91影视

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

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:

$N_m=\underset{n_x}{鈭�}\underset{n_y}{鈭�}\underset{n_z}{鈭�}{\stackrel{-}{n}}_{\text{P}1}\left(系\right)=\underset{n_x,n_y,n_z}{鈭�}{\stackrel{-}{n}}_{\text{Pl}}\left(系\right)$

but,

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

where,

$系=\frac{p^2}{2m}$

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

$p=\frac{h}{位}=\frac{hn}{2L}$

therefore,

$系=\frac{h^2{n}^2}{8m{L}^2}$

substitute into the above equation to get:

$N_m=\underset{n_z,n_y,n_z}{鈭�}\frac{1}{e^{h^2{n}^2/8m{L}^{2}kT}鈭�1}$

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

$N_m={鈭�}_0^{蟺/2}d\text{桅}{鈭�}_0^{蟺/2}\sin \left(胃\right)d胃{鈭�}_0^{\mathrm{鈭�}}\frac{n^2}{e^{h^2{n}^2/8m{L}^{2}kT}鈭�1}dn$

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

localid="1650306816172" $N_m=\frac{蟺}{2}{鈭�}_0^{\mathrm{鈭�}}\frac{n^2}{e^{h^2{n}^2/8m{L}^{2}kT}鈭�1}dn$

now let,

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

thus,

localid="1650306839587" $N_m=\frac{蟺}{2}{鈭�}_0^{\mathrm{鈭�}}\frac{1}{e^x鈭�1}\sqrt{\frac{2m{L}^{2}kT}{h^2}}\frac{8m{L}^{2}kTx}{h^2}\frac{1}{\sqrt{x}}dx$

localid="1650306845592" $N_m=2蟺{\left(\frac{2m{L}^{2}kT}{h^2}\right)}^{3/2}{鈭�}_0^{\mathrm{鈭�}}\frac{\sqrt{x}}{e^x鈭�1}dx$

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

localid="1650306857173" $N_m=2蟺V{\left(\frac{2mkT}{h^2}\right)}^{3/2}{鈭�}_0^{\mathrm{鈭�}}\frac{\sqrt{x}}{e^x鈭�1}dx$

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

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

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

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

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

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

solve for $T_0$, we get:

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

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

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

$T_0={\left(\frac{1}{2\left(2.3152\right)蟺}\frac{\left(6.022脳{10}^{23}\right)}{\left(7.11脳{10}^{鈭�6}{\text{m}}^3\right)}\right)}^{2/3}\left(\frac{{\left(6.626脳{10}^{34}\text{J}鈰�\text{s}\right)}^2}{2\left(1.24脳{10}^{29}\text{kg}\right)\left(1.38脳{10}^{鈭�23}\text{J}/\text{K}\right)}\right)$

$T_0=4152K$

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

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:

$U=\underset{n_x}{鈭�}\underset{n_y}{鈭�}\underset{n_z}{鈭�}系{\stackrel{-}{n}}_{\text{Pl}}\left(系\right)=\underset{n_z,n_y,n_z}{鈭�}系{\stackrel{-}{n}}_{\text{Pl}}\left(系\right)$

but,

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

where,

$系=\frac{p^2}{2m}$

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

$p=\frac{h}{位}=\frac{hn}{2L}$

therefore,

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

substitute into the above equation to get:

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

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

localid="1650306926387" $U=\frac{h^2}{8m{L}^2}{鈭�}_0^{蟺/2}d\text{桅}{鈭�}_0^{蟺/2}\sin \left(胃\right)d胃{鈭�}_0^{\mathrm{鈭�}}\frac{n^4}{e^{h^2{n}^2/8m{L}^{2}kT}鈭�1}dn$

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

localid="1650306941435" $U=\frac{h^2}{8m{L}^2}\frac{蟺}{2}{鈭�}_0^{\mathrm{鈭�}}\frac{n^4}{e^{h^2{n}^2/8m{L}^{2}kT}鈭�1}dn$

now let,

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

thus,

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

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

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

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

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

we can write the heat capacity as follow:

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

where

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

but,

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

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:

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

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

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

$N_m=\underset{n_z,n_y}{鈭�}\frac{1}{e^{系/kT}鈭�1}$

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

$N_m=\frac{蟺}{2}{鈭�}_0^{\mathrm{鈭�}}\frac{n}{e^{系/kT}鈭�1}dn$

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

$N_m鈭�{鈭�}_0^{\mathrm{鈭�}}\frac{n}{e^x鈭�1}dx$

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

$\begin{array}{c}{N}_m鈭�{鈭�}_0^{\mathrm{鈭�}}\frac{1}{1+x鈭�1}dx\\ N_m鈭�{鈭�}_0^{\mathrm{鈭�}}\frac{1}{x}dx\end{array}$

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