What's Wrong With Landau's Theory of Phase Transitions

  • #1

Main Question or Discussion Point

What's "Wrong" With Landau's Theory of Phase Transitions

Every book under the sun tells you that the Landau method is wrong because it "fails to consider fluctuations" but I don't see how that's true. It's got a gradient term of the form [tex] (\nabla M )^2 [/tex] and one could then include higher order gradient terms if one wants. How are these gradient terms not a way of allowing fluctuations? I guess my question is if we include terms of all orders in space and temperature (all temperature terms and all gradient terms, to infinite order), where does the Landau approach go "wrong"? Why doesn't it give correct critical exponents?
 

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  • #2
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Every book under the sun tells you that the Landau method is wrong because it "fails to consider fluctuations" but I don't see how that's true. It's got a gradient term of the form [tex] (\nabla M )^2 [/tex] and one could then include higher order gradient terms if one wants. How are these gradient terms not a way of allowing fluctuations? I guess my question is if we include terms of all orders in space and temperature (all temperature terms and all gradient terms, to infinite order), where does the Landau approach go "wrong"? Why doesn't it give correct critical exponents?
Hi,

First the Landau method is found to be correct above a critical dimension (depending on the system) so it is wrong to state, as a principle, that the method is wrong. The only problem is that this critical dimension is for most of cases above 3 and so can't really be applied for real systems.
Second, the Landau method is a mean field method in the sense it confuses basically
exp(<X>) with <exp(X)> which is not at all the same in general.
the error that you do by applying such a rough approximation is given by the cumulant expansion whose second term corresponds to the "correctly computed" mean fluctuations.
Estimating the order of magnitude of these fluctuations is basically the idea of Ginsburg to explain the validity range of the Landau theory for phases transitions.
 
  • #3


Hi,

First the Landau method is found to be correct above a critical dimension (depending on the system) so it is wrong to state, as a principle, that the method is wrong. The only problem is that this critical dimension is for most of cases above 3 and so can't really be applied for real systems.
Second, the Landau method is a mean field method in the sense it confuses basically
exp(<X>) with <exp(X)> which is not at all the same in general.
the error that you do by applying such a rough approximation is given by the cumulant expansion whose second term corresponds to the "correctly computed" mean fluctuations.
Estimating the order of magnitude of these fluctuations is basically the idea of Ginsburg to explain the validity range of the Landau theory for phases transitions.
Where is the exp(X) that we're talking about? Are we talking about field theory or just the standard phenomenological landau?
 
  • #4
Physics Monkey
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Where is the exp(X) that we're talking about? Are we talking about field theory or just the standard phenomenological landau?
You may think of trying to calculate something like
[tex]
Z = \int DM \exp{(- \beta H[M] )}.
[/tex]

[tex] Z [/tex] is the partition function, [tex] \beta [/tex] is the inverse temperature, [tex] H[M] [/tex] is roughly the energy as a function of the field M e.g. magnetization, and [tex] \int DM [/tex] is the instruction to sum over all configurations of M.

Landau's theory attempts to approximate the free energy [tex] \beta F = - \log{Z} [/tex] by the classical minimum writing [tex] Z \approx \exp{(-\beta H[M_{\mbox{min}}])} [/tex]. Including fluctuations means including other configurations of M besides the one that minimizes the energy. These fluctuations are necessary to obtain a quantitative description of many critical points even if the non-fluctuating approximation captures some aspects of the phase transition.

Hope this helps.
 
  • #5


You may think of trying to calculate something like
[tex]
Z = \int DM \exp{(- \beta H[M] )}.
[/tex]

[tex] Z [/tex] is the partition function, [tex] \beta [/tex] is the inverse temperature, [tex] H[M] [/tex] is roughly the energy as a function of the field M e.g. magnetization, and [tex] \int DM [/tex] is the instruction to sum over all configurations of M.

Landau's theory attempts to approximate the free energy [tex] \beta F = - \log{Z} [/tex] by the classical minimum writing [tex] Z \approx \exp{(-\beta H[M_{\mbox{min}}])} [/tex]. Including fluctuations means including other configurations of M besides the one that minimizes the energy. These fluctuations are necessary to obtain a quantitative description of many critical points even if the non-fluctuating approximation captures some aspects of the phase transition.

