is important.) 

Since $\sigma$ is given by 
\begin{equation}\label{sigma-g}
\sigma = F(\chi) = \left\{ \begin{array}{rcrrrr}
\sinh\chi & & \rm for & k & = & -1 \\
        \chi & & \rm for & k & = &  0 \\
\sin \chi & & \rm for & k & = & +1 
\end{array}\right.,
\end{equation} 
an expression for $\chi(z)$ is sufficient for calculating the angular
size distance~$D$ (and of course the luminosity distance~$D^{\rm L}$
(via Eq.~(\ref{lumang-g})) and the `coordinate distance'~$\sigma$ (via
Eq.~(\ref{sigma-g})).  In general, $\sigma_{xy} \neq \sigma_{y} -
\sigma_{x}$; however, $\chi_{xy} = \chi_{y} - \chi_{x}$, so that 
\begin{equation}
\sigma_{xy} = F (\chi_{xy})
\end{equation}
where $F$ is given by Eq.~(\ref{sigma-g}).  Using Eq.~(\ref{dp-g}) one
can calculate 
\begin{equation}
\chi_{xy} = \frac{D^{P}}{R_{0}} = 
\frac{c}{H_{0}R_{0}}\int\limits_{z_{x}}^{z_{y}}\frac{{\rm d}z}{\sqrt{Q(z)}}.
\end{equation}
In the general case, Eq.~(\ref{dp-g}) can be solved by elliptic
integrals, as explained in Feige (\cite{BFeige92a}).  For the cases
$\lambda_{0}=0$ and $\Omega_{0}=0$ the formulae using elliptic integrals
break down; in these cases, easier analytic formulae, which fortunately
exist, can be used.  The case $\Omega_{0}=0$ has been discussed above. 
The case $\lambda_{0}=0$ will be discussed below. {\em Again, we stress
that the differential equation derived in Sect.~\ref{rainer} is
completely general and can be used in {\em all} cases}. 



\subsection{$\lambda_{0} = 0$}\label{lambdanull} 


For $\lambda_{0} = 0$, there is in general no simpler solution. This
case has been discussed by Dyer and Roeder for $\eta=0$
(\cite{CDyerRRoeder72a}) and for general $\eta$ values
(\cite{CDyerRRoeder73a}).  They point out the interesting result that
the maximum in the angular size distance from $z_{1}=0$ to $z_{2}$
increases monotonically from $1.25$ to $\infty$ as $\eta$ decreases from
$1$ to $0$.  See also the discussion (with a differing notation!) in
Sect.~4.5.3 in SEF.  However, some solutions exist for special values of
$\Omega_{0}$ and $\eta$.  The case $\Omega_{0}=0$ has been discussed in
Sect.~\ref{omeganull} above; the value of $\eta$ is of course irrelevant
in this case.   With the exception of Sect~\ref{mattig} below, in the
following we simply quote results from SEF in our notation. 



