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A question about RG scaling of masses..

by jys34
Tags: masses, scaling
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jys34
#1
Nov24-13, 07:28 PM
P: 7
Hi, I'm studying about Renormalization group.

I have a question about mass beta-function.

Usually, when we perform one-loop calculation to get counter-term coefficients,
resulting RG coefficient for mass scaling is given by

[itex]\mu \frac{d m}{d \mu} = a_1 (e ) m [/itex]

and [itex]a_1(e)[/itex] is come from divergent loop diagrams (with 1/epsilon divergence)

but mass itself is dimensionful, and I thought that it must scale not only at quantum-level but also tree-level, too.

so I expect that

[itex]\mu \frac{d m}{d \mu} = m + a_1 (e ) m [/itex]

because mass has +1 mass dimension and irrelevant under momentum scaling.

but, on the other hand, it is weird that masses in non-interacting theory also should be scaled.


how do we treat RG scaling (or calculation of beta-functions) of dimensionful parameters in dimensional regularization? (I think that I have certain mistakes in above procedure..)




p.s.
Is it possible to change the result if we consider about some derivative-type background field

[itex]A_\mu^{ext} = \partial_\mu \phi(x) [/itex]

instead of mass? (here phi(x) is some dimensionless non-dynamical parameter)
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andrien
#2
Nov26-13, 08:04 AM
P: 1,020
you can use minimal subtraction scheme in which you can define your unrenormalized parameters mass,coupling and field by a laurent series.For mass,you can write
m0=μ[1+Ʃn=1(mn(g)/εn],now these unrenormalied parameters should be independent of subtraction point i.e. ∂m0/∂μ=0.With the help of it you can get recursion relations.You can relate them to original renormalization group parameters.You have to assume that μ(∂m/∂μ) should be an analytic function of ε.You can use the same procedure for coupling strength and field to get the recursion relation.you have to however multiply by the corresponding dimension like με in coupling strength.
jys34
#3
Nov27-13, 05:38 AM
P: 7
Thanks for comments, but I'm still confused because many references have two different answers for mass running..

In Shankar's famous RMP (http://rmp.aps.org/abstract/RMP/v66/i1/p129_1)(or cond-mat/9307009)
the derivation of mass (or chemical potential) scaling at tree-level can be found, given by
[itex]\mu \frac{dm}{d \mu} = -m [/itex]
and at one-loop, given by
[itex]\mu \frac{dm}{d \mu} = -m + \frac{e^2}{8 \pi^2} m [/itex]

the derivation via Wilsonian RG give the scaling even at tree-level.


However, In QED, mass scaling up to one-loop, given by
[itex]\mu \frac{dm}{d \mu} = \frac{e^2}{8 \pi^2} m [/itex]
in some other well-known references(Srednicki, Peskin & Schroeder)

and redefined dimensionless parameter [itex] m = \mu \tilde{m} [/itex] scales as (up to one-loop)
[itex]\mu \frac{d \tilde{m}}{d \mu} = -\tilde{m} + \frac{e^2}{8 \pi^2} \tilde{m} [/itex]

here, there is no tree-level scaling for dimensionful parameter [itex]m [/itex] ,
but there is tree-level (and non-interacting limit) scaling for dimensionless parameter [itex] \tilde{m} [/itex]


I've been confused due to the disagreement between these references..
There are many other references, but both of them have been widely used.. how can we intepret this difference? or, did I have some misunderstandings?

andrien
#4
Nov27-13, 08:40 AM
P: 1,020
A question about RG scaling of masses..

can you point out exactly where there is a tree level correction giving μ(dm/dμ) in the reference you have given,I cannot just seem to find it.Also the result you wrote down are for one loop correction to coupling constant rather than mass.For mass,μ(∂m/∂μ)=γmm,where γm is anomalous dimension.Also you don't need to convert m to any other dimensionless parameter,with the approach I have mentioned above you can get the correction by knowing the original renormalization parameters.
jys34
#5
Nov27-13, 09:14 AM
P: 7
I mean that the momentum scaling [itex] p → bp, x → x/b [/itex] in action (without quantum correction) can give scaling contribution (as eq (63)-(69), (91) in Shankar's review)

and with quantum correction (upto one-loop), we can get additional term as eq (97) in Shankar's review.

However, It seems to be different from other(Peskin, Srednicki) results, in which there is no free field scaling.

