I have read that little group for the massive particles is $SO(3)$ and for the massless particles is $E(2)$ in 4 dimensions. How does one take zero mass limits for the representations and show that it becomes $E(2)$, because explicitly in the representations for the massive case, I mean in the matrices, the mass $m$ won't appear in the representations, how does one then take the zero mass limit?

I know it's related to contraction of groups but is a fairly non-trivial task for any limit (like mass going to zero or radius going to infinity).

Also if I have a spin-zero state for massive particle, then will the zero mass limit still have spin zero?

  • $\begingroup$ WP. $\endgroup$ – Cosmas Zachos Feb 20 at 16:55
  • 2
    $\begingroup$ The simplest review is the one by Wigner & Kim JMP 1990 where they write down the explicit little subalgebras of the Lorentz group, 4x4 matrices, for the massive and massless case. They are, respectively, SO(3) and E2, whose geometries are the sphere and the plane: the group contraction of the sphere to the plane in the infinite radius limit is self-explanatory. What more are you after? The spineless massive particle will go to the 0-helicity light like particle (no spin for it). $\endgroup$ – Cosmas Zachos Feb 21 at 0:06
  • $\begingroup$ @CosmasZachos thanks for the links they are really helpful. As I was reading up I had a question come up, the noether’s theorem gives momentum, angular momentum and boost operators for a given lagrangian in QFT, they are also obey the same algebra as the poincare group, is there a connection when I study the zero mass limit for these operators ( which are also algebraic expresssions) and the representations, using contractions ? $\endgroup$ – user183683 Feb 21 at 1:56
  • $\begingroup$ I'm not sure what you are after. All these subalgebras are sets of 4x4 matrices. The masses are properties of the representations, in this case 4-momenta, or suitable quantum fields. As I indicated, contractions are most useful in intuiting the geometries associated with the subalgebras. $\endgroup$ – Cosmas Zachos Feb 21 at 2:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy