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\begin{document}

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

\begin{enumerate}

\item{\bf An astrophysical observation (redshift):} %\smallskip

\noindent The equation for the redshift is
\begin{eqnarray} \nonumber
  \lambda'\ =\ \lambda\,\sqrt{\frac{1+\beta}{1-\beta}}\ &\Rightarrow&\
  \left(\frac{\lambda'}{\lambda}\right)^2\ =\ {\frac{1+\beta}{1-\beta}}
  \\ \nonumber
  \left(\frac{\lambda'}{\lambda}\right)^2 -\beta\,
  \left(\frac{\lambda'}{\lambda}\right)^2 &=& 1+\beta
  \\ \nonumber
  \left(\frac{\lambda'}{\lambda}\right)^2 - 1\ =\ \beta\,\left[ 1
  + \left(\frac{\lambda'}{\lambda}\right)^2 \right]\ &\Rightarrow&\
  \beta\ =\ \frac{(\lambda'/\lambda)^2-1}{(\lambda'/\lambda)^2+1}\,.
\end{eqnarray}
With $\lambda'=(729.2\,[nm])\,m^2/(m^2-4)$ given and from Quantum 
Mechanics books $\lambda=(364.56\,[nm]) \,m^2/(m^2-4)$ we find
$\lambda'/\lambda=2$ and, hence,
$$ \beta\ =\ \frac{v}{c}\ =\ \frac{3}{5}\ =\ 0.6\,.$$
\medskip


\item{\bf Light signals and travel in two inertial frames:}

\begin{figure}[th] % From EMscript/figures_plt/minkowski_space_sol.
\centerline{\hbox{ \psfig{figure=figures/1msol,width=12cm} }}
\caption{Minkowski space in which observer~B is at rest and
observer~A moves with speed $4/5$ in negative $x$ direction. 
\label{fig_mid1mink} } \end{figure}

We use natural units, $c=1$, and express everything in seconds,
omitting $[s]$ in the following.

\begin{enumerate}

\item In the rest frame $S$ of A the coordinates of $A_1$ are $(t,x)
=(15,0)$. In the rest frame $S'$ of B we have $15=\tau=\sqrt{t'^2-x'^2}$ 
for the proper time of A at this space-time point and $x'=-4t'/5$. 
Putting these equations together, we find:
$$ 15 = \sqrt{(t')^2-(4t'/5)^2} = 3\,t'/5~~\Rightarrow~~
   t'=25~~\Rightarrow~~(t',x')=(25,-20)\,.$$
Alternatively, we may perform the Lorentz transformations
$$ t' = \gamma\,t - \beta\gamma\,x\,~~~
   x' = \gamma\,x - \beta\gamma\,t\,.$$
Now, $\beta=4/5$ and $\gamma=1/\sqrt{1-(4/5)^2}=5/3$, 
$\beta\gamma=4/3$. Hence, $(t,x)=(15,0)$ transforms into
$$ t' = 5t/3 = 25\,,~~x'=-4t/3=-20\,~~{\rm i.e.,}~~
   (t',x')=(25,-20)\,. $$


\item In $S$ the coordinates of $B_0$ are the meeting point of the 
straight lines $x=4t/5$ and $x=t-15$ for $t\ge 15$. The solution is 
$(t,x)=(75,60)$ and the proper time of B at this point is $\tau=
\sqrt{t^2-x^2}=45$. In the $S'$ frame we have by its definition 
$x'=0$ for B and the space-time point $B_0$ becomes $(t',x')=(45,0)$.

\item In $S$ the space-time position $A_2$ is at $(t,x)=(135,0)$. We
can deal with this like before with $(t,x)=(15,0)$: 
$$ 135 = \sqrt{(t')^2-(4t'/5)^2} = 3\,t'/5~~\Rightarrow~~t'=225\,,~~x'
   = -4\,t'/5 = -180\,.$$
Alternatively, we may perform the Lorentz transformations. Now,
$$ t' = 5t/3 = 225\,,~~x'=-4t/3=-180\,~~{\rm i.e.,}~~
   (t',x')=(225,-180)\,. $$

\item See the figure.
\bigskip

\item{\bf Space travel (Rindler problem 2.9):} %\smallskip

$$ \triangle x = \beta\,\triangle t = 8\,,$$
$$ \triangle\tau = 8 = \frac{\triangle t}{\gamma} = \triangle t\,
   \sqrt{1-\beta^2} = 8\,\frac{\sqrt{1-\beta^2}}{\beta}\,, $$
Therefore,
$$ 1 = \frac{\sqrt{1-\beta^2}}{\beta} = \frac{1-\beta^2}{\beta^2}\ 
   \Rightarrow\ 2\,\beta^2 = 1\ \Rightarrow\ \beta = 
   \frac{1}{\sqrt{2}} = 0.7071\dots \,.$$

\end{enumerate}

\end{enumerate} 
\end{document}
