# Problem with Curie's temperature

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## Main Question or Discussion Point

According to Curie-Weiss law, the magnetic susceptibility obeys the equation $\chi=\frac{C}{T-T_c}$ where $T_c$ is the Curie temperature. People say this implies that the material is ferromagnetic for $T<T_c$,i.e. has non-zero magnetization in the absence of external magnetic field and then loses this property for $T>T_c$. I have 2 questions now:
1) What is the domain of applicability of the mentioned equation?
2) A material is ferromagnetic if $\chi \to \infty$. But the mentioned equation says that this only happens for $T\to T_c$ and below and above $T_c$, there is no ferromagneticity for the material. But in all references, it is said that below $T_c$ there is ferromagneticisty for the material. How can I solve this contradiction?
Thanks

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The Curie-Weiss law is valid only above T_c

Gold Member
The Curie-Weiss law is valid only above T_c
So what about below $T_c$?
My problem is, because we have $\vec M=\chi \vec H$, we can have non-zero magnetization with zero magnetic field, only if $\chi\to \infty$ and because below $T_c$, the material is ferromagnetic, it seems that for all $T<T_c$, we should have $\chi\to\infty$. But it seems wrong to me. What is wrong with my reasoning? Can you give some reference that treats this?
In dielectrics, we have $\varepsilon=\frac{1+\frac{8\pi}{3}\sum N_i \alpha_i}{1-\frac{4\pi}{3}\sum N_i \alpha_i}$ where $\varepsilon$ is the dielectric constant,$N_i$ is the number density of type i atoms and $\alpha_i$ is the polarizability of type i atoms. Now here, for $\sum N_i \alpha_i=\frac{3}{4\pi}$, we have $\varepsilon\to \infty$ which means we have ferroelectricity. I wanna know is there analogous calculations for ferromagnetism where we have the singularity in terms of the properties of the material only and not in terms of temperature?

Last edited:
Matterwave
Gold Member
The equation $\vec{M}=\chi \vec{H}$ is a linear relation between magnetization and the external field. This is only approximately true for some materials (e.g. para magnets). Ferro magnets will exhibit hysteresis and as such the magnetization is not a one-to-one function of the external field. You may want to look here: http://en.wikipedia.org/wiki/Magnetic_hysteresis

Once you get into ordered phases like ferro-, ferri- and antiferromagnets the physics becomes much more complicated. The response is not linear anymore. In many cases it depends on the history of the sample (e.g. hysteresis loops as mentioned above), and there may be threshold fields above which phase transitions to a different ordered state occur. For practical applications you also have to worry about the formation of domains, which is yet another can of worms.

Even in the para- or diamagnetic state $\vec{M} = \chi \vec{H}$ is an approximation that assumes that the material is isotropic. In anisotropic materials $\chi$ is a rank-2 tensor.

DrDu
The Curie Weiss law can be obtained for an Ising model in the mean field approximation:
http://en.wikipedia.org/wiki/Mean_field_theory

From the expression found, you can derive the linear response corresponding to the Curie Weiss law for T>Tc and spontaneous magnetization below.

According to Curie-Weiss law, the magnetic susceptibility obeys the equation $\chi=\frac{C}{T-T_c}$ where $T_c$ is the Curie temperature. People say this implies that the material is ferromagnetic for $T<T_c$,i.e. has non-zero magnetization in the absence of external magnetic field and then loses this property for $T>T_c$. I have 2 questions now:
1) What is the domain of applicability of the mentioned equation?
2) A material is ferromagnetic if $\chi \to \infty$. But the mentioned equation says that this only happens for $T\to T_c$ and below and above $T_c$, there is no ferromagneticity for the material. But in all references, it is said that below $T_c$ there is ferromagneticisty for the material. How can I solve this contradiction?
Thanks
There is no contradiction. The Curie-Weiss law is an oversimplification of the behavior of ferromagnets in the region near Tc.