Hope this helps.
Thanks that helps a lot. Ok, so I understand from a field theory perspective where the short coming is. How does one quantitatively point out the shortcoming in the non-field theory form. The form where we just have some free energy expanded in powers about the critical point with a gradient term. Is it something of the nature that a gradient term is only the continuum approximation of an interaction such as

[tex]\frac{S_{i+1} - S_i}{a} {lim}_{a \rightarrow 0} [/tex]


and thus really only considers nearest neighbour (and not next and next-next nearest and so on up to the correlation length)? Is this correct? Or is it something else that is wrong.
 
  • #6
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Thanks that helps a lot. Ok, so I understand from a field theory perspective where the short coming is. How does one quantitatively point out the shortcoming in the non-field theory form. The form where we just have some free energy expanded in powers about the critical point with a gradient term. Is it something of the nature that a gradient term is only the continuum approximation of an interaction such as

[tex]\frac{S_{i+1} - S_i}{a} {lim}_{a \rightarrow 0} [/tex]


and thus really only considers nearest neighbour (and not next and next-next nearest and so on up to the correlation length)? Is this correct? Or is it something else that is wrong.
If you consider a real discrete Ising model then you can apply the standard mean field approximation to solve it (this could be done by using the Gibbs-Bogoliubov inequality for instance). You will end up with a uniform order parameter M that satisfies a non linear implicit equation.
Close to a critical point, this equation can be expanded in powers of M (usually we stop at M to the third) and it can be shown that it is the extremum of a Landau free energy expanded in powers of M up to M to the fourth.

Now, if you are interested in spatial correlations in your system or if there can exist inhomogeneities in your system you need more than a description with only a global variable i.e. you need a field description.
Close to a critical point the correlation length is expected to diverge and as a consequence the spatial variations of the field M(x) should be very smouth (it is not that straightforward to link the two statements but roughly speacking this is it) which is the reason why people only consider powers in gradients of the field.
 
  • #7


If you consider a real discrete Ising model then you can apply the standard mean field approximation to solve it (this could be done by using the Gibbs-Bogoliubov inequality for instance). You will end up with a uniform order parameter M that satisfies a non linear implicit equation.
Close to a critical point, this equation can be expanded in powers of M (usually we stop at M to the third) and it can be shown that it is the extremum of a Landau free energy expanded in powers of M up to M to the fourth.

Now, if you are interested in spatial correlations in your system or if there can exist inhomogeneities in your system you need more than a description with only a global variable i.e. you need a field description.
Close to a critical point the correlation length is expected to diverge and as a consequence the spatial variations of the field M(x) should be very smouth (it is not that straightforward to link the two statements but roughly speacking this is it) which is the reason why people only consider powers in gradients of the field.
But a gradient term isn't a global term, it's a local one. If you imagine something like a 2d ferromagnetic system as a taut sheet of rubber with local peaks and valleys and for each deviation from a flat surface you penalize the free energy to infinite order in gradients, how is that an incorrect description of the system? What is missing?
 
  • #8
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But a gradient term isn't a global term, it's a local one. If you imagine something like a 2d ferromagnetic system as a taut sheet of rubber with local peaks and valleys and for each deviation from a flat surface you penalize the free energy to infinite order in gradients, how is that an incorrect description of the system? What is missing?
Exactly a gradient is not a global quantity but a local one (but its integral is a global quantity) and hence if you are only interested in the magnetization and its corresponding susceptibility (uniform in the bulk) you normally don't need such local description.
However, if you are not in bulk and have to deal, say, with an interface or if you want to describe the spatial correlations between two points in your system then you will need a field M(x) and some of its gradients in the description.

The reason for uniformity a priori when you are only interested in the value of M is that by symmetry you can't discriminate a specific lattice site from another and therefore the magnetization at each point shall be the same.
 
  • #9


Exactly a gradient is not a global quantity but a local one (but its integral is a global quantity) and hence if you are only interested in the magnetization and its corresponding susceptibility (uniform in the bulk) you normally don't need such local description.
However, if you are not in bulk and have to deal, say, with an interface or if you want to describe the spatial correlations between two points in your system then you will need a field M(x) and some of its gradients in the description.

The reason for uniformity a priori when you are only interested in the value of M is that by symmetry you can't discriminate a specific lattice site from another and therefore the magnetization at each point shall be the same.
I'm confused here, Landau's theory (at least in the form I'm concerned with) DOES include gradient terms, potentially to infinite order. My question is what is WRONG with Landau's theory (when it has gradient terms). I've seen in textbooks where they keep up to like 4th order in T and 1st order in gradients and show that it only gives mean field results but what happens if you kept all orders in both. What is "wrong" with that sort of Landau theory? What is it failing to include?
 
  • #10
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I'm confused here, Landau's theory (at least in the form I'm concerned with) DOES include gradient terms, potentially to infinite order. My question is what is WRONG with Landau's theory (when it has gradient terms). I've seen in textbooks where they keep up to like 4th order in T and 1st order in gradients and show that it only gives mean field results but what happens if you kept all orders in both. What is "wrong" with that sort of Landau theory? What is it failing to include?
Can you give an example of the type of Landau free energy you are talking about 'maybe I am also confusing with something else)?

Considering what is wrong in Landau approach in principle even if one includes all the gradient he can? The problem is that it confuses the free energy of the system with the most probable action (i.e. the energy of the system expressed in term of the order parameter).

Whatever the number of terms in your guess for the action, it won't become more accurate than a mean field approximation anyway (in the best cases you can include non linearities and/or non localities but it is still mean field).

To see how rough is a mean field approximation in general just imagine that classical trajectories for a particle are actually a mean field approximation of the quantum description where actually the system can take any path to go from a point to another (even though it's weighted).
 
  • #11


Can you give an example of the type of Landau free energy you are talking about 'maybe I am also confusing with something else)?

Considering what is wrong in Landau approach in principle even if one includes all the gradient he can? The problem is that it confuses the free energy of the system with the most probable action (i.e. the energy of the system expressed in term of the order parameter).

Whatever the number of terms in your guess for the action, it won't become more accurate than a mean field approximation anyway (in the best cases you can include non linearities and/or non localities but it is still mean field).

To see how rough is a mean field approximation in general just imagine that classical trajectories for a particle are actually a mean field approximation of the quantum description where actually the system can take any path to go from a point to another (even though it's weighted).
I'm talking about a Landau expansion of the form

[tex] \int dx -h m(x) + b m(x)^2 + c m(x)^3 + d m(x)^4 + \ldots + e (\nabla m(x))^2 + f (\nabla m(x))^4 + \ldots [/tex]

Pretty much every textbook under the sun, before introducing any field theory (i.e. discussion of functional integrals and classical actions), before introducing renormalization groups, discards this approach by saying it is "only a mean field approach and fails to include fluctuations". My question is, with infinite order in gradients, what are these fluctuations (without any reference to field theory) that are being neglected? Why do these books say that an expansion such of this, taken to infinite order, will not yield the correct critical exponent? (and then they often "prove" this by showing that if you only take it to 4th order you don't get the correct exponents as if that closes the matter). What physics is happening in this system at the level of discussion of pre-field theory pre-RG that is being missed by a phenomenological expansion in both T AND position of the free energy?
 
  • #12
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I'm talking about a Landau expansion of the form

[tex] \int dx -h m(x) + b m(x)^2 + c m(x)^3 + d m(x)^4 + \ldots + e (\nabla m(x))^2 + f (\nabla m(x))^4 + \ldots [/tex]

Pretty much every textbook under the sun, before introducing any field theory (i.e. discussion of functional integrals and classical actions), before introducing renormalization groups, discards this approach by saying it is "only a mean field approach and fails to include fluctuations". My question is, with infinite order in gradients, what are these fluctuations (without any reference to field theory) that are being neglected? Why do these books say that an expansion such of this, taken to infinite order, will not yield the correct critical exponent? (and then they often "prove" this by showing that if you only take it to 4th order you don't get the correct exponents as if that closes the matter). What physics is happening in this system at the level of discussion of pre-field theory pre-RG that is being missed by a phenomenological expansion in both T AND position of the free energy?
Ok we speak about the same things :smile:.

If you consider pre-field theory and pre-RG understanding well I can't really tell. I guess that first it failed to predict the good exponents and second, again in principle there is no real reason for the free energy to obey such an expansion.

About the gradients terms I will just repeat myself I am afraid. Since they are local quantities you don't need them (they are just useless) if you have a uniform external field or if you are not looking at the correlations.

