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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