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I would like to ask if the nuclear and sub nuclear particles react to heat energy supplied to them. For example the quark triads in protons and neutrons?

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closed as unclear what you're asking by John Rennie, John Duffield, CuriousOne, Kyle Kanos, ACuriousMind Mar 25 '16 at 11:57

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    $\begingroup$ How do you define "temperature" in this case? $\endgroup$ – CuriousOne Mar 23 '16 at 19:10
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    $\begingroup$ Same question... what is "heat" energy supposed to be in this case? The closest you can come to a thermodynamic scenario is a quark-gluon plasma. Is that what you are asking? $\endgroup$ – CuriousOne Mar 23 '16 at 19:14
  • $\begingroup$ no i am asking about the scenerio in which neutrons and protons are fused due to great heat and pressure. I would like to know the situation of the quarks present inside them. $\endgroup$ – user111211 Mar 23 '16 at 19:20
  • $\begingroup$ It's similar to what happens to nuclei during chemical reactions, they are mostly unaffected. Fusion doesn't change the number of neutrons and protons. $\endgroup$ – CuriousOne Mar 23 '16 at 20:02
  • $\begingroup$ Closely related: physics.stackexchange.com/q/114120/44126 $\endgroup$ – rob Mar 24 '16 at 23:10
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Thermal energy affects condensed matter, molecules, atoms, and nuclei by exciting atomic transitions with energies of roughly $E=kT$ and below. At room temperature $T=300\,\rm K$, $kT = 25\,\rm meV$ you mostly excite vibrational states in solids, whence the expression "heat is jiggling atoms." At room temperature you can generally make molecules rotate, but not vibrate. I don't have vibrational energies at hand, but you start to make electronic excitations in atoms when you get $kT \approx 1\,\rm eV$, corresponding to $T\approx 12\,000\,\rm K$. Since molecules are held together by electronic interactions, this is the sort of temperature where you can reliably dissociate molecules and you expect to find most atoms not in the ground state.

Generally the first excited state in a nucleus is a couple of MeV above the ground state. So from the perspective of a nucleus any temperature below hundreds of millions of kelvin is "absolute zero," in the sense that you are astoundingly unlikely to have a thermal excitation drive the nucleus away from the ground state. Ordinary temperatures do not have enough energy to thermally excite nuclei.

With quarks bound into nucleons the situation is more extreme. The "first excited state" of the nucleon is the $\Delta$ baryon with mass around 1200 MeV. You won't find thermally-produced $\Delta$s until $kT$ approaches 200 MeV.

Ordinarily you can't talk about temperature unless you are pretty close to a thermal equilibrium; by definition, almost, all the nuclear reactions we observe are far from equilibrium. However, "temperature" turns out to be a pretty good way to describe the energy density in a strongly interacting fluid like the quark-gluon plasma, which was the state of matter for the first minute or so after the Big Bang.

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