The Lorentz transformation

Consider two Cartesian frames $S(x,y,z,t)$ and $S'(x',y',z',t')$ in the standard configuration, in which $S'$ moves in the $x$-direction of $S$ with uniform velocity $v$, and the corresponding axes of $S$ and $S'$ remain parallel throughout the motion, having coincided at $t=t'=0$. It is assumed that the same units of distance and time are adopted in both frames. Suppose that an event (e.g., the flashing of a light-bulb, or the collision of two point particles) has coordinates ($x$, $y$, $z$, $t$) relative to $S$, and ($x'$, $y'$, $z'$, $t'$) relative to $S'$. The ``common sense'' relationship between these two sets of coordinates is given by the Galilean transformation:

$$x' = x - v t,$$ (1324)

$$y' = y,$$ (1325)

$$z' = z,$$ (1326)

$$t' = t.$$ (1327)

This transformation is tried and tested, and provides a very accurate description of our everyday experience. Nevertheless, it must be wrong! Consider a light wave which propagates along the $x$-axis in $S$ with velocity $c$. According to the Galilean transformation, the apparent speed of propagation in $S'$ is $c-v$, which violates the relativity principle. Can we construct a new transformation which makes the velocity of light invariant between different inertial frames, in accordance with the relativity principle, but reduces to the Galilean transformation at low velocities, in accordance with our everyday experience?

Consider an event $P$, and a neighbouring event $Q$, whose coordinates differ by $dx$, $dy$, $dz$, $dt$ in $S$, and by $dx'$, $dy'$, $dz'$, $dt'$ in $S'$. Suppose that at the event $P$ a flash of light is emitted, and that $Q$ is an event in which some particle in space is illuminated by the flash. In accordance with the laws of light-propagation, and the invariance of the velocity of light between different inertial frames, an observer in $S$ will find that

$$dx^2 + dy^2 + dz^2 -c^2 dt^2 = 0$$ (1328)

for $dt>0$, and an observer in $S'$ will find that

$$dx'^2 + dy'^2 + dz'^2 -c^2 dt'^2 = 0$$ (1329)

for $dt'>0$. Any event near $P$ whose coordinates satisfy either (1328) or (1329) is illuminated by the flash from $P$, and, therefore, its coordinates must satisfy both (1328) and (1329). Now, no matter what form the transformation between coordinates in the two inertial frames takes, the transformation between differentials at any fixed event $P$ is linear and homogeneous. In other words, if

$$x' = F(x, y, z, t),$$ (1330)

where $F$ is a general function, then

$$dx' = \frac{\partial F}{\partial x} dx + \frac{\partial F... ...artial F}{\partial z} dz + \frac{\partial F}{\partial t} dt.$$ (1331)

It follows that

$$dx'^2 + dy'^2 + dz'^2 -c^2 dt'^2 = a dx^2 + b dy^2 + c dz^2 + d dt^2 + g dx dt + h dy dt$$

$$+ k dz dt + l dy dz + m dx dz + n dx dy,$$ (1332)

where $a$, $b$, $c$, etc. are functions of $x$, $y$, $z$, and $t$. We know that the right-hand side of the above expression vanishes for all real values of the differentials which satisfy Eq. (1328). It follows that the right-hand side is a multiple of the quadratic in Eq. (1328): i.e.,

$$dx'^2 + dy'^2 + dz'^2 -c^2 dt'^2 = K (dx^2 + dy^2 + dz^2 -c^2 dt^2),$$ (1333)

where $K$ is a function of $x$, $y$, $z$, and $t$. [We can prove this by substituting into Eq. (1332) the following obvious zeros of the quadratic in Eq. (1328): $(\pm 1, 0, 0, 1)$, $(0,\pm 1, 0, 1)$, $(0, 0, \pm 1, 1)$, $(0, 1/\sqrt{2}, 1/\sqrt{2}, 1)$, $(1/\sqrt{2}, 0, 1/\sqrt{2}, 1)$, $(1/\sqrt{2}, 1/\sqrt{2}, 0, 1)$: and solving the resulting conditions on the coefficients.] Note that $K$ at $P$ is also independent of the choice of standard coordinates in $S$ and $S'$. Since the frames are Euclidian, the values of $dx^2 + dy^2 +dz^2$ and $dx'^2+dy'^2 + dz'^2$ relevant to $P$ and $Q$ are independent of the choice of axes. Furthermore, the values of $dt^2$ and $dt'^2$ are independent of the choice of the origins of time. Thus, without affecting the value of $K$ at $P$, we can choose coordinates such that $P=(0,0,0,0)$ in both $S$ and $S'$. Since the orientations of the axes in $S$ and $S'$ are, at present, arbitrary, and since inertial frames are isotropic, the relation of $S$ and $S'$ relative to each other, to the event $P$, and to the locus of possible events $Q$, is now completely symmetric. Thus, we can write

$$dx^2 + dy^2 + dz^2 -c^2 dt^2 = K (dx'^2 + dy'^2 + dz'^2 -c^2 dt'^2),$$ (1334)

in addition to Eq. (1333). It follows that $K=\pm 1$. $K= -1$ can be dismissed immediately, since the intervals $dx^2 + dy^2 + dz^2 -c^2 dt^2$ and $dx'^2 + dy'^2 + dz'^2 -c^2 dt'^2$ must coincide exactly when there is no motion of $S'$ relative to $S$. Thus,

$$dx'^2 + dy'^2 + dz'^2 -c^2 dt'^2 =dx^2 + dy^2 + dz^2 -c^2 dt^2.$$ (1335)

Equation (1335) implies that the transformation equations between primed and unprimed coordinates must be linear. The proof of this statement is postponed until Sect. 10.7.

