From the two-dimensional disk we are moving to three-dimensional space-time. We will meet Einstein-Poincare-Minkowski special relativity, though in a baby version, with x and y, but without z in space. It is not too bad, because the famous Lorentz transformations, with length contraction and time dilation happen already in two-dimensional space-time, with x and t alone. We will discover Lorentz transformations today. First in disguise, but then we will unmask them.
First we recall, from The disk and the hyperbolic model, the relation between the coordinates on the Poincare disk , and on the unit hyperboloid
We have the group SU(1,1) acting on the disk with fractional linear transformations. With and in SU(1,1)
the fractional linear action is
By the way, we know from previous notes that is in SU(1,1) if and only if
Having the new point on the disk, with coordinates we can use Eq. (1) to calculate the new space-time point coordinates This is what we will do now. We will see that even if depends on in a nonlinear way, the space-time coordinates transform linearly. We will calculate the transformation matrix and express it in terms of and We will also check that this is a matrix in the group SO(1,2).
The program above involves algebraic calculations. Doing them by hand is not a good idea. Let me recall a quote from Gottfried Leibniz, who, according to Wikipedia
He became one of the most prolific inventors in the field of mechanical calculators. While working on adding automatic multiplication and division to Pascal’s calculator, he was the first to describe a pinwheel calculator in 1685 and invented the Leibniz wheel, used in the arithmometer, the first mass-produced mechanical calculator. He also refined the binary number system, which is the foundation of virtually all digital computers.
I used Mathematica as my machine. The same calculations can be certainly done with Maple, or with free software like Reduce or Maxima. For those interested, the code that I used, and the results can be reviewed as a separate HTML document: From SU(1,1) to Lorentz.
Here I will provide only the results. It is important to notice that while the matrix has complex entries, the matrix is real. The entries of depend on real and imaginary parts of and
Here is the calculated result for :
In From SU(1,1) to Lorentz it is first verified that the matrix is of determinant 1. Then it is verified that it preserves the Minkowski space-time metric. With defined as
Since the transformation preserves the time direction. Thus is an element of the proper Lorentz group .
Remark: Of course we could have chosen with on the diagonal. We would have the group SO(2,1), and we would write the hyperboloid as It is a question of convention.
In SU(1,1) straight lines on the disk we considered three one-parameter subgroups of SU(1,1):
We can now use Eq. (7) in order to see which space-time transformations they implement. Again I calculated it obeying Leibniz and using a machine (see From SU(1,1) to Lorentz).
Here are the results of the machine work:
The third family is a simple Euclidean rotation in the plane. That is why I denoted the parameter with the letter In order to “decode” the first two one-parameter subgroups it is convenient to introduce new variable and set The group property is then lost, but the matrices become evidently those of special Lorentz transformations, transforming and , leaving unchanged, and transforming and leaving unchanged (though with a different sign of ). Taking into account the identities
In the following posts we will use the relativistic Minkowski space distance on the hyperboloid for finding the distance formula on the Poincare disk.
3 thoughts on “From SU(1,1) to the Lorentz group”
Thanks! Fixewd. Also in the database.
Still “” in the middle.
Thx. Not any more.