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How can electron spins in Iron at room temperature have ferromagnetic order even though they are travelling at very high speeds?

One could argue that spin and motion are completely uncorrelated and hence you can have superfast electrons that still somehow manage to orient their spins - but then how do you explain domains?

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This is a nice question and I am curious if we can attract a good answer. Edmund C. Stoner needed 43 pages in 1937 JSTOR link. You might want to start with understanding what electrons are doing in a metal instead of being local to a single atom. –  Alexander May 13 '13 at 20:51
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2 Answers

In ferromagnetic materials there is an unpaired electron in the outermost orbital, giving an overall magnetic moment equal to one electron spin to the atom. In a ferromagnetic bulk crystal, these orbitals can overlap between neighbouring atoms which causes the spontaneous magnetisation through the exchange interaction. This interaction is incredibly short range, just a few Angstrom (approx nearest neighbour atoms only) and so cannot be entirely responsible for domain structure. The important aspect of the exchange interaction for domain structure is an effective restoring torque on the spins between neighbouring atoms. Imagine the spins as rungs on a rope-ladder, if you try to misalign one rung the ones next to it will move slightly and if you let go, the torque will restore alignment. (A similar torque exists for the anisotropy in the crystal).

The next important contribution for domains is how the overall magnetic order in the material aligns. The internal magnetisation $M$ must terminate at the surfaces of the material, so if we take the usual $B = \mu_0 [H+M]$, and take the divergence:

$\nabla\cdot B = \mu_0 [\nabla\cdot H + \nabla \cdot M] $

so: $ \nabla \cdot H = -\nabla \cdot M$

All this means is that all magnetisation which terminates at the surface must produce a stray field, $H$. In short, the production of the stray field corresponds to a cost in energy $\frac{1}{2\mu_0}\int H^2 dV$ which can be minimised by reducing the amount of $M$ which terminates at the surface. The consequence is that the minimum energy state for the magnetic order is certainly not for all the spins to align with one-another in most cases. This is why we see twists and turns in the magnetic order, 'Domain structure'.

So basically, the exchange interaction provides a method for spins to align with one another on the short scale and the minimisation of the stray field shapes the larger structure across whole domains. (Other contributions can come from external magnetic fields and magneto-crystalline anisotropy).

I may have gone off topic there but I hope it answers some part of your question :)

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Wow very interesting points! Thanks for putting it together. I'm still digesting parts of that - but there is one question that just occurred to me: as far as ferromagnetism in Iron is concerned, is 1) the Nearly Free Electron model or 2) the Tight-Binding model the more useful intuitive picture? –  Shreyas Patankar May 17 '13 at 0:12
    
Good question.. I don't know. I havn't studied the band structure of ferromagnets in any detail, but there's a good review on the subject of domain formation and polarized currents that may be of interest to you (if you can access it): C.H. Marrows "Spin Polarised Currents and Magnetic Domain Walls". Advanced in Physics '54'(8) 2005. link –  StickyCube May 17 '13 at 12:06
    
This answer is a good description of ferromagnetism but doesn't help to reconcile that behaviour with the common conception of a metal. If it's simply the case that electrons moving between domains quickly reorient to align with the new net moment, can we perhaps make an order of magnitude estimate for the rate at which domain boundaries fluctuate? –  Eric Thewalt May 21 '13 at 0:36
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It's because the electrons in the conduction band are correlated, motion and spin are not uncorrelated since pauli principle is acting, if the spins are opposite the motion can be more "free", but if they point towards the same direction they can be closer (conterintuitevely) Exchange interaction

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so does that mean that the spins are constrained by the exchange interaction to be stationary? then how is a ferromagnet still a metal? –  Shreyas Patankar May 13 '13 at 6:09
    
You are assuming that all the electrons are perfectly aligning towards a fixed direction, I don't think this is true, it's more like a tendency, the magnetization reach its maximum at zero degrees continuosly, it's a second order phase transition –  Ikiperu May 13 '13 at 6:11
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