Why to change from momentum space integrals to spherical coordinate ones?

[Breo]

So I was asked to compute loop contributions to the Higgs and compute the integrals in spherical coordinates, I gave a look to Halzen book but did not found anything. Why, when and how to make that change?

 

[Orodruin]

I suspect they mean spherical coordinates in momentum space ...

 

[Breo]

I suspect they mean spherical coordinates in momentum space ...

"[...]Change the momentum integrals to spherical coordinates[...]"

 

[Orodruin]

So what is the problem? Going to spherical coordinates is a standard procedure from multivariable calculus.

 

[ChrisVer]

$d^3 p \rightarrow p^2 dp d \Omega$
is quite useful ...

 

[Breo]

But why? because I solved loop integrals using dim regularization and other tricks and never used the change of variables. That is why I asked when to use the change of variables instead to keep using my previous methods (Used in many loop computations by some textboks like Peskin's)

 

[Orodruin]

But why? because I solved loop integrals using dim regularization and other tricks and never used the change of variables.

Did you perhaps use the standard forms of the integrals that you end up with in dim-reg which you can find, eg, in the end of Peskin? How do you think these integrals are evaluated?

 

[Breo]

oh!

On the unit sphere area?

 

[ChrisVer]

why "unit" sphere area?

 

[Breo]

why "unit" sphere area?

I am still wondering haha

Do not know for now... tip for clairvoyance?

 

[Orodruin]

The entire point with dim reg is to rewrite the integral in 4-d as a limit of an integral in d "dimensions". Interpreting it as the area of a d-1-dimensional unit sphere and an integral where $p^{d-1}\, dp$ has replaced the volume $d^dp$.

 

[ChrisVer]

Let's try it this way... How would you solve this integral:
$\int \frac{d^3 p}{(2 \pi)^3} \frac{a }{|p|^2+m^2} e^{-i\vec{p} \cdot \vec{x}}$
?

 

[ChrisVer]

area of a d-1-dimensional unit sphere

Why saying a unit sphere? This doesn't make sense, exactly because the $d^d p$ is a volume. The $d^d p = |p|^{d-1} dp d\Omega_{d-1}$ is a volume element of an "hypersphere" between radius $p$ and $p+dp$.

edit: I think I figured it out

 

[Orodruin]

Why saying a unit sphere? This doesn't make sense, exactly because the $d^d p$ is a volume. The $d^d p = |p|^{d-1} dp d\Omega_{d-1}$ is a volume element of an "hypersphere" between radius $p$ and $p+dp$.

Yes, exactly. I took one step extra and integrated the $d\Omega$. This gives you the area of the d-1-dimensional unit sphere.