# The Lorentz Transformations

*Michael Fowler, UVa Physics. 2/26/08*

### Problems with the Galilean Transformations

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,

_{}

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* (*x*, *y*, *z*, *t*)
in frame* S*, for example, a small bomb explosion, as functions of the
coordinates (*x ,y, z, t*) 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′ * =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′ * at which the bomb exploded in her frame, *O′* could determine the distance of the point (*x′*, *y′*, *z′*)
from the 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′ *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′ *-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 _{} 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′ 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).

### Deriving the Lorentz Transformations

Let us now suppose that *O′* and her crew observe a small bomb to
explode in *S′* at (*x′, 0, 0, t′*). In this section, we shall find the space
coordinates and time (*x*, *y*, *z*, *t*) 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′ ,z′ * zero because they transform
trivially—there is no Lorentz contraction perpendicular to the motion, so *y* = *y′ * and *z* = *z′ *.

First, we consider at what *time* the bomb explodes as measured by *O*. *O′’*s crew found the
bomb to explode at time *t′ * as measured by a local clock, that is,
one located at the site of the explosion, *x′ .* Now, as observed by *O* from frame *S*, *O′’*s clock at *x′ * is *not* synchronized with *O′’*s
clock at the origin *S′.* When the bomb explodes and the clock at *x′ *reads *t′ ,* *O* will see *O′’*s origin clock to read _{} 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 _{} 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 _{} he will find that
the true time* t* in his frame is equal to this appropriately scaled to
allow for time dilation, that is,

_{}

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′ *meters
beyond *O′*, as measured by *O′*, but of course *O* will see that distance *x′ *as
contracted to _{} since it’s
in a moving frame.

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

_{}

This can be written as an equation for *x* in terms of *x′ , t′ * by substituting for* t* using the first Lorentz transformation above, to give

_{}

Therefore, we have found the Lorentz transformations expressing the
coordinates (*x, y, z, t*) of an event in frame *S* in terms of the
coordinates (*x′, y′, z′, t′*) of the same event
in frame *S′*:

_{}

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 (*x*,*y*,*z*,*t*) to (*x′*, *y′*, *z′*, *t′ *)) has
exactly the same form as that given above with *v* replaced by -*v*.

### Spheres of Light

Consider now the following scenario: suppose that as* O′* passes *O* (the instant both of them agree is at time *t′ * = *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′*,
then, *O′* (or, to be precise, her local observers out there in the
frame) will see a shell of light of radius *ct′*, that is to say,
they will see the light to have reached all points (*x′*, *y′*, *z′ *) on the surface

_{}

*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′ *can be written

_{}

and can be thought of as *a surface in the four dimensional* (*x′,
y′, z′, t′*) *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 (*x*, *y*, *z*, *t*) space, all we have to do is to change from one set of
variables to the other using the Lorentz transformations:

_{}

On putting these values of (*x′, y′, z′, t′*)
into_{} we find that the corresponding surface of events in (*x*, *y*, *z*, *t*) space is:

_{}

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′ *she sees a sphere of radius *r′*, in particular she sees the
light to have reached the spots +*r′ *and –*r′* on
the *x′ * axis. But from *O’*s point
of view the expanding light sphere does *not* reach the point +*r′* at the same time it reaches –*r′*! (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′ around O′ at time t′ corresponds in S to a continuum of
different events happening at different times*.

### Lorentz Invariants

We found above that for an event (*x′, y′, z′, t′*)
for which _{} the coordinates
of the event (*x*, *y*, *z*, *t*) as measured in the other
frame *S* satisfy _{} The
quantity _{} 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 (*x*,*y*) has coordinates (*x′*, *y′ *)
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′ *^{2} + *y′ *^{2}, so for
the transformation from coordinates (*x*, *y*) to (*x′*, *y′ *), *x*^{2} + *y*^{2} is an invariant. Similarly, if a point *P*_{1} has coordinates (*x*_{1}, *y*_{1}) and (*x*_{1}′, *y*_{1}′) and another point *P*_{2} has
coordinates (*x*_{2}, *y*_{2}) and (*x*_{2}′, *y*_{2}′) then clearly the two points are the same distance
apart as measured with respect to the two sets of axes, so

_{}

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

_{}

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

This *s* is sometimes called the “space-time
interval”. *The big
difference from the two-dimensional rotation case is that s ^{2} can be
positive or negative*. For

*s*negative just taking

^{2}*s*as its square root the “distance” itself would be imaginary—so we need to think this through carefully. As we shall see below, it’s clear enough what to do in any particular case—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*_{1}′
= *t*_{2}′. They
will not be simultaneous in frame *S*, but they *will* satisfy

_{}

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—now 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, _{} Then in
frame *S*:

_{}

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

### The Light Cone

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

_{}

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

_{}

Now think about this as a surface in the three-dimensional space (*x*, *y*, *t*). 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* (*x*, *y*, *z*, *t*) from the origin is spacelike if _{}and timelike if _{} It is
said to be *lightlike* if_{} 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 in 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.