What Is the Relationship Between Free Energy and Magnetisation?

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SUMMARY

The relationship between free energy (F) and magnetisation (M) is defined by the equation M = (-∂F/∂H)ₜ, where H represents the magnetic field. For a system of N independent spin-1/2 paramagnets, the partition function Z is given by Z = 2ⁿ coshⁿ(βμH), leading to the free energy expression F = -kTN.log[2cosh(βμH)]. The susceptibility is derived as χ = ∂M/∂H = (Nμ²/kT) sech²(μH/kT), which simplifies to (Nμ²/kT) in low field or high temperature limits. The derivation of these relationships is crucial for understanding magnetic systems.

PREREQUISITES
  • Understanding of Helmholtz free energy and its differential form.
  • Familiarity with partition functions in statistical mechanics.
  • Knowledge of magnetisation and susceptibility in magnetic systems.
  • Basic concepts of thermodynamics, particularly Legendre transformations.
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  • Study the derivation of Helmholtz free energy in magnetic systems.
  • Learn about the Legendre transformation and its applications in thermodynamics.
  • Explore the concepts of Gibbs free energy and its relationship to magnetisation.
  • Investigate the role of statistical mechanics in deriving physical properties of paramagnets.
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Physicists, particularly those specializing in statistical mechanics and thermodynamics, as well as students studying magnetic properties of materials.

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Homework Statement



1. State a relationship between the free energy, F, and the magnetisation, M.

2. State a partition function for the case of a system of N independent spin-1/2 paramagnets in a field, and derive an expression for its susceptibility.

The Attempt at a Solution



(1) Looking in my notes, I see that dF = - MdH - SdT (*). It follows, then, that

M = \left(-\frac{\partial F}{\partial H}\right)_{T}

That's the question answered, but I wonder how one arrives at this expression for F.

I know that F = U - TS and dW = - HdM, so it would seem that

dF = TdS - dW - TdS - SdT = -dW - SdT = HdM - SdT

ie. not MdH. Could someone show me how (*) is derived?

(2) For the whole system, I find that

Z = 2^{N}cosh^{N}(\beta\mu H) (H is my field, \mu is the Bohr magneton )

I thus find the free energy to be

F = -kTN.log[2cosh(\beta\mu H)]

And thus, by the relationship stated above,

M = N\mu tanh(\frac{\mu H}{kT})

The susceptibility, I presume, is \frac{\partial M}{\partial H}\right). It comes out for me as

\frac{N\mu^{2}}{kT} sech^{2}(\frac{\mu H}{kT})

(which, I find, tends to \frac{N\mu^{2}}{kT} in the limit of low field or high temperature.

Does that seem sensible?

Cheers!
 
Last edited:
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Your post contains some confusion. First, your equation (*) for Helmholtz free energy should be written with dF rather than F.

Here's a quick trick that may be helpful for treating the magnetization. How can we decide without derivation whether dW is HdM or MdH? Well, the differential energy dF is extensive (depends on volume), and only HdM depends differentially on a volume-dependent quantity M. Accordingly

dF=HdM-SdT. This should replace your (*).

From this we can say that H=(\partial F/\partial M)_{T}.

To get M, we need to define a new quantity, usually called Gibbs free energy,

G=F-HM. By Legendre transformation,

dG=-SdT-MdH

and

M=-(\partial G/\partial H)_T.

So I don't get your result for part 1...

EDIT: oops, corrected signs
 
Last edited:
marcusl said:
Your post contains some confusion. First, your equation (*) for Helmholtz free energy should be written with dF rather than F.

Sorry. That was a typo.

Here's a quick trick that may be helpful for treating the magnetization. How can we decide without derivation whether dW is -HdM or -MdH? Well, the differential energy dF is extensive (depends on volume), and only -HdM depends differentially on a volume-dependent quantity M. Accordingly

dF=-HdM-SdT. This should replace your (*).

Hmmm... I find in 'Statistical Mechanics, A Survival Guide', by Glazer and Wark, that the result dF = - MdB - SdT (the one I have written down) is stated as a known result on pg. 34 (without proof: "from our knowledge of thermodynamics we know that for a magnetic system..."), followed by the expression

M = \left(-\frac{\partial F}{\partial B}\right)_{T}

(I'm using H, of course, rather than B)

But I'm not sure how they arrived at that expression for dF...
 
Last edited:
Looking around online I found this
http://www.cmp.caltech.edu/refael/phys127/notes10new.pdf"
where Eqs. (12)-(13) seem to match if you replace my G with his \mathcal{F}.

Anyone else?

EDIT: removed comment about free energy.
 
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