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bhobba submitted a new PF Insights post
Why Renormalisation Needs a Cutoff
Continue reading the Original PF Insights Post.
Why Renormalisation Needs a Cutoff
Continue reading the Original PF Insights Post.
nrqed said:This is the "old" approach to renormalization (pre Ken Wilson, say). The modern point of view is that the cutoff should not be taken to infinity. But then one must treat the theory as and effective field theory and there is an infinite of terms to be included in the lagrangian. This is for another post. But my point here was to convey that the cutoff does not go away even in renormalizable theories if we don't take the limit cutoff goes to infinity.
nrqed said:Instead, one gets typically something of the form [itex] \ln( (\Lambda^2+k_1^2)/k_2^2) [/itex].
Jimster41 said:But regardless, do I understand correctly that this is saying that G(x,λ) can be decomposed into a linear combination of functions [itex]{ G }_{ i }(x)[/itex] multiplied by powers of λ (That just what the power series expansion technique)?
Jimster41 said:Do I understand correctly that this just normalizes (scales) the "power series representation of G(x,λ)" to the difference between the first to constants of expansion of G(x,λ)?
Jimster41 said:This is because powers of small numbers go to zero in the limit, correct?
Jimster41 said:I can declare something "Dimensional" to be suddenly "Dimensionless", change length into Btu's or whatever). After all it's just a computer, I can make it do whatever I want. But it seems telling to me that without instructions for how/when/where to do this, the computer can't "automatically" do so .
Thanks for the direction Bill.bhobba said:Its a Taylor series expansion - in applied math you generally assume you can do that.
Not quite - because of the division it creates something dimensionless - its different to a rescaling which would simply be a change of units.
"Suppose λ is small, then F(x) = λ, F has the dimensions of λ, so is dimensionless"
No. Its because F must be dimensionless - but the expansion says it isn't. This is an inconsistency - to accommodate it, it must be infinity or a constant - if it actually depended on x it woul not be dimensionless.
I am sorry - but you can't do that. Its modelling something - nothing you can do can change what its modelling.
Thanks
Bill
bhobba said:No. Its because F must be dimensionless - but the expansion says it isn't. This is an inconsistency - to accommodate it, it must be infinity or a constant - if it actually depended on x it woul not be dimensionless.
Hi Bill,bhobba said:But for large Λ its the same. I have gone through the exact calculations for the messon/meeson scattering in my original paper, and it, even without taking a large number approximation you get [itex] \ln( (\Lambda^2/k_2^2) [/itex] . Are you sure you are talking about the large energy approximation I am using in the paper?
Thanks
Bill
You are absolutely correct, stevendaryl. I was going to make the same remark. The higher Fs come from a Taylor expansion and the terms in a Taylor expansion (the terms that multiply the powers of the variable in which we expand) all have different dimensions since they come from derivatives with respect to a dimensional quantity! So there is no problem with the dimensions of any of the F's. We cannot conclude anything one way or another from the existence of the Taylor expansion.stevendaryl said:I might be just being stupid, but I don't understand this point. You have a dimensionless function of x, [itex]F(x)[/itex]. It can be written as a power series in x, as follows:
[itex]F(x) = F_0 + x F_1 + x^2 F_2 + ...[/itex]
If x is small, then we can approximate F by just the first two terms, so:
[itex]F(x) = F_0 + x F_1[/itex]
I don't understand why you say that if [itex]F[/itex] actually depended on x, it would not be dimensionless. What it seems to me is that [itex]F[/itex] is dimensionless, and so is [itex]F_0[/itex], but [itex]F_1[/itex] has the dimensions of [itex]\frac{1}{D}[/itex], where [itex]D[/itex] is the dimensions of x.
I agree that if you want all the [itex]F_i[/itex] to be dimensionless, then you can't have an expansion in [itex]x[/itex], you have to have an expansion in [itex]\frac{x}{\Lambda}[/itex] where [itex]\Lambda[/itex] has the same dimensions as [itex]x[/itex]. But saying that [itex]F[/itex] is dimensionless doesn't imply that [itex]F_1[/itex] is dimensionless.
stevendaryl said:I don't understand why you say that if [itex]F[/itex] actually depended on x, it would not be dimensionless.
bhobba said:I simply expanded F2 to make it clearer what's going on. You can argue it has dimension F2(X) - whatever F2 is - if its squared it's dimensions x^2, if its √ it has dimensions x^1/2 etc. If its constant then its dimensionless. It looks obvious to me from what dimensions means in dimensional analysis.
nrqed said:I was talking about a general calculation, so I am not assuming that the external momenta are much larger than other physical scales (like the masses of the particles in the loops). My point is that in general, if one does a one loop QFT calculation
stevendaryl said:I guess it's obvious that if [itex]F(x)[/itex] is dimensionless, but [itex]x[/itex] is not, then [itex]F(x)[/itex] can be rewritten as a function of [itex]\frac{x}{\Lambda}[/itex], where [itex]\Lambda[/itex] is a scale factor with the same dimensions as [itex]x[/itex]. But I don't see what that shows about the interpretation of [itex]\Lambda[/itex] as a cutoff.
bhobba said:Hi Patrick.
I am sure you are correct in general - but I was not considering the general case - I was considering the high energy scale of a one parameter theory. The example you gave has two parameters K1 and K2.