I insist on the fact that the terms in gradient in the Landau expansion are not fluctuations but merely spatial variations and as such are meaningless if you are in the bulk without any inhomogeneities induced by the presence of non uniform external field.

If you don't agree with that statement (which is ok with me) can you give an example of uniform magnetization whose value would depend on the gradient terms in the free energy plz?
 
  • #13


Ok we speak about the same things :smile:.

I insist on the fact that the terms in gradient in the Landau expansion are not fluctuations but merely spatial variations and as such are meaningless if you are in the bulk without any inhomogeneities induced by the presence of non uniform external field.

Perhaps this is where my problem is. What are the fluctuations that are being alluded to? When we talk about the fluctuations of the mean field I have always understood this to mean the spatial variations from the mean value that relate to random non-mean-field domains. If this is what is meant by fluctuations then surely an infinite gradient expansion should capture this physics to infinite accuracy. But is this not what is meant by "fluctuations"?
 
  • #14
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@maverick_starstrider:

Fluctuations doesn't refer to spatial fluctuations and what you are writing down IS actually a field theory. But if you don't like to think in terms of field theory, you don't have to.

In statistical mechanics the partition function is calculated by summing an exponential function over ALL configurations (or states) possible [tex]Z = \sum_i e^{-\beta S_i}[/tex] (this is what Physics Monkey write in a path-integral representation in the field theoretical case). A stationary phase, or mean-field approximation is essentially what you get if instead of summing over all configurations, you only consider the configuration that contributes the most (minimizes [tex]S_i[/tex]). This configuration need NOT be spatially uniform, for example this is not the case in a type II superconductor because of http://www.tkm.uni-karlsruhe.de/lehre/TopTKM.WS1011/Talks/Abrikosovvortices.pdf". Using this approximation you lose the fluctiations around the mean-field, meaning that you calculate [tex]Z[/tex] without the contribution of all the other configurations of the system.

Now, if you knew the microscopic theory you could do all of this "easily". Landau's method on the other hand is a matter of, when knowing the order parameter, to write the most general form [tex]S_i[/tex] can have instead of appealing to the microscopics. Since one usually minimize [tex]S_i[/tex], only one configuration is considered and therefore fluctuations are neglected.

I hope this helps.

Update: I see that Zacku has already said most of what I write here.
 
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  • #15
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If this is what is meant by fluctuations then surely an infinite gradient expansion should capture this physics to infinite accuracy.
This is not really the topic of the question, but I think its still important to comment on. If you expand a function or a functional, it is NOT correct that you get infinite accuracy if you keep all terms! This depends on radius of convergence, since the expansion need not be convergent everywhere! For example if you use perturbation theory to solve a complicated model, you don't (in most cases) get the exact result even if you include all infinite orders! Therefore perturbation theory doesn't work for phase transitions, since then observables are not analytic.
 
  • #16
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A stationary phase, or mean-field approximation is essentially what you get if instead of summing over all configurations, you only consider the configuration that contributes the most (minimizes [tex]S_i[/tex]). This configuration need NOT be spatially uniform, for example this is not the case in a type II superconductor because of http://www.tkm.uni-karlsruhe.de/lehre/TopTKM.WS1011/Talks/Abrikosovvortices.pdf".
I have to say I didn't know about these vortices. I am not sure I understand the gradient term in Landau-Ginsburg equation 1 in you link or at least I am not sure that it disagrees with my previous statement.
Since it is a term in [tex] \left|\left(-i \hbar \vec{\nabla} - \vec{A}\right) \psi \right|^2 [/tex]
it is just proportional to the mean kinetic energy of an effective particle that belong to the calculated current and hence I can't exactly rely on symmetry arguments to guess any uniformity of the field itself since one of the order parameters in one sense is related to a current or is the current itself (I imagine that a non zero current in a given direction in the limit where the magnitude of the external field tends to zero could be a definition of what is superconductivity).

In summary I tend to claim that the order parameter in bulk should be uniform as long as
the phase transition is not characterized by a non zero current.

Would you mind to tell me if I am off base or not in my "intuitive" reasoning?
 