\subsubsection{$\lambda_{0} = 0$ and $\eta = 0$} 


As discussed above, for $\eta=0$ Eq.~(\ref{RKeqG}) is effectively a
first order equation for $D'$.  For $\lambda_{0}=0$ $Q(z)$ is
sufficiently simplified to allow an analytic solution.  Recalling
Eq.~(\ref{domega-g}), 
\begin{displaymath}
D_{xy} = \frac{\rm c}{{\rm H}_0}(1 + z_{x})(\omega(z_{y})-\omega(z_{x})), 
\end{displaymath}
Eq.~(\ref{omega-g}) simplifies to 
\begin{equation}
\omega(z) = \int\limits_{0}^{z} \frac{{\rm d}z'}
                      {(1 + z')^{3}\sqrt{\Omega_{0}z'+1}},
\end{equation}
which has the solution:
\begin{equation}
\omega(z) =
\left\{\begin{array}{lr}
\frac{3\Omega_{0}^{2}}{4(\Omega_{0}-1)^\frac{5}{2}} 
\arctan(\psi) + \frac{3}{4(\Omega_{0}-1)^{2}} \quad \times & \\ 
\left(
\frac{\left(\Omega_{0}z+\frac{5\Omega_{0}}{3}-\frac{2}{3}\right)
\sqrt{\Omega_{0}z+1}}{(1+z)^{2}}-\frac{5\Omega_{0}}{3}+\frac{2}{3}\right)
                                             &  (\rm I)\\
\quad & \\
\frac{2}{5}\left(1 - (\Omega_{0}z+ 1 )^{-\frac{5}{2}}\right)
                                             &  (\rm II) \\
\quad & \\
\frac{3\Omega_{0}^{2}}{4(1-\Omega_{0})^\frac{5}{2}} 
\arctanh(\psi) + \frac{3}{4(1-\Omega_{0})^{2}} \quad \times & \\ 
\left(
\frac{\left(\Omega_{0}z+\frac{5\Omega_{0}}{3}-\frac{2}{3}\right)
\sqrt{\Omega_{0}z+1}}{(1+z)^{2}}-\frac{5\Omega_{0}}{3}+\frac{2}{3}\right)
                                             &  (\rm III)
\end{array}\right.
\end{equation}
with
\begin{equation}
\psi =
\left\{ \begin{array}{lr}
 \left(\frac{\sqrt{\Omega_{0}-1}(1 + \sqrt{\Omega_{0}z+1})}
{\Omega_{0}-1+\sqrt{\Omega_{0}z+1}}\right)
                                                &(\rm I) \\&  \\
\left(\frac{\sqrt{1-\Omega_{0}}(1 + \sqrt{\Omega_{0}z+1})}
{\Omega_{0}-1+\sqrt{\Omega_{0}z+1}}\right)
                                                & (\rm III)
\end{array}\right.
\end{equation}
and
\begin{equation}
\begin{array}{lr}
\mbox{case I:}                      &      \Omega_{0} > 1 \\
\mbox{case II:}                     &      \Omega_{0} = 1 \\
\mbox{case III:}                    &  0 < \Omega_{0} < 1 
\end{array}.
\end{equation}
Note that in SEF, the text at the top of page~137 is unclear---the
expression in parentheses in the denominator of the first term
($\Omega-1$) for the $\Omega>1$ case has to be replaced with
($1-\Omega$) as well for $\Omega<1$.  Note also that
$\Omega\equiv\Omega_{0}$ and that after page 131 $\lambda_{0}=0$ is
always assumed. 



\subsubsection{$\lambda_{0} = 0$ and $\Omega_{0} = 1$} 


For $\Omega_{0}=1$ and $\lambda_{0}=0$ (the Einstein-de~Sitter model) we 
have the solution 
\begin{equation}
D_{xy}=\frac{c}{H_{0}}\frac{1}{2\beta}\left(\frac{(1+z_{y})^{\beta-\frac{5}{4}}}
{(1+z_{x})^{\beta+\frac{1}{4}}}-\frac{(1+z_{x})^{\beta-\frac{1}{4}}}
{(1+z_{y})^{\beta+\frac{5}{4}}}\right),
\end{equation}
where 
\begin{equation}
\beta:=\frac{1}{4}\sqrt{25 - 24\eta}.
\end{equation}



