*Michael Fowler, University of Virginia*

We have already seen that Newtonian mechanics is invariant under the *Galilean*
transformations relating two inertial frames moving with relative speed $v$ in the* **$x$** *-direction,

$\begin{array}{l}x={x}^{\prime}+v{t}^{\prime},\\ y={y}^{\prime},\\ z={z}^{\prime},\\ t={t}^{\prime}.\end{array}$

However, these transformations presuppose that *time* is a well-defined
universal concept, that is to say, it’s the same time everywhere, and all
observers can agree on what time it is. Once
we accept the basic postulate of special relativity, however, that the laws of
physics, including Maxwell’s equations, are the same in all inertial frames of
reference, and consequently the speed of light has the same value in all
inertial frames, then as we have seen, observers in different frames do *not*
agree on whether clocks some distance apart are synchronized. Furthermore, as we have discussed,
measurements of moving objects are compressed in the direction of motion by the
Lorentz-Fitzgerald contraction effect. Obviously, the above equations are too naïve! We must think more carefully about time and
distance measurement, and construct new transformation equations consistent
with special relativity.

Our aim here, then, is to find a set of equations analogous to those above
giving the coordinates of an *event* $\left(x,y,z,t\right)$ in frame* S*, for example, a small bomb
explosion, as functions of the coordinates $\left({x}^{\prime},{y}^{\prime},{z}^{\prime},{t}^{\prime}\right)$ of the same event measured in the parallel
frame *S′* which is moving at speed $v$ along the $x$ -axis of frame *S*. Observers *O* at the origin in frame *S*
and *O′* at the origin in frame *S *synchronize their clocks at $t={t}^{\prime}=0,$ at the instant they pass each other, that is,
when the two frames coincide . (Using our previous notation, *O* is Jack
and *O′* is Jill.)

To determine the time ${t}^{\prime}$ at which the bomb exploded in her frame, *O′*
could determine the distance of the point $\left({x}^{\prime},{y}^{\prime},{z}^{\prime}\right)$ from her origin, and hence how long it would
take light from the explosion to reach her at the origin. A more direct
approach (which is helpful in considering transformations between different
frames) is to imagine *O′* to have
a multitude of helpers, with an array of clocks throughout the frame, which
have all been synchronized by midpoint flashes as described in the previous
lecture. Then the event—the bomb
explosion—will be
close to a clock, and that local clock determines the time ${t}^{\prime}$ of the event, so we do not need to worry about
timing a light signal.

In frame *S′*, then, *O′* and her crew have clocks all along the ${x}^{\prime}$ -axis (as well as everywhere else) and all
synchronized:

Now consider how this string of clocks appears as viewed by *O* from
frame *S*. First, since they are all moving at speed $v,$ they will be registering time more slowly by
the usual time dilation factor $\sqrt{1-{v}^{2}/{c}^{2}}$ than *O’*s own physically identical
clocks. Second, they will not be synchronized. From the clocks on a train argument in the
last lecture, if the clocks are* **$L$*apart as measured by *O′*, successive
clocks to the right (the direction of motion) will be behind by $Lv/{c}^{2}$ as observed by *O*.

It should be mentioned that this lack of synchronization as viewed from
another frame only occurs for clocks *separated in the direction of relative
motion*. Consider two clocks some distance apart *on the **${z}^{\prime}$
axis* of *S′. * If they are synchronized in *S′* by both being started by a flash of light from
a bulb half way between them, it is clear that as viewed from *S* the
light has to go the same distance to each of the clocks, so they will still be
synchronized (although they will start later by the time dilation factor).

Let us now suppose that *O′* and her crew observe a small bomb to
explode in *S′* at $\left({x}^{\prime},0,0,{t}^{\prime}\right).$ In this section, we shall find the space
coordinates and time $\left(x,y,z,t\right)$ of this
event as observed by *O* in the frame *S*. (As above, *S′* moves relative to *S*
at speed $v$ along the $x$ -axis). In other words, we shall derive the Lorentz
transformations—which are
just the equations giving the four coordinates of an event in one inertial
frame in terms of the coordinates of the same event in another inertial frame. We take ${y}^{\prime},{z}^{\prime}$ zero because they transform trivially—there is
no Lorentz contraction perpendicular to the motion, so $y={y}^{\prime}$ and $z={z}^{\prime}.$

