Question

A ferromagnet is a material (like iron) that magnetizes spontaneously, even in the absence of an externally applied magnetic field. This happens because each elementary dipole has a strong tendency to align parallel to its neighbors. At $t=0$the magnetization of a ferromagnet has the maximum possible value, with all dipoles perfectly lined up; if there are $N$atoms, the total magnetization is typically$~2\mu eN$, where µa is the Bohr magneton. At somewhat higher temperatures, the excitations take the form of spin waves, which can be visualized classically as shown in Figure $7.30$. Like sound waves, spin waves are quantized: Each wave mode can have only integer multiples of a basic energy unit. In analogy with phonons, we think of the energy units as particles, called magnons. Each magnon reduces the total spin of the system by one unit of $\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$21r$}\right.$and therefore reduces the magnetization by $~2\mu e$. However, whereas the frequency of a sound wave is inversely proportional to its wavelength, the frequency of a spin-wave is proportional to the square of $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$.. (in the limit of long wavelengths). Therefore, since$\in =hf$and $p=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$.. for any "particle," the energy of a magnon is proportional

In the ground state of a ferromagnet, all the elementary dipoles point in the same direction. The lowest-energy excitations above the ground state are spin waves, in which the dipoles precess in a conical motion. A long-wavelength spin wave carries very little energy because the difference in direction between neighboring dipoles is very small.

to the square of its momentum. In analogy with the energy-momentum relation for an ordinary nonrelativistic particle, we can write $\in =\raisebox{1ex}{$p^2$}\!\left/ \!\raisebox{-1ex}{$2pm$}\right.$*, where$m$* is a constant related to the spin-spin interaction energy and the atomic spacing. For iron, m* turns out to equal $1.24\times {10}^{29}kg$, about$14$times the mass of an electron. Another difference between magnons and phonons is that each magnon ( or spin-wave mode) has only one possible polarization.

(a) Show that at low temperatures, the number of magnons per unit volume in a three-dimensional ferromagnet is given by

$\frac{Nm}{V}=2\pi {\left(\frac{2m\times kT}{h^2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx.$

Evaluate the integral numerically.

(b) Use the result of part ($a$) to find an expression for the fractional reduction in magnetization, $\left(M\right(O)-M(T\left)\right)/M\left(O\right).$Write your answer in the form ${(T/To)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$, and estimate the constant$T_0$for iron.

(c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low temperature. You should find $Cv/Nk={(T/Ti)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$, where $Ti$differs from To only by a numerical constant. Estimate$Ti$for iron, and compare the magnon and phonon contributions to the heat capacity. (The Debye temperature of iron is $470k$.)

(d) Consider a two-dimensional array of magnetic dipoles at low temperature. Assume that each elementary dipole can still point in any (threedimensional) direction, so spin waves are still possible. Show that the integral for the total number of magnons diverge in this case. (This result is an indication that there can be no spontaneous magnetization in such a two-dimensional system. However, in Section $8.2$we will consider a different two-dimensional model in which magnetization does occur.)

Short answer

(a). The required condition of this part is proved.

(b). The result is $\frac{M\left(0\right)-M\left(T\right)}{M\left(0\right)}=2\left(2.3152\right)\pi \frac{V}{N}{\left(\frac{2mkT}{h^2}\right)}^{3/2}$

(c). The heat capacity due to magnetic excitations in a ferromagnet at low temperature is $C_V=5\pi \left(1.7833\right)V{\left(\frac{2m}{h^2}\right)}^{3/2}{k}^{5/2}{T}^{3/2}$.

(d). The required condition of this part is proven.

Step-by-step solution

Part (a) step 1: Given Information 

We need to find the number of magnons per unit volume in a three-dimensional ferromagnet is given.

