parameter μ dimensional regularization qed


by paolorossi
Tags: μ, dimensional, parameter, regularization
paolorossi
paolorossi is offline
#1
Nov12-12, 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?
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xepma
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#2
Nov12-12, 04:42 PM
P: 527
They are absorbed into the renormalized coupling constants and field renormalization.
paolorossi
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#3
Nov12-12, 05:10 PM
P: 22
a book to view this???

andrien
andrien is offline
#4
Nov13-12, 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.
paolorossi
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#5
Nov13-12, 08:32 AM
P: 22
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.
I speak about the parameter μ that we introduct in the extension of the 4 dimensional minkowsky space to a D dimensional space. This parameter is such that the Lagrangian density of interaction has the correct physical dimension MD in D dimensions.

But when we return to 4 dimension some terms depend on μ. xepma say that
They are absorbed into the renormalized coupling constants and field renormalization.
, but how I can see that? can somebody suggest a book or paper?

(sorry for my english)
andrien
andrien is offline
#6
Nov14-12, 07:00 AM
P: 977
I speak about the parameter μ that we introduct in the extension of the 4 dimensional minkowsky space to a D dimensional space. This parameter is such that the Lagrangian density of interaction has the correct physical dimension MD in D dimensions.
So what exactly is μ.
paolorossi
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#7
Nov14-12, 08:20 AM
P: 22
So what exactly is μ.
Is it a question?
μ is the arbitrary parameter, with mass dimension, which appears in the Lagrangian density of interaction in D dimesion

LI = e μ 2-D/2 [itex]\bar{\psi}[/itex][itex]\gamma[/itex][itex]\mu[/itex][itex]\psi[/itex] A[itex]\mu[/itex]

so that LI has the correct dimension MD (note that the fields have the dimension [ A[itex]\mu[/itex] ] = MD/2-1/2 , [ [itex]\psi[/itex] ] = MD/2-1 ).
My question is: if the counter-terms 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?
andrien
andrien is offline
#8
Nov15-12, 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)
paolorossi
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#9
Nov15-12, 08:37 AM
P: 22
Quote Quote by andrien View Post
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)
yes, what you say is correct, in fact, the bare interaction Lagrangian density will not have to depend on the parameter μ (in the limit D=4) ... so if we express the lagrangian in terms of physical quantity (renormalizated charge, mass , ... ) then these physical parameters must transform adequately under a change of the parameter μ, so that the theory (ie the amplitudes calculated) are independent from the choice of μ. that is, if we make the transformation

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 so-called " renormalization group" (RG). thanks anyway for answer!


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