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Helicity states degrees of freedom

  1. Oct 10, 2009 #1

    1)Why does the number of helicity states depend on space-time dimension ?

    Masselss graviton in 4 dimensional space has 2 helicity states ( 2 degrees of freedom). In 5 dimensional space it has (still massless in 5 D) 5 helicity states (5 degrees of freedom) ....

    In 6 dimesional space according to formula d(d-3)/2 it has 9 helicity states .... but beeing a particle of spin s=2 it has 2s+1=5 states so where from are the additional 4 helicity states...

  2. jcsd
  3. Oct 10, 2009 #2
    If you recall that in 4d for the photon only 2 transverse modes which can be picturized to be the left and right circularly polarized. For d-dimensions there are d-2 transverse directions and so according to me the number of degrees of freedom i.e. the number of helicities ought to be 2 * (no. of polarizations mixing two transverse directions) = 2* (d-2)(d-3)/2= (d-2)(d-3). Can you refer me the text/paper from where you quoted the helicities = d*(d-3)/2 formula.
  4. Oct 10, 2009 #3
    Tomas Ortin gravity and strings page 295
  5. Oct 10, 2009 #4
    Ah I see that now. Thats just the number of independent components (polarizations) of an (p+1)-antisymmetric tensor in (d-2) transverse dimensions i.e. (d-2) C _(p+1) = (d-2)!/(p+1)!(d-p-3)! while the massive case you can have longitudinal modes so you count the number of components of an antisymmetric tensor with (p+1) indices in (d-1) spatial directions i.e. (d-1)C_(p+1).
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