First, we consider at what *time* the bomb explodes as measured by *O*.
*O′’*s crew found the bomb to
explode at time ${t}^{\prime}$ as measured by a local clock, that is, one
located at the site of the explosion, ${x}^{\prime}.$ Now, as observed by *O* from frame *S*,
*O′’*s clock at ${x}^{\prime}$ is *not* synchronized with *O′’*s clock
at her origin*.* When the bomb
explodes and the clock at ${x}^{\prime}$ reads ${t}^{\prime},$ *O* will see *O′’*s origin clock to
read ${t}^{\prime}+v{x}^{\prime}/{c}^{2}.$ What
does *O′*s own clock read at this point? Recall that *O*, *O′*
synchronized their origin clocks at the moment they were together, at $t={t}^{\prime}=0.$ Subsequently, *O* will have observed *O′’*s
clock to be running slowly by the time-dilation factor. Therefore, when at the
instant of the explosion he sees *O′’*s origin clock to be reading ${t}^{\prime}+v{x}^{\prime}/{c}^{2},$ he will find that the true time* **$t$** *in his frame is equal to this
appropriately scaled to allow for time dilation, that is,

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

This is the first of the Lorentz transformations.

The second question is: *where* does *O* observe the explosion to
occur?

Since it occurs at time $t$ after *O′* passed *O*, *O′* is $vt$ meters beyond *O* at the time of the
explosion. The explosion takes place ${x}^{\prime}$ meters beyond *O′*, as measured by *O′*,
but of course *O* will see that distance ${x}^{\prime}$ as contracted to ${x}^{\prime}\sqrt{1-{v}^{2}/{c}^{2}}$ since it’s in a moving frame.

Therefore *O* observes the explosion at point $x$ given by

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

This can be written as an equation for $x$ in terms of ${x}^{\prime},{t}^{\prime}$ by substituting for* **$t$** *using the first Lorentz transformation
above, to give

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

Therefore, we have found the Lorentz transformations expressing the
coordinates $\left(x,y,z,t\right)$ of an event in frame *S* in terms of the
coordinates $\left({x}^{\prime},{y}^{\prime},{z}^{\prime},{t}^{\prime}\right)$ of the same event in frame *S′*:

$\begin{array}{l}x=\frac{{x}^{\prime}+v{t}^{\prime}}{\sqrt{1-{v}^{2}/{c}^{2}}},\\ y={y}^{\prime},\\ z={z}^{\prime},\\ t=\frac{{t}^{\prime}+v{x}^{\prime}/{c}^{2}}{\sqrt{1-{v}^{2}/{c}^{2}}}.\end{array}$

Notice that nothing in the above derivation depends on the $x$ -velocity $v$ of *S′* relative to *S* being
positive. Therefore, the inverse transformation—from $\left(x,y,z,t\right)$ to $\left({x}^{\prime},{y}^{\prime},{z}^{\prime},{t}^{\prime}\right)$ --has exactly the same form as that given
above with $v$ replaced by $-v.$

Consider now the following scenario: suppose that as* O′* passes *O*
(the instant both of them agree is at time ${t}^{\prime}=t=0$ ) *O′* flashes a bright light, which she
observes to create an expanding spherical shell of light, centered on herself
(imagine it’ s a slightly foggy day, so she can see how the ripple of light
travels outwards). At time ${t}^{\prime},$ then, *O′* (or, to be precise, her local
observers out there in the frame) will see a shell of light of radius $c{t}^{\prime},$ that is to say, they will see the light to
have reached all points $\left({x}^{\prime},{y}^{\prime},{z}^{\prime}\right)$ on the surface

${{x}^{\prime}}^{2}+{{y}^{\prime}}^{2}+{{z}^{\prime}}^{2}={c}^{2}{{t}^{\prime}}^{2}.$

*Question*: how do *O* and his observers stationed throughout the
frame *S* see this light as rippling outwards?

To answer this question, notice that the above equation for where the light
is in frame *S′* at a particular time ${t}^{\prime}$ can be written

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

and can be thought of as *a surface in the four dimensional* $\left({x}^{\prime},{y}^{\prime},{z}^{\prime},{t}^{\prime}\right)$ *space*, the totality of all the “events”
of the light reaching any particular point. Now, to find the corresponding
surface of events in the four dimensional $\left(x,y,z,t\right)$ space, all we have to do is to change from one
set of variables to the other using the Lorentz transformations:

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

On putting these values of $\left({x}^{\prime},{y}^{\prime},{z}^{\prime},{t}^{\prime}\right)$ into ${{x}^{\prime}}^{2}+{{y}^{\prime}}^{2}+{{z}^{\prime}}^{2}-{c}^{2}{{t}^{\prime}}^{2}=0,$ we find that the corresponding surface of events in $\left(x,y,z,t\right)$ space is:

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

This means that at time $t,$ *O* and his observers in frame *S*
will say the light has reached a spherical surface centered on *O*.

How can *O′* and *O*, as they move further apart, possibly both be
right in maintaining that at any given instant the outward moving light pulse
has a spherical shape, each saying it is centered on herself or himself?

Imagine the light shell as *O′ *sees it—at the
instant ${t}^{\prime}$ she sees a sphere of radius ${r}^{\prime},$ in particular she sees the light to have
reached the spots $+{r}^{\prime}$ and $-{r}^{\prime}$ on the ${x}^{\prime}$ -axis. But from *O′’*s point of view the
expanding light sphere does *not* reach the point $+{r}^{\prime}$ at the same time it reaches $-{r}^{\prime}$!
(This is just the old story of
synchronizing the two clocks at the front and back of the train one more time.)
That is why *O* does not see *O′’*s
sphere: *the arrival of the light at
the sphere of radius **${r}^{\prime}$** around O′ at time **${t}^{\prime}$** corresponds in S to a continuum of different
events happening at different times*.

We found above that for an event $\left({x}^{\prime},{y}^{\prime},{z}^{\prime},{t}^{\prime}\right)$ for which ${{x}^{\prime}}^{2}+{{y}^{\prime}}^{2}+{{z}^{\prime}}^{2}-{c}^{2}{{t}^{\prime}}^{2}=0,$ the coordinates of the event $\left(x,y,z,t\right)$ as measured in the other frame *S*
satisfy ${x}^{2}+{y}^{2}+{z}^{2}-{c}^{2}{t}^{2}=0.$ The
quantity ${x}^{2}+{y}^{2}+{z}^{2}-{c}^{2}{t}^{2}$ is said to be a *Lorentz invariant*: it doesn’t vary on going from one frame to
another.

A simple two-dimensional analogy to this invariant is given by considering
two sets of axes, *Oxy* and *Ox′ y′ * having the same origin *O*, but the axis *Ox′
*is at an angle to *Ox*, so one
set of axes is the same as the other set but rotated. The point *P* with coordinates $\left(x,y\right)$ has coordinates $\left({x}^{\prime},{y}^{\prime}\right)$ measured on the *Ox′ y′ * axes. The square of the distance of the point *P*
from the common origin *O* is ${x}^{2}+{y}^{2}$ and is also ${{x}^{\prime}}^{2}+{{y}^{\prime}}^{2},$ so for the transformation from coordinates $\left(x,y\right)$ to $\left({x}^{\prime},{y}^{\prime}\right),$ ${x}^{2}+{y}^{2}$ is an
invariant. Similarly, if a point *P*_{1}
has coordinates $\left({x}_{1},{y}_{1}\right)$ and $\left({{x}^{\prime}}_{1},{{y}^{\prime}}_{1}\right)$ and another point *P*_{2} has
coordinates $\left({x}_{2},{y}_{2}\right)$ and $\left({{x}^{\prime}}_{2},{{y}^{\prime}}_{2}\right)$ then clearly the two points are the same
distance apart as measured with respect to the two sets of axes, so

${\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}={\left({{x}^{\prime}}_{1}-{{x}^{\prime}}_{2}\right)}^{2}+{\left({{y}^{\prime}}_{1}-{{y}^{\prime}}_{2}\right)}^{2}.$

This is really obvious: the distance between two points in an ordinary plane can’t depend on the angle at which we choose to set our coordinate axes.

