5.5 The lorentz transformation  (Page 3/12)

Using the lorentz transformation for length

Solution

1. Identify the known: $L=100\text{m};v=0.20c;\text{Δ}\tau =0.$
2. Identify the unknown: $L\prime \text{.}$
3. Express the answer as an equation. The surveyor in frame S has measured the two ends of the stick simultaneously, and found them at rest at $x_2$ and $x_1$ a distance $L=x_2-x_1=100\text{m}$ apart. The spaceship crew measures the simultaneous location of the ends of the sticks in their frame. To relate the lengths recorded by observers in $\text{S}\prime$ and S, respectively, write the second of the four Lorentz transformation equations as:
   
   $\begin{array}{cc}\hfill x{\prime }_2-x{\prime }_1& =\frac{x_2-vt}{\sqrt{1-v^2\text{/}{c}^2}}-\frac{x_1-vt}{\sqrt{1-v^2\text{/}{c}^2}}\hfill \\ & =\frac{x_2-x_1}{\sqrt{1-v^2\text{/}{c}^2}}=\frac{L}{\sqrt{1-v^2\text{/}{c}^2}}.\hfill \end{array}$
4. Do the calculation. Because $x{\prime }_2-x{\prime }_1=100\text{m},$ the length of the moving stick is equal to:
   
   $\begin{array}{cc}\hfill L\prime & =\left(100\text{m}\right)\sqrt{1-v^2\text{/}{c}^2}\hfill \\ & =\left(100\text{m}\right)\sqrt{1-{\left(0.20\right)}^2}\hfill \\ & =98.0\text{m.}\hfill \end{array}$
   
   Note that the Lorentz transformation gave the length contraction equation for the street.

Lorentz transformation and simultaneity

Solution

1. Identify the known: $\text{Δ}t=0.$
   Note that the spatial separation of the two events is between the two lamps, not the distance of the lamp to the passenger.
2. Identify the unknown: $\text{Δ}t\prime =t_2^{\prime }-t_1^{\prime }.$
   Again, note that the time interval is between the flashes of the lamps, not between arrival times for reaching the passenger.
3. Express the answer as an equation:
   
   $\text{Δ}t=\frac{\text{Δ}t\prime +v\text{Δ}x\prime \text{/}{c}^2}{\sqrt{1-v^2\text{/}{c}^2}}.$
4. Do the calculation:
   
   $\begin{array}{ccc}\hfill 0& =\hfill & \frac{\text{Δ}t\prime +\frac{c}{2}\left(26\text{m}\right)\text{/}{c}^2}{\sqrt{1-v^2\text{/}{c}^2}}\hfill \\ \hfill \text{Δ}t\prime & =\hfill & -\frac{26\text{m/s}}{2c}=-\frac{26\text{m/s}}{2\left(3.00\times {10}^8\text{m/s}\right)}\hfill \\ \hfill \text{Δ}t\prime & =\hfill & \mathrm{-4.33}\times {10}^{\mathrm{-8}}\text{s.}\hfill \end{array}$

Significance