Why do some papers use binomial expansion in dimensional regularization?

In summary, the conversation discusses a paper that uses dimensional regularization to derive an expression involving an integral. The rightmost factor in the expression is expanded using binomial expansion, but there is confusion about whether this is valid due to the behavior of the factor when x approaches 1. The paper in question is the second one in a list of papers by Donoghue and Holstein, and the relevant equation can be found on page 49 (or 50) of the scanned version. The listener expresses surprise at the techniques used in the paper and notes that they have not seen them in other QFT books.
  • #1
RedX
970
3
I was looking at a paper that used dimensional regularization and the following expression was derived:

[tex]\int dx \mbox{ }[p^2(1-x)^2-\lambda^2(1-x)]^{\epsilon} [/tex]

Factoring out [tex]p^2(1-x)^2 [/tex]:

[tex]\int dx \mbox{ }[p^2(1-x)^2]^{\epsilon}[1-\frac{\lambda^2}{p^2(1-x)}]^{\epsilon} [/tex]

The part that I don't understand is that they expanded the rightmost factor in binomial expansion. [tex]\lambda^2[/tex] is smaller than [tex]p^2 [/tex] (in fact [tex]\lambda^2=p^2-m^2 [/tex]), but the 1/(1-x) changes all that when x approaches 1, making [tex]\frac{\lambda^2}{p^2(1-x)} [/tex] much greater than 1.

Is it okay to expand the rightmost factor in binomial expansion because when x goes to 1, the factor on the left [tex][p^2(1-x)^2]^{\epsilon} [/tex] goes to zero, so it doesn't matter what's on the rightmost side at that point? If so, isn't there an error term that needs to be calculated?
 
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  • #2
RedX said:
I was looking at a paper that used dimensional regularization and the following expression was derived:

[tex]\int dx \mbox{ }[p^2(1-x)^2-\lambda^2(1-x)]^{\epsilon} [/tex]

Factoring out [tex]p^2(1-x)^2 [/tex]:

[tex]\int dx \mbox{ }[p^2(1-x)^2]^{\epsilon}[1-\frac{\lambda^2}{p^2(1-x)}]^{\epsilon} [/tex]

The part that I don't understand is that they expanded the rightmost factor in binomial expansion. [tex]\lambda^2[/tex] is smaller than [tex]p^2 [/tex] (in fact [tex]\lambda^2=p^2-m^2 [/tex]), but the 1/(1-x) changes all that when x approaches 1, making [tex]\frac{\lambda^2}{p^2(1-x)} [/tex] much greater than 1.

Is it okay to expand the rightmost factor in binomial expansion because when x goes to 1, the factor on the left [tex][p^2(1-x)^2]^{\epsilon} [/tex] goes to zero, so it doesn't matter what's on the rightmost side at that point? If so, isn't there an error term that needs to be calculated?

Could be. Could you give us some more context? Do you have the citation for the paper? Cheers,

Adam
 
  • #3
olgranpappy said:
Could be. Could you give us some more context? Do you have the citation for the paper? Cheers,

Adam

There's a lot of pages, so I'll just link to a link of it:

http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=FIND+A+DONOGHUE+AND+HOLSTEIN+and+robinett&FORMAT=www&SEQUENCE= [Broken]

It's the second paper in the list, and you can download a copy by clicking on Scanned Version (KEK).

The part I'm referring to is on page 49, equation (A1).

It's an older paper, so they probably do things a little differently, but still, the result should be the same as with today's techniques, so I'm quite surprised: I haven't seen any of these techniques tried before in QFT books on QED.

*actually, it's page 50 if you click on the scanned images (Tiff and Gif). On the actual paper, it is page 49.
 
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What is dimensional regularization and how does it work?

Dimensional regularization is a technique used in theoretical physics and mathematics to handle divergent integrals in calculations. It involves extending the number of dimensions in which the calculation is performed, and then taking the result in the physical number of dimensions as the final result. This allows for a more general and consistent way of dealing with infinities in calculations.

Why is dimensional regularization important in theoretical physics?

Dimensional regularization is important in theoretical physics because it allows for the calculation of physical quantities that would otherwise be infinite or undefined. This technique is particularly useful in quantum field theory, where infinities often arise due to the nature of the theory.

What are the advantages of using dimensional regularization?

One advantage of dimensional regularization is that it maintains the symmetries of the underlying theory, unlike other regularization techniques which can break these symmetries. Additionally, it can be applied to a wide range of theories and calculations, making it a versatile tool in theoretical physics.

Are there any limitations to dimensional regularization?

While dimensional regularization is a powerful tool, it does have some limitations. It may not always work for non-renormalizable theories, and it can introduce artificial poles in the calculation. It is also important to carefully consider the choice of dimensional regularization scheme, as different schemes can lead to different results.

How is dimensional regularization related to renormalization?

Dimensional regularization is closely related to the process of renormalization, which is used to remove infinities from physical calculations. In fact, dimensional regularization is often used as a tool in the renormalization process, as it allows for the calculation of finite and well-defined quantities in theories that would otherwise be plagued by infinities.

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