ferromagnetism at the curie temperature tc, the magnetism m becomes zero. tc is mainly determined by...
Post on 21-Dec-2015
218 views
TRANSCRIPT
Ferromagnetism
• At the Curie Temperature Tc, the magnetism M becomes zero.
• Tc is mainly determined by the exchange J.
• As T approaches Tc, M approaches zero in a power law manner (critical behaviour).
M
Tc
Coercive behaviour
• Hc, the coercive field, is mainly determined by the anisotropy constant (both intrinsic and shape.)
Hc
Coherent rotation model of coercive behaviour
• E=-K cos2 ()+MH cos(-0).
• E/=0; 2E/2=0.• E/= K sin 2()-MH
sin(-0).
• K sin 2=MH sin(-0).
• 2E/2=2K cos 2()-MH cos(-0).
• 2K cos 2=MH cos(-0).
Coherent rotation
• K sin 2=MHc sin(-0).
• K cos 2=MHc cos(-0)/2.
• Hc(0)=(2K/ M)[1-(tan0)2/3 +(tan0)4/3 ]0.5 / (1+(tan0)2/3).
Special case: 0=0
• Hc0=2K/M.
• This is a kind of upper limit to the coercive field. In real life, the coercive field can be a 1/10 of this value because the actual behaviour is controlled by the pinning of domain walls.
Special case: 0=0, finite T, H<Hc
• Hc=2K/M.
• In general, at the local energy maximum, cos m=MH/2K.
• Emax= -K cos2 m +MH cos m= (MH)2/4K.
• E0=E(=0)=-K+MH
• For Hc-H=, U=Emax-E0=NM22/4K.
• Rate of switching, P = exp(-U/kBT) where is the attempt frequency
Special case: 0=0, Hc(T)
• Hc0=2K/M.
• For Hc0-H=, U=Emax-E0=NM22/4K.
• Rate of switching, P = exp(-U/kBT).
• Hc(T) determined by P ¼ 1. We get Hc(T)=Hc0-[4K kB T ln()/NM2]0.5
• In general Hc0-Hc(T)/ T. For 0=0, =1/2; for 0 0, =3/2
Non-uniform magnetization: formation of domains due to the
dipolar interaction• Edipo=(0/8) s d3R d3R’ M(R)M(R’) iajb(1/|
R-R’|).
• After two integrations by parts and assuming that the surface terms are zero, we get
• Edipo=(0/8) s d3R d3R’M(R)M(R’)/|R-R’| where the magnetic charge M=r ¢ M.
Non-uniform magnetization: formation of domains
• For the case of uniform magnetization in fig. 1, there is a large dipolar energy proportional to the volume V
• For fig. 2, the magnetic energy is very small. But there is a domain wall energy / V2/3.
M Fig. 1
Fig. 2
Uniform magnetization: magnetic energy
• The magnetic charge at the top is M(z=L/2)=Md (z-L/2)/dz=M (z-L/2); similarly M(z=-L/2)=-M (z+L/2). The magnetic energy is 0 M2 AL/80 M2 V/8
M Fig. 1
Fig. 2
L/2-L/2
Magnetic charge density is small for closure domains
• For the closure domain, as one crosses the domain boundary, the magnetic charge density is M=dMx/dx +dMz/dz=-M+M=0. Thus the magnetic energy is small.
M
Fig. 1
x
z
Domain walls
• Bloch wall: the spins lie in the yz plane. The magnetic charge is small.
• Neel wall: the spins lie in the xz plane. The dipolar energy is higher because the magnetic charge is nonzero here.
z
xy
Domain wall energy
• Because the exchange J is largest, first neglect the dipolar copntribution.
• Assume that the angle of orientation changes slowly from spin to spin.
• The exchange energy is approximately Js (d/dx)2
Domain wall energy
• Energy to be minimized: U=J s (d/dx)2-Ks cos2().
• Minimizing U, we get the equation • –Jd2/dx2+2K sin(2)=0. This can be written as• -d2/dt2+2 sin (2)=0 where t=x/l; the magnetic
length l=(J/K)0.5.
This looks like the same equation for the time dependence of a pendulum in a gravitational field : m d2y/dt2-m g sin y=0.
Domain wall energy
• From the ``conservation of energy’’, we obtain the equation (d/dt)2+ cos (2)=C where C is a constant.
• From this equation, we get s d /[C-cos(2)]0.5= t. To illustrate, consider the special case with C=1, then we get the equation s d/sin()=t. Integrating, we get ln|tan()|=2t; =2 tan-1 exp(2t).
• t=-1, =0; t=1, =.
Non-uniform magnetization: Spin wave
Rate of change of angular momentum, ~ dSi/dt is equal to the torque, [Si, H]/i where H is the Hamiltonian, the square bracket means the commutator.
• Using the commutation relationship [Sx,Sy ]=iSz
: [S, (S¢ A)]=iA£ S. For
example x component [Sx, SyAy+SzAz] =iSzAy-iAzSy
• We obtain ~ dSi/dt=2J Si£ Sj+
Ferromagnetic spin waves
• Consider a ferromagnet with all the spins line up in equilibrium. Consider small deviation from it. Write Si=S0+ Si, we get the linearized equation
• We get ~ d Si/dt=-2J S0£ ( Si- Si+ )
Ferromagnetic spin waves:
• ~ d Si/dt=-2J S0£ ( Si- Si+ ) • Write Si=Ak exp(ik t-k r), we obtain the
equation• i ~ k Ak =CAk £ S0; C= 2J(1-eik ). In
component form (S0 along z): i ~ k Akx
=CAky , i ~ k Aky =-CAkx
• For S0 along z, Ak=A(1, i, 0) and ~k = 2J|S0| (1-cos{k }). For k small, k~Dk2 where D=JzS02.
Spin wave energy gap
• At k=0, k=0.
• Suppose we include an anisotropy term Ha=-
(K/2)i Siz2=-(K/2)i[S-( S)
2]. In terms of Fourier transforms Ha=(K/2)k ( Sk)2+constant.
• i ~ k Ak =C’Ak £ S0; C’= 2J(1-eik )+K.
• ~k = 2J|S0| (1-cos{k })+K.
k=0=K. This is usually measured by FMR