# 5.5 The lorentz transformation  (Page 3/12)

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## Using the lorentz transformation for length

A surveyor measures a street to be $L=100\phantom{\rule{0.2em}{0ex}}\text{m}$ long in Earth frame S. Use the Lorentz transformation to obtain an expression for its length measured from a spaceship $\text{S}\prime ,$ moving by at speed 0.20 c , assuming the x coordinates of the two frames coincide at time $t=0.$

## Solution

1. Identify the known: $L=100\phantom{\rule{0.2em}{0ex}}\text{m};v=0.20c;\text{Δ}\tau =0.$
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 $L={x}_{2}-{x}_{1}=100\phantom{\rule{0.2em}{0ex}}\text{m}$ 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 $x{\prime }_{2}-x{\prime }_{1}=100\phantom{\rule{0.2em}{0ex}}\text{m},$ the length of the moving stick is equal to:
$\begin{array}{cc}\hfill L\prime & =\left(100\phantom{\rule{0.2em}{0ex}}\text{m}\right)\sqrt{1-{v}^{2}\text{/}{c}^{2}}\hfill \\ & =\left(100\phantom{\rule{0.2em}{0ex}}\text{m}\right)\sqrt{1-{\left(0.20\right)}^{2}}\hfill \\ & =98.0\phantom{\rule{0.2em}{0ex}}\text{m.}\hfill \end{array}$

Note that the Lorentz transformation gave the length contraction equation for the street.

## Lorentz transformation and simultaneity

The observer shown in [link] standing by the railroad tracks sees the two bulbs flash simultaneously at both ends of the 26 m long passenger car when the middle of the car passes him at a speed of c /2. Find the separation in time between when the bulbs flashed as seen by the train passenger seated in the middle of the car.

## 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:
$\begin{array}{ccc}\hfill 0& =\hfill & \frac{\text{Δ}t\prime +\frac{c}{2}\left(26\phantom{\rule{0.2em}{0ex}}\text{m}\right)\text{/}{c}^{2}}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}\hfill \\ \hfill \text{Δ}t\prime & =\hfill & -\frac{26\phantom{\rule{0.2em}{0ex}}\text{m/s}}{2c}=-\frac{26\phantom{\rule{0.2em}{0ex}}\text{m/s}}{2\left(3.00×{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)}\hfill \\ \hfill \text{Δ}t\prime & =\hfill & -4.33×{10}^{-8}\phantom{\rule{0.2em}{0ex}}\text{s.}\hfill \end{array}$

## Significance

The sign indicates that the event with the larger ${x}_{2}\prime ,$ namely, the flash from the right, is seen to occur first in the $S\prime$ frame, as found earlier for this example, so that ${t}_{2}<{t}_{1}.$

## Space-time

Relativistic phenomena can be analyzed in terms of events in a four-dimensional space-time . When phenomena such as the twin paradox, time dilation, length contraction, and the dependence of simultaneity on relative motion are viewed in this way, they are seen to be characteristic of the nature of space and time, rather than specific aspects of electromagnetism.

In three-dimensional space, positions are specified by three coordinates on a set of Cartesian axes, and the displacement of one point from another is given by:

$\left(\text{Δ}x,\text{Δ}y,\text{Δ}z\right)=\left({x}_{2}-{x}_{1},{y}_{2}-{y}_{1},{z}_{2}-{z}_{1}\right).$

The distance $\text{Δ}r$ between the points is

$\text{Δ}{r}^{2}={\left(\text{Δ}x\right)}^{2}+{\left(\text{Δ}y\right)}^{2}+{\left(\text{Δ}z\right)}^{2}.$

The distance $\text{Δ}r$ is invariant under a rotation of axes. If a new set of Cartesian axes rotated around the origin relative to the original axes are used, each point in space will have new coordinates in terms of the new axes, but the distance $\text{Δ}r\prime$ given by

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