I Lorentz invariant phase space and cross section

[simonjech]

TL;DR Summary

I did not understand one step in the QFT and Standard Model book from Matthew D. Schwartz.

Can someone please explain to me how can we obtain this integral in eq. 5.27 from eq. 5.26? I quite do not understand how is it possible to make this adjustment and why the (p_(f))^2 appeared there in the numerator and also why a solid angle appeared there suddenly.

 

[malawi_glenn]

What is p_f ? is it the modulus of p3 (and/or p4)? I suspect "f" means final here...

If that is the case, then the reason why there is a solid angle measure is because any three-dimensional vector can be represented in spherical coordinates as modulus and solid angle.

 

[simonjech]

Yes, f means final. Can you please be more specific?

 

[malawi_glenn]

I assume that if you are studying QFT you must have done integrations of the form $\int \mathrm{d}x\mathrm{d}y\mathrm{d}z$ by transforming to spherical coordinates thousands times before? If not, here is a reference https://tutorial.math.lamar.edu/classes/calciii/tisphericalcoords.aspx

 

[vanhees71]

The idea is to get rid of the "energy-momentum conserving" $\delta$-distribution. To that end you first integrate over $\vec{p}_4=-\vec{p}_f=\vec{p}_3$. This gives you (putting all the factors $(2\pi)$ and the factor 1/4 together and use $E_{\text{CM}}=E_1+E_2$
$$\mathrm{d} \Pi_{\text{LIPS}}=\frac{1}{16 \pi^2} \delta(E_{\text{CM}}-E_3-E_4) \frac{\mathrm{d}^3 p_3}{E_3 E_4}.$$
Now you have to get rid of the energy-conserving $\delta$-function. For this you note that
$$E_3^2=m_3^2+p_f^2, \quad E_4^2=m_4^2+p_f^2.$$
This means it's a good idea to introduce first spherical coordinates for the integration over $\vec{p}_f$. The $\delta$ function can obviously be integrated out by the integral over $p_f=|\vec{p}|_f$, and the final result is only to be integrated over the angles, which is denoted by the "solid angle", $\mathrm{d} \Omega=\mathrm{d} \vartheta \mathrm{d} \varphi \sin \vartheta$, where $\vartheta$ and $\varphi$ are the usual spherical angular coordinates of $\vec{p}_f$.

So finally end indeed up with Eq. (5.27) by using $\mathrm{d}^3 p_f =\mathrm{d} p_f \mathrm{d} \Omega p_f^2$ and then integrating over $p_f$ by substitution as explained in the next steps of the book.

 

[simonjech]

Thank you for your respond. It was helpful to me.

The idea is to get rid of the "energy-momentum conserving" $\delta$-distribution. To that end you first integrate over $\vec{p}_4=-\vec{p}_f=\vec{p}_3$. This gives you (putting all the factors $(2\pi)$ and the factor 1/4 together and use $E_{\text{CM}}=E_1+E_2$
$$\mathrm{d} \Pi_{\text{LIPS}}=\frac{1}{16 \pi^2} \delta(E_{\text{CM}}-E_3-E_4) \frac{\mathrm{d}^3 p_3}{E_3 E_4}.$$
Now you have to get rid of the energy-conserving $\delta$-function. For this you note that
$$E_3^2=m_3^2+p_f^2, \quad E_4^2=m_4^2+p_f^2.$$
This means it's a good idea to introduce first spherical coordinates for the integration over $\vec{p}_f$. The $\delta$ function can obviously be integrated out by the integral over $p_f=|\vec{p}|_f$, and the final result is only to be integrated over the angles, which is denoted by the "solid angle", $\mathrm{d} \Omega=\mathrm{d} \vartheta \mathrm{d} \varphi \sin \vartheta$, where $\vartheta$ and $\varphi$ are the usual spherical angular coordinates of $\vec{p}_f$.

So finally end indeed up with Eq. (5.27) by using $\mathrm{d}^3 p_f =\mathrm{d} p_f \mathrm{d} \Omega p_f^2$ and then integrating over $p_f$ by substitution as explained in the next steps of the book.