# Lifetime of the magnetization of a ferromagnet

In a ferromagnetic material, there's spontaneous magnetisation, meaning in the absence of a $\mathbf{B}$ field $\mathbf{M}\neq 0.$ This means that there may be different domains in the sample where all magnetic moments are pointing in the same direction at the equilibrium state. Some everyday example of such materials that occur naturally would be iron, nickel, cobalt, and so on. Such magnets create a non-zero field $\mathbf{H}$ outside the sample. Which is responsible for example for two magnets that interact.

• If the magnet is left on its own, it keeps generating a magnetic field in its vicinity even if no other magnetizable material is around (so no interaction), so in a long enough amount of time, does it lose its magnetization? I mean it cannot provide a magnetic field indefinitely, something has to be providing for the energy.

Admittedly the title of the post is rather dubious, because the net magnetisation of a ferromagnet at $\mathbf{B}=0$ is ultimately there because it corresponds to the lowest energy state of the sample at hand, but it is still radiating a magnetic field around itself.

• Now suppose we bring a paramagnet into the picture, then some of the magnetic moments in this paramagnet will align themselves with the magnetic field created by the ferromagnet ($\mathbf{B}=\mu_0 \chi \mathbf{M}$). Does this interaction cause the ferromagnet to fall into a lower magnetization state?
• physics.stackexchange.com/questions/14667/… – mmesser314 Jan 16 '15 at 11:04
• Creating and maintaining a field is not an efford as such. Think of a charge - it creates an electric field around it. Or any object - it creates a gravitational field around it. The field itself is not energy demanding to maintain. But when something interacts with this field, then energy transfers start – Steeven Jan 16 '15 at 16:05

Strictly speaking, a magnetized ferromagnet (FM) is not in its lowest energy state. $\mathbf{M}=0$ would be the lowest energy. However, the FM can not get there from a magnetized state, because that would involve moving domain boundaries around, which requires energy the magnet does not normally have. Therefore it sits there magnetized, in a metastable state. You can think of it as a local free energy minimum. To get to the global minimum ($\mathbf{M}=0$ state), which is the lowest energy, an energy barrier must be overcome. The ferromagnet will stay magnetized unless something supplies enough energy to hop over the barrier. That something is often the thermal bath that keeps your magnet at a fixed temperature. The higher the temperature, the quicker this will occur.
Realize that a magnetized FM is not in equilibrium. At least formally we can say that there is a current associated with magnetization: $\nabla \times \mathbf{M}=\mathbf{J}$, and where there is any current, there is no equilibrium. Now, it may happen suddenly that there is just enough energy to move a little bit of a domain wall, because there are energy fluctuations whenever you are in thermal contact with a heat bath (heat is basically random motion of atoms, particles, etc.). Then you can reduce the overall magnetization a little. So, if you wait, than you could see the magnetization creeping down a little (it is called creep). This is in principle. In practice, I am not sure the rate of such creep can be measured anywhere but extremely close to the Curie point of a typical FM like iron. So in practice we don't worry about it. However, for spin glass the creep is measurable, basically because the energy barriers involved are very small and it is easier to find enough thermal energy to surmount them.
• Sure, you can edit it yourself, just find the edit button below your answer, and replace  \nabla \cdot \mathbf{M}=\mathbf{J} with  \nabla \times \mathbf{M}=\mathbf{J} to get the cross prod. Don't forget the two \$ signs around each math expression (use them like parentheses). – Phonon Jan 16 '15 at 19:25