As a result, my question is the reason why
there is no scaling dimensions in Peskin, Srednicki and [itex] \mu \frac{dm}{d \mu} = 0 [/itex] in free theory( [itex] e=0[/itex]) , However, in Shankar's review, [itex] \mu \frac{dm}{d \mu} = -m [/itex] in free theory( [itex] e=0[/itex]).

Are their masses physically not same?

p.s.
I mentioned about dimensionless parameter, because there is free action scaling in it. (I heard that these setup is essential to find RG fixed point for dimensionful parameters (of Relevant, or Irrelevant operators), for example, Wilson-fisher fixed point structure, and I did want to know about the physical intepretations of [itex] m, \tilde{m} [/itex] in Peskin, Srednicki, and [itex] m [/itex] in Shankar's review.

and, as you know, if we follow methods in Srednicki or Peskin, we can get [itex] γ_m = -\frac{3e^2}{8 \pi^2 } + O(e^4) [/itex] (sorry for mistakes in numerical constant, but it is not important in this question;;)
andrien
#6
Nov28-13, 03:37 AM
P: 1,020
Ok,so if you are not confused between renormalization scale and some scale parameter to obtain the behaviour of 1PI at higher momenta,then as you know the RG eqn is written for 1PI as,
[μ(∂/∂μ)+β(∂/∂er)+mrγm(∂/∂mr)]ζ(pi,mr,er,μ)=0............(1),where I have omitted some terms in between and ζ(pi,mr,er,μ) is the renormalized vertex.
Now under a scale transformation pi→tpi,we can obtain another eqn by considering a scale transformation pi,t-1mr,t-1μ→tpi,mr,μ.Under this transformation the vertices will transform like
ζ(tpi,mr,er,μ)=tdζ(pi,t-1mr,er,t-1μ),where d is the dimension of the vertex.On differentiating it with respect to t,you end up with
[t(∂/∂t)+mr(∂/∂mr)+μ(∂/∂μ)-d]ζ(tpi,mr,er,μ)=0,you can now eliminate μ(∂/∂μ) between this and eqn. (1),which will give
[t(∂/∂t)-β(∂/∂er)-mrm-1)(∂/∂mr)-d]ζ(tpi,mr,er,μ)=0............(2),now you can solve it for a particular choice of renormalization scale μ=μ0,you can look for a solution of the form
ζ(tpi,mr,er0)=λ(t)ζ(pi,m'(t),e'(t),μ0) with the boundary condition of λ(1)=1,m'(1)=mr0),e'(1)=er0),you can now differentiate this eqn. w.r.t. t which will give you
t(∂/∂t)ζ(tpi,mr,er0)=[t/λ(dλ/dt)+t(∂m'/∂t)(∂/∂m')+t(∂e'/∂t)(∂/∂e')]ζ(tpi,mr,er0),where mr,er are evaluated at μ0.
On comparing this eqn to (2) above you will see that compatibility is provided
t(∂m'/∂t)=mrm-1) with other condition on β function and λ.
jys34
#7
Nov28-13, 07:10 PM
P: 7
Thanks for your help, andrien. I did understand your mention.

Can I ask some more questions?

the final equation for [itex] t \frac{\partial m'}{\partial t} = t \frac{\partial (t^{-1} m_r )}{\partial t} [/itex] means that [itex] t \frac{\partial m_r}{\partial t} = \gamma_{m} m_r [/itex],
and under momentam scaling [itex] p \rightarrow tp [/itex] in free field action, [itex] m_r [/itex] itself is not scaled. Am I right?

then, I wonder whether the mass parameter [itex] r_0 = m^2 [/itex] in Sankar's review (which scales as [itex] r_0 \rightarrow s^{-2} r_0 [/itex] under momentum scaling, even in free action) corresponds to [itex] m' [/itex].


(I think that scaled mass [itex] m' [/itex] is some kind of effective mass, which feels momentum scaling effect, so, physically measured value of particle's mass is [itex] m_r [/itex] itself and it is not changed without quantum correction, but if momentum become small under scaling transformation t<1, then, effectively particle become highly massive.. then we can see these effect from '-1' term in RGE for m' .. Is my understanding correct?;;)


actually, I thought that their different results in (Peskin, Srednicki, and others) and (Shankar, Fradkin, and others) are come from the assumption that

Shankar : every terms in action is invariant under momentum scaling. (then, mass parameter is scaled in momentum scaling, even without quantum corrections.)

Peskin, Srednicki : relevant, irrelevant terms in action in not invariant under momentum scaling, the magnitude of these terms becomes small and grows, respectively.

