91影视

Number of particles =

Here,

$g\left(蔚\right)$=density of states,

$k$= Boltzmann's constant,

$T$= temperature.

The energy levels in the 3-dimensional harmonic oscillator,

$蔚=nhf$

Here,

$n$is any nonnegative integer and $f$is the classical oscillation frequency, and $h$is the Planck's constant

$n=\frac{蔚}{hf}$

The degeneracy level of $n$=

$g\left(n\right)dn=\frac{1}{2}(n+1)(n+2)$

Assuming $n>>1$$g\left(n\right)dn=\frac{n路n}{2}dn$,

$g\left(n\right)dn=\frac{n^2}{2}dn$

Differentiating the equation $蔚=nhf$on both side

$d蔚=dnhf$

$dn=\frac{d蔚}{h{f}^{\text{'}}}$

Substituting the value of $dn=\frac{d蔚}{h{f}^{\text{'}}}$and $n=蔚/hf$in the equation $g\left(n\right)dn=\frac{n^2}{2}dn$,

$g\left(蔚\right)d蔚=\frac{1}{2}{\left(\frac{蔚}{hf}\right)}^2\frac{d蔚}{hf}$

$=\frac{1}{2}\frac{{蔚}^2}{(hf{)}^3}d蔚$

Thus, Number of density states,$g\left(蔚\right)=\frac{1}{2}\frac{{蔚}^2}{(hf{)}^3}$

Substituting the value of $渭=0,N_0=0,g\left(蔚\right)=\frac{1}{2}\frac{{蔚}^2}{{\left(h{f}^{\text{'}}\right)}^3}$

$N={鈭�}_0^{鈭�}\frac{1}{2}\frac{{蔚}^2}{(hf{)}^3}\frac{d蔚}{e^{\frac{蔚}{kT}}-1}$

Let ,$x=\frac{蔚}{kT}\text{and}dx=\frac{d蔚}{kT}$

$=\frac{1}{2}{\left(\frac{k{T}_C}{hf}\right)}^3{鈭�}_0^{鈭�}\frac{x^{2}dx}{e^x-1}$

${\left(\frac{k{T}_C}{hf}\right)}^3=\frac{2N}{{鈭�}_0^{鈭�}\frac{x^{2}dx}{e^x-1}}$ (Equation-2)

As we know,

${鈭�}_0^{鈭�}\frac{x^2}{e^x-1}dx=螕\left(3\right)味\left(3\right)$

$=2!(1.202)$

$=2.404$

Substituting the value of ${鈭�}_0^{鈭�}\frac{x^2}{e^x-1}dx=2.404$in Equation-2

${\left(\frac{k{T}_C}{hf}\right)}^3=\left(\frac{2N}{2.404}\right)$

$T_C=\frac{hf}{k}{\left(\frac{N}{1.202}\right)}^{\frac{1}{3}}$

Thus, the condensate temperature for this system=$\frac{hf}{k}{\left(\frac{N}{1.202}\right)}^{\frac{1}{3}}$

We have,

The expression for potential energy

$E_{\mathrm{pot}}=\frac{C{L}^2}{2}$

$k{T}_C=\frac{C{L}^2}{2}$

Here,

$C$= spring constant

$L$= distance from the equilibrium position.

Angular frequency of system, $蝇=\sqrt{\frac{C}{m}}$

$C=m{蝇}^2$

Substituting the value of C in equation $k{T}_C=\frac{C{L}^2}{2}$

$k{T}_c=\frac{m{蝇}^2{L}^2}{2}$

$\frac{k{T}_C}{{蝇}^2}=\frac{m{L}^2}{2}$

$\frac{{蝇}^2}{k{T}_c}=\frac{2}{m{L}^2}$

Multiplying and dividing left side with $h^2$,

$\frac{{\mathrm{鈩�}}^2{蝇}^2}{{\mathrm{鈩�}}^{2}k{T}_C}=\frac{2}{m{L}^2}$

$\frac{(hf{)}^2}{{\mathrm{鈩�}}^{2}k{T}_c}=\frac{2}{m{L}^2}$

Applying square on both sides

${\left(k{T}_C\right)}^2=(hf{)}^2{\left(\frac{N}{1.202}\right)}^{\frac{2}{3}}$

$k{T}_C=\frac{(hf{)}^2}{k{T}_C}{\left(\frac{N}{1.202}\right)}^{\frac{2}{3}}$

$={\mathrm{鈩�}}^2\left(\frac{(hf{)}^2}{{\mathrm{鈩�}}^{2}k{T}_c}\right){\left(\frac{N}{1.202}\right)}^{\frac{2}{3}}$

Substituting the value of $\frac{2}{m{L}^2}\text{=}\frac{(hf{)}^2}{{\mathrm{鈩�}}^{2}k{T}_C}$

$k{T}_c={\mathrm{鈩�}}^2\frac{2}{m{L}^2}{\left(\frac{N}{1.202}\right)}^{\frac{2}{3}}$

$={\left(\frac{h}{2蟺}\right)}^2\frac{2}{m{L}^2}{\left(\frac{N}{1.202}\right)}^{\frac{2}{3}}$

Substituting $V^{1/3}\text{=}L$

$k{T}_C=\frac{1}{蟺}\left(\frac{h^2}{2蟺m}\right){\left(\frac{N}{V}\right)}^{\frac{2}{3}}{\left(\frac{1}{1.202}\right)}^{\frac{2}{3}}$

$=0.318\left(\frac{h^2}{2蟺m}\right){\left(\frac{N}{V}\right)}^{\frac{2}{3}}$

Therefore, the expression obtained was roughly equal to the condensate temperature of bosons confined inside a box with rigid walls.