Special theory of relativity: Time dilation in an airplane

Question:
Physicists conduct an experiment to measure time dilation of an atomic clock in a flying airplane. If the velocity of the airplane is 960 km/h, find the time dilated of the atomic clock from the laboratory flame. (You do not need to consider the effect from the general theory of relativity.)

Answer:
We consider one direction of motion, say, $x$ direction. The Lorentz transformations for the displacement and time are
\begin{eqnarray}
x' &=& \frac{-vt+x}{\sqrt{1-\beta^2}}  \\
t' &=& \frac{t-(v/c^2)x}{\sqrt{1-\beta^2}}
\end{eqnarray}
where $\beta=v^2/c^2$. Suppose that the clock is placed at the origin of moving flame, $S'$. Then, one measures the time in laboratory flame, $S$. The origin in $S'$ indicates that $x'=0$. Thus, we have
\begin{equation}
0 = \frac{-vt+x}{\sqrt{1-\beta^2}} \quad \rightarrow \quad x = vt
\end{equation}
Plug it in the transformation for time.
\begin{equation}
t' = \frac{t-(v/c^2)(vt)}{\sqrt{1-\beta^2}}=\frac{t(1-\beta^2)}{\sqrt{1-\beta^2}}=\sqrt{1-\beta^2}t
\end{equation}
Since $v\ll c \rightarrow \beta \ll 1$, we can use the approximated expression with a Taylor expansion;
\[
t' = \left(1-\frac{1}{2}\beta^2\right)t
\]
Now, convert the velocity of the airplane into m/s. Note that 1 km = 1000 m and 1 hour = 3600 s.
\[
960 \ \mathrm{km/h} = 960 \times 1000 \div 3600 = 266.7 \ \mathrm{m/s}
\]
Therefore, we have the time dilation compared with the lab frame.
\[
t'= \left\{1-\frac{1}{2}\left(\frac{266.7}{3.00\times 10^8}\right)^2\right\}t=(1-3.95\times 10^{-13})t
\]
This means: While one second elapses in lab frame, the clock in the airplane only elapses $1-3.95\times 10^{-13}$ seconds.

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Special theory of relativity: Velocity addition

Question:
The speed of light is measured in a liquid that has refractive index of $n$. If the liquid is moving at $v$, what is the speed of light in the liquid detected in the laboratory frame?

Answer:
The addition of two velocities from the lab frame in terms of relativity is
\[
v_{\mathrm{added}} = \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}
\]
The speed of light in liquid is given by
\[
v' = \frac{c}{n}
\]
In this case, $v_1=\frac{c}{n}$ and $v_2=v$.
\begin{eqnarray}
v_{\mathrm{added}} &=& \frac{\frac{c}{n}+v}{1+\frac{\frac{c}{n}v}{c^2}} \\
  &=& \frac{c/n+v}{1+v/nc}  \\
  &=& \frac{c(vn+c)}{nc+v}
\end{eqnarray}

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Lorentz contraction

Question:
A 2.00-m stick is moving in one direction, and the length is observed to be 1.00 m from the laboratory frame. How fast is the stick moving?

Answer:
This kinematics is based on the special theory of relativity. Due to time dilation, the original length in motion is observed shorter. Suppose $L_0$ and $L$ are the original length and the contracted length, respectively. From the insight of relativity, we have
\[
L = L_0\sqrt{1-\frac{v^2}{c^2}}
\]
where $v$ and $c$ are speed of the stick and speed of light. Let us solve for the speed of the stick.
\begin{eqnarray*}
\left(\frac{L}{L_0}\right)^2 &=& 1 - \frac{v^2}{c^2}   \\
\frac{v^2}{c^2} &=& 1 - \left(\frac{L}{L_0}\right)^2   \\
v^2 &=& c^2\left[ 1 - \left(\frac{L}{L_0}\right)^2 \right]  \\
v &=& \sqrt{c^2\left[ 1 - \left(\frac{L}{L_0}\right)^2 \right]}
\end{eqnarray*}
Then, plug in the numbers.
\[
v = \sqrt{(3.0\times 10^8)^2 \left[ 1 - \left(\frac{1.00}{2.00}\right)^2 \right]}
   = 2.60 \times 10^8 \quad \mathrm{m/s}
\]

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