       Re: Brillouin function for a Ferromagnet

• To: mathgroup at smc.vnet.net
• Subject: [mg111493] Re: Brillouin function for a Ferromagnet
• From: DrHasenbein <fstromberg at gmx.de>
• Date: Tue, 3 Aug 2010 06:38:36 -0400 (EDT)
• References: <i293l9\$m0s\$1@smc.vnet.net>

```On 22 Jul., 11:39, Alexei Boulbitch <alexei.boulbi... at iee.lu> wrote:
> First I have to apologize if this is too basic for this forum... I have some temperature dependent magnetization data from a FeNi-alloy. The measurement was done from RT up to the Curie temperature T_c of the alloy. I want to compare my data to aBrillouinfunction with different angular momenta J. Basically I have to solve
>
> M/M_0=B_J[(T_c/T)(M/M_0)],
> with
> B_J=(2J+1/J)coth[(2J+1/2J)x]-(1/2J)coth[x/2J],
> where
> M_0 is the saturation magnetization
>
> How do I basically solve this and in parallel fit it to my data? In principle I want to obtain M_0 and T_c. My data consists of M(T).
>
> Frank
>
> I think, Frank, that I will not make a mistake, if I write that at this site you should better write your functions in
> the Mathematica notations, so that we copy-paste them into our notebooks. This belongs to a basic politeness. However,
> let us try it like it is.
>
> First of all, your equation looks somewhat strange (comparehttp://en.wikipedia.org/wiki/Brillouin_function).
> Your notations are not quite clear.  If I get it right in your case
>
> x=(T_c/T)(M/M_0), is it?
>
> In this case something is wrong with your notations, since your equation MUST have a bifurcation at T=Tc, i.e. Tc/T=1.
> I made a simple check with J=1 and have got Tc/T=3/2 (note that I used the traditionalBrillouinfunction, different
> from the one you write). Here Tc/T is the parameter you used in the above equation and thus,
> Tc is not the Curie temperature, but your notation. The relation Tc/T=3/2 must be the one for determination the Curie
> temperature T=2Tc/3. If you were right however, it would give T=Tc. So, there is a mistake in your formulation and
> you should first fix it.
>
> Second, equation you are looking at is very well known, and it is well-known that it has no analytic solution, like M=M(J,T,M0).
> In contrast, the problem of finding the bifurcation point, Tc, HAS the analytic solution.
> It enables one to establish the Curie point, Tc=Tc(J). You may find details in a number of popular textbooks on magnetism,
> As soon as you have this solution, you get the experimental Tc value, Tc(exp) from your experimental data, and write down
> the equation Tc(J)=Tc(exp). This gives you J value. You cannot get more from the combination of this equation and the
> experimental data, since like this you already fix all available parameters, and the theory you are using offers you
> no degree of freedom more to play with.
>
> I do not understand, why do you need to go beyond this simple procedure and take troubles to make fitting etc. However,
> if you have a strong reason, this problem may be approached from several directions.
>
> First, one can try to solve equation numerically. It is only possible, if parameters
> (such as Tc/T and J) have some fixed numerical values. Then FindRoot is the operator
> that will make the job for you. In principle, you may try to think about a routine that will scan
> the parameter space, numerically retrieve the theoretical dependence and compare it with your experimental data.
> In this case it is up to you, what routine to write. It depends upon details of your problem,
> and after all, it is your problem.
>
> Second, one may think of what I call a "semi-analytic" solution. In this case you start with separating out
> a region in the space of parameters (i.e., J, Tc/T), where you can expect them to take values. Then you take
> a set of points (e.g. a lattice) within this region and solve your equation numerically in each point of this lattice.
> You further, look for an analytic function, M/M0=f[J, Tc/T,a,b,...], (where a, b,... are some fitting parameters) which
> would approximate your numeric solution. This function is a product of your guess, and should be as simple as compatible
> with the precision level you require. For example, it could be a polynomial. The function f[J,Tc/T,a,b...]
> should be fitted to your numeric solutions. On this step the fitting parameters are a, b,..., rather than J, and thus,
> a, b,... will be fixed. For the purposes of fitting you may use Fit and/or FindFit operations. There is also a couple of
> other fitting operators that you could use in complex cases.
> After the first fitting procedure is finished you have an analytical function, f[J,Tc/T], approximating your theoretical
> results. Now you may fit this analytical function to your experimental data using J as the fitting parameter.
> You may again use Fit and/or FindFit for this purpose. It is this approach that I would chose myself, if I need to
> attack this problem.
>
> Third, one can try some analytic approximation to the solution, say asymptotic expansion. This is again up to you,
> which approximation to choose. As soon as you have such an approximation, you may again use Fit and/or FindFit to
> make fitting of the analytic function in hand to your experimental data.
>
> In any case you should familiarize yourself with Mathematica first. This pays off.
>
> Finally, let me make a guess: since the theory you used is much too rough, the solution far from the Curie point will
> badly (if at all) fit to experiment, independently of the approach you choose, or the Curie point got from the fitting
> function will be away from the experimental one.
>
> Alexei
>
> --
> Alexei Boulbitch, Dr. habil.
> Senior Scientist
> Material Development
>
> IEE S.A.
> ZAE Weiergewan
> 11, rue Edmond Reuter
> L-5326 CONTERN
> Luxembourg
>
> Tel: +352 2454 2566
> Fax: +352 2454 3566
> Mobile: +49 (0) 151 52 40 66 44
>
> e-mail: alexei.boulbi... at iee.lu
>
> www.iee.lu
>
> --
>
> This e-mail may contain trade secrets or privileged, undisclosed or
> otherwise confidential information. If you are not the intended
> recipient and have received this e-mail in error, you are hereby
> notified that any review, copying or distribution of it is strictly
> prohibited. Please inform us immediately and destroy the original

First of all thanks for your explicit answer. I must be honest that I
did forget to check back here for answers. I also had some trouble
with how to use this group etc...
And yes, I am still starting to use Mathematica, but as usual one
wants to take too many steps at a time.

Sincerely,
Frank

```

• Prev by Date: Re: ReplaceAll and ReplaceRepeated Strange Behavior
• Next by Date: Re: ReplaceAll and ReplaceRepeated Strange Behavior
• Previous by thread: StringForm v.s Row
• Next by thread: More streamline with StreamDensityPlot