Bondi, Lorentz, k-calc, k-calculus, Minkowskian geometry, relativity,anti-relativity
This is a method of deriving the Lorentz transform equations was invented by Hermann Bondi. It uses space time diagrams like this
My diagram (fig 2) represents the space time diagram drawn from the point of view of observer B and shows A observing two events separated by a time T as measured by observer B.
The description of the k-calculus I am using is by Ray D'Inverno who says in his book that he spent his years as a research student at King's College London in the era of Bondi. Ray D'Inverno does not include the diagram of fig 2, but makes an addition to his diagram giving fig 3. He then simplifies this by moving the top pair of parallel lines down until they meet the other pair on B's world line and then moves the world lines of A and B together so that they now meet at the first event. This reduces the length of the lower of each pair of parallel lines to zero and they disappear giving the diagram in fig 4.
The mathematical fiddle lies in the geometry of fig 3. If we look at fig 2, we find that if B sees the time interval as T, then A will see the same time interval as kT. If B measured the time interval as kT then A would observe this to be k2T. In his diagram (fig 4) D'Inverno inserts this result, but the geometry of his diagram no longer supports the result. The result which is written in fig 4 fits the geometry of similar triangles.
The problem comes from the fact that it is impossible to draw a space time diagram which gives equal status to all observers. We might draw a diagram in which A and B have equal status, but it would always be possible to draw a vertical line and call it observer C. Whoever is represented by the vertical line is the privileged observer because their world line is draw at 45° to the lines representing light rays. Figs 2 and 4 both put A in the position of privilege contrary to the fundermental assertion that all observers have equal status. To give both observers equal status, we must assert that figs 1 and 2 are equally valid. but how do we then combine them? It is impossible to do so in a way which gives equal status and allows us to derive the required result.
When we apply the assumptions of special relativity to fig 5 we require that BAC = OBA=45° and that ABC=90°. This can only be so if OC is parallel to OB which makes a nonsense of the diagram.
The k-calculus is a fiddle.