Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Some problems from Kittel Thermal Physics, chapter II

Posted by peeterjoot on January 10, 2013

[Click here for a PDF of this post with nicer formatting]


Some review from the ancient second (or third?) year thermal physics course I took.


Energy and temperature [1] problem 2.1

Suppose g(U) = C U^{3N/2}, where C is a constant and N is the number of particles. This form of g(U) actually applies to an ideal gas.

Show that U = 3 N t/2

Show that (\partial^2 \sigma/\partial U^2)_N is negative.



We’ve got

\begin{aligned}\frac{1}{{\tau}}= \frac{\partial {\sigma}}{\partial {U}}= \frac{\partial {}}{\partial {U}} \left( \ln C + \frac{3N}{2} \ln U \right)= \frac{3N}{2} \frac{1}{{U}},\end{aligned} \hspace{\stretch{1}}(1.2.1)


\begin{aligned}U = \frac{3N}{2} \tau.\end{aligned} \hspace{\stretch{1}}(1.2.2)

Second derivative of entropy

From above

\begin{aligned}\frac{\partial^2 \sigma}{\partial U^2}= -\frac{3N}{2} \frac{1}{{U^2}}.\end{aligned} \hspace{\stretch{1}}(1.2.3)

This doesn’t seem particularly suprising if we look at the plots. For example for C = 1 and 3N/2 = 1 we have (Fig 1)


Fig1: Plots of entropy and its derivatives for this multiplicity function

The rate of change of entropy with energy decreases monotonically and is always positive, but always has a negative slope.

Paramagnetism [1] problem 2.2

Find the equilibrium value at temperrature \tau of the fractional magnetization

\begin{aligned}\frac{M}{N m} = \frac{2 \left\langle{{s}}\right\rangle}{N}\end{aligned} \hspace{\stretch{1}}(1.2.4)

of the system of N spins each of magnetic moment m in a magnetic field B. The spin excess is 2 s. Take the entropy as the logarithm of the multiplicity g(N, s) as given in (1.35):

\begin{aligned}\sigma(s) \approx \ln g(N, 0) - \frac{2 s^2}{N},\end{aligned} \hspace{\stretch{1}}(1.2.5)

for {\left\lvert{s}\right\rvert} \ll N. Hint: Show that in this approximation

\begin{aligned}\sigma(U) = \sigma_0 - \frac{U^2}{2 m^2 B^2 N},\end{aligned} \hspace{\stretch{1}}(1.2.6)

with \sigma_0 = \ln g(N, 0). Further, show that 1/\tau = - U/ (m^2 B^2 N), where U denotes \left\langle{{U}}\right\rangle, the thermal average energy.


I found this problem very hard to interpret. What exactly is being asked for? Equation (1.35) in the text was

\begin{aligned}g(N, s) \approx g(N, 0) e^{-\frac{2s^2}{N}}\end{aligned} \hspace{\stretch{1}}(1.0.7a)

\begin{aligned}g(N, 0) \approx \sqrt{ \frac{2}{\pi N}} 2^N,\end{aligned} \hspace{\stretch{1}}(1.0.7b)

from which we find the entropy 1.2.5 directly after taking logarithms. The temperature is found directly

The magnetization, for a system that has spin excess 2 s was defined as

\begin{aligned}U = -2 s m B \equiv - M B\end{aligned} \hspace{\stretch{1}}(1.0.8)

and we can substitute that for s

\begin{aligned}\sigma(U) = \sigma_0 - \frac{U^2}{2 m^2 B^2 N},\end{aligned} \hspace{\stretch{1}}(1.0.9)

and take derivatives for the temperature

\begin{aligned}\frac{1}{{\tau}}= \frac{\partial {\sigma}}{\partial {U}}= \frac{\partial {}}{\partial {U}}\left(\sigma_0 - \frac{U^2}{2 m^2 B^2 N} \right)=- \frac{U}{m^2 B^2 N}\end{aligned} \hspace{\stretch{1}}(1.0.10)

This gives us a relation between temperature and the energy of the system with spin excess 2 s, and we could write

\begin{aligned}\frac{M}{N m} = -\frac{U}{B N m} = \frac{m B}{\tau}.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Is this the relation that this problem was asking for?

