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I would like to get an understanding of the value of the Hagedorn temperature and the units this temperature can be given in. Is the Hagedorn temperature the maximum temperature, just as $0\ K$ is the lowest temperature? What happens to matter when the Hagedorn temperature is reached?

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Jens Vigen
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    Wellcome to PSE. Why don't you search the Web first ??? – Frobenius Jun 30 '20 at 06:38
  • There is a formal (mathematical) condition for the existence of the Hagedorn temperature in the system, it must be  a system with exponential growth of the density of states. Then at a certain temperature the partition function of the system diverges. In this sense, Kittel’s simplest statphysical model (molecular zipper) can apparently serve as a toy model illustrating this process. – Aleksey Druggist Jul 01 '20 at 15:55
  • I need to be educated on "Kittel’s simplest statphysical model (molecular zipper) " as such model was not defined in the classroom I teach and the only google hit is this comment. Moreover, I do remember that Hagedorn temperature was associated with exponential growth of mass spectrum. Singularity of deensity of states is a microcanonical feature associated in general with phase transitions. These two things seem different to me. – JohannR Jul 03 '20 at 14:05
  • Why not to browse wiki pointed in OP question, topics "History" and "Technical explanation" :" increase the entropy of the system rather than the temperature" and "HT is the temperature above which the partition sum diverges in a system with exponential growth in the density of states" ...Singularity of deensity of states* - this phrase is completely incomprehensible to me – Aleksey Druggist Jul 03 '20 at 15:57
  • Reading Wiki is not always illuminating: many inputs addressing also general public. The gen-ed physics CERN Courier article https://cerncourier.com/a/the-tale-of-the-hagedorn-temperature/ explains just your Q in fig 1: within a specific model (not in gen) called SBM (statitstical bootstrap model) with point particles there is correspondance of singularities, but not for fin size. Hagedorn explains this also himself in video available on YouTube https://www.youtube.com/playlist?list=PLCVlSP-AskCfm0L23K9-4Vggmqp35gyl7 . Thus even with exp. mass spectrum FS hadrons can dissolve into QGP. – JohannR Jul 05 '20 at 12:33
  • I agree, Wikipedia does not always clarify the situation, let's turn to the original article by Hagedorn http://cds.cern.ch/record/346206/files/CM-P00057114.pdf , see formulas (30) and (31) on page 19, which demonstrate the main idea - postulating the exponential dependence of the density of states on energy (with some complication in the form preexponential factor $E^{\alpha -1}$. This leads to the divergence of the partition function at a certain critical temperature (for a purely exponential density, we must put $\alpha=1$. The math is identic as I noted in my initial comment – Aleksey Druggist Jul 06 '20 at 00:46
  • Hagedorn 1965 was for point particles, these had common singular point in mass spectrum and density of states in SBM. But by 1979 we allowed finite hadron volume which is widely accepted - earliest mention is here http://cds.cern.ch/record/134674/files/CM-P00055555.pdf. (reprinted here https://link.springer.com/chapter/10.1007%2F978-3-319-17545-4_23). In the model published you never reach singular point of density of states as hadrons melt.Singular density of states with compressible hadrons disappears, phase transformation (not transition) to QGP while keeping exp-mass spectrum is OK. – JohannR Jul 06 '20 at 04:53
  • @JohannR, let me express my deep respect to you, I'm completely ignorant of particle physics, I was just attracted by the formal similarity of the behavior of the partition function in two different models (it seems to me that one can add the Peierls droplet model here) – Aleksey Druggist Jul 06 '20 at 12:33

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The Hagedorn temperature is a maximum temperature in the same sense that the boiling point of water is a maximum temperature: namely, it's the maximum temperature for a particular "phase" of matter, and going beyond that temperature requires a "phase transition". In the case of water, this phase transition is evaporation (aka boiling); once the water has evaporated, it can begin to reach higher temperatures. In the case of hadronic matter, the phase transition is the quark-hadron phase transition; above the Hagedorn temperature, quark matter takes the form of a "quark-gluon plasma".

probably_someone
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Hagedorn Temperature is the temperature at which the strong force gluing quarks in hadrons is exceeded by quarks' vibration energy, thus making ordinary matter disintegrate into quark matter. After matter conversion into quark plasma, this plasma can be heated further until Planck temperature is reached. So Hagedorn Temperature is not highest possible particle ensemble temperature, but just the temperature of phase transition $T_{\text{matter} \to \text{quark plasma}} $. Maximum possible temperature vs. Hagedorn Temperature ratio is $$\frac {T_{\text{Planck}}}{T_{\text{Hagedorn,Quarks}}} \approx 10^{20}$$ In String theory, a separate Hagedorn Temperature can be defined, this time not for hadrons, but for strings. In effect this means that quark plasma is disintegrated further into "String Plasma", i.e. interaction forces between Strings are exceeded by String energy, thus breaking matter further into a soup of strings. This string phase transition temperature is very near Planck scale: $$ \frac {T_{\text{Planck}}}{T_{\text{Hagedorn,Strings}}} \approx 100 $$ So it's not likely that this can be tested in laboratories in the near future.

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Agnius Vasiliauskas
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