\subsubsection{$\lambda_{0} = 0$ and $\eta = 1$}\label{mattig}


For $\eta=1$ the special case of the expression for $\chi(z)$ for
$\lambda_{0}=0$ is (Feige~\cite{BFeige92a}) 
\begin{equation}
\chi(z) = 
-2\left\{ \begin{array}{lr}
\arcsin\left(\sqrt{\frac{\Omega_{0}-1}{\Omega_{0}(1+z)}}\right)
                                              &   (  \Omega_{0} > 1 ) \\
&  \\
\sqrt{\frac{1}{1+z}}
                                            &   (  \Omega_{0} = 1 ) \\
&  \\      
\arcsinh\left(\sqrt{\frac{1-\Omega_{0}}{\Omega_{0}(1+z)}}\right)
                                              & (0 < \Omega_{0} < 1 )
\end{array}\right.,
\end{equation}
where $\chi_{xy} = \chi(y) - \chi(x)$. 
(It is obvious that in the case $\lambda_{0}=\Omega_{0}=0$
Eq.~(\ref{omeganull-g}) should be used.)  From this, it is possible to
obtain a general expression for the angular size distance (see, e.g.,
SEF): 
\begin{equation}
D_{xy} = \frac{c}{H_{0}}\frac{2}{\Omega_{0}^{2}}(1+z_{x})\left(
R_{1}(z_{y})R_{2}(z_{x}) - R_{1}(z_{x})R_{2}(z_{y})\right),
\end{equation}
with
\begin{equation}
R_{1}(z) = \frac{\Omega_{0}z - \Omega_{0} + 2}{(1+z)^{2}}
\end{equation}
and
\begin{equation}
R_{2}(z) = \frac{\sqrt{\Omega_{0}z+1}}{(1+z)^{2}}.
\end{equation}
For $z_{x}=0$ and $z_{y}=z$ one gets for the angular size distance
\begin{eqnarray}
\label{msef-g}
D(z)& = & \frac{c}{H_{0}}
\frac{2}{\Omega_{0}^{2}(1+z)^{2}} \qquad\times\nonumber \\
 & &
\left(
\Omega_{0}z - (2 - \Omega_{0})\left(\sqrt{\Omega_{0}z+1}-1\right)\right).
\end{eqnarray}
valid for $\Omega_{0}>0$.  For $\Omega_{0}=0$ one obtains
\begin{equation}
\label{msef0-g}
D = \frac{c}{H_{0}}\frac{z\left(1 + \frac{z}{2}\right)}{\left(1 + z\right)^{2}}
\end{equation} 
(Multiplying Eq.~(\ref{msef-g}) or Eq.~(\ref{msef0-g}) with
$(1+z)/R_{0}$ results in the respective expression for $\sigma$ as a
function of redshift as first derived by Mattig (\cite{WMattig58a}). See
also Sandage (\cite{ASandage95a}), Sect.~1.6.3). In this case, the
volume element given by Eq.~(\ref{dvolume-g}) reduces to 
\begin{equation}
\D V = 16\pi R_{0}^{3}
\frac{(\Omega_{0}z -(2-\Omega_{0})(\sqrt{\Omega_{0}z+1}-1)^{2})}
{\Omega_{0}^{4}(1+z)^{3}\sqrt{\Omega_{0}z+1}}
\end{equation}
Of course, for the physical, as opposed to comoving, density, an {\em
additional\/} factor of $(1+z)^{3}$ must be added to the denominator. 

\subsubsection{$\lambda_{0} = 0$ and $\eta = \frac{2}{3}$} 


For $\eta=\frac{2}{3}$ and $\lambda_{0}=0$ there is also an analytic
solution (see SEF): 
\begin{eqnarray}
D_{xy} & = &
\frac{c}{H_{0}}\frac{2}{3\Omega_{0}^{2}}(1+z_{x})\qquad\times\nonumber\\
 & &
\left(R_{1}(z_{x})R_{2}(z_{y}) - R_{2}(z_{x})R_{1}(z_{y})\right),
\end{eqnarray}
with
\begin{equation}
R_{1}(z) = \frac{1}{(1+z)^{2}}
\end{equation}
and
\begin{equation}
R_{2}(z) = \frac{\sqrt{\Omega_{0}z+1}(\Omega_{0}z+3\Omega_{0}-2)}{(1+z)^{2}}.
\end{equation}



\subsection{Other cases}


We can offer no proof that no other easier solutions, either reducing
Eq.~(\ref{RKeqG}) to a more easily (numerically) integrated form or even
to an analytic solution, exist.  This is left as an exercise to the
interested reader.  The authors are of course interested in such
solutions and are willing to verify them.  As far as we know,
Eq.~(\ref{RKeqG}) must be used except in the special cases mentioned in
this appendix. 



\subsection{Light travel time}\label{time}


Feige (\cite{BFeige92a}) not only gives the distance but also the light
travel time by means of elliptic integrals.  As for the distance, and
for the same reasons, simple analytic formulae can and must be used for
the special cases $\Omega_{0}=0$ and $\lambda_{0}=0$.  For $k=0$, an
analytic expression for the light travel time exists, although the
elliptic integrals can also be used in this case.  For completeness, we
give these special cases here for the light travel time $t_{xy} = t_{x}-t_{y}$. 

