4
$\begingroup$

In deriving the rate of emission of arbitrary numbers of soft photons in a general QED process, Weinberg performs the following integral (equations 13.2.8-9):

$$-\pi(\vec{p}_m\cdot \vec{p}_n)\int_{\lambda\leq|\vec{q}|\leq\Lambda}\frac{d^3\vec{q}}{|\vec{q}|^3(E_n-\hat{q}\cdot\vec{p}_n)(E_m-\hat{q}\cdot\vec{p}_m)}=\frac{2\pi^2}{\beta_{nm}}\ln\left(\frac{1+\beta_{nm}}{1-\beta_{nm}}\right)\ln\left(\frac{\Lambda}{\lambda}\right)$$

where $$\beta=\sqrt{1-\frac{m_n^2m_m^2}{(\vec{p}_n\cdot\vec{p}_m)^2}}.$$

I am trying to compute this integral for myself but am having some trouble with the angular integral. Would someone mind giving me some assistance?

$\endgroup$

1 Answer 1

8
$\begingroup$

I think I solved it; unfortunately I don't have time to do all the manipulations after the integral is computed.

Let's change to spherical coordinates $$I=-\pi(\vec{p}_m\cdot\vec{p}_n)\int \text{d}q\text{d}\phi \sin(\phi) \text{d}\theta\frac{1}{q}\frac{1}{(E_n-p_n \cos(\phi))(E_m-p_m \cos(\phi))}$$

It follows that: $$I=-\pi(\vec{p}_m\cdot\vec{p}_n)\ln\left(\frac{\Lambda}{\lambda}\right)2\pi \int_0^{\pi}\text{d}\phi \frac{\sin(\phi)}{(E_n-p_n \cos(\phi))(E_m-p_m \cos(\phi))}$$

Now, let's change variables $$x=\cos(\phi) \\ \text{d}x=-\sin(\phi)\text{d}\phi$$ so we have $$I=-2\pi^2(\vec{p}_m\cdot\vec{p}_n)\ln\left(\frac{\Lambda}{\lambda}\right) \int_{-1}^{1}\text{d}x\frac{1}{(E_n-p_n x)(E_m-p_m x)}$$

Now, we have to make the following step: $$\frac{A}{(E_n-p_n x)}+\frac{B}{(E_m-p_m x)}$$ you find that $$A=\frac{-p_n}{E_n p_m-E_m p_n} \\ B=\frac{p_m}{E_np_m-E_mp_n}$$

Then, it's easy to see that $$I=2\pi^2(\vec{p}_m\cdot\vec{p}_n)\ln\left(\frac{\Lambda}{\lambda}\right)\left[\frac{A}{p_n}\ln\left(E_n-p_nx\right)+\frac{B}{p_m}\ln\left(E_m-p_mx\right)\right]^1_{-1}$$ by making $A$ and $B$ explicit we get: $$ I=2\pi^2\frac{(\vec{p}_m\cdot\vec{p}_n)}{E_np_m-E_np_n}\ln\left(\frac{\Lambda}{\lambda}\right) \left[-\ln\left(E_n-p_nx\right)+\ln\left(E_m-p_mx\right)\right]^1_{-1}$$ therefore $$I=2\pi^2\frac{(\vec{p}_m\cdot\vec{p}_n)}{E_np_m-E_np_n}\ln\left(\frac{\Lambda}{\lambda}\right)\left[\ln\left(\frac{E_m-p_m}{E_n-p_n}\right)+\ln\left(\frac{E_n+p_n}{E_m+p_m}\right)\right]$$ and then it's should just be rearranging the factors in a convenient way. I hope that helped!

$\endgroup$
1
  • $\begingroup$ Why should the angle between q and pm, and q and pn be the same (phi)? $\endgroup$
    – nox
    Oct 9, 2021 at 5:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.