# Will my idea for simple-minded renormalization work?

+ 1 like - 3 dislike
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This is a simpleminded approach to dealing with
divergences.

We are trying to calculate some Feynman diagrams and there are
a finite number of divergent integrals. Presumably the high energy
behavior of the theory is not being handled properly.

We generally believe in the theory so we believe that if it properly
handled the high energy behavior the integrals would turn out finite.
Since we can't calculate the integrals we simply treat them as
additional parameters to be determined by observation.
So e.g. instead of $\{e,m\}$, we have $\{e,m,I_1,I_2,I_3\}$. 
We then proceed to calculate. Things are very simple. We don't have to
worry about overlapping divergences. There are no infinities at all.
We don't have to prove that the integrals can be absorbed by the
original parameters. If they can be it will appear in the calculations.
If they can't  it's not fatal. 

Will this work?

Closed as per community consensus as the post is not appropriate for the Q&A section of PO
recategorized Jun 7, 2015

Hi ockeghem,

this looks rather like an abstract of some paper, the question if what you are doing will work can not be answered without looking at your specific calculations. I therefore vote to close it here. The @downvoters may probably also want to follow the link to my closevote above and upvote it there ...

But you can request the correspoding paper to be imported into our reviews section, to obtain a review and further discussion. This will most probably also answer the question if your approach works.

In the context of a proper renormalization group analysis, I think your terms $I_1$, $I_2$, $I_3$ could in principle correspond to some operators (or terms in the action, Lagrangian) that are only important at certain high energy scales (corresponding to new physics or effects kicking in).

But it is really difficult to tell without knowing in some more detail what exactly you are doing...

Do I get it right, that you declare the results of the integrals as constants to be determined experimentally? That doesn't explain why the integral diverges, and that's why you need renormalisation.

Everyone misunderstood my question! I wasn't explicit enough.

1. I am presenting a simple general approach to dealing with

divergences in field theory. This is not an abstract! I have no

particular problem in mind and am not looking for help solving it!

2. Consider QED. Everyone believes that it is mostly valid and

very successful, but has problems at high energies that are not

understood and cause divergences. Renormalization does not

explain these divergences but only evades them.

3. The usual (textbook) way of dealing with these divergences is

complicated  and involves counterterms and dealing with

overlapping divergences. There are mysterious invisible infinities.

4. Instead I propose to simply consider the divergent integrals as

additional  parameters --- like masses or coupling constants.

The justification for this is that if we did understand the high

energy behavior we could do the integrals and they would come

out finite.

No further thought is required --- we don't have to prove anything more.

We simply calculate. If the new parameters $I_i$  can be absorbed

into the old ones $e,m$  it will show up when the diagrams are

calculated. Otherwise there are additional parameters that have to be fitted

to observed data.

5. Is this renormalization? I don't think so. It's logically simpler.

If it gives the same results then the answer to my question is 'yes'

-- it works!

6. The point of all this is not necessarily practical calculation but

to show that a theory with a finite number of divergent integrals is

calculable without any appeal to special renormalization techniques.

I hope this clarifies things. I'm having trouble with the editor. Why

is this doublespaced?

@ockeghem It's clear enough. It's just nonsensical.

@dimension10 Please allow him the chance to explain himself.

@ockeghem Try to place you thoughts in the context of scalar field theory. Preferably clearly detailing the steps. I can definitely say I don't understand your motives.

@Prathyush as already said, the post seems to be the abstract of a paper. So he should submit this paper to our Reviews section, where we can review and discuss it.

In case there are not yet any calculations done to support the conclusions and claims made in the above "question", neither the Q&A nor the Reviews section of PO would be an appropriate place for it.

Simply rewriting this Q&A question does not help, the missing information that must be added to judge if the above approach works is substantial and worth a whole paper. It should therefore definitively go into the reviews section.

+ 0 like - 2 dislike

Will this work?

You want us to do this research for you and give you the answer?

We generally believe in the theory so we believe that if it properly
handled the high energy behavior the integrals would turn out finite.

How can you "generally believe in theory" if the high energy behavior is certainly handled in a wrong way and you know it?

Yes, "properly handled the high energy behavior" may help, I call it "reformulation", but for that you have to understands why you currently handle it in a wrong way. Replacing divergent integrals with cut-off integrals and considering the cut-off theory as a "fitting function" is the actual renormalization approach, so you propose nothing new. And this is fatal! You did not get the right physics!

answered Jun 7, 2015 by (112 points)
edited Jun 7, 2015
###### (Oops, the original revision of my above comment, saying that there was something right with the answer, was posted despite the fact that I edited it once I understood what exactly the OP was trying to say. The final revision was:)

Erm, no - this is not renormalisation at all. The OP is basically saying that we should consider the results of divergent integrals as something to be experimentally verified. However, this does not actually explain anything, on the contrary - renormalisation does explain everything.

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