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\begin{document}

\centerline{\bf Special and General Relativity PHZ 4601 Fall 2018: 
Test on Homework Solutions.} \bigskip 

\begin{enumerate}

{\bf \item \bf Twin travel in 2D Minkowski space.} % \bigskip

\begin{enumerate}

\item $\beta_B^1 = 4/5$.
\smallskip

\item The proper time of $B$ at $B_0$, the end of the first part 
of its travel, is $$\sqrt{(5)^2-(4)^2}=3\,.$$
\item $2\times 5=10$ is the time on the clock of $A$ when $A$ meets
$B$ again.
\smallskip

\item $2\times 3=6$ is the time on the clock of $B$ when $A$ meets
$B$ again.
\smallskip

\item We have $\beta_A^2=-\beta_B^1=-4/5$ for the velocity of $A$
in the inertial frame $K_2$.
\smallskip

\item The coordinates of $B_0$ in $K_2$ are $\pmatrix{3\cr 0}$.
\bigskip

\end{enumerate}


\item {\bf Radar Distance.} % \bigskip

Let the coordinate times for light-signals from $A$ to $B$ and $B$ to 
$A$ be $\triangle t_{A\to B}$ and $\triangle t_{B\to A}$, respectively. 
The coordinate round-trip time $\triangle t$ for a radio echo is the 
same at $A$ and $B$: $\triangle t = \triangle t_{A\to B} + \triangle 
t_{B\to A}$. But the radar distance is determined with standard clocks:
Radar distance measured at~$A$: $L_A=(c/2)\,\triangle \tau_A =
(c/2)\,\exp(\Phi_A/c^2)\,\triangle t$ and, similarly, $L_B=(c/2)\,
\triangle \tau_B = (c/2)\,\exp(\Phi_B/c^2)\,\triangle t$. Therefore,
$$ \frac{L_A}{L_B} = \frac{\exp(\Phi_A/c^2)}{\exp(\Phi_B/c^2)}\,. $$
\bigskip

\item{\bf Schwarzschild Metric.} %\smallskip

We choose the identification 
\begin{eqnarray} \nonumber
  d\vec{s}\,^2 &=& a\,(dx^1)^2+b\,(dx^2)^2+(dx^3)^2+(dx^4)^2\\ \nonumber
               &=& a\,(dt)^2+b\,(dr)^2+c\,(d\theta)^2+d\,(d\phi)^2\,,
\end{eqnarray} 
respectively, with $a=\exp[+A(r)]$, $b=-\exp[+B(r)]$, $c=-r^2$, 
$d=-r^2\sin^2\theta$. The non-zero derivatives are the 
$$ '= \frac{\partial~}{\partial\,x^2}=\frac{\partial}{\partial\,r}
   ~~{\rm  and,\ on}\ d\ {\rm only,}\ \frac{\partial~}{\partial\,x^3} 
   = \frac{\partial~}{\partial\,\theta}\,.$$
 Therefore, the non-zero contributions to the Ricci tensor are
$$ R_{tt} = R_{11} = \beta\,a_{22} - \beta\,a_2\,(\alpha\,a_2+\beta\,b_2
                   -\gamma\,c_2-\delta_2\,d_2)\,, $$
where $\alpha$, $\beta$, $\gamma$, $\delta$ are defined by $\alpha=
1/2a$, $\beta =1/2b$, $\gamma=1/2c$, $\delta=1/2d$. We have
$$ a'=A'\exp(+A)\,,~~b'=-B'\exp(+B)\,~~c'=-2r\,~~d'=-2r\,\sin^2\theta$$
and $a''=A''\exp(A)+(A')^2\exp(A)$. Inserting these gives
\begin{eqnarray} \nonumber
 R_{tt}\ &=& - \exp(A-B)\,\frac{1}{2}\,\left(A''+{A'}^2\right)\\
 \nonumber &~& + \exp(A-B)\,\frac{1}{2}\,A'\,\left(\frac{1}{2}\,A'
           +\frac{1}{2}\,B' -\frac{1}{r}-\frac{1}{r} \right)\\ \nonumber 
        &=& - \exp(A-B)\,\left(\frac{1}{2}\,A''-\frac{1}{4}\,A'\,B'
            +\frac{1}{4}\,{A'}^2+A'/r\right)\,.
\end{eqnarray} 

\end{enumerate} 
\end{document}


