New Reply

Wigner's little group, massless case

 
Share Thread Thread Tools
Jul5-12, 05:02 AM   #1
 

Wigner's little group, massless case


Hello,

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
[tex]
S^\mu~_\nu(\alpha,\beta) =

\begin{bmatrix}
1 & 0 & -\alpha & \alpha\\
0 & 1 & -\beta & \beta\\
\alpha & \beta & 1-\zeta & \zeta\\
\alpha & \beta & -\zeta & 1+\zeta\\
\end{bmatrix},
\\
[/tex]
Then he says that R and S separately are Abelian subgroups of the little group
[tex]R(\theta_1)R(\theta_2)=R(\theta_1+\theta_2)\\
S(\alpha_1,\beta_1)S(\alpha_2,\beta_2)=S(\alpha_1+\alpha_2,\beta_1+\bet a_2)[/tex]
and S is also an invariant subgroup
[tex]R(\theta)S(\alpha,\beta)R^{-1}(\theta)=S(\alpha\cos\theta+\beta\sin\theta,-\alpha\sin\theta+\beta\cos\theta)[/tex]
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).
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> King Richard III found in 'untidy lozenge-shaped grave'
>> Google Drive sports new view and scan enhancements
>> Researcher admits mistakes in stem cell study
Jul5-12, 09:54 AM   #2
 
Blog Entries: 27
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor
Hello Mesmerized!
Quote by Mesmerized View Post
From these relations he concludes that the S subgroup represents the translations which seems logical.
no, he says (bottom of page 70) …
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 …

so the line in a particular direction represents all possible frequencies of light in that direction …

and a lorentz transformation of rapidity α in say the x-direction increases the frequency of every ray in the x-direction by eα (in actual space), corresponding simply to a translation by α along the x-direction in the "graph" …

then the little group (leaving the z-direction invariant) consists of all redshifts in the x-y plane, and rotations in the x-y plane
Jul6-12, 04:25 AM   #3
 
thank you for such a detailed answer.
New Reply

Tags
little group, weinberg
Thread Tools


Similar Threads for: Wigner's little group, massless case
Thread Forum Replies
are photons massless or practically massless High Energy, Nuclear, Particle Physics 10
Massless representations of the Poincare group Quantum Physics 2
Group Theory - specific non-abelian case Calculus & Beyond Homework 4
Isometry sub-group of the gauge group & the center of structural group Differential Geometry 0
Learn the general case first, or the special case first? General Math 6