Part (a) step 2: Simplify

The number of magnons equal the sum of Planck distribution all over the modes $n_x,n_y$and $n_z$multiplied by the factor of $1$because we have two polarization modes, that is:

$N_m=\sum _{n_x}\sum _{n_y}\sum _{n_z}{\stackrel{-}{n}}_{\text{P}1}\left(ϵ\right)=\sum _{n_x,n_y,n_z}{\stackrel{-}{n}}_{\text{Pl}}\left(ϵ\right)$

but,

${\stackrel{-}{n}}_{\text{Pl}}\left(ϵ\right)=\frac{1}{e^{ϵ/kT}-1}$

where,

$ϵ=\frac{p^2}{2m}$

consider we have a cubic box with volume of and side width of , the allowed momenta for magnons is given by:

$p=\frac{h}{\lambda }=\frac{hn}{2L}$

therefore,

$ϵ=\frac{h^2{n}^2}{8m{L}^2}$

substitute into the above equation to get:

$N_m=\sum _{n_z,n_y,n_z}\frac{1}{e^{h^2{n}^2/8m{L}^{2}kT}-1}$

now we need to change the sum to an integral in spherical coordinates, by multiplying this by the spherical integration factor $n^2\sin \left(\theta \right)$so we get:

$N_m={\int }_0^{\pi /2}d\text{Φ}{\int }_0^{\pi /2}\sin \left(\theta \right)d\theta {\int }_0^{\infty }\frac{n^2}{e^{h^2{n}^2/8m{L}^{2}kT}-1}dn$

the first two integrals are easy to evaluate, and they give a factor of $\pi /2$so:

localid="1650306816172" $N_m=\frac{\pi }{2}{\int }_0^{\infty }\frac{n^2}{e^{h^2{n}^2/8m{L}^{2}kT}-1}dn$

now let,

localid="1650306825074" $x=\frac{h^2{n}^2}{8m{L}^{2}kT}\to n=\sqrt{\frac{8m{L}^{2}kTx}{h^2}}\to dn=\sqrt{\frac{2m{L}^{2}kT}{h^2}}\frac{1}{\sqrt{x}}dx$

thus,

localid="1650306839587" $N_m=\frac{\pi }{2}{\int }_0^{\infty }\frac{1}{e^x-1}\sqrt{\frac{2m{L}^{2}kT}{h^2}}\frac{8m{L}^{2}kTx}{h^2}\frac{1}{\sqrt{x}}dx$

localid="1650306845592" $N_m=2\pi {\left(\frac{2m{L}^{2}kT}{h^2}\right)}^{3/2}{\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx$

but the volume is given as localid="1650306851531" $V=L^3$therefore:

localid="1650306857173" $N_m=2\pi V{\left(\frac{2mkT}{h^2}\right)}^{3/2}{\int }_0^{\infty }\frac{\sqrt{x}}{e^x-1}dx$

using we can find the integral numerically, the code is shown in the following picture and the value of the integral is localid="1650306864643" $2.3152$, therefore:

localid="1650306870482" $N_m=2\left(2.3152\right)\pi V{\left(\frac{2mkT}{h^2}\right)}^{3/2}$

Part (b) step 1: Given Information 

We need to find an expression for the fractional reduction in magnetization.

Part (b) step 2: Simplify 

The total magnetization at the temperature of $T=0$is $2{\mu }_{B}N$each magnon reduces the total magnetization by a value of $2{\mu }_B$then the fractional reduction in the magnetization is:

$\frac{M\left(0\right)-M\left(T\right)}{M\left(0\right)}=\frac{2{\mu }_B{N}_m}{2{\mu }_{B}N}=\frac{N_m}{N}$

substitute from the result of part (a) to get:

