I am having some troubles trying to classify a modified photon gas into which phase transition it belongs (if it really is a phase transition).

First of all, it is a photon gas that interacts with an external electric field, so the q-potential (from the grand canonical ensemble) depends on the external field and the temperature alone. The potential is continuous but not analytic only when the external field is varied (if the external field is maintained fixed it might be on one side or the other of the critical point, independently of the temperature). Because the density of energy is the derivative with respect to $\beta$, it has the same qualitative properties of the q-potential.

Now, I've found two definitions for phase transitions, Ehrenfest's one and the "modern" one. Following Ehrenfest's classification I found that it's a first order transition since the discontinuity appears in the first derivative of the potential with respect to a thermodynamic variable, the electric field. But, following the "modern" definition, I found that it is a second order phase transition, or, in the modern classification, a continuous phase transition, because the entropy doesn't have a discontinuity and thus the system doesn't require a latent heat to perform the transition.

My questions are, have I well understood the definitions? Is the system really having a phase transition? And, if it does, which one is it having or it is of a third kind I didn't mention?


  • $\begingroup$ What's your Hamiltonian? It's hard to comment on this unless we know formally what you are doing. $\endgroup$ – webb May 13 '14 at 22:30
  • $\begingroup$ @webb The energy of the photon is given, in terms of it's momentum and the external field, by $\varepsilon=\frac{\sqrt{\left(1+aE^{2}\right)\{a\vec{p}\cdot\vec{E}-p^{2}\}}}{1+aE^{2}}$, where $a$ is a constant. $\endgroup$ – Pierre May 13 '14 at 22:42
  • $\begingroup$ Why? Where did you get this energy from? $\endgroup$ – webb May 13 '14 at 23:00
  • $\begingroup$ @webb It's a made-up system, it's a toy model... $\endgroup$ – Pierre May 13 '14 at 23:20
  • $\begingroup$ I'm not sure why you'd expect intuitive behavior from an unphysical model. To spare us the work, could you show the details of your calculation that you've done so far? $\endgroup$ – webb May 13 '14 at 23:49

Your Answer

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

Browse other questions tagged or ask your own question.