Two things I don’t understand from this problem:

  • Where does 2 \left\langle{{s}}\right\rangle/N come from? If we calculate the expectatation of the spin excess, we find that it is zero

    \begin{aligned}\left\langle{{2 s}}\right\rangle = \frac{ \sqrt{ \frac{2}{\pi N}} 2^N \int_{-\infty}^\infty ds 2 se^{-\frac{2s^2}{N}}}{2^N}=0.\end{aligned} \hspace{\stretch{1}}(1.0.12)

  • If 2 \left\langle{{s}}\right\rangle has a non-zero value, then doesn’t that make \left\langle{{U}}\right\rangle also zero? It seems to me that U in 1.0.10 is the energy of a system with spin excess s, and not any sort of average energy?

Quantum harmonic oscillator [1] problem 2.3


Find the entropy of a set of N oscillators of frequency \omega as a function of the total quantum number n. Use the multiplicity function (1.55) and make the Stirling approximation \ln N! \approx N \ln N - N. Replace N - 1 by N.

Planck Energy

Let U denote the total energy n \hbar \omega of the oscillators. Express the entropy as \sigma(U, N). Show that the total energy at temperature \tau is

\begin{aligned}U = \frac{N \hbar \omega}{\exp\left( \hbar \omega/\tau \right) -1}\end{aligned} \hspace{\stretch{1}}(1.13)

This is the Planck result; it is derived again in Chapter 4 by a powerful method that does not require us to find the multiplicity function.



The multiplicity was found in the text to be

\begin{aligned}g(N, n) = \frac{(N + n -1)!}{n! (N-1)!}\end{aligned} \hspace{\stretch{1}}(1.0.14)

I wasn’t actually able to follow the argument in the text, and found the purely combinatoric wikipedia argument [3] much clearer. A similar diagram and argument can also be found in [2] section 3.8.

Taking logarithms and applying the Stirling approximation, our entropy is

\begin{aligned}\begin{aligned}\sigma&= \ln g \\ &= \ln (N + n - 1)! - \ln (N - 1)! - \ln n! \\ &\approx(N + n - 1) \ln (N + n - 1) - (N + n - 1)- (N - 1) \ln (N - 1) + (N - 1)- n \ln n + n \\ &=(N - 1) \ln \frac{N + n - 1}{N - 1}+ n \ln \frac{N + n - 1}{n}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.15)

Planck Energy

Now we make the N - 1 \rightarrow N replacement suggested in the problem (ie. assuming N \gg 1), for

\begin{aligned}\begin{aligned}\sigma &\approx N \ln \frac{N + n}{N}+ n \ln \frac{N + n}{n} \\ &= (N + n)\ln (N + n) - N \ln N - n \ln n \\ &= \left(N + \frac{U}{\hbar\omega} \right)\ln \left(N + \frac{U}{\hbar\omega} \right) - N \ln N - \frac{U}{\hbar\omega} \ln \frac{U}{\hbar\omega}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.16)

With (x \ln x)' = \ln x + 1, we have

\begin{aligned}\frac{1}{{\tau}}= \frac{\partial {\sigma}}{\partial {U}}=\frac{1}{{\hbar\omega}} \left( \ln \left(N + \frac{U}{\hbar\omega} \right) - 1 -\ln \frac{U}{\hbar\omega} + 1 \right),\end{aligned} \hspace{\stretch{1}}(1.0.17)


\begin{aligned}U e^{\frac{\hbar \omega}{\tau}} = N \hbar \omega + U.\end{aligned} \hspace{\stretch{1}}(1.0.18)

A final rearrangement gives us the Planck result.


[1] C. Kittel and H. Kroemer. Thermal physics. WH Freeman, 1980.

[2] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

[3] Wikipedia. Einstein solid — Wikipedia, The Free Encyclopedia, 2012. URL [Online; accessed 2-January-2013].


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