The linearity of the transformation allows the coordinate axes in the two frames to be orientated so as to give the standard configuration mentioned earlier. Consider a fixed plane in $S$ with the equation $l x + m y + n z + p=0$. In $S'$, this becomes (say) $l (a_1 x' + b_1 y' +c_1 z' +d_1 t' + e_1) + m (a_2 x' +\cdots) +n (a_3 x'+\cdots) + p =0$, which represents a moving plane unless $l d_1 +m d_2 + n d_3=0$. That is, unless the normal vector to the plane in $S$, $(l, m, n)$, is perpendicular to the vector $(d_1, d_2, d_3)$. All such planes intersect in lines which are fixed in both $S$ and $S'$, and which are parallel to the vector $(d_1, d_2, d_3)$ in $S$. These lines must correspond to the direction of relative motion of the frames. By symmetry, two such frames which are orthogonal in $S$ must also be orthogonal in $S'$. This allows the choice of two common coordinate planes.

Under a linear transformation, the finite coordinate differences satisfy the same transformation equations as the differentials. It follows from Eq. (1335), assuming that the events $(0,0,0,0)$ coincide in both frames, that for any event with coordinates $(x, y, z, t)$ in $S$ and $(x', y', z', t')$ in $S'$, the following relation holds:

$$x^2 + y^2 + z^2 -c^2 t^2 = x'^2 + y'^2 + z'^2 - c^2 t'^2.$$ (1336)

By hypothesis, the coordinate planes $y=0$ and $y'=0$ coincide permanently. Thus, $y=0$ must imply $y'=0$, which suggests that

$$y' = A y,$$ (1337)

where $A$ is a constant. We can reverse the directions of the $x$- and $z$-axes in $S$ and $S'$, which has the effect of interchanging the roles of these frames. This procedure does not affect Eq. (1337), but by symmetry we also have

$$y = A y'.$$ (1338)

It is clear that $A=\pm 1$. The negative sign can again be dismissed, since $y=y'$ when there is no motion between $S$ and $S'$. The argument for $z$ is similar. Thus, we have

$$y' = y,$$ (1339)

$$z' = z,$$ (1340)

as in the Galilean transformation.

Equations (1336), (1339) and (1340) yield

$$x^2 - c^2 t^2 = x'^2 - c^2 t'^2.$$ (1341)

Since, $x'=0$ must imply $x=v t$, we can write

$$x' = B (x-v t),$$ (1342)

where $B$ is a constant (possibly depending on $v$). It follows from the previous two equations that

$$t' = C x + D t,$$ (1343)

where $C$ and $D$ are constants (possibly depending on $v$). Substituting Eqs. (1342) and (1343) into Eq. (1341), and comparing the coefficients of $x^2$, $x t$, and $t^2$, we obtain

$$B=D = \frac{1}{\pm (1-v^2/c^2)^{1/2}},$$ (1344)

$$C = \frac{- v/c^2}{\pm(1-v^2/c^2)^{1/2}}.$$ (1345)

We must choose the positive sign in order to ensure that $x'\rightarrow x$ as $v/c\rightarrow 0$. Thus, collecting our results, the transformation between coordinates in $S$ and $S'$ is given by

$$x' = \frac{x-v t}{(1-v^2/c^2)^{1/2}},$$ (1346)

$$y' = y,$$ (1347)

$$z' = z,$$ (1348)

$$t' = \frac{t-v x/c^2}{(1-v^2/c^2)^{1/2}}.$$ (1349)

This is the famous Lorentz transformation. It ensures that the velocity of light is invariant between different inertial frames, and also reduces to the more familiar Galilean transform in the limit $v \ll c$. We can solve Eqs. (1346)-(1349) for $x$, $y$, $z$, and $t$, to obtain the inverse Lorentz transformation:

$$x = \frac{x'+v t'}{(1-v^2/c^2)^{1/2}},$$ (1350)

$$y = y',$$ (1351)

$$z = z',$$ (1352)

$$t = \frac{t'+v x'/c^2}{(1-v^2/c^2)^{1/2}}.$$ (1353)

Not surprizingly, the inverse transformation is equivalent to a Lorentz transformation in which the velocity of the moving frame is $-v$ along the $x$-axis, instead of $+v$.