Electron and graviton degrees of freedom

unih
Messages
27
Reaction score
0
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!
 
Physics news on Phys.org
It's all in the Hamiltonian formulation. The nr. of DOF is derived there by merely counting constraints.
 
unih said:
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

unih said:
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.

unih said:
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.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top