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

1. May 16, 2015

### 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?

2. May 16, 2015

### Orodruin

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

3. May 16, 2015

### Breo

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

4. May 16, 2015

### Orodruin

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

5. May 16, 2015

### ChrisVer

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

6. May 16, 2015

### 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)

7. May 16, 2015

### Orodruin

Staff Emeritus
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?

8. May 16, 2015

### Breo

oh!

On the unit sphere area?

9. May 16, 2015

### ChrisVer

why "unit" sphere area?

10. May 16, 2015

### Breo

I am still wondering haha

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

11. May 16, 2015

### Orodruin

Staff Emeritus
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 $d^dp$.

12. May 16, 2015

### ChrisVer

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

13. May 16, 2015

### ChrisVer

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

14. May 16, 2015

### Orodruin

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