5
$\begingroup$

Consider the Lorentz group $SO(2,1)$ in 2+1 spacetime dimensions. It's little group for massless particles should be just "$SO(1)$", which is just a trivial group with an identity element. So I expect that the massless particles should not have any propagating degrees of freedom in 2+1 dimensions. In fact, its already true for gravity in 2+1 dimensions that it doesn't have any propagating wave solutions.

But we also know that in 2+1 electromagnetism, the tensor $F_{\mu\nu}$ has 3 independent components, which is in contradiction to what I said earlier. Can someone help me correct my understanding? I am sure I am thinking incorrectly, but precisely can't put finger on what's wrong.

$\endgroup$
3

1 Answer 1

7
$\begingroup$

In 1+2D EM, the field has one degree of freedom. Indeed, $F$ has three real components, but you have two dynamic equations (first order in time) and one constraint (Gauss' law). Phase space is therefore 2D, so one degree of freedom. This checks out physically, as there is only one transverse mode for a given non zero light-like wavevector (only one normal direction).

You can make the result consistent with the little group argument by using the entire little group $O(1)\sim\mathbb Z_2$ (i.e. add parity and time reversal). You now have irreps that are 1D where the action is non trivial, but still abelian.

Hope this helps.

$\endgroup$
2
  • $\begingroup$ This 1 degree of freedom means that there is not even E and M, only a single field? $\endgroup$
    – peterh
    Apr 22 at 17:46
  • $\begingroup$ No, you still have two electric components and one magnetic component (eg $E_x,E_y,B_z$). It’s like in 1+3D, you have 6 components but two degrees of freedom (two possible polarizations) $\endgroup$
    – LPZ
    Apr 22 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.