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\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions} \bigskip

{\bf Set 1:} \bigskip

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% \end{figure} \centerline{Potential as function of $z$.}

\begin{enumerate}

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\item Non-relativistic distance measurement \smallskip

With $x=vt$ the first event is conveniently chosen $x_0=0$ for $t=0$.

In the non-relativistic limit $u'=u-v$ and the elastic bounce back 
velocity in the frame $K'$ is $-u'$. So it is in frame $K$ 
$$ -u'+v=-u+2v\,.$$
Hence, $u>2v$ is required for the ball to come back at all. Assume 
the ball is emitted at time $t^e$. It will hit $O'_0$ at a time $t_1$ 
determined by
$$ x_1 = v\,t_1 = u\,(t_1-t^e)\ \Rightarrow\
   t_1 = \frac{u\,t^e}{u-v}~~{\rm and}~~
   x_1 = v\,t_1 = \frac{u\,v\,t^e}{u-v}\,.$$
With $t^r_{1}$ defined as the time at which the ball is received back 
by $O_0$, % the time to travel back is determined by the equation
$$ (u-2v)\,(t^r_1-t_1) = x_1 = v\,t_1\,, $$
$$ (u-2v)\,t^r = (u-2v)\,t_1+v\,t_1 = (u-v)\,t_1= u\,t^e\,. $$
Introducing the time difference $\triangle t=t^r-t^e$ between received 
and emitted,
$$ u\,(t^r-t^e)=u\,\triangle t=2v\,t^r=2v\,(t^e+\triangle t)\,,$$
with the final result
\begin{eqnarray} \label{v_from_u}
  v = \frac{u\,\triangle t}{2\,(t^e+\triangle t)}\ \to\ \frac{u}{2}
  \ {\rm for}\ \triangle t\to\infty\,.
\end{eqnarray}
This ought to be compared with the equation for a light signal, which 
is for the situation $x=v\,t$ (i.e., $x_0=0$) derived from
$$ x_1 = v\,(t^e+\triangle t/2) = \frac{c\,\triangle t}{2}\ 
   \Rightarrow\ v = \frac{c\,\triangle t}{2\,(t^e+\triangle t/2)}\
   \to\ c\ {\rm for}\ \triangle t\to\infty\,.$$
Due to the asymmetry of the outward and return travel in the 
non-relativistic case, equation (\ref{v_from_u}) is more difficult
to derive than the one for light. Besides, there is a crucial
difference by a factor of two for $\triangle t$ in the denominator.
A return signal is only received for $u>2v$, whereas $c>v$ is 
sufficient for light signals.
\bigskip

\item Galilei transformations \smallskip

\noindent The equation 
$$ c^2 t^2 - \vec{x}^{\,2} = 0 ~~~{\rm in}\ K $$
is derived from 
$$ \vec{x} - \vec{c}\,t = 0 $$
which holds in $K$ for the propagation of the light in the direction 
$\vec{c}$. This equation transforms into
$$  \vec{x}^{\,'} - \vec{v}\,t - \vec{c}\,t = 0 ~~~{\rm in}\ K'\ .$$
Hence, in $K'$
$$  \vec{x}^{\,'} - \vec{c}^{\,'} t = 0~~~{\rm with}~~~
    \vec{c}^{\,'} = \vec{c} + \vec{v}\, .$$
In $K'$ the speed of light $c'=\sqrt{c'^{\,2}}$ is no longer a constant, 
but $c'^{\,2}=(\vec{c}+\vec{v}\,)^2$ depends on the angle between 
$\vec{c}$ and $\vec{v}$.
\bigskip

\item Contractions.  \smallskip

${\rm (1),\ (2)\ and\ (3)}\!:\ \ 25-1-4-9=11$;~~

${\rm (4)\ and\ (5)}\!:\ \ 25-9-16=0$; \smallskip

${\rm (6)}\!:\ \ 25-9-16=-4$; \smallskip

${\rm (7)}\!:\ \ 25-0-9-16=0$.
\bigskip


\item Time and relativistic distance measurements by light signals.
\smallskip

\noindent
A sufficiently accurate approximation of The elapsed time is given by
$$ \frac{2,757,790}{9,193}\,[s] \approx 300\,[s]\,.  $$ 

\noindent
(1) The relation $x=\triangle t/2$ gives for all three times
$$ 10^{-3}\times 3\times 10^5\,[km]\ =\ 300\,[km]\,. $$
Therefore, $O_2$ is at rest with respect to $O_1$.
\smallskip

\noindent
(2) We find the following positions at the following times:
\begin{eqnarray} \nonumber
  x_1=300\,[km]\ &{\rm at}&\ t_1=1.001\,[s]\,, \\ \nonumber
  x_2=600\,[km]\ &{\rm at}&\ t_2=2.002\,[s]\,, \\ \nonumber
  x_3=900\,[km]\ &{\rm at}&\ t_3=3.003\,[s]\,.
\end{eqnarray} 
This gives the velocities
\begin{eqnarray} \nonumber
  v_{21} &=& \frac{x_2-x_1}{t_2-t_1}\ =\ \frac{300\,[km]}
  {(2.002-1.001\,[s]}\ =\ \frac{300\,[km]}{1.001\,[s]}
  \ \approx\ 299.7\,[km/s]\,, \\ \nonumber
  v_{32} &=& \frac{x_3-x_2}{t_3-t_2}\ =\ \frac{300\,[km]}
  {(3.003-2.002)\,[s]}\ =\ \frac{300\,[km]}{1.001\,[s]}
  \ \approx\ 299.7\,[km/s]\,. 
\end{eqnarray} 
So, the results are consistent with the idea that $O_2$ is at rest
in an inertial frame, which moves with about 299.7$\,[km/s]$ with 
respect to the inertial frame of $O_1$. The position of $O_2$
with respect to $O_1$ is then given by
$$ x(t)\ =\ x_0 + v\,t\ =\ \frac{300\,[km]}{1.001\,[s]}\,t\,,$$
where $x_0=0$ follows from $x(t_1)=300\,[km]$.
\smallskip

\noindent
(3) We find again the positions $x_1=300\,[km]$ at $t_1=1.001\,[s]$
and $x_2=600\,[km]$ at $t_2=2.002\,[s]$, which gives again $v_{21}
=(300/1.001)\,[km/s]$, but now
\begin{eqnarray} \nonumber
  x_3=1200\,[km]\ &{\rm at}&\ t_3=3.004\,[s]\,,
\end{eqnarray} 
which gives 
\begin{eqnarray} \nonumber
  v_{32} &=& \frac{x_3-x_2}{t_3-t_2}\ =\ \frac{600\,[km]}{1.002\,[s]}
  \ \approx\ 588.2\,[km/s]\,. 
\end{eqnarray} 
As $v_{21}$ and $v_{32}$ disagree, $O_2$ cannot be at rest in an
inertial frame.

Note: If one wants to find suitable $t_2$, $t_1$, $\triangle t_2$
and $\triangle t_1$ values for a given speed $v$, this can be
done using the formula
\begin{eqnarray} \nonumber
  t_2- t_1\ =\ \frac{c}{2}\,\frac{\triangle t_2 - \triangle t_1}{v}\,.
\end{eqnarray} 
There many solutions. For instance with, besides $v$, also $\triangle 
t_2 -\triangle t_1>0$ given, any $t_2-t_1$ difference that matches will
do. For instance, for $\triangle t_2 - \triangle t_1 = 2\times 10^{3}
[s]$ and $t_1=1.001\,[s]$, $t_2-t_1 = 1\,[s]$ (not $1.001\,[s]$) is
needed to get precisely $300\,[km/s]$. Means starting time for $t_2$
at $1.999\,[s]$, receiving time at $2.003\,[s]$..
\bigskip

\end{enumerate}
\end{document}