$\begin{array}{c}\frac{M\left(0\right)-M\left(T\right)}{M\left(0\right)}=2\left(2.3152\right)\pi \frac{V}{N}{\left(\frac{2mkT}{h^2}\right)}^{3/2}=\left(\frac{T}{T_0}\right)\\ 2\left(2.3152\right)\pi \frac{V}{N}{\left(\frac{2mkT}{h^2}\right)}^{3/2}={\left(\frac{T}{T_0}\right)}^{3/2}\end{array}$

solve for $T_0$, we get:

$$\begin{gathered} \frac{T}{T_0}={\left(2\left(2.3152\right)\pi \frac{V}{N}\right)}^{2/3}\left(\frac{2mkT}{h^2}\right) \\ {T}_0={\left(\frac{1}{2\left(2.3152\right)\pi }\frac{N}{V}\right)}^{2/3}\left(\frac{h^2}{2mk}\right) \end{gathered}$$

now we need to find this value for iron, the volume of one mole of iron is

$V=7.11\times {10}^{-6}{\text{m}}^3$and the number of atoms in one mole of iron is $N=6.022\times {10}^{23}$the mass of the iron atom is $m=1.24\times {10}^{29}\text{kg}$substituted with these values to get:

$T_0={\left(\frac{1}{2\left(2.3152\right)\pi }\frac{\left(6.022\times {10}^{23}\right)}{\left(7.11\times {10}^{-6}{\text{m}}^3\right)}\right)}^{2/3}\left(\frac{{\left(6.626\times {10}^{34}\text{J}\cdot \text{s}\right)}^2}{2\left(1.24\times {10}^{29}\text{kg}\right)\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)}\right)$

$T_0=4152K$

Part (c) step 1: Given Information 

We need to find the magnon and phonon contributions to the heat capacity.

Part (c) step 2: Simplify

To find the heat capacity we first need to find the total energy, in the same way of finding $N_m$,but we must multiply the distribution with, that is:

$U=\sum _{n_x}\sum _{n_y}\sum _{n_z}ϵ{\stackrel{-}{n}}_{\text{Pl}}\left(ϵ\right)=\sum _{n_z,n_y,n_z}ϵ{\stackrel{-}{n}}_{\text{Pl}}\left(ϵ\right)$

but,

${\stackrel{-}{n}}_{\text{Pl}}\left(ϵ\right)=\frac{1}{e^{ϵ/kT}-1}$

where,

$ϵ=\frac{p^2}{2m}$

consider we have a cubic box with volume of and side width of , the allowed momenta for magnons is given by:

$p=\frac{h}{\lambda }=\frac{hn}{2L}$

therefore,

localid="1650306911538" $ϵ=\frac{h^2{n}^2}{8m{L}^2}$

substitute into the above equation to get:

localid="1650306915837" $U=\frac{h^2}{8m{L}^2}\underset{n_x,n_y,n_z}{}\frac{n^2}{e^{h^2{n}^2/8m{L}^{2}kT}-1}$

now we need to change the sum to an integral in spherical coordinates, by multiplying this by the spherical integration factor localid="1650306922250" $n^2\sin \left(\theta \right)$so we get:

localid="1650306926387" $U=\frac{h^2}{8m{L}^2}{\int }_0^{\pi /2}d\text{Φ}{\int }_0^{\pi /2}\sin \left(\theta \right)d\theta {\int }_0^{\infty }\frac{n^4}{e^{h^2{n}^2/8m{L}^{2}kT}-1}dn$

he first two integrals are easy to evaluate, and they give a factor of localid="1650306933758" $\pi /2$, so:

localid="1650306941435" $U=\frac{h^2}{8m{L}^2}\frac{\pi }{2}{\int }_0^{\infty }\frac{n^4}{e^{h^2{n}^2/8m{L}^{2}kT}-1}dn$

now let,

localid="1650306949525" $x=\frac{h^2{n}^2}{8m{L}^{2}kT}\to n=\sqrt{\frac{8m{L}^{2}kTx}{h^2}}\to dn=\sqrt{\frac{2m{L}^{2}kT}{h^2}}\frac{1}{\sqrt{x}}dx$

