# Electron and graviton degrees of freedom

## Main Question or Discussion Point

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!!!

Related Quantum Physics News on Phys.org
dextercioby
Homework Helper
It's all in the Hamiltonian formulation. The nr. of DOF is derived there by merely counting constraints.

tom.stoer
a) 1) We have a spin 1 particle that should have 2s+1=3 spin state. So DOF=3.
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

2) We have 4 Aμ guys. One is out because of gauge invariance. So we have 3 left.
Does EOM gives us smth?
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.

What about photon in 1+D dim? Where and how zero photon mass get into play?
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.