Temperature dependence of magnetic domains

Does the size of magnetic domains depend on temperature ? Not able to find any papers on this subject, maybe because there is no such dependence...

• I would think that there is generally a large energy barrier for splitting magnetic domains or combining them, so I wouldn't be surprised that the temperature dependence is very small to nil. – user93237 Dec 20 '15 at 22:04
• Generally, when a substance gets colder, the magnetic domains become harder to change, and when it gets hotter it gets easier. Above the substance's Curie Temperature, the magnetic domains randomize and it loses any permanent magnetism. – CoilKid Dec 20 '15 at 22:44
• Let me put it another way. Let's take a piece of non magnetized iron at T1 below its Curie T. Let's check the number of domains in it = n1. Then let's change its temperature to T2, again below its Curie T and check the number n2 of domains in it. Is n1 = n2 ? n1 > n2 ? n1 < n2 ? – Anarchasis Dec 20 '15 at 23:00
• Answer is yes, domain size is temperature dependent. Especially near transition point (Curie temperature). I would expect that a valid answer could use Ising model as an example, and would use terms like 'critical exponent', 'transition point' and 'correlation length' (domain size). – Mikael Kuisma Dec 22 '15 at 3:21

As Samuel and CoilKid suggested, it seems that there is not really any temperature dependance of the number of magnetic domains in a bulk material as long as you stay away from the Curie temperature, because the energy scale associated to the creation/destruction of a domain wall is far away from any temperature scale. Let us detail this a little bit more :

What we call domain wall is the region where the magnetization varies continuously from one domain to another. Below is depicted a (Bloch) domain wall separating two domains for which magnetization rotates from the $+\hat{z}$ direction to the $-\hat{z}$ direction.

One can compute the static energy associated to the creation of such domain wall. There are two terms to be taken into account : the magnetocrystalline anisotropy term which is minimum when the magnetization is along the direction of the easy axis, and the exchange term which represents the local interaction due to the presence of a gradient in magnetization.

Let us considere a 1D domain wall. For simplicity, we assume that :

• The magnetocrystalline anisotropy is uniaxial, and the associated easy axis is $\hat{u}_K=+\hat{z}$.
• We define the local magnetization $\textbf{M}=M_s\textbf{m}$, so that $\|\textbf{m}\|=1$. We assume that the magnetization will stay in the $(\hat{y},\hat{z})$ plan along the domain wall.
• We note $\theta$ the angle between the local magnetization and the $+\hat{z}$ axis, such that : $$\textbf{m}= \begin{pmatrix} 0\\[1mm] \sin\theta(x)\\[1mm] \cos\theta(x) \\ \end{pmatrix}$$
• Let be $S$ the transverse surface of the domain wall in the $(\hat{y},\hat{z})$ plan, and $\Delta$ the width of the domain wall.

Thus, the exchange energy takes the form : $$E_\text{ex}=A\int_{-\Delta/2}^{+\Delta/2} \mathrm{d}S\,\mathrm{d}x\,(\nabla\textbf{m})^2=AS\int_{-\Delta/2}^{+\Delta/2}\mathrm{d}x\,(\partial_x\theta)^2$$ Moreover, the anisotropy energy reads : $$E_\text{anis}=K\int_{-\Delta/2}^{+\Delta/2} \mathrm{d}S\,\mathrm{d}x\,\left[1-\hat{u}_K\cdot\hat{z}\right]=KS\int_{-\Delta/2}^{+\Delta/2}\mathrm{d}x\,\sin^2\theta(x)$$

The equilibrium magnetization profile

The first thing to do is to calculate the equilibrium magnetization profile which minimizes its total energy $E=E_\text{ex}+E_\text{anis}$, following the optimization : $$\delta E=E(\theta+\delta\theta)-E(\theta)=0$$ One can show that it is equivalent to the condition : $$\ln\tan\left(\frac{\theta}{2}\right)=x\sqrt{\frac{K}{A}}\quad\text{with}\quad\sqrt{\frac{A}{K}}=\Delta$$ or equivalently : $$\theta(x)=2\arctan\exp\left(\frac{x}{\Delta}\right)$$

The total energy of a static domain wall

Now it is possible to perform the integrations for $E_\text{ex}$ and $E_\text{anis}$ and we would obtain : $$E=4S\sqrt{AK}$$ Now we can do some numerics. Let's take for example the cobalt for which : $$A=10^{-11}\,J.m^{-1}\qquad\text{and}\qquad K=1,25.10^{6}\,J.m^{-3}$$ at room temperature $T=300\,K$.

Let's take $S\sim\ell^2$ where $\ell$ is the typical size of a magnetic domain,i.e. typically $\ell\sim 10\,\mu m$, then we obtain : $$E=1,41.10^{-12}\,J\gg k_B T$$ at room temperature $k_B T\sim 4,1.10^{-21}\,J$.

Conclusion

If you are far from the Curie temperature, it is not possible to create or destroy a domain wall since the energy barrier is much higher than any thermal energy scale. Thus, changing the temperature does not modify the number of magnetic domains in the material.

However, it is well known that if you increase the temperature higher than the Curie temperature, the material loses its ferromagnetism and become paramagnetic by undergoing a phase transition. Such phase transition occurs when the temperature is such that : $$E\sim k_B T$$ i.e. when basically any thermal fluctuation can create a domain wall and kill the long range order of the spins.

• OK thanks a lot for your answer. This is what I needed. Isn't there a unit problem ? 4S.Sqrt (AK) isn't it an energy per unit length ? – Anarchasis Dec 21 '15 at 23:16
• Yeah the unit of $A$ was wrong. See the edit, thanks for pointing it out! – dolun Dec 22 '15 at 1:33