There are two different ways of representing the time dimension. We can also choose units so that the speed of light is 1 leading to a simplification of the terms Here I have used c = 1 and real time.
[L] = where =
giving = =
(Mathcad responds to the use of the hyphen key by placing brackets around the expression, so x' cannot be used)
For v = the matrix transform = =
The transform for v is
Those with Mathcad 6.0 can download the Mathcad file and verify that
So we are confident that this is indeed a Lorentz transform.
If we now consider three observers in inertial reference frames S_0, S_1 and S_2 such that at some moment, their axes are coincident and that they zero their clocks at that moment. If the velocity of S_1 in S_0 is u along the x axis and the velocity of S_2 in S_1 is the vector v defined above, we can apply first a Lorentz from S_0 to S_1 and then a Transform from S_1 TO S_2.
It can be seen that the second column remains unchanged by the post multiplication. If the Lorentz group exists then we should be able to equate this to a single tranform:
But this requires us to equate the terms of the second columns
We note that these equations are independent of u and conclude that no solution is possible which will be valid for all u less than c. Therefore, we are unable to show that there is a Lorentz transform from S_0 to S_2.
However, the condition that = is satisfied.
This can be verified by doing the sums on a separate Mathcad sheet. (It has a habit of filling its memory and crashing if too much is done in one sheet)