The Lorentz analog of this, dropping the $y,z$ coordinates, can be written

${c}^{2}{\left({t}_{1}-{t}_{2}\right)}^{2}-{\left({x}_{1}-{x}_{2}\right)}^{2}={c}^{2}{\left({{t}^{\prime}}_{1}-{{t}^{\prime}}_{2}\right)}^{2}-{\left({{x}^{\prime}}_{1}-{{x}^{\prime}}_{2}\right)}^{2}={s}^{2},$

say, where ${s}^{2}$ is some sort of measure of the “distance” between the two events $\left({x}_{1},{t}_{1}\right)$ and $\left({x}_{2},{t}_{2}\right).$

This ${s}^{2}$ is sometimes called the “space-time interval”.
*The big difference from the
two-dimensional rotation case is that, despite the notation, s ^{2} can be positive or negative*.
The
cases of spacelike and timelike separated events are best dealt with
separately, at least to begin with.

Consider first two events simultaneous in frame *S′*, so ${{t}^{\prime}}_{1}={{t}^{\prime}}_{2}.$ They
will not be simultaneous in frame *S*, but they *will* satisfy

${\left({x}_{1}-{x}_{2}\right)}^{2}-{c}^{2}{\left({t}_{1}-{t}_{2}\right)}^{2}>0.$

We say the two events are* spacelike separated*. This means that they are sufficiently removed
spatially that a light signal could not have time between them to get from one
to the other, so one of these events could not be the *cause* of the
other. The sequence of two events can be
different in different frames if the events are spacelike separated. Consider again the starting of the two clocks
at the front and back of a train as seen from the ground: the back clock starts
first. Now imagine viewing this from a
faster train overtaking the clock train—from this
view, the front clock will be the first to start. The important point is that although these
events appear to occur in a different order in a different frame, neither of
them could be the cause of the other, so *cause
and effect are not switched around*.

Consider now two events which occur at the same place in frame *S′* at
different times, $\left({{x}^{\prime}}_{1},\text{\hspace{0.17em}}{{t}^{\prime}}_{1}\right)\text{and}\left({{x}^{\prime}}_{2},\text{\hspace{0.17em}}{{t}^{\prime}}_{1}\right).$ Then in
frame *S*:

${c}^{2}{\left({t}_{1}-{t}_{2}\right)}^{2}-{\left({x}_{1}-{x}_{2}\right)}^{2}={c}^{2}{\left({{t}^{\prime}}_{1}-{{t}^{\prime}}_{2}\right)}^{2}>0.$

These events are said to be* timelike* separated. There is no frame in which they are
simultaneous. “Cause and effect” events
are timelike separated.

Let us try to visualize the surface in four-dimensional space described by the outgoing shell of light from a single flash,

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

It is helpful to think about a simpler situation, the circular ripple
spreading on the surface of calm water from a pebble falling in. Taking $c$ here to be the speed of the water waves, it is
easy to see that at time* **$t$** *after the splash the ripple is at

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

Now think about this as a surface in the three-dimensional space $\left(x,y,t\right).$ The
plane corresponding to time $t$ cuts this surface in a circle of radius $ct.$ This means the surface is a cone with its
point at the origin. (The
four-dimensional space flash-of-light surface is not so easy to visualize, but
is clearly the higher-dimensional analog: the plane surface corresponding to
time $t$ cuts it in a sphere instead of a circle.) This surface is called the *lightcone*.

We have stated above that the separation of a point *P* $\left(x,y,z,t\right)$ from the origin is *spacelike* if ${x}^{2}+{y}^{2}+{z}^{2}-{c}^{2}{t}^{2}>0,$ and *timelike
*if ${x}^{2}+{y}^{2}+{z}^{2}-{c}^{2}{t}^{2}<0.$

It is said to be *lightlike* if ${x}^{2}+{y}^{2}+{z}^{2}-{c}^{2}{t}^{2}=0.$

Points on the light cone described above are lightlike separated from the
origin. To be precise, the points
corresponding to an outgoing shell of light from a flash at the origin at $t=0$ form the *forward light cone*. Since the
equation depends only on ${t}^{2},$ there is a solution with $t$ negative, the “backward light cone”, just the
reflection of the forward light cone in the plane $t=0.$

Possible causal connections are as follows: an event at the origin (0, 0, 0, 0) could cause an event inside or on the forward light cone: so that is the “future”, as seen from the origin. Events in the backward light cone—the “past”—could cause an event at the origin. There can be no causal link between an event at the origin and an event outside the light cones, since the separation is spacelike: outside the light cones is “elsewhere” as viewed from the origin.