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Narrow Resonace of e+eby forceface
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#1
Sep413, 01:46 PM

P: 9

I was reading the paper about the discovery of the first e+e narrow resonancehttp://prl.aps.org/abstract/PRL/v33/i23/p1406_1 and one question that came to my mind, which was why was this discovery important. In my naive understanding I would believe that this pair annihilation could happen anytime, where at low energies only photons would be produced, but at high energies more massive particles has the possibility of being produced and the latter is what is happening in the paper. In my mind this production of particles would be represented by something more steady than a sudden resonance. As the energy increases I would think the cross section for scattered hardons would increase a related rate.



#2
Sep413, 02:30 PM

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PF Gold
P: 2,602

The fact that the crosssection vs centerofmass energy has peaks near the masses of unstable resonances can be understood fairly simply. We have a collection of processes ##e^+e^ \rightarrow X \rightarrow \mathrm{anything}##, where ##X## is one of the unstable particles that can be produced. The amplitude for a specific process looks like (we are assuming ##X## is a boson for simplicity)
$$ A = c_1 \frac{1}{p^2  M^2 + i M\Gamma} c_2.$$ Here ##M## is the mass of the resonance ##X## and ##\Gamma## is its decay rate. The factor in the middle is the propagator for ##X## at 4momentum ##p##. The factor ##c_2## describes the details of the interaction between ##X## and the ##e^+e^## pair, while ##c_1## describes the interaction of ##X## with a specific collection of particles in the end state. The inclusive cross section would involve summing over all possible final states. The crosssection depends on the modulus square of the amplitude and we can imagine expanding the result in terms of the ratio of the spatial components of the ##X## momentum with respect to the energy of the particle. For centerofmass energy ##E## close to the mass of the resonance, most of the energy must go into producing the particle, so ##E\sim M## and the spatial momenta must be very small. If we neglect these corrections, we can write the corresponding probability in the form $$ P \sim \frac{k}{(E^2 M^2)^2 + (M\Gamma)^2},$$ which is known as the BreitWigner distribution. We use the variable ##k## to contain all of the precise details of the interactions involved in a specific process. You can verify that this function has the form of a curve with peak centered at ##E=M##. The width of the peak is related to the decay rate: the larger the decay rate, the broader the peak. Conversely, the smaller the decay rate (or more stable the particle is), the narrower the peak is. We conclude from this that, whatever the detailed form of the crosssection for ##e^+e^ \rightarrow \mathrm{anything}##, we should find peaks around the masses of any resonances that are permitted to be produced at the given energy. At any given energy, there is a background corresponding to the events where we produce resonances with particles much lower than the c.o.m. energy, but whenever there is a stable enough resonance, there will be a welldefined peak in the spectrum that sticks out of the background. To be totally convincing, one would have to do the hard work of actually computing the details of the amplitudes involved, or better yet, actually make the measurements and verify that the resonance peaks do in fact stand out from the background events. The BreitWigner distribution is an approximate description that actually turns out to be very useful to discuss certain properties of the observations. 


#3
Sep613, 01:40 AM

P: 9

So in other words this is the J/Psi particle.



#4
Sep613, 02:24 AM

Mentor
P: 12,113

Narrow Resonace of e+e
That was the first observed resonance involving charm quarks, right.



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