Physics 512 (Spring 2018)

HOMEWORK 4: solutionsdue 8 February 2018

(1) Exact solution of the Weiss modelWe can set external applied field H = 0 and use Weiss molecular field to

find exact solutions. Written dimensionless magnetic moment, m = MMs, and

introducing

y =gµBJλM

kBT

x = exp (y)

the exact solution for the paramagnetic state is:

m = BJ (y)

(Tc = nµ2Bλ/kB and Ms = ngµBJ)For J = 1/2, express T/Tc as a function of m ONLY.Specifically, write down an expression containing only m on one side and

only T/Tc on the other and no other parameters involved.

Solution:For J = 1/2, g = 2 and

BJ (y) = tanh y =ey − e−yey + e−y

=x− 1

x

x+ 1x

=x2 − 1x2 + 1

= m

sox2 =

1 +m

1−mTaking ln of both sides, we have:

2 lnx = 2y = ln1 +m

1−m

with our definitions (and gJ = 1),

y =nµ2Bλ

kB

(gJ)2

T

M

ngµBJ=TcTm

therefore

2TcTm = ln

1 +m

1−mor

T

Tc=

2m

ln (1 +m)− ln (1−m)

1

(2) Helical magnetic orderSome materials have distinct layered magnetic structure. For example Dy.

Consider a situation when magnetic moments are ordered ferromagnetically ineach atomic plane, but the planes may have angle θ between directions of themagnetic moments in them. See the illustration below.

The energy can be described by nearest and next nearest neighbour exchangeconstants, J1 and J2 , respectively.

E = −2NS2 (J1 cos θ + J2 cos 2θ)

where N is the number of atoms in each plane, each atom having spin S.(a) Find equilibrium angle(s) θ corresponding to possible ground states in

terms of J1 and J2.(b) Which of the following states possible? ferromagnetic (FM), antiferro-

magnetic (AFM) or helical?(c) Draw the "phase diagram" of possible magnetic states on a (J1, J2) plane.

2

Solutions(a) To find possible solution, we need to minimize E with respect to θ, so

dE

dθ= 2NS2 (J1 sin θ + 2J2 sin 2θ) = 0

withsin 2θ = 2 sin θ cos θ

the possible solutions are found from:

sin θ (J1 + 4J2 cos θ) = 0

The solutions are:

sin θ = 0

sin θ = π

cos θ = − J14J2

(b) Therefore, all three states are possible: θ = 0 - FM, θ = π - AFM and

θ = arccos(− J14J2

)- helimagnetic order.

(c) Clearly, the following phase diagram shows possible magnetic states inthis model:

3

4