parameter μ dimensional regularization qedby paolorossi Tags: μ, dimensional, parameter, regularization 

#1
Nov1212, 02:42 PM

P: 22

in the dimensional regularization of the QED we introduct an arbitrary parameter μ with the dimension of a mass... so there are finite terms that are function of μ... so they are arbitrary??? how we can fix this parameter? what is the physical meaning of μ?




#2
Nov1212, 04:42 PM

P: 527

They are absorbed into the renormalized coupling constants and field renormalization.




#3
Nov1212, 05:10 PM

P: 22

a book to view this???




#4
Nov1312, 06:27 AM

P: 977

parameter μ dimensional regularization qed
Do you mean for photon mass which is definitely taken zero at the end of calculation because the integrals are divergent without it and photon should not have any mass.




#5
Nov1312, 08:32 AM

P: 22

But when we return to 4 dimension some terms depend on μ. xepma say that (sorry for my english) 



#6
Nov1412, 07:00 AM

P: 977





#7
Nov1412, 08:20 AM

P: 22

μ is the arbitrary parameter, with mass dimension, which appears in the Lagrangian density of interaction in D dimesion L_{I} = e μ^{ 2D/2} [itex]\bar{\psi}[/itex][itex]\gamma[/itex]^{[itex]\mu[/itex]}[itex]\psi[/itex] A_{[itex]\mu[/itex]} so that L_{I} has the correct dimension M^{D} (note that the fields have the dimension [ A_{[itex]\mu[/itex]} ] = M^{D/21/2} , [ [itex]\psi[/itex] ] = M^{D/21} ). My question is: if the counterterms are such that only the divergent terms are "eliminated" in the corrections to higher orders (propagators and vertex) , then there are finite terms that are function of μ. But μ is arbitrary. So how do you solve this? maybe there is a way to fix μ? 



#8
Nov1512, 06:54 AM

P: 977

it is clear that when D=4(as is the case) ,μ does not play any roll after all.But only a tool to handle regularization in arbitrary dimension.(I thought it to be some photon mass type thing which is taken zero at end)




#9
Nov1512, 08:37 AM

P: 22

μ > μ' then, for example, the charge change as e(μ) > e(μ')=e' so that, for example, the vertex (plus radiative corrections) remains unchanged. I saw that the set of this transformation forms the socalled " renormalization group" (RG). thanks anyway for answer! 


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