An example is the scattering amplitude from the Φ^4 theory which is what I considered in the first paper. From Zee page 145 - its
i*λ + 1/(2*(2π)^4) ∫ dk^4 1/((k^2 - m^2)((K - k)^2 - m^2)).
For simplicity we go to the high energy regime so m can be neglected
i*λ + 1/(2*(2π)^4) ∫ dk^4 1/(k^2*(K - k)^2)
Zee claims its i*λ + i*C*log(Λ^2/K^2) which is exactly the form I came up with.
However I have to say there were a number of approximations made in deriving it (Zee did it on page 151 and 152 - band its rather tricky). I don't know if that's a factor.
Thanks
Bill
nrqed said:Your demonstration worked only because you were taking the high energy limit but it does not follow in general, which was my point.
Hi Bill,bhobba said:Good point.
But what do you think of the following argument - take the integral before:
∫ dk^4 1/((k^2 - m^2)((K - k)^2 - m^2)).
We divide the integral into two bits - the sum of a finite integral bit and two high energy bits that are of the form ∫ dk^4 1/k^4 because k swamps the other terms - one from - ∞ the other to +∞. Then by subtracting from the finite integral ∫ dk^4 1/k^4 over that finite region, the improper integrals become ∫ dk^4 1/k^4 for -∞ to ∞. This has the form Limit Λ → ∞ C*log (Λ^2). In this case it has the form for my original argument to be applicable. Of course approximations are used - but what Zee used was full of them as well.
This was like in your example where Λ was large.
Thanks
Bill
nrqed said:there is no justification for dropping all masses and then, as I said, the cutoff does not cancel out unless we take it to infinity.
Using dimensional regularization would mean not introducing the cutoff and the integral would be quite different.bhobba said:Hi Guys
Just a quick question how one actually does integrals like that. Zee used a Pauli-Villiars approximation and even then you end up with
C* ∫ log (Λ^2/(m^2 - α(1-α)*K^2) dα where the integral is from 0 to 1
Do you do it in dimensional regularisation?
Thanks
Bill
nrqed said:You can use the online Mathematica integrator, for example.
Ah, your question was about what people in the field prefer to use to regularize their integrals? Then yes, dimensional regularization is almost always used, in practice. In an abelian gauge theory like QED, one may use a Pauli-Villars regularization but in non abelian gauge theories, one uses dimensional regularization and then, by habit, people use dim reg everywhere (after introducing the gauge fixing terms, Fadeed-Popov ghosts, etc, in the lagrangian)bhobba said:Do - yes - its actually in a table of integrals - plug and chug.
Thanks
Bill
nrqed said:Ah, your question was about what people in the field prefer to use to regularize their integrals?
Jimster41 said:Is "the cutoff" whatever it needs to be, dimensionally, and in value, depending on the specific problem being solved?
Or, is there a canonical "maximal/minimal" cutoff. For some reason I was thinking it was thelinit given by the Planck constant.
Is it correct to say that the cutoff is really along the "probability" axis regardless of the dimension in question? That renormalization is drawing a boundary around the peak of the "probability" wave. Even though the wave keeps going, the probability of observation is always? (or "generally")decreasing, so introducing a "cutoff" is practical, and is just about how infinitesimal a degree of uncertainty is tolerable.
If this is the case, is it true, do I understand correctly that "many body" case is more worrisome?
Jimster41 said:Is "the cutoff" whatever it needs to be, dimensionally, and in value, depending on the specific problem being solved?
Jimster41 said:Is it correct to say that the cut-off is really along the "probability" axis regardless of the dimension in question?
atyy said:For example, in quantum electrodynamics, the "absolutely necessary" (nonsharp) cutoff is given by the Landau pole and lies above the Planck scale. However, reality may intervene way before that, and cause our theory to be false. In particular, we expect quantum gravity to render QED already false below the Planck scale.
Jimster41 said:I was associating the "cutoff" with a boundary on the space of "possible paths", limiting the integration to run only over the ones with some minimum probability, rather than an infinite number of them.
Renormalization is a mathematical technique used in theoretical physics to remove infinities that arise in certain calculations. It is necessary because in some quantum field theories, the equations become infinite when certain parameters, such as energy or momentum, approach zero. The cutoff is a mathematical tool used to regulate these infinities, allowing for sensible calculations to be made.
The cutoff is a maximum value that is imposed on certain parameters in the equations. This prevents the equations from becoming infinite and allows for finite calculations to be made. The cutoff is usually removed at the end of the calculation, resulting in a finite and physically meaningful result.
Renormalization is a crucial tool in theoretical physics because it allows for the prediction of physical phenomena at extremely small scales, such as in quantum mechanics. Without renormalization, many important calculations would be impossible or would yield meaningless results.
While renormalization with a cutoff is a powerful technique, it does have its limitations. For example, it cannot be used in all quantum field theories and may not always give physically meaningful results. Additionally, the choice of cutoff can affect the final result, leading to some degree of uncertainty.
Renormalization with a cutoff is used in a wide range of fields, including particle physics, condensed matter physics, and cosmology. It has been successfully applied to explain phenomena such as the behavior of quarks and gluons in the strong nuclear force, the behavior of electrons in superconductors, and the evolution of the universe in the early stages of the Big Bang.