This review covers some of the basics
of special relativity and derives the Lorentz
Transform needed to relate events observed in inertial frames
moving at relative speeds $v$ approaching the speed of light $c$.

We
begin with the assumption that the
same physical laws are valid in all inertial frames. Consider two
inertial coordinate frames $S$ and the "primed" frame $S'$ moving with
a constant velocity $v$ along their common $x$-axis. The planes $y = 0$
and $z =
0$ always coincide with the planes $y' = 0$ and $z' = 0$, and we can
set the zero points on the clocks so that $t = t' = 0$ at the instant
when $x = x'$.

Thus $(x,y,z,t)$ might be coordinates
in the rest frame of the
Galaxy and $(x', y', z', t')$ an inertial frame moving with the
instantaneous velocity of a (non-inertial) cosmic-ray electron. An
event is something that occurs at one point $(x', y', z', t')$
in
space and time, such as firing the flash on a camera at rest in the
$S'$ frame.

According to the "intuitively obvious"
Galilean relativity,
the event coordinates in the unprimed frame $S$ are:

$$ x = x' +vt'
\qquad y = y' \qquad z = z' \qquad t = t' $$

Unfortunately, our
intuition is shaped by experiences with objects moving at speeds much
less than the speed of light in a vacuum, and it does not correspond to
reality at speeds approaching $c$. We "know" by experience that time is
absolute; that is, $t = t'$ in all inertial frames. Galilean relativity
implies that parallel velocities simply add. The speed of a photon
emitted by the flash in the $+x'$ direction will be seen as

$$c_{\rm x}
= {d x \over d t} = {d (x' + vt') \over d t'} = { dx' \over d t'} + v =
c_{\rm x}' + v$$

by an observer in the frame $S$. Thus Galilean
relativity is inconsistent with both observation and Maxwell's
equations, which correctly predict that the speed of light in a vacuum
is the same for all observers in all inertial frames ($c_{\rm x} =
c_{\rm x}'$), regardless of their relative velocities $v$. We
therefore drop the assumption that time is absolute.

The Lorentz transform is the only
coordinate transform consistent
with
both relativity (equivalence of inertial frames) and the existance of
some still-unspecified invariant speed (for example, the speed of
light, or even $\infty$). Note that the latter assumption is much
weaker than assuming $t = t'$ because if the invariant speed is
infinite, then the assumption has no consequences and the Lorentz
transform reduces to the Galilean transform. The Lorentz transform
can be derived with two reasonable assumptions (see Rindler's *
Essential Relativity*, p. 39
or Rindler's * Introduction to Special Relativity*, p. 12). We
will assume that space is both homogeneous
and isotropic:
that is, the
laws
of physics do not change from one place to another, or with orientation
of the coordinate frames.

Isotropy implies that the observers
moving with the two frames agree
on their relative speed $\vert v \vert$ since a 180 degree coordinate
rotation
exchanges the roles of the two frames. That rotation should have no
effect if space is isotropic. The assumption of homogeneity implies
that any transformation from one inertial frame to another is linear;
e.g., $y' = Ay + B$ where $A$ and $B$ are constants. Any nonlinear
terms (e.g., $y' = Ay + B + Cy^2$) would cause the transform itself to
vary under coordinate
translations and $C$ must therefore be zero in a homogeneous space.

Since
we can choose coordinate frames so $y = 0$ when $y' = 0$, linearity
requires that $y' = Ay$, where $A$ is some constant scale factor.
Reversing the
coordinate directions by 180 degrees reverses the roles of $S$ and
$S'$, so $y = A'y'$. Only $A = A' = \pm1$ is consistent with isotropy.
We can reject the negative solution $A = -1$ because it implies $y =
-y'$
when $v = 0$, so the Lorentz transforms for the $y$ and $z$ coordinates
are

$$y = y' {\rm \quad~and~likewise~\quad} z = z'$$

in agreement with the Galilean transform.

We can proceed in a similar way with
the
$x$-coordinate. Linearity
requires

$$x = \gamma'(x' + vt') {\rm ~and~} x' = \gamma(x - vt)$$

where $\gamma'$ and $\gamma$ are (still unspecified) constant scale
factors.
Reversing the directions of $S$ and $S'$ gives

$$ x = \gamma'(x' - vt')
{\rm ~and~} x' = \gamma(x + vt)~.$$

Reversing the roles of the two
frames gives

$$ x = \gamma(x' - vt') {\rm ~and~} x' = \gamma'(x +
vt)~.$$

These two pairs of equations imply $\gamma = \gamma'$; that is,
the observers in $S$ and $S'$ also agree on the scale factor $\gamma$
associated with the relative velocity $v$.

Suppose that there is some
speed $c$ which is the same in all inertial frames. (We already know
from
Maxwell's equations and by experiment that $c$ is the speed of light in
a
vacuum, but for this argument, it could be any speed, even $c =
\infty$,
in which case the Lorentz transform would end up being identical to the
Galilean transform.) Then $x = ct$ implies $x' = c t'$ and

$$ct =
\gamma t'(c + v) {\rm ~and~} ct' = \gamma t (c - v)$$

The product of
these these two equations is

$$ c^2 t t' = \gamma^2 t t' (c + v) (c -
v)~.$$

We solve for $\gamma$, which is called the Lorentz
factor:

$$\gamma = \biggl( 1 - {v^2 \over c^2} \biggr)^{-1/2}$$

Again, the negative solution to this
equation can be rejected as
unphysical. The $x$-coordinate transform is

$$ x = \gamma (x' + vt')
{\rm ~~and~~} x' = \gamma(x - vt)$$

Eliminating $x$ from this pair of
equations yields

$$t = \gamma (t' + v x' / c^2)$$

The Lorentz transform of special relativity is thus:

$$ x = \gamma (x' + vt') \qquad y = y' \qquad z = z' \qquad t = \gamma (t' + \beta x' / c) $$ $$ x' = \gamma (x - vt) \qquad y' = y \qquad z' = z \qquad t' = \gamma (t - \beta x / c) $$

where $\beta \equiv v/c$. Note that
the Galilean transform is just
the limit of the Lorentz transform as $c \rightarrow \infty$.
If $(\Delta x, \Delta y, \Delta z, \Delta t)$ and $(\Delta x',
\Delta y',
\Delta z', \Delta t')$ are the coordinate differences between two
events, the differential form of the (linear) Lorentz transform is:

$$\Delta x = \gamma(\Delta x' + v
\Delta t') \qquad \Delta y =
\Delta y' \qquad \Delta z = \Delta z' \qquad \Delta t = \gamma (\Delta
t' + \beta \Delta x' / c) $$

$$\Delta x' = \gamma(\Delta x - v \Delta
t) \qquad \Delta y' = \Delta y \qquad \Delta z' = \Delta z \qquad
\Delta t' = \gamma (\Delta t - \beta \Delta x / c) $$