Take this form of the QCD Phase Diagram for example:

QCD Phase Diagram

This baryon density is a number density - i.e. number of baryons in some volume.

Why are baryon density and temperature regarded as independent parameters in the QCD Phase Diagram?

The following qualitative reasoning is the source of the trouble:

For conventional matter, the number density and temperature are clearly not independent. If you compress a fixed number of molecules of a gas, thereby increasing the number density, molecules would be closer together, more collisions and hence, higher temperature.

What is the physical reason/ justification for the same, in the context of QCD?

Is it related to the time-scale of a high energy collision (a.k.a. the lab for these sort of things) - i.e. produced particles can't diffuse out fast enough and hence, can be simultaneously hot and dense, at least for the time scale of the collision? (Clearly, this rough hand-waving argument can't be the true reason. So, what is it?)

(Image Source = This )


2 Answers 2


Usually phase diagrams, e.g. of water, are shown as pressure vs temperature. However, we could just as well write a phase diagram as temperature vs density. This is because the equation of state of matter is a relation between pressure, density and temperature. The above phase diagram should be interpreted as showing: what is the state of matter at a certain density and temperature? For example, for high densities (particles heavily packed together) and high temperatures (large average kinetic energy), which occurs for example at beginning of the universe, the state of matter is expected to be a quark-gluon plasma.

A simple way to see why your reasoning fails is to consider a gas which satisfies $p=n kT$. At high densities (interparticle distance small) and but very low temperatures (small average kinetic energies), the pressure is very low. At low densities (interparticle distance large) and very high temperatures (large average kinetic energy), the pressure is high.

  • $\begingroup$ Re: Para 2 - So? $\endgroup$
    – 299792458
    Sep 25, 2014 at 15:39
  • $\begingroup$ I didn't want to sound blunt, but isn't that $n$ actually $n/V$? So, reading this as a number density, how do you arrive at the deductions? And what does that establish? (Sorry if I'm being a dimwit.) $\endgroup$
    – 299792458
    Sep 25, 2014 at 15:43
  • $\begingroup$ Note that equation of state for quark soups are a little bit different than the equation of state for an ideal gas. $\endgroup$
    – Kyle Kanos
    Sep 25, 2014 at 16:32
  • $\begingroup$ The n=N/V is the density, so the number of particles N over the volume V. Similarly, you wrote in your question in the QCD phase diagram you have the baryon density, which is the number of baryons in some volume. Also Kyle: I simply gave the ideal gas law as an example as the equation of state of quarks is still actively researched and wanted to avoid confusion, however, the arguments still hold. $\endgroup$
    – Jasper
    Sep 25, 2014 at 20:41

I think you are right to be confused by the nuclear phase diagram, but I will rephrase your question to show you why there is no contradiction in "hot and dense".

I believe we all understand that the density can be regarded as a proxy for the specific volume for a fixed number of particles. So I will take it from there.

Below is a typical P-V-T phase diagram. The three parameters P-V-T are not independent, but any two are.

p–v–T 3D diagram for fixed amount of pure material - From Wikipedia: Phase diagram

And here cuts perpendicular to the T-axis (constant temperature):

PV diagrams under constant temperature

The 3D phase diagram therefore shows you 1) that it is possible to vary the specific volume (read: density) without changing the temperature. 2) what pressure to apply in order to achieve a given $(V,T)$ combination.

In your example: You start with some $(P_0,V_0,T_0)$; you apply pressure getting to $P_1$, in order to decrease the specific volume down to some desired value $V_1$; the value of temperature $T_1$ which you reach will depend on both $V_1$, $T_1$. The phase diagram gives you that dependence.

As regards the nuclear phase diagram, it leaves unclear what happens to the pressure. More precisely, it leaves out the third axis, whatever it is. That would be a different question.

[Edit] I should have added a V-T diagram, to make the analogy with you QCD phase diagram more clear - I am not allowed to post one now, but I am sure you can find one easily.


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