Let $s,t,u$ be the usual Mandelstam variables defined by $s = (p_1 + p_2)^2$, $t = (p_1-p_3)^2$ and $u = (p_1 - p_4)^2 = 4m^2 - t -s$. I am trying to convice myself of that $stu \approx -s^2 t$ for large $s$. But for me it is not so clear why the term $st^2$ can be neglected, i.e. why $t \ll s$ would hold.
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asked Oct 19, 2021 at 9:13
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$\begingroup$ $\varphi \varphi \rightarrow \varphi \varphi$ with $\varphi$ a scalar field of mass $m$ $\endgroup$– Mathphys meisterCommented Oct 20, 2021 at 8:20
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$\begingroup$ Masses go to zero at high energies, and all 4-momenta are light like. But... at right angle scattering, your statement fails, of course: $t\approx -s/2$. You want a toy counterexample? $\endgroup$– Cosmas ZachosCommented Oct 20, 2021 at 12:59
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Suppose in the CM frame particle $1$ has $3$-momentum $p\hat{e}_i,\,p\hat{e}_j$ before and after the interaction, with $\hat{e}_i\cdot\hat{e}_i=1,\,\hat{e}_1\cdot\hat{e}_2=\cos\theta$. Then$$s=4(m^2+p^2),\,t=-2p^2(1-\cos\theta),\,u=-2p^2(1+\cos\theta).$$Since $-\frac{u}{s}=\frac{\cos^2(\theta/2)}{1+m^2/p^2}$, you need to convince yourself $\theta\to0$ as $s\to\infty$.
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$\begingroup$ Do you mean $\theta \rightarrow \pi$? $\endgroup$ Commented Oct 20, 2021 at 12:30
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$\begingroup$ @Mathphysmeister That would make $-u/s$ small, but you need it to approximate $1$. $\endgroup$– J.G.Commented Oct 20, 2021 at 12:50