I would like to ask the following question for which I cannot find a definite answer in the literature.
Of what ORDER is the phase transition leading to Bose-Einstein condensation for a ideal and real Bose gases?
Many thanks!
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2$\begingroup$ Second order, Ising universality. $\endgroup$– ThomasOct 27, 2015 at 19:06
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$\begingroup$ @Thomas: Do you have a particular system in mind? Naively, I would consider all kinds of transitions where a scalar order parameter picks up an expectation value (in the zero momentum mode) as an example of Bose-Einstein condensation. Therefore, I can think of it happening at first-order or second-order. $\endgroup$– SivaOct 27, 2015 at 21:32
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$\begingroup$ @Thomas: Cunfusingly, one can find e.g. this books.google.co.uk/…. So is the author correct? $\endgroup$– HamurabiOct 28, 2015 at 10:07
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2$\begingroup$ @Hamurabi: This is about the non-interacting Bose gas, which is a little bit of an artificial limit (if the gas actually is non-interacting, then it cannot thermalize). However, it does imply that for a weakly interacting Bose the critical regime is expected to be very small. (And the transition remains continuous.) $\endgroup$– ThomasOct 28, 2015 at 16:14
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2$\begingroup$ @Silva: I think one can never completely exclude that a first order transition ``intervenes'', for reasons unrelated to the symmetries of the order parameter. I don't know of an example of this, however. All the standard cases of BE condensation (liquid Helium, ultracold atomic Bose gases, excitons, kaon condensation, .. ) conform to the standard universality arguments. $\endgroup$– ThomasOct 28, 2015 at 16:19
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2 Answers
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The order parameter in the Bose-Einstein condensation is the "macroscopic wave function" or rather the square root of the single-particle reduced density matrix. The broken symmetry is usually said to be gauge symmetry.
Carter states that the presence of singularities in the heat capacity is a measure of the existence of a phase transition. The heat capacity is $$ C_V =k_B\beta^2 \dfrac{\partial^2 \ln(Z)}{\partial \beta^2} = \dfrac{k_Bc_d}{4} \int_{0}^{\infty} \dfrac{\epsilon^{d/2+1}d\epsilon}{\sinh^2(\beta(\mu-\epsilon)/2)} \ . $$
When μ=0, the singularity of the CV's integrand becomes $ϵ^{d/2}+1/ϵ^2=ϵ^{d/2−1}$. For $d=(1,2,3)$, the integrand's singularity equals ($ϵ^{−1/2}$,$ϵ^0$,$ϵ^{1/2}$), which all are convergent integral singularities.
trying to understand Bose-Einstein Condensate (BEC)
"It turns out that, as you cool a gas of bosons, you will eventually reach a point where the gas suddenly "condenses" into a state with nearly all of the particles occupying a single state, generally the lowest-energy available state. This happens with material particles because the wave-like character of the bosons becomes more and more pronounced as you lower the temperature. The wavelength associated with them, which at room temperature is many times smaller than the radius of the electron orbits eventually becomes comparable to the spacing between particles in the gas. When this happens, the waves associated with the different particles start to overlap, and at some point, the system "realizes" that the lowest-energy state would be for all the particles to occupy a single energy level, triggering the abrupt transition to a BEC.
This transition is a purely quantum effect, though, and has nothing to do with chemical bonding"
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$\begingroup$ Sorry, but the question was of what order it is actually. $\endgroup$– HamurabiOct 28, 2015 at 10:07
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$\begingroup$ Chemical bonding is a quantum effect, though. $\endgroup$ Mar 11, 2016 at 15:44
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Hamurabi, I assume you are asking for the order of the wave equation.
Try thinking about intramolecular vs. intermolecular forces. A wave equation is necessary to describe the energy of any quantum occurrence, so far as I recall.
It may take an approximation for any non-ideal system, but exact or not the wave equation describes the moment in time when the field harmonizes giving the energy at a transition. These wave equations commonly fit different basic "orders" or forms.
