Problem with Dimensional Regularization

In summary, we have a person who is trying to apply dimensional regularization to a divergent integral involving a term with two indices, l_{\mu}l_{\nu}. They are new to this and have a few questions, including how to apply Wicks rotation and whether it would work to rewrite the term in terms of the metric tensor and dimension. They also mention a method for determining integrals using a shift in variables.
  • #1
baranas
14
0
Good day to everyone. I am trying to apply dimensional regularization to divergent integral
[tex] \int\frac{d^{4}l}{\left(2\pi\right)^{4}}\frac{4\, l_{\mu}l_{\nu}}{\left[l^{2}-\triangle+i\epsilon\right]^{3}}.[/tex]
I am very new to these thing. The first question is how should i apply Wicks rotation to the term [tex]l_{\mu}l_{\nu}[/tex]As i understand it should be done before going to d dimensions. I need to avoid substitution [tex]l_{\mu}l_{\nu}\to\frac{1}{4}g_{\mu \nu}l^2[/tex]
Would it work to rewrite
[tex]l_{\mu}l_{\nu}=\frac{g_{\mu \nu}}{g_{\mu \nu}g^{\mu \nu}}g^{\mu \nu}l_{\mu}l_{\nu}[/tex]
Which gives in d dimensions
[tex]\frac{g_{\mu\nu}}{d}l^2[/tex]
I would appreciate any help.
 
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  • #2
The way to usually determine these integrals is to start with the result for

[tex]\int \frac{d^N l}{(l^2 + a^2)^A} = \frac{\Gamma(A-N/2)}{\Gamma(A)} \frac{\pi^{N/2}}{(a^2)^{A-N/2}}[/tex]

shift [itex]l=l'+p[/itex], so that

[tex]\int \frac{d^N l'}{((l')^2 + 2 p\cdot l' + a^2+p^2)^A} = \frac{\Gamma(A-N/2)}{\Gamma(A)} \frac{\pi^{N/2}}{(a^2+p^2)^{A-N/2}}.[/tex]

Now we can generate factors of [itex]l'_\mu[/itex] in the numerator by differentiating with respect to p, setting [itex]p=0[/itex] at the end as needed.
 

Related to Problem with Dimensional Regularization

What is dimensional regularization?

Dimensional regularization is a technique used in theoretical physics to handle integrals that would otherwise be divergent in higher dimensions. It involves treating the number of dimensions as a continuous parameter, which allows for the calculation of integrals that would otherwise be undefined.

Why is dimensional regularization useful?

Dimensional regularization is useful because it allows for the calculation of integrals that would otherwise be impossible to solve. It also helps to remove the dependence of results on arbitrary scale parameters, making calculations more physically meaningful.

What are the limitations of dimensional regularization?

One of the limitations of dimensional regularization is that it can only be used for certain types of integrals, specifically those that are related to scalar fields. It also does not work for integrals that are inherently divergent, such as those that arise in quantum field theory.

How does dimensional regularization differ from other regularization methods?

Dimensional regularization differs from other regularization methods in that it treats the number of dimensions as a continuous parameter, whereas other methods involve introducing a cutoff or modifying the integrand. It also has the advantage of being more mathematically rigorous and physically meaningful.

Are there any drawbacks to using dimensional regularization?

One potential drawback of dimensional regularization is that it can produce non-integer values for physical quantities, which may be difficult to interpret in some cases. It also requires a good understanding of the underlying theory and careful handling of divergences, which can be challenging for some researchers.

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