# Reverse engineering relativity?

1. Mar 15, 2005

Staff Emeritus
Is it possible to start with the Lorentz transformations - I mean the whole Poincare group SO(1,3) - and derive that there exists an invariant speed? Is this a known derivation? I suppose we could assume a "flat" manifold with metric Diag(1,-1,-1,-1,).

2. Mar 15, 2005

### marlon

I am just wondering, wouldn't it be one heck of a coincidence that you start with "such" transformations ? I don't see the point in doing so, to be honest. However, isn't this quite easy to prove ? Just apply those transformations to two reference frames...

marlon

3. Mar 15, 2005

### marlon

Besides, velocity being absolute is an inherent property of the Lorentz transformations, right ? They are constructed starting from the universal constant c and the homogenity of space...So if you start from the L transformations, you already have that property

marlon

4. Mar 15, 2005

### dextercioby

It would be interesting SA.But it would be cheating.You cannot teach that.Students might ask you:"Why SO(3,1)"??What would you say ?What would be the reasoning behind not follwing the Einstein approach,the aximatical one?

It sorta reminds of the "deriving SchrÃ¶dinger equation" threads.

Daniel.

5. Mar 15, 2005

### robphy

Suggestion: Find the eigenvectors. (They should lie on a cone through the origin.)
It's probably easiest to leave off the translations.
To see that this is plausible, try the usual Lorentz Transformations in 2D Minkowski.

Edit: Furthermore, one should be able to show that if the eigenvalue corresponding to an eigenvector is not 1 or -1, then that eigenvector must have zero norm [with respect to the metric preserved by the transformation].

Last edited: Mar 15, 2005
6. Mar 17, 2005