Feature request: Triaxial anisotropy

[borisgross]

Several magnetic system show biaxial crystalline anisotropy, such as e.g. the 2D magnets CrSBr or CrPS. Adding this functionality to mumax+ would enlarge the amount of magnetic systems that can be modeled.

[ilateur]

Do you have an expression and a source for biaxial crystalline anisotropy?

[borisgross]

I had implemented it in mumax3 using

Msat = 200000;
Ka := 100000:
u_a := ConstVector(1,0,0);
anisField := Mul(Const(2Ka/Msat), Mul(Dot(u_a, m),u_a));
anisEdens := Mul(Const(-0.5Msat), Dot(anisField, m));
AddFieldTerm(anisField);
AddEdensTerm(anisEdens);

I'll dig for the paper or source, that I had used for this.

[borisgross]

Ah now I remember, I had just added a second uniaxial term (as you see above) for the second crystal axis. For the first one I used the existing implementation in mumax3. However, there is a formulation for biaxial anisotropy in the thesis of Jonas De Clercq entitled 'Modelling antiferromagnetic interfaces with MuMax3' from Ghent University.

[ilateur]

Your implementation seems like two uniaxial anisotropies added together
$\varepsilon = - K_1 (c_{1} \cdot m) - K_2 (c_{2} \cdot m)$
while the implementation from Jonas De Clercq is a different single term
$\varepsilon = K_c (c_1 \cdot m)^2 (c_1 \cdot m)^2$

Taking $K_1 = K_2 = K_c = 1$ and $c_1$ and $c_2$ to be the x- and y-axes, we get the following energies on the unit sphere, with black being favored low energy and yellow being unfavored high energy.

These seem to behave very differently... Am I missing something?

If you want two uniaxial anisotropies, which happen to be perpendicular to each other ($c_1 \cdot c_2 = 0$) with exactly the same anisotropy constant ($K_1 = K_2$), then this is no different from a single uniaxial anisotropy with a negative sign up to a constant. For example:
$- K_u m^2_x - K_u m^2_y = - K_u (1 - m^2_z)$
And so this is already possible.

[borisgross]

No, you do not miss anything. I need to trace back what I had done in the past, and what I actually need now (maybe I did something wrong, or maybe I just do not remember the process to get to what I actually needed). Sorry for the confusion and thanks for your help. I'll get back to you once I am clear (which might be in 2 or 3 weeks after my vacation).

[borisgross]

I checked, and what I need (and also had used in the past) is two uniaxial anisotropies with unequal anisotropy constants. I wrongly pointed you to the formulation in Jonas' thesis, this is where I made a mistake, his formulation is different to what I need.

Since there is the possibility to set one uniaxial anisotropy axis already, my wish would indeed be a formulation as I had done for mumax3, adding a second uniaxial anisotropy axis, as shown further above. (I still need to research for a proper literature source for this.)

[jplauzie]

Hi,

With the term 'biaxial', one needs to be a bit careful, because it can refer to different things and people are not consistent about it. People often call it 'biaxial' because there are two axes, but two uniaxial anisotropies are not equivalent to a 'true' biaxial anisotropy (worse, sometimes people will call a cubic anisotropy "biaxial" because it looks biaxial if you only do in-plane loops. So there are three different things people will often use the name for 😆). The version in De Clercq is 'true' biaxial. Usually the give away for which type people are talking about is to look for things like two fold vs four-fold symmetry, 2nd degree/4th degree, or 2nd/4th power in the direction cosines ('true' biaxial being the latter of those terms).

If you have two uniaxial anisotropies of different strengths, you can represent this as a single uniaxial anisotropy (See Cullity section 7.11, Eq 7.66). If your anisotropy energies are

$E_a= K_a * sin^2(theta)$
$E_b= K_b * sin^2(90- theta)$

You can rewrite E_a+E_b as

$K_b+(K_a-K_b)*sin^2(theta)$

Dropping the constant term since only differences matter, you are left with:

$(K_a-K_b)*sin^2(theta)$

Which is just a normal uniaxial anisotropy with strength K_a-K_b. (This is basically what Ian did above, just with K_a != K_b. Notably, if K_a=K_b, there is no anisotropy in the plane. And while there is no anisotropy in the plane, there is an out of plane anisotropy- in his example, the m_z. The above equations are omitting the second angle, but you can carry it through easily enough)

That said, it can be convenient to just be able to add arbitrary uniaxial anisotropies, like how mumax3 does with Custom Fields. Also it's not so simple if you start including higher order terms, it does not simply so easily, as discussed here (and those higher terms do look like a true biaxial, in the sense that they are 4th order in the direction cosines)

Cullity also has the form for a true biaxial anisotropy (Eq 7.11). (This should be equivalent to the one used by De Clercq, if I'm recalling correctly).

(If you want to generalize, there is also a triaxial anisotropy, where all 3 axis strengths differ, and that is different from the above cases. I've only ever seen one or two people ask for it on the mumax3 boards though, it's pretty rare. As a side note, I believe CrSBr is actually triaxial, see e.g. here. The form of the triaxial energy looks like this. Despite the name, you can reduce it down to two parameters from 3 using the $m_x^2+m_y^2+m_z^2=1$ constraint. It ends up functionally looking like 2 uniaxials)

[borisgross]

Hi,
thank you for the clarification on the terminology, I have a déjà-vu on having this discussing before. Then I used with 'biaxial anisotropy' the wrong terminology for what I am look for, it actually seems to be triaxial anisotropy!

[borisgross]

Is there any chance that triaxial anisotropy will be implemented at any time soon? Otherwise, at least for my current project, it makes most sense for me to switch to mumax3.

[lamoreel]

The addition of this new feature has been placed on top of our TODO list, which we juggle alongside our own projects. Without making any promises, I suspect it will be available on our dev branch within one (or two) month(s). If this should seem too late, then perhaps switching to mumax3 might be the more fruitful option for you for the time being.