- #36
Haelfix
Science Advisor
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Reread the section again, you missed the point or don't understand what physicists mean.. UV finiteness, by original classical definition, is one possible outcome of a specific set of procedures physicists do in a quantum theory.
If you do not follow the procedure, it makes no sense to say xyz is UV finite unless you redefine it that way. It is NOT a statement that says divergences are absent in some intermediate step in a calculation. It says roughly that all counterterm coefficients possible in the most general effective lagrangian (preserving all the original symmetries) after an *arbitrary* regularization and under a massless renormalization scheme like msbar are zero and the regularizer drops out *always*. In short no renormalization is necessary or even possible.
Note how this is absolutely NOT what LQG are doing.. The symmetries of the original system are traded off early on, and a very specific set of regularization rules are imposed (ones that by necessity are compatible with diffeomorphism invariant states), the original desired symmetries must then for consistency reemerge much later in a limit (this is the famous task of trying to get poincare invariance to flow out of the theory).
Its very much analogous to picking a particular path along a renormalization group orbit in a conventional QFT and looking at the divergence structure thereof. Eg you might see something very singularity free along that particularly trajectory, but you are very far from proving finiteness for the original thing. You essentially have to hope (or guess right) that the particular trajectory you picked flows to an attractor. In LQG's case this ambiguity or choice is traded into uncertainties in the hamiltonian constraint, but does NOT mean the original EH action that is being loop quantized is UV finite.
Anyway, different definitions.
If you do not follow the procedure, it makes no sense to say xyz is UV finite unless you redefine it that way. It is NOT a statement that says divergences are absent in some intermediate step in a calculation. It says roughly that all counterterm coefficients possible in the most general effective lagrangian (preserving all the original symmetries) after an *arbitrary* regularization and under a massless renormalization scheme like msbar are zero and the regularizer drops out *always*. In short no renormalization is necessary or even possible.
Note how this is absolutely NOT what LQG are doing.. The symmetries of the original system are traded off early on, and a very specific set of regularization rules are imposed (ones that by necessity are compatible with diffeomorphism invariant states), the original desired symmetries must then for consistency reemerge much later in a limit (this is the famous task of trying to get poincare invariance to flow out of the theory).
Its very much analogous to picking a particular path along a renormalization group orbit in a conventional QFT and looking at the divergence structure thereof. Eg you might see something very singularity free along that particularly trajectory, but you are very far from proving finiteness for the original thing. You essentially have to hope (or guess right) that the particular trajectory you picked flows to an attractor. In LQG's case this ambiguity or choice is traded into uncertainties in the hamiltonian constraint, but does NOT mean the original EH action that is being loop quantized is UV finite.
Anyway, different definitions.