The idea behind renormalization group.

[mhill]

What is the idea behind renormalization group ??

i believe you begin with an action $$S[\phi] =\int d^{4}x L(\phi , \partial _{\mu} \phi )$$

then you expand the fields into its Fourier components upto a propagator..

$$\phi (x) =C \int_{ \Lambda}d^{4}x e^{i \vec p \vec x} + c.c$$

but then i do not more, i know that every quantity measured by the QFT theory will have the form:

$$m(\Lambda) = alog(\Lambda) + \sum_{n=1}^{\infty} b_{n} \Lambda ^{n}$$

and that from the definition of propagator we should ask for 'conformal invariance' but i do not know how this theory helps to find finite (regularized) value of the quantities.

the idea i am looking for, it is to know if using renormalization group we could find a relation between the bare quantities $$m^{(0)}$$ and the renormalized ones $$m^{(R)}$$ via some differential or integral equation.

another mor informal question, if we found a method to obtain finite values for integrals.

$$\int_{0}^{T}dx x^{n}$$ as T-->oo and for every positive 'n'

would the problem of renormalization be solved ??

 

[kdv]

What is the idea behind renormalization group ??

i believe you begin with an action $$S[\phi] =\int d^{4}x L(\phi , \partial _{\mu} \phi )$$

then you expand the fields into its Fourier components upto a propagator..

$$\phi (x) =C \int_{ \Lambda}d^{4}x e^{i \vec p \vec x} + c.c$$

but then i do not more, i know that every quantity measured by the QFT theory will have the form:

$$m(\Lambda) = alog(\Lambda) + \sum_{n=1}^{\infty} b_{n} \Lambda ^{n}$$

and that from the definition of propagator we should ask for 'conformal invariance' but i do not know how this theory helps to find finite (regularized) value of the quantities.

the idea i am looking for, it is to know if using renormalization group we could find a relation between the bare quantities $$m^{(0)}$$ and the renormalized ones $$m^{(R)}$$ via some differential or integral equation.

another mor informal question, if we found a method to obtain finite values for integrals.

$$\int_{0}^{T}dx x^{n}$$ as T-->oo and for every positive 'n'

would the problem of renormalization be solved ??

I will meep itshort just in case you are not around anymore or not interested in this issue anymore.

The renormalization group is simply the implementation of the fact that changing the scale at which the renormalization procedure is applied should not change the physics. It's neat because even if you do the calculation at a specific order, imposing the renormalizatipn scale invariance allows you to determine the leading log correction of the next orders and solving the renormalization group differential equation essentially sums up those leading logs.
I could say much more about your questions but will wait to see if you are still interested.

 

[mhill]

yes, thank you very mach KDV for your response. The main problem is that i only know a bit about Lie Group theory , i know about the Beta function (how is defined) and the Callan-Szymanzik equation, but i do not know how could be RG used to obtain finite results in the renormalization process.

 

[kdv]

yes, thank you very mach KDV for your response. The main problem is that i only know a bit about Lie Group theory , i know about the Beta function (how is defined) and the Callan-Szymanzik equation, but i do not know how could be RG used to obtain finite results in the renormalization process.

group theory is really a completely separate concept. We can discuss renormalization without ever talking about group theory. Of course, if you have a gauge theory, some group theory stuff will creep in the discussion but it has nothing to do with renormalization per se.

Second, the renormalization group equations don't have anything to do with obtaining finite results in the renormalization process! You can renormalize a theory without ever talking about the renormalization group. If you want to discuss renormalization, that's a separate issue. It would be important for you to first understand renormalization before tackling the renormalization group.

Regards