Lorentz Transformation

Derivation of Lorentz transformation equations using orthogonal transformations

Let us consider two uniformly moving frames of reference where origins coincides at t=0. let the source of light is fixed at unprimed frame of reference and emits a pulse of light. The observer fixed in the unprimed frame of reference will observe a spherical wave-front propagating with the speed of light c , whose equation can be written as

x^2+y^2+z^2 = c^2 t^2                                (1)

Since speed of light is invariant according to special theory of relativity so the observer fixed in the primed frame of reference would also observe the light pulse propagating as the spherical wave-front from his own origin whose equation is

x'^2+y'^2+'z^2 = c^2 t'^2                                (2)

We now need appropriate transformation equations because here we note that time is no more scalar invariant. The transformations should be such that the light pulse viewed from both the frames of reference must be in form that wave pulse is propagating as simultaneously concentric spheres in both the systems. This reveals that

x^2+y^2+z^2 - c^2 t^2 = x'^2+y'^2+'z^2 = c^2 t'^2 (3)

or in terms of x_1,x_2,x_3 and x_4 where x_4=ict we have

x_1^2 + x_2^2 + x_3^2 + x_4^2 = x'_1^2 + x'_2^2 + x'_3^2 + x'_4^2


\sum\limits_{\mu = 1}^4 {x_\mu ^2} = \sum\limits_{\mu = 1}^4 {x'_\mu ^2} (4)

Thus square of radius vector is invariant under a transformation of co-ordinates in four dimensional space. We can thus say that equation (4) represents the orthogonal transformation of vectors in four dimensional Minkowiski space. Thus Lorentz transformations can be represented as
x{'_\mu } = \sum\limits_{\mu  = 1}^4 {{a_{\nu \mu }}{x_\nu }}


is a linear transformation matrix. Now here frame S' is moving w.r.t. S with velocity v in the positive x_3 direction and all the axis of S' are parallel to that of S. We will now obtain the matrix element a_{\mu\nu} of transformation between x and x' for pure Lorentz transformation.
x{'_\mu } = \sum\limits_{\mu  = 1}^4 {{a_{\nu \mu }}{x_\nu }}                            (5)
matrix equation2




x'_{4}=a_{41}x_{1}+a_{42}x_{2}+a_{43}x_{3}+a_{44}x_{4}                                (equations 6)

As the primed system is taken to be moving along x_3 axis , then x_1 and x_2, being in a direction perpendicular to that system , motion will remain unaffected by the transformation i.e.,



as the components x_3 and x_4 are the components to undergo transformation so components x_1 and x_2 should not appear in the transformation matrix of x'_3 and x'_4. Thus above equation 6 becomes




x'_{4}=a_{43}x_{3}+a_{44}x_{4}                                (equations 7)

Therefore the transformation matrix is

\begin{pmatrix} 1 &0 &0 &0 \\ 0& 1 & 0 &0 \\ 0& 0 &a_{33} &a_{34} \\ 0 & 0 & a_{43} &a_{44} \end{pmatrix}                                                       (8)

Further since these matrix elements must obey the orthogonality condition that is,

\sum_{\nu}a_{\mu \nu}a_{\lambda\nu}=\delta_{\mu\nu}                         (9)

it is =1 if \mu=\lambda and it is =0 if \mu\neq \lambda

putting \mu=\lambda=3 , and first \nu=3 and second \nu=4 we get

a_{33}^{2}+a_{34}^{2}=1                            (10)

Also with \mu=\lambda=4,  \nu=3 and \nu=4 we get

a_{43}^{2}+a_{44}^{2}=1                                (11)

a_{33}a_{43}+a_{34}a_{44}=0                                          (12)

Thus the orthogonality condition furnishes the above three equations connecting the four matrix elements.

The four unknown elements can be determined uniquely when the fourth relation between them is provided. We know that the origin of primed frame is (x'_{3}=0) is moving uniformly along x_3 axis; thus after time t it’s x_3 co-ordinate will be vt, i.e.,

x_3=vt=v\frac{x_4}{ic}=-\frac{iv}{c}x_4=-i\beta x_4

which, with matrix relation for x'_3 gives

x'_{3} = a_{33} x_{3} +a_{34} x_{4}

x'_{3}=(a_{34}-i\beta a_{33})x_{4}=0

a_{34}=i\beta a_{33}

which when substituted with equation 10 gives


so that,


On solving equation 11 and equation 12 it is quite easy to show that

a_{44}=\frac{1}{\sqrt{1-\beta^2}} and a_{43}=-\frac{i\beta}{\sqrt{1-\beta^2}}

We can now write the Lorentz transformation as

\begin{pmatrix} x'_1\\ x'_2\\ x'_3\\ x'_4 \end{pmatrix}=\begin{pmatrix} 1 &0 &0 &0 \\ 0 & 1& 0 &0 \\ 0 & 0 & \frac{1}{\sqrt{1-\beta^2}} &\frac{i\beta}{\sqrt{1-\beta^2}} \\ 0& 0 & -\frac{i\beta}{\sqrt{1-\beta^2}} & \frac{1}{\sqrt{1-\beta^2}} \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{pmatrix}


\begin{pmatrix} x'_1\\ x'_2\\ x'_3\\ ict' \end{pmatrix}=\begin{pmatrix} 1 &0 &0 &0 \\ 0 & 1& 0 &0 \\ 0 & 0 & \frac{1}{\sqrt{1-\beta^2}} &\frac{i\beta}{\sqrt{1-\beta^2}} \\ 0& 0 & -\frac{i\beta}{\sqrt{1-\beta^2}} & \frac{1}{\sqrt{1-\beta^2}} \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ x_3\\ ict \end{pmatrix}


x'=x\\  y'=y\\  z'=\frac{z-vt}{\sqrt{1-\beta^2}}\\  ict'=z\left ( -\frac{i\beta}{\sqrt{1-\beta^2}} \right ) +\frac{ict}{\sqrt{1-\beta^2}}                     (13)


t'=\frac{t-\left ( \frac{v}{c^2} \right )z}{\sqrt{1-\beta^2}}

equations 13 are known as Lorentz transformation equations. Lorentz transformation to real co-ordinate is possible when \beta<1 indicating that one can not have relative velocity greater than c, the sped of light.