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Gibbs vs Helmoltz potential in magnetic systems |
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| Apr17-09, 05:55 AM | #1 |
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Gibbs vs Helmoltz potential in magnetic systems
Hi all,
I am a bit confused about the relation between the thermodynamic potentials and the partition function. Usually one has [tex]Z = e^{-\beta A}, \qquad A = U-TS[/tex] right? Here A is the Helmoltz free energy. Now, when addressing magnetic system I see [tex]Z = e^{-\beta G}[/tex] where G is the Gibbs potential, whose "natural variables" are T (temperature, intensive) and H (magnetic field, intensive). That is, G is the analogue of the usual G(T,P) defined for non magnetic system, linked to the internal energy through two Legendre transforms. What's the justification of this? I mean, G = A - PV. In a magnetic system G = A+MH, where M is the magnetization. How can Z be the exponential of both A and G? Also, I read equations like [tex]U=G-T \frac{\partial G}{\partial T}[/tex] (K. Huang, Statistical Mechanics pag 393). But this seems to me the relation between U and A. It seems to me that a PV (or -MH for magnetic systems ) term is missing in the case of G. Is the second equation for Z obtained by changing some of the assumptions that result in the first equation according to the "usual derivation"? In some sense a different canonical ensemble? Thanks a lot for any insight F |
| Apr17-09, 10:21 AM | #2 |
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Could the sometimes-missing [itex]PV[/itex] term indicate that the authors consider [itex]P-V[/itex] work (i.e., thermal expansion work) to be negligible compared to magnetic work? (Or alternatively, that they assume the system to be in a vacuum?) |
| Apr17-09, 11:08 AM | #3 |
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Recognitions:
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| Apr17-09, 11:21 AM | #4 |
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Gibbs vs Helmoltz potential in magnetic systems
Hey, thanks a lot for your time
Sorry about that. Shouldn't that be [tex] Z = e^{-\beta (G+MH)} [/tex] Why the magnetic work term seems to be ignored? Also, the equation linking [itex]U[/itex] and [itex]G[/itex] still seems weird to me. I mean, ignoring the PV terms as much as one usually ignores the MH terms, it seems to me that one should have [tex] G - T \frac{\partial G}{\partial T} = G + TS = A - MH + TS = U - MH [/tex] i.e. once again one work term too much (K. Huang has U instead of U-MH). Could this be related to the fact that [itex]H[/itex] is constant? I'm really confused about this. Thanks a lot again, F |
| Apr17-09, 01:52 PM | #5 |
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Let me try to tell you what I am getting from all of this. So when deriving the "usual" canonical ensemble one assumes that T is fixed and the system exchanges heat with a heat bath. The volume is fixed in that case, so the system does not exchange other forms of work. If the volume is allowed to change while keeping fixed the pressure, mechanical work is exchanged as well. Now I can believe that by going along the same lines of the derivation of the "usual canonical ensemble" one gets an additional work term in the exponential giving the probability of a microstate, other than the internal energy term. There are some signs and coefficients that I do not see right now, but now everything makes more sense (perhaps). I just see that the signs for the mechanic and magnetic work terms must be opposite, essentially because pressure reduces the volume while the magnetic field increases the magnetization. I'll try to work out the details. So the whole point seems to be that one wants to fix not only the temperature but also the external force (e.g. pressure or magnetic field). Could we say that Gibbs Canonical Ensemble is doubly canonical? Something I do not grasp completely is: why don't we include the additional term in the energy of the microstate? I'm thinking to particles with a magnetic moment, for instance... But perhaps I should understand completely the isobaric ensemble and draw an analogy with the magnetic system. Thanks again F |
| Apr17-09, 03:53 PM | #6 |
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It can't. This is the story : source of your ( and not only yours! ) confusion is that some authors insist that internal energy U must have all extensive variables ! The price they pay for such IMHO unreasonable approach is that in presence of field B they "redefine" internal energy as U' = U + M*B ! End result is that whenever magnetic field appears in thermodynamics of your problem, all formulas must be changed to preserve this "deep principle about natural variables of U", so you got two sets of formulas : magnetic ones and nonmagnetic ones. Difference between those is what confuses you. In my opinion, the best thing you can do is avoid this by accepting that field B ( H ) is "natural" variable of internal energy U ( named spectroscopic internal energy ) and only then will you have consistent relations where one doesn't have to worry is Z exponential of A or G this Monday =) |
| Apr20-09, 12:55 PM | #7 |
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thanks for your reply. I'm sort of glad of knowing that I'm not the only one who gets confused by this. Unfortunately at the moment I'm still struggling with all of this stuff. I sort of see the issues at stake, but I'm not able to "tie the loose ends". Meaning I think I have not really understood the problem yet. A part of the story seems to be changing the name of things. What people seems to call energy here is what one would usually call an enthalpy. Or is it? I think this has to do with your comment about the "natural variables" of the internal energy being extensive quantities. I see that they are all forms of energy, but this name-changing makes me loose my bearings with thermodynamic functions. Doesn't all this mess up the "thermodynamic squares"? Could you suggest a book or website where these things are discussed? So far the only place I've found is the MIT lecture notes suggested by atyy, but I'd like to see some examples, or perhaps a derivation... Anyway thanks again F |
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