# Detection of resonance particles

why we detect an increment in the number of events of the reactions for the energy of resonance particles? I read that cross section of this new unstable particle must be added to the cross section of the target particle but i cannot understand the reason. A little example of concepts i cannot understand, we have a process like

$$\begin{equation}\mathrm{e}^{-}+\mathrm{p} \longrightarrow \mathrm{e}^{-}+\Delta^{+}\end{equation}$$

where after $$\approx 10^{-23}$$ s we have

$$\begin{equation}\mathrm{e}^{-}+\mathrm{p} \longrightarrow \mathrm{e}^{-}+\pi^{+}+\mathrm{n}\end{equation}$$

via

$$\begin{equation}\Delta^{+} \longrightarrow \pi^{+}+\mathrm{n}\end{equation}$$

the compatible collisions must satisfy

$$\begin{equation}m_{\Delta} c^{2}=\sqrt{\left(E_{\pi}+E_{\mathrm{n}}\right)^{2}-\left(\mathbf{p}_{\pi}+\mathbf{p}_{\mathrm{n}}\right)^{2} c^{2}}\end{equation}$$

so if we detect and increment in the number of events in such collisions we can say that we have a resonant particle. where

$$\begin{equation}Z=\sqrt{\left(E_{\pi}+E_{\mathrm{n}}\right)^{2}-\left(\mathbf{p}_{\pi}+\mathbf{p}_{\mathrm{n}}\right)^{2} c^{2}}\end{equation}$$

I don't understand what is being displayed, do our detectors differentiate between a simple collision and one where the resonant particle intervenes? Is it possible that the peak detected is due to the fact that the detector is counting both processes if they occur? And finally, should we associate only the local maxima when representing number of events versus energy with resonant particles or can we also associate the valleys?

• Can you put a reference to the plot you're displaying or to the experiment? In this kind of experiment, you usually measure the same final states and you count the number of reactions that occurred. The resonance shows an increment of final state production wrt due to the production of a short living particle. I.e. you count the number of processes $e^-+p\rightarrow e^-+n+\pi^+$. – Lox Aug 24 at 7:50
• In the context of DIS, the differential cross-section $\frac{d\sigma}{d\Omega dE_3}$ ($E_3$ the energy of the outgoing electron) can be seen as a distribution of $E_3$ at fixed $\theta$ and the invariant mass of the $\pi^+ + n$, $m_X^2$, is a linear function of $E_3$ therefore you can relate the cross-section to the invariant mass of the final state. – Lox Aug 24 at 8:07
• @Lox I don't know how to put a image reference in physics.stackexchange.com but is from: Serway R.A., Moses C.J., Moyer C.A. Modern physics p. 567 – Elpis Aug 24 at 8:19
• Ok. In this case, the dashed curve is the distribution of produced $\pi^+ + n$ in the case where no $\Delta^+$ is produced. "I read that cross section of this new unstable particle must be added to the cross section of the target particle but i cannot understand the reason" it is not clear what you mean for cross-section of the target particle. You must add the cross-section of $\sigma(no\ \Delta^+)+\sigma(\Delta^+)$ to obtain the blue curve on the first resonance. Z is the invariant mass of the $\pi^+ + n$ system since you can produce it as a final state independently from the $\Delta^+$ – Lox Aug 24 at 8:50