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  • #17
Mute
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Why doesn't [Landau Theory] give correct critical exponents?
Ultimately, Landau theory doesn't give the correct critical exponents because it has thrown out the microscopic length scale of the system. The assumption that it is ok to do this because the length scale is much smaller than the correlation length (which diverges at the critical point) fails because even if [itex]\ell \ll \xi[/itex], setting [itex]\ell/\xi = 0[/itex] in a function [itex]F(\ell/\xi)[/itex] is only valid if F(0) isn't singular, which, as it turns out in critical phenomena, the functions one is interested in are usually singular. You can add all the higher order gradient terms to the landau free energy that you want, it won't change the fact that you've thrown away a length scale that's actually important even though by all rights it seems like it shouldn't be.

If you haven't already, I would recommend borrowing/buying a copy "Lectures on Phase Transitions and the Renormalization Group" by Goldenfeld. It talks at length about Landau theory, it's breakdown, anomalous dimensions, etc. (from a stat mech/condensed matter viewpoint).
 
  • #18


Ultimately, Landau theory doesn't give the correct critical exponents because it has thrown out the microscopic length scale of the system. The assumption that it is ok to do this because the length scale is much smaller than the correlation length (which diverges at the critical point) fails because even if [itex]\ell \ll \xi[/itex], setting [itex]\ell/\xi = 0[/itex] in a function [itex]F(\ell/\xi)[/itex] is only valid if F(0) isn't singular, which, as it turns out in critical phenomena, the functions one is interested in are usually singular. You can add all the higher order gradient terms to the landau free energy that you want, it won't change the fact that you've thrown away a length scale that's actually important even though by all rights it seems like it shouldn't be.

If you haven't already, I would recommend borrowing/buying a copy "Lectures on Phase Transitions and the Renormalization Group" by Goldenfeld. It talks at length about Landau theory, it's breakdown, anomalous dimensions, etc. (from a stat mech/condensed matter viewpoint).
Wow, this book looks great. My uni library doesn't have a copy but based on the table of content I just might have to buy it.
 
  • #19
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Ultimately, Landau theory doesn't give the correct critical exponents because it has thrown out the microscopic length scale of the system. The assumption that it is ok to do this because the length scale is much smaller than the correlation length (which diverges at the critical point) fails because even if [itex]\ell \ll \xi[/itex], setting [itex]\ell/\xi = 0[/itex] in a function [itex]F(\ell/\xi)[/itex] is only valid if F(0) isn't singular, which, as it turns out in critical phenomena, the functions one is interested in are usually singular. You can add all the higher order gradient terms to the landau free energy that you want, it won't change the fact that you've thrown away a length scale that's actually important even though by all rights it seems like it shouldn't be.

If you haven't already, I would recommend borrowing/buying a copy "Lectures on Phase Transitions and the Renormalization Group" by Goldenfeld. It talks at length about Landau theory, it's breakdown, anomalous dimensions, etc. (from a stat mech/condensed matter viewpoint).
I am not able to follow your reasoning. The essence of critical exponents is that they are universal, in particular they don't depend on the microscopic detail. Well, more precisely they only depend on microscopics through relevant couplings (in the RG sense) and therefore for many theories the mean-field results are reliable for higher dimensions (independent of microscopic details), ie above the http://en.wikipedia.org/wiki/Critical_dimension" [Broken]. For lower dimension fluctuations may dominate, and Landaus theory won't be reliable.

I feel like I am misunderstanding you.
 
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  • #20
Mute
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I am not able to follow your reasoning. The essence of critical exponents is that they are universal, in particular they don't depend on the microscopic detail. Well, more precisely they only depend on microscopics through relevant couplings (in the RG sense) and therefore for many theories the mean-field results are reliable for higher dimensions (independent of microscopic details), ie above the http://en.wikipedia.org/wiki/Critical_dimension" [Broken]. For lower dimension fluctuations may dominate, and Landaus theory won't be reliable.

I feel like I am misunderstanding you.
Critical exponents are universal in the sense that they don't depend on the exact details of the microscopic hamiltonian, beyond symmetries, length of interactions and dimension. The value of the microscopic length scale will not change the critical exponents, but the fact that there is one can. As I mentioned above, you can't set [itex]\ell/\xi[/itex] (microscopic length scale over the correlation length) to zero in a function unless the function is well-behaved in the limit [itex]\ell/\xi \rightarrow 0[/itex]. In dimensions above the upper critical dimension it turns out that's fine, but below the critical dimension this limit is singular and you get anomalous critical exponents, different from the ones Landau theory predicts (the mean field theory values).
 
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