For $\Omega_{0}=0$ we have:
\begin{equation}\label{tomeganull-g}
t(z) = \frac{1}{H_{0}\sqrt{|\lambda_{0}|}}
\left\{ \begin{array}{lr}
\arcsin(\psi)                         & \lambda_{0} < 0 \\&  \\
\frac{\sqrt{|\lambda_{0}|}}{(1+z)}   & \lambda_{0} = 0 \\&  \\
\arcsinh(\psi)                        & 0 < \lambda_{0} < 1 \\ & \\
-\sqrt{|\lambda_{0}|}\ln(1+z)         & \lambda_{0} = 1 \\&  \\
\arccosh(\psi)                        & \lambda_{0} > 1
\end{array}\right.,
\end{equation} 
where $\psi:=\frac{1}{(1+z)}\sqrt{\frac{|\lambda_{0}|}{|1-\lambda_{0}|}}$ 

For $\lambda_{0}=0$ we have:
\begin{equation}
t(z) = A \times
\left\{ \begin{array}{lr}
\frac{\sqrt{(\Omega_{0}z+1)(\Omega_{0}-1)}}{\Omega_{0}(1+z)} \quad - & \\
\arcsin\left(\sqrt{\frac{\Omega_{0}-1}{\Omega_{0}(1+z)}}\right)
                                                       & \Omega_{0} > 1  \\ 
\qquad\\
-\frac{\sqrt{\Omega_{0}-1}^{3}}{\Omega_{0}} 
\frac{2}{3}\left(\sqrt{\frac{1}{1+z}}\right)^{3} 
                                   & \Omega_{0} = 1                       \\
\qquad\\
\arcsinh\left(\sqrt{\frac{1-\Omega_{0}}{\Omega_{0}(1+z)}}\right) \quad - \\
\frac{\sqrt{(\Omega_{0}z+1)(1-\Omega_{0})}}{\Omega_{0}(1+z)}
                                      & 0 < \Omega_{0} < 1
\end{array}\right.,
\end{equation} 
where 
\begin{displaymath}
A = -\frac{\Omega_{0}}{H_{0}(\sqrt{|\Omega_{0}-1|})^{3}}
\end{displaymath}
(For $\Omega_{0}=0$ the appropriate case from Eq.~(\ref{tomeganull-g}) 
must be used.)  

For $k=0$ we have:
\begin{equation}
t(z) = \frac{2}{3H_{0}} \times
\left\{ \begin{array}{lr}
\frac{1}{\sqrt{\Omega_{0}-1}} 
\arcsin(\psi)
& \qquad     \Omega_{0} > 1 \\&  \\
\left(\sqrt{\frac{1}{1+z}}\right)^{3}       
   &    \qquad  \Omega_{0} = 1 \\ & \\
\frac{1}{\sqrt{1-\Omega_{0}}}
\arcsinh(\psi)
& \qquad 0 < \Omega_{0} < 1
\end{array}\right., 
\end{equation} 
where 
\begin{equation}
\psi = \left\{  \begin{array}{lr}
\sqrt{\frac{\Omega_{0}-1}{\Omega_{0}(1+z)^{3}}}
                   & \qquad\qquad \Omega_{0} > 1 \\&  \\
\sqrt{\frac{1-\Omega_{0}}{\Omega_{0}(1+z)^{3}}}
& \qquad\qquad 0 < \Omega_{0} < 1
\end{array}\right..
\end{equation} 
(For $\Omega_{0}=0$ the appropriate case from Eq.~(\ref{tomeganull-g}) 
must be used.)  



