Speculation on the Statistical Enhancement Factor For Bosons
Francesco R. Ruggeri
In a previous note (1), we argued that even though photons and particles with rest mass both have the same quantum free particle probability (wavefunction) exp(-iEt+ipx), they have different spins, spin 1 (photon) and spin ½ (fermion like nucleons, electron). We argued that one may find equations for spin matrices by identifying an inherent force associated with the object. For a particle with rest mass, we argued that one chooses rest mass, mo, which is proportional to the force, inertia. For a photon, one chooses the electric field. The key is then to find an equation which is linear in the inherent force and contains p, E (for special relativistic effects and geometry). For a particle with rest mass, the Lorentz invariant -EE + cc p dot p = -momo cccc is linearized as Dirac showed. For a photon, one seeks an equation linear in the electric field. One may linearize d/dt partial (.5eo El dot El + .5/uo B dot B) + 1/uo grad dot (El x B) = 0, where El=electric field and B=magnetic field. For a particle with rest mass, one must retain the identity of the rest mass when forces interact. If one has two identical mo’s with the same center-of-mass, one must somehow distinguish between the two to deliver an impulse hit to only one. This may be done by using a magnetic field and having one particle have spin up, and the other, spin down. This leads to the Pauli exclusion principle. Using the Pauli exclusion principle in a two body elastic scattering problem, one may easily establish time reversal balance by: p(e1)p(e2) {1-p(e3)}{1-p(e4)} = p(e3)p(e4) {1-p(e1)}{1-p(e2)} ((1)). From this one may obtain the Fermi-Dirac distribution. The question becomes: What does one do for photons? We argue that photons do not interact in the same manner as two particles with rest mass which elastically scatter. Instead we consider photons interacting with electrons in atoms. We note that in the case of particles with rest mass, there is a grouping of two energies, i.e. e1,e2 on the LHS side of ((1)), and e3,e4, on the RHS. We keep this notion of grouping, but note that e1, and e2 are not being converted into e3, e4 photons. Rather, all four types of photons are present and may interact with the electrons in atoms. Thus, we consider the sum of the probabilities for the e1,e2 grouping, namely (e1,e2) + (e1,e2) e3 + (e1,e2) e4 + (e1,e2) e3, e4. We argue that the sum of probabilities for this set of groupings should equal the sum of probabilities for the grouping of (e3,e4), i.e. (e3,e4)+ (e3,e4) e1 + (e3,e4) e2 + (e3,e4) e1,e2. This is equivalent to: p(e1)p(e2) {1+p(e3)}{1+p(e4)} = p(e3)p(e4) {1+p(e1)}{1+p(e2)}. From this, the Bose-Einstein distribution follows.
•Mar 4, 2026•0 citations