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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.

enter image description here

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?

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  • $\begingroup$ 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 $e^-+p\rightarrow e^-+n+\pi^+$. $\endgroup$ – Lox Aug 24 at 7:50
  • $\begingroup$ 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. $\endgroup$ – Lox Aug 24 at 8:07
  • $\begingroup$ @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 $\endgroup$ – Elpis Aug 24 at 8:19
  • $\begingroup$ 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 $\Delta^+$ $\endgroup$ – Lox Aug 24 at 8:50
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I don't understand what is being displayed, do our detectors differentiate between a simple collision and one where the resonant particle intervenes?

The experiment in your given plot, starts with a fixed number/Delta(E) which for some (confusing) reason they call Z. Usually the protons are stationary and the energy of the electrons is increased step by step and the number of events with three identified particles, e , pion, neutron are counted per bin.

The gray is the measured plot. The black line is the theoretical curve one would fit if there were no extra attraction ( resonance) between the electron and the proton but just the electromagnetic interaction. Then one counts how many standard deviations the peaks and valleys are from the theoretical smooth curve. If the difference is more than four standard deviations from the theoretical curve , one has a definite signal of an extra attractive force.

As the y axis is not given numerically, one trusts they have done a good job in their analysis and the blue curve shows the fit with the resonances and the expected background added.

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  • $\begingroup$ thank you, the sentence "Then one counts how many standard deviations .." was very clarifying for me. However you introduce a new concept that i don't know, that is "the attraction between the electron and the proton". Can we assume that the local maxima with respect to the theoretical curve correspond to an attractive interaction and the valleys to a repulsive interaction in the scattering? How does having attraction or repulsion affect the type of resonant particle formed? I have never interpreted these processes in this way so I would be grateful for the clarification $\endgroup$ – Elpis Aug 24 at 8:36
  • $\begingroup$ in electron proton interactions the first order feynman diagram is electromagnetic, with the exchange of a virtual electron, researchgate.net/figure/… . Then the proton can be excited into its components, the quarks, which allows new couplings bewtween quarks and that is how the resonances are formed. The word attractve i use comes from the 1/r attractive electromagnetic potential that enters to first order giving the virtual photon. $\endgroup$ – anna v Aug 24 at 12:56
  • $\begingroup$ extra because the excess must be due to something else that gives the statistically signivicant excess to the black curves number of events. $\endgroup$ – anna v Aug 25 at 3:39

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