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\begin{document}

\begin{center}
{\bf ASTR 5770 Cosmology Spring 2006.  Problem Set 1.  Due Mon 6 Feb}
\end{center}
\vskip4pt

\sect{Hubble Diagram}

Derive explicit expressions for the luminosity distance $d_L$
as a function of redshift $z$ for universes which are:
\\
(a) flat, matter-dominated, $\Omega_{\rm m} = 1$;
\\
(b) open, curvature-ominated, $\Omega_{\rm K} = 1$;
\\
(c) flat, vacuum-dominated, $\Omega_\Lambda = 1$.
\\
Use a computer to draw Hubble diagrams for each case,
that is, a graph of $\log d_L$ versus $\log z$.
\\{}
[Hint:
Luminosity distance $d_L$ is defined by
\begin{equation}
  F = {L \over 4 \pi d_L^2}
\end{equation}
where $F$ is the observed flux from the object,
and $L$ is its intrinsic luminosity.]

\sect{Tensor}

This is mostly a concept question, so think about it carefully.
In a general coordinate system $x^\mu$,
which of the following are tensors,
and if so, of what type?  Explain.
\\
(a) An infinitesimal coordinate interval $d x^\mu$;
\\
(b) The coordinates $x^\mu$;
\\
(c) The proper time $\tau$ along the worldline of an arbitrarily accelerating object.
\\{}
[Hint:
A ``tensor'' is a quantity
$t^{\lambda\kappa ...}_{\mu \nu ...}$
which transforms under coordinate transformations
$x^\mu \rightarrow x^{\prime \mu}$
as
\begin{equation}
  t^{\prime \kappa \lambda ...}_{\, \mu \nu ...} =
 {\partial x^{\prime \kappa} \over \partial x^\alpha}
 {\partial x^{\prime \lambda} \over \partial x^\beta}
 {\partial x^\gamma \over \partial x^{\prime \mu}}
 {\partial x^\delta \over \partial x^{\prime \nu}}
 \ 
 ...
 \ 
 t^{\alpha \beta ...}_{\gamma \delta ...}
 \ .
\end{equation}
A tensor with $n$ indices has rank $n$.
A scalar is a tensor of rank zero.
A 4-vector is a tensor of rank one.
A tensor with all indices up is said to be contravariant.
A tensor with all indices down is said to be covariant.
A tensor with $n_U$ indices up and $n_D$ indices down
is said to be mixed, with $n_U$ contravariant indices
and $n_D$ covariant indices.]

\sect{Equation of motion}

Variation of the action yields
Lagrange's equations for the
motion of a particle
in terms of its Lagrangian
${\cal L}(x^\mu, u^\mu)$
\begin{equation}
\label{eqmotLag}
  {d \over d \tau} {\partial {\cal L} \over \partial u^\lambda}
  =
  {\partial {\cal L} \over \partial x^\lambda}
\end{equation}
where $\tau$ is the particle's proper time,
and $u^\mu$ its 4-velocity
\begin{equation}
  u^\mu \equiv {d x^\mu \over d \tau}
  \ .
\end{equation}
The coordinates $x^\mu$ and velocities $u^\mu$ in this approach
are taken to be independent coordinates, so
the velocity partial derivatives $\partial / \partial u^\lambda$
in equation~(\ref{eqmotLag})
are to be interpreted as being done with the coordinates $x^\mu$ held fixed,
and conversely
the coordinate partial derivatives $\partial / \partial x^\lambda$
are done with the velocities $u^\mu$ held fixed.
In general relativity,
the effective Lagrangian of a free particle can be taken to be
(see Blau's 1999 lecture notes
\url{http://www.unine.ch/phys/string/lecturesGR.ps.gz})
\begin{equation}
  {\cal L}
  =
  g_{\mu \nu}(x) u^\mu u^\nu
\end{equation}
with the metric $g_{\mu\nu}$
being a function only of the coordinates $x^\mu$, not the velocities $u^\mu$.
Show that Lagrange's equations~(\ref{eqmotLag})
are equivalent to the usual equations of motion
\begin{equation}
\label{eqmotPE}
  {d u \over d \tau} = 0
  \ .
\end{equation}
[Hint:
First
show that $g^\lambda . g_\mu = \delta^\lambda_\mu$.
Then from
$\partial ( g^\lambda . g_\mu ) / \partial x^\nu = 0$
and from the definition of Christoffel symbols,
$\partial g_\mu / \partial x^\nu \equiv \Gamma^\lambda_{\mu\nu} g_\lambda$,
deduce that
$\partial g^\lambda / \partial x^\nu = - \Gamma^\lambda_{\mu\nu} g^\mu$.
Show that the equation of motion~(\ref{eqmotPE}),
with
$u = g^\lambda u_\lambda$,
becomes
$g^\lambda
\left( d u_\lambda / d \tau - \Gamma_{\mu\lambda\nu} u^\mu u^\nu \right) = 0$,
where $\Gamma_{\mu\lambda\nu} \equiv g_{\mu\kappa} \Gamma^\kappa_{\lambda\nu}$.
Now argue that
$\partial {\cal L} / \partial u^\lambda = 2 u_\lambda$,
and that
$\partial {\cal L} / \partial x^\lambda = 2 \Gamma_{\mu\nu\lambda} u^\mu u^\nu$.
To complete,
you'll need to invoke the no-torsion symmetry
$\Gamma_{\mu\nu\lambda} = \Gamma_{\mu\lambda\nu}$.
]

The great advantage of Lagrange's equations~(\ref{eqmotLag})
is that if, thanks to some symmetry,
the metric is independent of a coordinate $\chi$,
then the right hand side of equation~(\ref{eqmotLag}) vanishes,
and it is possible immediately to deduce that
\begin{equation}
  u_\chi = \mbox{constant}
\end{equation}
is a constant of the motion.

\sect{Electrically charged universe [no credit]}

Could the observed acceleration of the Universe
be caused by a uniform finite charge density in the Universe,
as opposed to by some mysterious gravitationally repulsive dark energy?
[Hint:
In a spatially homogeneous and isotropic universe,
which way can the electric field vector point?
In electromagnetism,
the energy density of the electric field is proportional to the
square of the electric field.]

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