Question

In Section 2.5 I quoted a theorem on the multiplicity of any system with only quadratic degrees of freedom: In the high-temperature limit where the number of units of energy is much larger than the number of degrees of freedom, the multiplicity of any such system is proportional to $U^{Nf/2}$, where$Nf$ is the total number of degrees of freedom. Find an expression for the energy of such a system in terms of its temperature, and comment on the result. How can you tell that this formula for$\Omega$ cannot be valid when the total energy is very small?

Short answer

The required expression is$U=\frac{f}{2}NkT$.

Step-by-step solution

Given

It is given that,

$\Omega \propto U^{\frac{Nf}{2}}$,

where,

$Nf$is the total number of degrees of freedom

Expression for multiplicity that is given as:

$\Omega =\frac{V^N(2\pi mU{)}^{\frac{3N}{2}}}{h^{3N}N!\left(\frac{3N}{2}\right)!}$

And for Einstein solid in the high temperature limit expression for multiplicity is given as:

$\Omega \approx {\left(\frac{qe}{N}\right)}^N$

Here, $N$is number of oscillator in solid, $q$is number of energy unit.

Calculation

Mathematically, temperature can be defined as:

$\frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)..........\left(1\right)$

Where,

$\partial S$is the change in entropy and $\partial U$is the change in the internal energy of the body.

The given system has two quadratic terms in its energy. One is potential energy $\frac{1}{2}k{x}^2$and another one is kinetic energy term $\frac{1}{2}m{v}^2$. Each of this is interpreted as degree of freedom.

So, multiplicity is written as:

localid="1646989414665" $$\begin{gathered} \Omega \propto U^{\frac{Nf}{2}} \\ \Omega =A{U}^{\frac{NF}{2}} \end{gathered}$$

Here, $A$is constant of proportionality.

Now, Entropy is given by

$S=k\ln \Omega$

Where, $k$is Boltzmann constant and $\Omega$is multiplicity.

By substituting the value of $\Omega$in the above, we get,

$S=k\ln \left(A{U}^{\frac{Nf}{2}}\right)$

But we know that,

$\ln \left(a^b\right)=b\ln \left(a\right),\ln \left(\frac{a}{b}\right)=\ln \left(a\right)-\ln \left(b\right)$and $\ln \left(a^b\right)=b\ln \left(a\right)$

Hence,

$S=k\ln \left(A\right)+\frac{Nfk}{2}\ln U$

Now, by substituting this value of $S$in the equation (1), we get,

localid="1646990034822" $$\begin{gathered} \frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)=\frac{\partial }{\partial U}\left(k\ln \left(A\right)+\frac{Nfk}{2}\ln U\right) \\ \frac{1}{T}=\frac{Nfk}{2U} \\ T=\frac{2U}{Nfk} \end{gathered}$$

By rearranging the terms, we get,

$U=\frac{f}{2}NkT$

Final answer

Hence, the required expression for the energy of the given system in terms of its temperature is$U=\frac{f}{2}NkT$.