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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..

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  • $\begingroup$ 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? $\endgroup$ – joigus Feb 17 at 11:33
  • $\begingroup$ 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. $\endgroup$ – user248824 Feb 17 at 11:37
  • $\begingroup$ 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... $\endgroup$ – Cosmas Zachos Feb 17 at 16:23

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