# Understanding the $Z$ production cross section in $e^{+}e^{-}\rightarrow Z\rightarrow \mu^{+}\mu^{-}$

In "Modern Particle Physics", Mark Thomson derives the total cross section for $$\sigma\left(e^{+}e^{-}\rightarrow Z^{0}\rightarrow\mu^{+}\mu^{-}\right)$$, cf. subchapter 16.1.1 on page 431 ff. He assumes that $$\sqrt{s}\sim m_{Z}$$, s.t. the QED Feynman diagram is in first approximation negligible.

Now, in the first Eq. on page 431 that does not have a number, i.e. $$\mathcal M_{fi} = - \frac{g_{Z}^{2}g_{\mu\nu}}{\left( s-m_{Z}^{2} + im_{Z}\Gamma_{Z}\right)}\dots,$$ I do not understand the term that Mark Thomson wrote down, which is supposed to be the propagagtor, I assume. Usually, we would write the $$W^{\pm}/Z^{0}$$ propagator as

$$\frac{-i\left( g_{\mu\nu} - q_{\mu}q_{\nu}/m_{Z}^{2}\right)}{\left(q^{2}-m_{Z}^{2}+im_{Z}\Gamma_{Z}\right)}.$$ In an $$s$$-channel Feynman diagram, $$s = q^{2}$$, but what happened to the term $$q_{\mu}q_{\nu}/m_{Z}^{2}$$? After all, in the following derivation, M. Thomson does not ignore the term $$s-m_{Z}^{2}$$ in the denominator..

• I'm not very fresh on this, but I've met similar apparently puzzling approx. in quantum statistical mechanics getting the $T=0$, $T=\infty$ limits. Example: $$\frac{e^{-\varepsilon}}{1-e^{-\varepsilon}}\underset{\varepsilon\rightarrow\infty}{\simeq}e^{-\varepsilon}$$ The lowdown is: using asymptotics is very sensitive to whether you do it in the numerator or the denominator. Could that be something like it? – joigus Feb 17 at 11:33
• Dear joigus, I believe in my case, it is for sure the nominator that we approx., as the denominator is left as it should be. – user248824 Feb 17 at 11:37
• I'm shadow-boxing here, lacking the book, but have you checked $q^\mu$ dotting on the initial and final spinors' bilinear does not resolve to $p\!/u(p)\sim m u(p)$s, and get suppressed by $m_Z$ in the denominator? The leptons are essentially massless compared to the the Z... – Cosmas Zachos Feb 17 at 16:23