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\begin{document}

\centerline{\bf Special and General Relativity (PHZ 4601/560 
Fall 2017) } \medskip 

\centerline{\bf Solutions Final December 14} \bigskip 

\begin{enumerate}

\item {\bf Schwarzschild effective potential.}

\begin{figure}[th] 
\centerline{\hbox{ \psfig{figure=V03.eps,width=14cm} }}
\caption{Newton and Schwarzschild potentials.\label{fig_v03} } 
\end{figure}

\begin{enumerate}

\item We have to calculate the zeros of the derivative of the effective 
potential
$$V'_{GR}(r)=-2\,\frac{h^2}{r^3}+\frac{2m}{r^2}+\frac{6m\,h^2}{r^4}\,.$$
After multiplying with $r^4$ we arrive at
$$ 0 = -2\,h^2\,r + 2m\,r^2 + 6m\,h^2~\Rightarrow~
   0 = r^2 - \frac{h^2}{m}\,r + 3h^2\,. $$
So, $p=h^2/m$, $q=3h^2$ and
$$r_{1,2}=\frac{h^2}{2m}\pm\sqrt{\left(\frac{h^2}{2m}\right)^2-3h^2}\,.$$

\item In terms of $h$ and $m$ the equation $q=(p/2)^2-(p/4)^2=3p^2/16$ 
reads
$$3h^2=\frac{3h^4}{16m^2}~\Rightarrow~16m^2=h^2~\Rightarrow~h=4m\,,$$
resulting in $r_{1,2}=(8\pm 4)\,m$ and
the Schwarzschild effective potential 
$$V_{GR}(r)=\frac{16m^2}{r^2}-\frac{2m}{r}-\frac{32m^3}{r^3}\,$$
where the first two terms are Newton's potential.

\item For the chosen parameters the Scharzschild and Newton's 
potential are drawn in the figure on page~1.

\end{enumerate}


\item {\bf Schwarzschild radius and density.}

\begin{enumerate}

\item The Scharzschild radius is $r_S=2m$ and
$$ m = \frac{4\pi}{3}\,r^3\,\rho\,.$$
Therefore
$$ R_0=\frac{8\pi}{3}\,R_0^3\,\rho~\Rightarrow~R_0^2=\frac{3}{8\pi\,\rho}$$
and $r_S>R$ for $R>R_0$, because $r_s$ increases $\sim R^3$, i.e., faster
than $R$.

\item In conventional units we have $m=G\,M/c^2$ and, therefore,
$$ R_0^2=\frac{3\,c^2}{8\pi\,G\,\rho}\,.$$
With the given numbers, and rounding to three digits,  we find $R_0=4
\times 10^{27}\,m = 42.3\times10^9\,ly$. This is larger than the size 
of the visible universe, which is approximately $15\times 10^9\,[ly]$.

\end{enumerate}
\bigskip

\item{\bf Einstein's equations.}

\begin{enumerate}

\item We have
$$ R^{\mu}_{~\mu}-\frac{1}{2}\,g^{\mu}_{~\mu}\,R = 
   -\kappa\,T^{\mu}_{~\mu}~\Rightarrow~R-2\,R = - 
   \kappa\,T~\Rightarrow~R = \kappa\,T$$
and, therefore,
$$ R_{\mu\nu}-\kappa\,\frac{1}{2}\,g_{\mu\nu}\,T = -\kappa\,T_{\mu\nu}
   ~\Rightarrow~ R_{\mu\nu} =-\kappa\,\left(T_{\mu\nu}-
    \frac{1}{2}\,g_{\mu\nu}\,T\right)\,.$$

\item In vacuum we have
$$ R^{\mu}_{~\mu}-\frac{1}{2}\,g^{\mu}_{~\mu}\,R + 
   g^{\mu}_{~\mu}\,\Lambda = 0~\Rightarrow~R-2\,R+4\,\Lambda = 0
   ~\Rightarrow~R=4\,\Lambda\,,$$
i.e., $n=4$.

\end{enumerate} 


\item{\bf Lorenz gauge with covariant derivatives.}

\noindent According to (10.55) of Rindler we have 
$$ \Phi^{\mu}_{~;\rho\sigma}-\Phi^{\mu}_{~;\sigma\rho} 
   = - \Phi^{\alpha}\,R^{\mu}_{~\alpha\rho\sigma}\,. $$
Therefore, the definition (10.68) of the Ricci tensor yields
$$ \Phi^{\mu}_{~;\rho\mu}-\Phi^{\mu}_{~;\mu\rho} = - 
   \Phi^{\alpha}\,R^{\mu}_{~\alpha\rho\mu} = -
   \Phi^{\alpha}\,R_{\alpha\rho}\,.  $$
In the Lorenz gauge $\Phi^{\mu}_{~;\mu\rho}=0$ holds and
we are left with
$$ \Phi^{\mu}_{~;\rho\mu}=-\Phi^{\alpha}\,R_{\alpha\rho}\,.$$

\end{enumerate} 
\end{document}
