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Q. 7.73

An Introduction to Thermal Physics
Found in: Page 325
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

Consider a gas of n identical spin-0 bosons confined by an isotropic three-dimensional harmonic oscillator potential. (In the rubidium experiment discussed above, the confining potential was actually harmonic, though not isotropic.) The energy levels in this potential are ε=nhf, where n is any nonnegative integer and f is the classical oscillation frequency. The degeneracy of level n is (n+1)(n+2)/2.

(a) Find a formula for the density of states, g(ε), for an atom confined by this potential. (You may assume n>>1.)

(b) Find a formula for the condensation temperature of this system, in terms of the oscillation frequency f.

(c) This potential effectively confines particles inside a volume of roughly the cube of the oscillation amplitude. The oscillation amplitude, in turn, can be estimated by setting the particle's total energy (of order kT ) equal to the potential energy of the "spring." Making these associations, and neglecting all factors of 2 and π and so on, show that your answer to part (b) is roughly equivalent to the formula derived in the text for the condensation temperature of bosons confined inside a box with rigid walls.

(a) Number of density states g(ε) = 12ε2(hf)3.

(b) The condensation temperature for this system =hfkN1.20213

(c) The expression obtained was roughly equal to the condensate temperature of bosons confined inside a box with rigid walls.

See the step by step solution

Step by Step Solution

Step 1. Given information

Number of particles =

N=0g(ε)dεe(ε-μ)kT-1 (Equation-1)


g(ε)=density of states,

k= Boltzmann's constant,

T = temperature.

Step 2.  (a) To find the formula for density state

The energy levels in the 3-dimensional harmonic oscillator,



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


The degeneracy level of n=


Assuming n>>1g(n)dn=n·n2dn,


Differentiating the equation ε=nhfon both side



Substituting the value of dn=dεhf'and n=ε/hf in the equation g(n)dn=n22dn,



Thus, Number of density states, g(ε)=12ε2(hf)3

Step 3. To find the condensate temperature of the system we have N=∫0∞g(ε)dεe(ε-μ)kT-1 

Substituting the value of μ=0, N0=0, g(ε)=12ε2hf'3


Let ,x=εkT and dx=dεkT



kTChf3=2N0x2dxex-1 (Equation-2)

As we know,


=2 !(1.202)


Substituting the value of 0x2ex-1dx=2.404 in Equation-2



Thus, the condensate temperature for this system=hfkN1.20213

Step 4. To find the condensation temperature of bosons confined inside a box with rigid walls

We have,

The expression for potential energy




C = spring constant

L = distance from the equilibrium position.

Angular frequency of system, ω=Cm


Substituting the value of C in equation kTC=CL22




Multiplying and dividing left side with h2,



Step 5.  The condensate temperature is kTC=hfN1.20213

Applying square on both sides




Substituting the value of 2mL2 = (hf)22kTC



Substituting V1/3 = L



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

Most popular questions for Physics Textbooks

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~2μeN, 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 h21 rand therefore reduces the magnetization by ~2μe. 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 1λ.. (in the limit of long wavelengths). Therefore, since=hf and p=hλ.. 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 =p22pm*, wherem* is a constant related to the spin-spin interaction energy and the atomic spacing. For iron, m* turns out to equal 1.24×1029kg, about14 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

NmV=2π2m×kTh232 0xex-1dx.

Evaluate the integral numerically.

(b) Use the result of part (a) to find an expression for the fractional reduction in magnetization, (M(O) - M(T))/M(O). Write your answer in the form (T /To)32, and estimate the constantT0 for iron.

(c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low temperature. You should find Cv / N k = (T /Ti)32 , where Ti differs from To only by a numerical constant. EstimateTi 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.)


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