thus,

localid="1650306955308" $\begin{array}{c}U=\frac{h^2}{8m{L}^2}\frac{\pi }{2}{\left(\frac{8m{L}^{2}kTx}{h^2}\right)}^2{\int }_0^{\infty }\frac{x^2}{e^x-1}\sqrt{\frac{2m{L}^{2}kT}{h^2}\frac{1}{\sqrt{x}}}dx\\ U=2\pi V{\left(\frac{2mkT}{h^2}\right)}^{3/2}\left(kT\right){\int }_0^{\infty }\frac{x^{3/2}}{e^x-1}dx\end{array}$

using we can find the integral numerically, the code is shown in the following picture and the value of the integral is localid="1650306961001" $1.7833$therefore:

localid="1650306967244" $U=2\pi \left(1.7833\right)V{\left(\frac{2m}{h^2}\right)}^{3/2}{k}^{5/2}{T}^{5/2}$

now to find the heat capacity we diffrentiate the total energy with respect to localid="1650306974153" $T$, that is:

localid="1650306980247" $C_V=\frac{\partial U}{\partial T}=\frac{\partial }{\partial T}\left[2\pi \left(1.7833\right)V{\left(\frac{2m}{h^2}\right)}^{3/2}{k}^{5/2}{T}^{5/2}\right]$

localid="1650306985176" $C_V=5\pi \left(1.7833\right)V{\left(\frac{2m}{h^2}\right)}^{3/2}{k}^{5/2}{T}^{3/2}$

we can write the heat capacity as follow:

$\frac{C_V}{Nk}={\left(\frac{T}{T_1}\right)}^{3/2}$

where

$\begin{array}{c}\frac{1}{T_1^{3/2}}=5\pi \left(1.7833\right)\frac{V}{N}{\left(\frac{2mk}{h^2}\right)}^{3/2}\\ \frac{1}{T_1}={\left(5\pi \left(1.7833\right)\frac{V}{N}{\left(\frac{2mk}{h^2}\right)}^{3/2}\right)}^{2/3}\\ \frac{1}{T_1}={\left(5\pi \left(1.7833\right)\frac{V}{N}\right)}^{2/3}\left(\frac{2mk}{h^2}\right)\\ T_1={\left(\frac{1}{5\pi \left(1.7833\right)}\frac{N}{V}\right)}^{2/3}\left(\frac{h^2}{2mk}\right)\end{array}$

but,

$T_0{\left(2\left(2.3152\right)\right)}^{2/3}={\left(\frac{1}{\pi }\frac{N}{V}\right)}^{2/3}\left(\frac{h^2}{2mk}\right)$

therefore:

$T_1={\left(\frac{2\left(2.3152\right)}{5\left(1.7833\right)}\right)}^{2/3}{T}_0$

substitute with $T_0$to get:

$\begin{array}{c}{T}_1={\left(\frac{2\left(2.3152\right)}{5\left(1.7833\right)}\right)}^{2/3}\left(4152\text{K}\right)=2682.5\text{K}\\ T_1=2682.5\text{K}\end{array}$

Part (d) step 1: Given Information 

We need to find the integral for the total number of magnons that diverge.

Part (d) step 2: Simplify

For two dimensional array, we can write the number as:

$N_m=\sum _{n_z,n_y}\frac{1}{e^{ϵ/kT}-1}$

now we need to change the sum to an integral in polar coordinates, by multiplying this by the polar integration factor $\left(nd\theta \right),\left(dn\right)$so we get:

$N_m=\frac{\pi }{2}{\int }_0^{\infty }\frac{n}{e^{ϵ/kT}-1}dn$

change the integral to be in $x$, that is

$N_m\propto {\int }_0^{\infty }\frac{n}{e^x-1}dx$

near $x=0$we can expand the exponential using $e^x=1+x$so we get:

$\begin{array}{c}{N}_m\propto {\int }_0^{\infty }\frac{1}{1+x-1}dx\\ N_m\propto {\int }_0^{\infty }\frac{1}{x}dx\end{array}$

but the integration of$1/x$is$\ln \left(x\right)$so the number diverge at the lower limit$\left(0\right)$.