%***********************************************************************
\section{Volume element}


Sometimes the distance is only a means of calculating the volume element
at a given redshift.  In the static Euclidean case the volume element is
of course 
\begin{equation}
\D V = 4\pi r^{2}dr.
\end{equation}
In the cosmological case, the volume element is, with $r=R_{0}\sigma$,
\begin{equation}
\D V = 4\pi r^{2}\D D^{\rm P} = 
4\pi r^{2}\frac{c}{H_{0}}\frac{{\rm d}z}{\sqrt{Q(z)}}.
\end{equation}
$R_{0}\sigma_{y}$ is, for $\eta=1$, simply $D_{y0}=(1 + y)D_{0y}$; see
Sect~\ref{eta1}.  Thus, the distance $D_{y0}$ is all that is needed to
calculate the volume; this first can be calculated by Eq.~(\ref{RKeqG})
with $\eta=1$ (This applies even if one would calculate {\em
distances\/} with another value of $\eta$ since the volume element is a
quantity related to the global geometry of the universe---alternatively,
one can use elliptic integrals, as in Sect.~\ref{eta1} and Feige
(\cite{BFeige92a}).) If one has an expression for $\sigma(z)$, then,
since 
\begin{equation}
\D D^{\rm P} = R_{0}\frac{\D \sigma}{\sqrt{1-k\sigma^{2}}},
\end{equation}
which follows directly from Eq.~(\ref{rwm-g}), one can write
\begin{equation}
\label{dvolume-g}
dV(\sigma) = 4\pi R_{0}^{3}
\int\limits_{0}^{\sigma}\frac{\sigma'^{2}\D\sigma'}{\sqrt{1-k\sigma'^{2}}}.
\end{equation}
where $R_{0}$ is given by Eq.~(\ref{rnull-g}) for the present values:
\begin{equation}
R_{0} = \frac{c}{H_{0}}\frac{1}{\sqrt{|\Omega_{0} + \lambda_{0} -1|}}.
\end{equation}
and 
\begin{equation}
\sigma = \frac{D_{y0}(1+z_{y})}{R_{0}}
\end{equation}
Integration gives
\begin{equation}
\label{volume-g}
V(\sigma) = \left\{
\begin{array}{lr}
2\pi r^{3}\left(\frac{\sqrt{1+\sigma^{2}}}{\sigma^{2}} -
\frac{\arcsinh \sigma}{\sigma^{3}}  
\right)
& \qquad k = -1 \\
\frac{4}{3}\pi r^{3}
& \qquad k = 0 \\
2\pi r^{3}\left(\frac{\arcsin \sigma}{\sigma^{3}} - 
\frac{\sqrt{1-\sigma^{2}}}{\sigma^{2}}\right)
& \qquad k = +1
\end{array}
\right.
\end{equation}
Thus, for $k=+1$, the total volume of the universe is
$2\pi^{2}R_{0}^{3}$. (See, e.g., Sandage (\cite{ASandage95a}),
Sect.~1.6.1; Sandage's $d$ is our $D^{\rm P}$ and his $r$ is our
$\sigma$.) Since 
\begin{displaymath}
\D\chi=\frac{\D\sigma}{\sqrt{1-k\sigma^{2}}}
\end{displaymath}
Eq.~(\ref{volume-g}) can also be written as 
(cf.~Feige (\cite{BFeige92a}), Eq.~(116); Feige's $r$ is our $\sigma$)
\begin{equation}
\label{volume2-g}
V(\chi) = 2\pi R_{0}^{3}\left\{
\begin{array}{lr}
\sinh(\chi)\cosh(\chi) - \chi
& \qquad k = -1 \\
\frac{2}{3}\chi^{3}
& \qquad k = 0 \\
\chi - \sin(\chi)\cos(\chi)
& \qquad k = +1
\end{array}
\right.
\end{equation}
Of course, all this refers to volumes {\em now\/} at the distance
corresponding to $z=y$.  If the volume at another time is important, say
at the time of emission of the light we see now---for instance if one is
concerned with the space density of some comoving objects---then the volume
element must be divided by $(1 + z)^{3}$. 




%***********************************************************************


\begin{thebibliography}{}
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\novolume{1972}{CDyerRRoeder72a}{\firstauthor{C.~C.}{Dyer}, %
\otherauthor{R.~C.}{Roeder}}%
{}
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%
%\bibitem{CDyerRRoeder73a}
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\otherauthor{R.~C.}{Roeder}}%
{}%
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%
%\bibitem{BFeige92a}
\article{1992}{BFeige92a}{\firstauthor{B.}{Feige}}%
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%
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%
\end{thebibliography}



%***********************************************************************


\end{document}
