# 5.5 The lorentz transformation  (Page 2/12)

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## The lorentz transformation equations

The Galilean transformation nevertheless violates Einstein’s postulates, because the velocity equations state that a pulse of light moving with speed c along the x -axis would travel at speed $c-v$ in the other inertial frame. Specifically, the spherical pulse has radius $r=ct$ at time t in the unprimed frame, and also has radius $r\prime =ct\prime$ at time $t\prime$ in the primed frame. Expressing these relations in Cartesian coordinates gives

$\begin{array}{ccc}\hfill {x}^{2}+{y}^{2}+{z}^{2}-{c}^{2}{t}^{2}& =\hfill & 0\hfill \\ \hfill {{x}^{\prime }}^{2}+{{y}^{\prime }}^{2}+{{z}^{\prime }}^{2}-{c}^{2}{{t}^{\prime }}^{2}& =\hfill & 0.\hfill \end{array}$

The left-hand sides of the two expressions can be set equal because both are zero. Because $y=y\prime$ and $z=z\prime ,$ we obtain

${x}^{2}-{c}^{2}{t}^{2}={{x}^{\prime }}^{2}-{c}^{2}{{t}^{\prime }}^{2}.$

This cannot be satisfied for nonzero relative velocity v of the two frames if we assume the Galilean transformation results in $t=t\prime$ with $x=x\prime +vt\prime .$

To find the correct set of transformation equations, assume the two coordinate systems S and $S\prime$ in [link] . First suppose that an event occurs at $\left(x\prime ,0,0,t\prime \right)$ in $S\prime$ and at $\left(x,0,0,t\right)$ in S , as depicted in the figure.

Suppose that at the instant that the origins of the coordinate systems in S and $S\prime$ coincide, a flash bulb emits a spherically spreading pulse of light starting from the origin. At time t , an observer in S finds the origin of $S\prime$ to be at $x=vt.$ With the help of a friend in S , the $S\prime$ observer also measures the distance from the event to the origin of $S\prime$ and finds it to be $x\prime \sqrt{1-{v}^{2}\text{/}{c}^{2}}.$ This follows because we have already shown the postulates of relativity to imply length contraction. Thus the position of the event in S is

$x=vt+x\prime \sqrt{1-{v}^{2}\text{/}{c}^{2}}$

and

$x\prime =\frac{x-vt}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}.$

The postulates of relativity imply that the equation relating distance and time of the spherical wave front:

${x}^{2}+{y}^{2}+{z}^{2}-{c}^{2}{t}^{2}=0$

must apply both in terms of primed and unprimed coordinates, which was shown above to lead to [link] :

${x}^{2}-{c}^{2}{t}^{2}=x{\prime }^{2}-{c}^{2}t{\prime }^{2}.$

We combine this with the equation relating x and $x\prime$ to obtain the relation between t and $t\prime :$

$t\prime =\frac{t-vx\text{/}{c}^{2}}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}.$

The equations relating the time and position of the events as seen in S are then

$\begin{array}{ccc}\hfill t& =\hfill & \frac{t\prime +vx\prime \text{/}{c}^{2}}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}\hfill \\ \hfill x& =\hfill & \frac{x\prime +vt\prime }{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}\hfill \\ \hfill y& =\hfill & y\prime \hfill \\ \hfill z& =\hfill & z\prime \text{.}\hfill \end{array}$

This set of equations, relating the position and time in the two inertial frames, is known as the Lorentz transformation    . They are named in honor of H.A. Lorentz (1853–1928), who first proposed them. Interestingly, he justified the transformation on what was eventually discovered to be a fallacious hypothesis. The correct theoretical basis is Einstein’s special theory of relativity.

The reverse transformation expresses the variables in S in terms of those in $S\prime .$ Simply interchanging the primed and unprimed variables and substituting gives:

$\begin{array}{ccc}\hfill t\prime & =\hfill & \frac{t-vx\text{/}{c}^{2}}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}\hfill \\ \hfill x\prime & =\hfill & \frac{x-vt}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}\hfill \\ \hfill y\prime & =\hfill & y\hfill \\ \hfill z\prime & =\hfill & z.\hfill \end{array}$

## Using the lorentz transformation for time

Spacecraft $S\prime$ is on its way to Alpha Centauri when Spacecraft S passes it at relative speed c /2. The captain of $S\prime$ sends a radio signal that lasts 1.2 s according to that ship’s clock. Use the Lorentz transformation to find the time interval of the signal measured by the communications officer of spaceship S .

## Solution

1. Identify the known: $\text{Δ}t\prime ={t}_{2}\prime -{t}_{1}\prime =1.2\phantom{\rule{0.2em}{0ex}}\text{s};\phantom{\rule{0.2em}{0ex}}\text{Δ}x\prime =x{\prime }_{2}-x{\prime }_{1}=0.$
2. Identify the unknown: $\text{Δ}t={t}_{2}-{t}_{1}.$
3. Express the answer as an equation. The time signal starts as $\left(x\prime ,{t}_{1}\prime \right)$ and stops at $\left(x\prime ,{t}_{2}\prime \right).$ Note that the $x\prime$ coordinate of both events is the same because the clock is at rest in $S\prime .$ Write the first Lorentz transformation equation in terms of $\text{Δ}t={t}_{2}-{t}_{1},$ $\text{Δ}x={x}_{2}-{x}_{1},$ and similarly for the primed coordinates, as:
$\text{Δ}t=\frac{\text{Δ}t\prime +v\text{Δ}x\prime \text{/}{c}^{2}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}.$

Because the position of the clock in $S\prime$ is fixed, $\text{Δ}x\prime =0,$ and the time interval $\text{Δ}t$ becomes:
$\text{Δ}t=\frac{\text{Δ}t\prime }{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}.$
4. Do the calculation.
With $\text{Δ}t\prime =1.2\phantom{\rule{0.2em}{0ex}}\text{s}$ this gives:
$\text{Δ}t=\frac{1.2\phantom{\rule{0.2em}{0ex}}\text{s}}{\sqrt{1-{\left(\frac{1}{2}\right)}^{2}}}=1.6\phantom{\rule{0.2em}{0ex}}\text{s.}$

Note that the Lorentz transformation reproduces the time dilation equation.

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