I'm not sure that they really assume in this way,

but I want to ask that which statement is correct?

or Both of them can be right? If it is possible, then, what is the criterions for determining scaling behaviors of dimensionful parameters.. Is it system dependent?

Of course, I think
if the mass parameter [itex] r_0 = m^2 [/itex] in Sankar's review really corresponds to [itex] m' [/itex], then [itex] m_r [/itex] itself is not scaled under [itex] p \rightarrow tp [/itex] ,
and non-marginal terms really non-invariant under [itex] p \rightarrow tp [/itex] scaling.

Sorry for many questions, because I'm still confused about RG behaviors of non-marginal term and dimensionful parameter.


p.s.
I think that I've misunderstood about quantum correction and momentum scaling, which are driven by other parameter, [itex] \mu [/itex] and t , respectively.
andrien
#8
Nov29-13, 04:09 AM
P: 1,020
Quote Quote by jys34 View Post
I think that scaled mass [itex] m' [/itex] is some kind of effective mass, which feels momentum scaling effect, so, physically measured value of particle's mass is [itex] m_r [/itex] itself and it is not changed without quantum correction, but if momentum become small under scaling transformation t<1, then, effectively particle become highly massive.. then we can see these effect from '-1' term in RGE for m' .. Is my understanding correct?;;)
This sort of explanation will never come out right,as you see that m' is a function that is to be determined from boundary condition and has the role that renormalized mass evaluated at particular subtraction point will determine that initial value.And hence as you have noted m'=t-1mr with μ=tμ0.I cannot go into much discussion now,but you can consult book 'Renormalization' by Collins.
jys34
#9
Nov29-13, 11:07 PM
P: 7
honestly, I don't understand what you are saying because your statement imply [itex]m'(t) = m_r (t^{-1} \mu_{0})[/itex], then, m'(t) is just renormalized mass at some particular scale [itex]t \mu_{0}[/itex].

However, in your last comment, I thought that you mean both of them ([itex] m'(t) [/itex] and [itex] m_r(t \mu_{0}) [/itex]) are physical, but they are non-invariant under not only quantum correction but also momentum scaling (because [itex] t\frac{\partial m'}{\partial t} = m' ( \gamma_{m} - 1 ) [/itex] )

actually, measured electron mass itself (0.5MeV) is not scaled in this way.
(As I know, electron mass is 0.5MeV at [itex] \sqrt{s} = 1eV [/itex] and [itex] \sqrt{s} = 1GeV [/itex] as well. there is only very small [itex] \gamma_{m} [/itex] contributions.)
(m'(t) is doubled if t become 1/2)

then, I thought we can conclude that [itex] m'(t) ( = m_r (t \mu_{0}) \ )[/itex] is not actual value but some kind of screened mass.

p.s.
actually, I should have considered that one of t and [itex] \mu [/itex] can be arsorbed in another. but It is not clear for me that which behavior ([itex] \mu \frac{\partial m_r}{\partial \mu} = m_{r} \gamma_{m} [/itex] or [itex] t \frac{\partial m'(t)}{\partial t} = m_{r} ( \gamma_{m} - 1 ) [/itex]) is related to measured physical value.

I'm sorry, My understanding about RG isn't good until now -_-;;
andrien
#10
Nov30-13, 06:56 AM
P: 1,020
Quote Quote by jys34 View Post
honestly, I don't understand what you are saying because your statement imply [itex]m'(t) = m_r (t^{-1} \mu_{0})[/itex], then, m'(t) is just renormalized mass at some particular scale [itex]t \mu_{0}[/itex].

However, in your last comment, I thought that you mean both of them ([itex] m'(t) [/itex] and [itex] m_r(t \mu_{0}) [/itex]) are physical, but they are non-invariant under not only quantum correction but also momentum scaling (because [itex] t\frac{\partial m'}{\partial t} = m' ( \gamma_{m} - 1 ) [/itex] )

actually, measured electron mass itself (0.5MeV) is not scaled in this way.
(As I know, electron mass is 0.5MeV at [itex] \sqrt{s} = 1eV [/itex] and [itex] \sqrt{s} = 1GeV [/itex] as well. there is only very small [itex] \gamma_{m} [/itex] contributions.)
(m'(t) is doubled if t become 1/2)
Well,the electron mass which you are measuring is varied with μ(renormalization scale),not with the scaling property of momentum.So when you are measuring electron mass,only γm will contribute according to μ(∂m/∂μ)=γmm.In any case,you should not consider m' as physical mass,it is mr.


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