We can also choose units so that the speed of light is 1 leading to a simplification of the terms. There are two different ways of representing the time dimension. Here I have used c = 1 and imaginary time.
In the matrix multiplications which follow, the rules of multiplication of matrices over the complex plane are not adhered to correctly. They are defined to preserve the geometry of the Argand plane and require the complex conjugates of complex elements of the right hand matrix to be used. Mathcad does not know this rule and gives the correct results for the geometry of Minkowski space.
[L] = where =
giving = =
(Mathcad respond 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 can download the Mathcad file and attempt to 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)