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Electron and graviton degrees of freedom

  1. Oct 11, 2012 #1
    Hi dear all
    Please explain to a stupid dummy a very simple thing.
    Take an a photon in 1+3 dimensions. How DOF it has? We all know that 2. How we calculate it?
    a) 1) We have a spin 1 particle that should have 2s+1=3 spin state. So DOF=3.
    2) We have 4 Aμ guys. One is out because of gauge invariance. So we have 3 left.
    Does EOM gives us smth?
    3) Electromagnetic tensor has 4×4 components. Its anti-symmetric so 4*3/2=6 components left. Bianci identities F(αβ;γ)=0 gives us C43=4!/3!1!= 4 conditions (is it true? Or it is 6eq? Its the same as a symmetry conditions???), so we left with 2 DOF.
    4) Electric and magnetic fields have three components each so we have 6. (4 Maxwell equations? ->2 )
    What about photon in 1+D dim? Where and how zero photon mass get into play?

    b) For graviton in 1+3.
    1) hαβ has 16 components. The matrix is symmetric so 4*5/2=10 components left. Bianci gives us h[αβ;γ]=0 - 4 equations. The gauge conditions (i.e. coordinate transformations with non zero Jacobian) gives me 4 tranformation (1 for each coordinate). Left with 2 - Fine
    2. In the 1+4 the same calculation give
    5*6/2=15 (symetric matrix)
    15-5=10 (because of gauge)
    Bianci gives us how?
    How DOF in the 5dim case????
    Help!!!
     
  2. jcsd
  3. Oct 11, 2012 #2

    dextercioby

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    It's all in the Hamiltonian formulation. The nr. of DOF is derived there by merely counting constraints.
     
  4. Oct 11, 2012 #3

    tom.stoer

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    For massless particles the Poincare group rep. is not a vector rep.; it's not about spin but about the helicity and we are left with two helicity states

    It's most easily seen in the A°=0 gauge. A° is a Lagrangian multiplier, not a d.o.f. b/c there is no canonical conjugate momentum due to the missing ∂°A° term. So you eliminate one d.o.f. by A°=0 and a second one by the Gauß law constraint; 4-1-1 = 2.

    In one spatial dimension there is no dyn. d.o.f. for the el.-mag field, only a single zero mode with ∂1A1 = 0 which is solved by A1 = const. survives; this corresponds to 2 components of Aμ and 2-1-1 = 0 as in 4 dim.
     
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