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Can there only be a critical temperature if there is some natural unit for an observable in the model, i.e. if there is a natural scale for something? Otherwise I don't see how for a system there could be a rule how the value of $T_c$ gets actually terminated. And do this transitions vanish from the model, if one does a limit where these units get irrelevant?

Is it a priori arbitrary which quantity (lenght, energy, charge,...) has to have the specified unit? And/Or have there to be more such units to make up a critical temperature? Lastly, can there only be a phase transition in a system if there is an associated critical temperature $T_c$?

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Ultimately for those kinds of systems you are taking $T_c$ and other thermodynamical properties are functions of the intermolecular potential. That is it is rather a whole function than a single energy value (scale). Thermodynamic properties depend on it via complex multidimensional integrals. – Yrogirg Jan 30 '12 at 17:20
For example, for Lennard-Jones model potential ( $T_c = 1.326 \varepsilon$ where $\varepsilon$ is the depth of the potential. – Yrogirg Jan 30 '12 at 17:27
@Yrigurg: Okay, so the answer to the first question is yes, there have to be some kind of parameters with certain values, which fixate the specific model? – NikolajK Jan 30 '12 at 17:39
The model can be defined with infinite number of numbers. For example if we are talking about simplest pair potential $f(r)$ it is a function $f : [0, \infty) \to \mathbb R$ --- that's a whole energy curve. Though I guess some rough estimation of $T_c $can be made based on a single energy value --- for example the depth of the potential. – Yrogirg Jan 30 '12 at 17:56
You're talking about postulating a whole space dependent function? In any case, the answer to my comment question is yes, right? – NikolajK Jan 30 '12 at 18:04

Phase transitions are abundant and occur also in athermal systems. For example, a collection of hard spheres will undergo a phase transition from a liquid-like state to a solid-like state when the volume ratio of the spheres is larger than some value. A graph with random links will undergo a phase transition if the average degree of a node is more than 1. Another good example is percolation.

So clearly, phase transitions can occur also in systems where energy plays no role. However, when a phase transition occurs as function of temperature, it is clear, on physical grounds, that the critical temperature will scale as some basic energy scale of the system. The problem is that usually you have quite a few energy scales.

As to your last question - phase transitions might occur when you change any of the control parameters. Take the liquid-vapor-solid phase transition for example. The phase space is plotted in this image, courtesy of wikipedia:

enter image description here

According to the current parameters $(T,P)$, your system is at a given point in this phase-space. When you change either of these parameters, you might cross a boundary between two phases - this is a phase transition.

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thx for the response. In "it is clear, on physical grounds, that the critical temperature will scale as some basic energy scale of the system", what does "the critical temperature will scale" mean? My main question if the value where phase transitions occur due to some other distinguished scale (such as atom radius). – NikolajK Jan 22 '12 at 13:42
If you set $k_b=1$, and there's no reason not to do so, then temperature is measured energy units. So clearly, if you multiply ALL energy scales in your system by 2, then $T_c$ will also be multiplied by 2. This is what I meant. I can't quite understand your question beyond that. – yohBS Jan 22 '12 at 15:29
I mean that if there is no smallest natural unit (not just some random energy unit for counting, but a physical scale), then I don't see why such a thing as a critical temperature could even emerge. I ask therefore if a critical temperature can only exist in a model with some natural unit (atom radius was my example) which makes up the formula for $T_c$. – NikolajK Jan 22 '12 at 16:10
I cannot imagine such a case. If the phase transition occurs as a function of $T$, this means that you're working with a Hamiltonian, and average expressions of the form $e^{-\beta H}$. And if you have a Hamiltonian, you have an energy scale. – yohBS Jan 22 '12 at 18:21
"And if you have a Hamiltonian, you have an energy scale" Why, necessarily? – NikolajK Jan 22 '12 at 19:12

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