Lorentz Transform


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 $x = x'$.


Drawing of an event in two frames


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:

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