|Jul5-12, 05:02 AM||#1|
Wigner's little group, massless case
I'm reading Weinberg's vol.1 on Quantum Theory of Fields and stuck on the following problem. In the massless case Wigner's little group is the group of Lorentz transformations that keep the vector (0,0,1,1) invariant. (I'm going with Wigner's notations, where the vector is denoted (z,y,x,t) and the metric tesor has (1,1,1,-1) on the main diagonal). He shows, that all the matrices that keep (0,0,1,1) invariant are of the form [itex]S(\alpha,\beta)R(\theta)[/itex], where R are rotations around z axis and the S matrix has the following form
1 & 0 & -\alpha & \alpha\\
0 & 1 & -\beta & \beta\\
\alpha & \beta & 1-\zeta & \zeta\\
\alpha & \beta & -\zeta & 1+\zeta\\
Then he says that R and S separately are Abelian subgroups of the little group
and S is also an invariant subgroup
From these relations he concludes that the S subgroup represents the translations which seems logical.
My question is: translations in which plane? I assume that a general translation should act on a 4-vector as: TX=(X+a), where 'X' and 'a' are 4-vectors, 'T' is the translation operator. With the S matrix above I can see that it keeps (0,0,1,1) vector invariant, but can't see how it is a translation. I can't find a particular vector on which it will act as SX=(X+a).
|Jul5-12, 09:54 AM||#2|
The reader will recognise these multiplication rules as those of the group ISO(2), consisting of translations … and rotations … in two dimensions.… he's simply pointing out a familiar version of the same group!
if you want a "physical" representation, it would be to plot all possible rays of light onto a 3D "graph" in which direction is represented in the usual way, but distance represents the log of red-shift …
|Jul6-12, 04:25 AM||#3|
thank you for such a detailed answer.
|little group, weinberg|
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