8
$\begingroup$

Why do phase transitions even exist? Why not smooth density change curves? What properties of matter, quantum or otherwise, predicts that matter will undergo phases at different pressures and temperatures?

Some materials have all phases, others are missing some.

If this were to be researched from very little existing knowledge, a great place to start would be by examining the differences between materials that are missing some phases, and comparing them with ones that aren't.

Improve this question
$\endgroup$
1
6
$\begingroup$

Phase transitions are a many-body effects. You can not generate sharp transition with a finite number of degrees of freedom (or particles). However as you add particles the features of the system may become sharper. In the limit of infinitely many particles (thermodynamic limit) you get a truly discontinuous transition.

In practice nothing is infinite. The typical number of atoms in normal matter is however 10^(23) which is indistinguishable from infinity in the sense that phase transitions appear perfectly sharp.

The mechanism from which the transition occurs depends on the particular system and transition that you consider. For example, in the case of water freezing, you have a competition in between disorder (temperature) and atom-atom interaction. The atoms want to stick together in ordered pattern (crystal) but the temperature wants them to have random position. There is a critical temperature above which the disorder wins and below which it is the potential energy that is stronger.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Would it be valid to say that the 1) bond length, and; 2) the distance at which molecules can rigidly hold each other being rather short and abrupt; are the reasons for such a sharp disjoint between randomness and order? One could imagine a really long bond length, or low kinetic energy resulting in a much denser liquid, with much more viscosity if molecules had much larger distances for still being in "order" rather than random. $\endgroup$ – ahnbizcad Sep 14 '14 at 22:32
  • $\begingroup$ Intuitively, it seems like when the kinetic energy of molecules in the liquid phase become low enough, some kind of attraction force abruptly becomes way more dominant, leading to an abrupt phase change. If we can pinpoint what type of force it is that dominates, perhaps we can attribute a certain force becoming dominant in the gas-liquid transition. (dipole-dipole, electronegativity, etc) $\endgroup$ – ahnbizcad Sep 14 '14 at 22:38
  • $\begingroup$ I think this answer is spot on, but I think there would be a more specific answer... like the typical escape velocity being small, or the "escape distance" of molecules being very short... and why that is so... like the charge of an electron, or the force being proportional to the inverse of the radius to some exponential factor. $\endgroup$ – ahnbizcad Sep 14 '14 at 22:46
  • 1
    $\begingroup$ Microscopic parameters such as bond length, etc. can be used to determine the value of the critical temperature. However, the sharpness of the transition should be thought of as the result of an instability. If the temperature is above the critical temperature (T_c) and the system is ordered, then a tiny perturbation will grow without bounds and affect the whole system to break the crystal ordering. Below T_c the perturbation decreases and simply dies out. The crystal is preserved. $\endgroup$ – Steven Mathey Sep 14 '14 at 22:52
1
$\begingroup$

It's not the details of the interaction that provide the clue, it's the cooperativity of the effect. For example, in a ferromagnet, there's no point to any one spin lining up unless its neighbors are aligned. That leads to the abrupt change from no long-range magnetic order to the onset of some long-range magnetic order starting at a well-defined temperature in an infinite system. Likewise, there's no point to a water molecule lining up in a particular way if its neighbors are tumbling around. Again, that cooperative effect leads to the abruptness of the transition from liquid to solid. So it takes not only the above-mentioned infinite number of degrees of freedom but also some cooperativity among them to generate a phase transition.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.