Taking e^- e^+ →e^- e^+to be the s channel process, verify that s=4(k^2+m^2) t=-2 k^2

Step 1: Define the Mandelstam variables. The Mandelstam variables ( s ), ( t ), and ( u ) are de

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Taking $\mathrm{e}^{-} \mathrm{e}^{+} \rightarrow \mathrm{e}^{-} \mathrm{e}^{+}$to be the $s$ channel process, verify that $$ \begin{aligned} & s=4\left(k^2+m^2\right) \\ & t=-2 k^2(1-\cos \theta) \\ & u=-2 k^2(1+\cos \theta) \end{aligned} $$ where $\theta$ is the center-of-mass scattering angle and $k=\left|\mathbf{k}_i\right|=\left|\mathbf{k}_j\right|$, where $\mathbf{k}_i$ and $\mathbf{k}_f$ are, respectively, the momenta of the incident and scattered electrons in the center-of-mass frame. Show that the process is physically allowed provided $s \geq 4 m^2, t \leq 0$, and $u \leq 0$. The physical region is shown shaded on Fig. 4.7. Note that $t=0(u=0)$ corresponds to forward (backward) scattering.

   Taking $\mathrm{e}^{-} \mathrm{e}^{+} \rightarrow \mathrm{e}^{-} \mathrm{e}^{+}$to be the $s$ channel process, verify that
$$
\begin{aligned}
& s=4\left(k^2+m^2\right) \\
& t=-2 k^2(1-\cos \theta) \\
& u=-2 k^2(1+\cos \theta)
\end{aligned}
$$
where $\theta$ is the center-of-mass scattering angle and $k=\left|\mathbf{k}_i\right|=\left|\mathbf{k}_j\right|$, where $\mathbf{k}_i$ and $\mathbf{k}_f$ are, respectively, the momenta of the incident and scattered electrons in the center-of-mass frame. Show that the process is physically allowed provided $s \geq 4 m^2, t \leq 0$, and $u \leq 0$. The physical region is shown shaded on Fig. 4.7. Note that $t=0(u=0)$ corresponds to forward (backward) scattering.

Quarks and leptons: introductory course in modern particle physics

Francis Halzen, Alan… 1st Edition

Chapter 4, Problem 6

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Step 1

The Mandelstam variables $ s $, $ t $, and $ u $ are defined as follows for any two-body scattering process: \[ s = (p_1 + p_2)^2 \] \[ t = (p_1 - p_3)^2 \] \[ u = (p_1 - p_4)^2 \] where $ p_1 $ and $ p_2 $ are the four-momenta of the incoming particles, Show more…

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Taking $\mathrm{e}^{-} \mathrm{e}^{+} \rightarrow \mathrm{e}^{-} \mathrm{e}^{+}$to be the $s$ channel process, verify that $$ \begin{aligned} & s=4\left(k^2+m^2\right) \\ & t=-2 k^2(1-\cos \theta) \\ & u=-2 k^2(1+\cos \theta) \end{aligned} $$ where $\theta$ is the center-of-mass scattering angle and $k=\left|\mathbf{k}_i\right|=\left|\mathbf{k}_j\right|$, where $\mathbf{k}_i$ and $\mathbf{k}_f$ are, respectively, the momenta of the incident and scattered electrons in the center-of-mass frame. Show that the process is physically allowed provided $s \geq 4 m^2, t \leq 0$, and $u \leq 0$. The physical region is shown shaded on Fig. 4.7. Note that $t=0(u=0)$ corresponds to forward (backward) scattering.

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