For a centre-of-mass energy of $3\,\mathrm{GeV}$ is it possible to make a charm quark?

Question

This is not strictly a homework question (but I have added the tag anyway). I am using an answer from a given question which states you cannot make a charm quark if given a centre-of-mass energy $3\,\mathrm{GeV}$ as evidence that is contradicatory to another source which says you can make a charm quark with a centre-of-mass energy $3\,\mathrm{GeV}$.

Here is the question from the first source (Imperial College London):

with the given data required.

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I will typeset the correct answers to (i), (ii) and (iii):

$(\mathrm{i})$ Each quark type has a production cross section proportional to $$(ee_q)^2=e^4\left(\frac{e_q}{e}\right)^2=e^4 {Q_q}^2=\alpha^2 {Q_q}^2$$ Any quark type produced gives hadrons so the total hadron production cross section is $$\sigma(\mathrm{Hadrons})\propto \sum_q {Q_q}^2$$ where the sum is over all types of quarks which can be produced. Hence, the ratio of the cross sections is simply $$R=\frac{\sigma(\mathrm{Hadrons})}{\sigma(\mu^+\mu^-)}=\sum_q {Q_q}^2=3\sum_f {Q_f}^2$$ where the latter sum above is over quark flavours and the factor of three arises as each exists with one of three colours.

At $6\,\mathrm{GeV}$, no bottom or top quark involved, so ratio becomes $$\sum_f {Q_f}^2=\left(+\frac{2}{3}\right)^2+\left(+\frac{2}{3}\right)^2+\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^2=2\times \frac49 +2\times \frac19=\frac{10}{9}\approx 1.1$$ for the $u$, $c$, and $d$, $s$ quarks respectively. With colour, the sum is three times larger, i.e $\color{blue}{\dfrac{10}{3}\approx 3.3}$

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$(\mathrm{ii})$ At $35\,\mathrm{GeV}$ all quarks but the top quark will be involved. In the absence of colour, the sum over the quark flavours gives $$\sum_f {Q_f}^2=\left(+\frac{2}{3}\right)^2+\left(+\frac{2}{3}\right)^2+\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^2=2\times \frac49 +3\times \frac19=\frac{11}{9}\approx 1.2$$ for the $u$, $c$, and $d$, $s$, $b$ quarks respectively. With colour, the sum is three times larger, i.e $\dfrac{11}{3}\approx 3.7$

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$(\mathrm{iii})$ At $3\,\mathrm{GeV}$, one is close to twice the $c$ quark mass so would not expect asymptotic freedom to hold so the above formula does not apply, and in fact would expect to see resonant behaviour.

I have no idea what the above answer means and I have already read this post on Asymptotic Freedom and don't really understand it.

But I thought the correct answer should be $$R=3\left(2\cdot\frac29+2\cdot\frac19\right)=\frac83 + \frac23=\color{blue}{\frac{10}{3}\approx 3.3}$$ as before with the $6\,\mathrm{GeV}$ case in part $(\mathrm{i})$.

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I have done some research and found out that I am not the only one who thinks this is the case; see page $3$ of this pdf from Southhampton University or below:

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Obviously both universities cannot be right, so the obvious question is; Which one is correct?

If you believe (or know) Southhampton university is wrong and Imperial College London is correct then could you please explain why asymptotic freedom would have to hold in order for $$R=\frac{\sigma(\mathrm{Hadrons})}{\sigma(\mu^+\mu^-)}=\sum_q {Q_q}^2=3\sum_f {Q_f}^2$$ to apply and what is meant by "resonant behaviour"?

Answer 1

There are a couple of things gong on here. One is in the value of the quark masses that you are given, and the other is related to threshold behavior.

1. If the quark mass you are given is taken from a constituent quark model if may be low. (But the PDB says $m_c = 1.0$–$1.6\,\mathrm{GeV}$, so not by much.)

2. The actual cross-section for reactions like this depends on the phase space for the products, and while that is roughly equal for the numerator and denominator in a most cases, it isn't true near threshold for a particular particle. At CoM energies just slightly larger than $2m_c$, the phase space for charm production is very much smaller than than for production of lighter quarks and hadrons and the expression for $R$ lies a little: the real value is smaller than expected.

   So while you "can" produce charm-anticharm pairs at threshold it actually happens so rarely as to be experimentally moot. In fact, experimental measurement of production thresholds are very difficult for exactly this reason and masses are usually measured in more subtle ways.

Answer 2

The plot in the answer of Luc J. Bourhis shows that the disagreement is a matter of context.I am referring to that plot in this complimentary answer.

Your question

"resonant behaviour"?

Resonant behavior in scattering particles,e+e- in this case, indicates a particle with the invariant mass of the two input particles, in this case a Ψ, but different quantum numbers in the decay products .

Looking at the plot provided:

At the Ψ resonance the existence of new quarks was established, and the need of 3 GeV energy is supported .

The creation of charm quantum numbers in separate charmed hadrons in order to see the charges contributing needs more energy . The creation of D mesons almost with a mass close to 2 GeV, in the plot, become evident in the ratio at around 4 GeV .

At 3GeV, one is close to twice the c quark mass so would not expect asymptotic freedom to hold so the above formula does not apply, and in fact would expect to see resonant behaviour.

Asymptotic freedom is the assumption of QCD that at very high energies the quarks will be free , and the paragraph uses it for distinguishing between constituent mass and current mass. Current mass is the mass that quarks have in QCD calculations and the one shown in the standard model . Quarks are "dressed" with virtual particles within hadrons and the charge argument cannot be used. They are handwaving the argument that at the resonance of Ψ it is a bad approximation to assume that the e+e- generate just the current quarks, the extra quarks-antiquarks-gluons in a hadron will dress the charmed quark into its constituent mass . They assume that current quarks are involved when there are no hadronic resonances to mask the charges. They equate the use of the current quarks masses to asymptotic freedom.

(in calculations with Feynman diagrams always the current masses are used)

In summary:

If one assumes the Ψ peak as a clear indication of new quarks needed to interpret it, one goes to 3 GeV . If one wants to measure the charge and color content the plot shows that one needs 4 GeV and more

Answer 3

Physics is an experimental science, so let's see some actual measurements. Here is the ratio $R$ plotted at low energy, from [BP11]. So clearly, the formula

$$R=\frac{\sigma(\mathrm{Hadrons})}{\sigma(\mu^+\mu^-)}=3\sum_f {Q_f}^2$$

starts to be valid only above 5 GeV. At 3 GeV, it is clearly falsified.

[BP11] H. Burkhardt and B. Pietrzyk. Recent bes measurements and the hadronic contribution to the qed vacuum polarization. Phys. Rev. D, 84:037502, Aug 2011.