This Quantum World/Appendix/Relativity/Lorentz transformations

Lorentz transformations (general form)Edit

We want to express the coordinates   and   of an inertial frame   in terms of the coordinates   and   of another inertial frame   We will assume that the two frames meet the following conditions:

  1. their spacetime coordinate origins coincide (  mark the same spacetime location as  ),
  2. their space axes are parallel, and
  3.   moves with a constant velocity   relative to  

What we know at this point is that whatever moves with a constant velocity in   will do so in   It follows that the transformation   maps straight lines in   onto straight lines in   Coordinate lines of   in particular, will be mapped onto straight lines in   This tells us that the dashed coordinates are linear combinations of the undashed ones,


We also know that the transformation from   to   can only depend on   so       and   are functions of   Our task is to find these functions. The real-valued functions   and   actually can depend only on   so   and   A vector function depending only on   must be parallel (or antiparallel) to   and its magnitude must be a function of   We can therefore write     and   (It will become clear in a moment why the factor   is included in the definition of  ) So,


Let's set   equal to   This implies that   As we are looking at the trajectory of an object at rest in     must be constant. Hence,


Let's write down the inverse transformation. Since   moves with velocity   relative to   it is


To make life easier for us, we now chose the space axes so that   Then the above two (mutually inverse) transformations simplify to


Plugging the first transformation into the second, we obtain


The first of these equations tells us that


The second tells us that


Combining   with   (and taking into account that  ), we obtain  

Using   to eliminate   we obtain   and  

Since the first of the last two equations implies that   we gather from the second that  

  tells us that     must, in fact, be equal to 1, since we have assumed that the space axes of the two frames a parallel (rather than antiparallel).

With   and     yields   Upon solving   for   we are left with expressions for   and   depending solely on  :


Quite an improvement!

To find the remaining function   we consider a third inertial frame   which moves with velocity   relative to   Combining the transformation from   to  


with the transformation from   to  


we obtain the transformation from   to  :


The direct transformation from   to   must have the same form as the transformations from   to   and from   to  , namely


where   is the speed of   relative to   Comparison of the coefficients marked with stars yields two expressions for   which of course must be equal:


It follows that   and this tells us that


is a universal constant. Solving the first equality for   we obtain


This allows us to cast the transformation


into the form


Trumpets, please! We have managed to reduce five unknown functions to a single constant.