The Sommerfeld expansion is an expansion in powers of kTεF, which is assumed to be small. In this section I kept all terms through order kTεF2, omitting higher-order terms. Show at each relevant step that the term proportional to localid="1650117451748" T3 is zero, so that the next nonvanishing terms in the expansions forlocalid="1650117470867" μ and localid="1650117476821" Uare proportional to localid="1650117458596" T4. (If you enjoy such things, you might try evalu...

Q. 7.30 - An Introduction to Thermal Physics

Q. 7.30

Question

The Sommerfeld expansion is an expansion in powers of $\frac{k T}{ε_{F}}$ , which is assumed to be small. In this section I kept all terms through order ${(\frac{k T}{ε_{F}})}^{2}$ , omitting higher-order terms. Show at each relevant step that the term proportional to $T^{3}$ is zero, so that the next nonvanishing terms in the expansions for $μ$ and $U$ are proportional to $T^{4}$ . (If you enjoy such things, you might try evaluating the $T^{4}$ terms, possibly with the aid of a computer algebra program.)

Step-by-Step Solution

Verified

Answer

At each step the term proportional to $T^{3}$ is zero and the next nonvanishing terms in the expansion for $μ$ and $U$ are proportional to $T^{4}$ in the proof of expression $U \approx \frac{2}{5} g_{0} [μ^{\frac{5}{2}} + \frac{5 π^{2}}{8} {(k T)}^{2} μ^{\frac{1}{2}} + \frac{7 π^{4}}{384} {(k T)}^{4} μ^{\frac{- 3}{2}}]$

1Step 1: Given information

We have been given that the Sommerfeld expansion is an expansion in powers of $\frac{k T}{ε_{F}}$ , which is assumed to be small and we kept all terms through order ${(\frac{k T}{ε_{F}})}^{2}$ , omitting higher-order terms.

We need to show at each relevant step that the term proportional to $T^{3}$ is zero, so that the next nonvanishing terms in the expansions for $μ$ and $U$ are proportional to $T^{4}$ .

2Step 2: Simplify

The total energy given by the integral $7.54$ is:

$U = \int_{0}^{\infty} ε g (ε) {\bar{n}}_{F D} (ε) d ε$

$U = g_{0} \int_{0}^{\infty} ε^{\frac{3}{2}} {\bar{n}}_{F D} (ε) d ε$

Integrating by parts, we get:

$U = \frac{2}{5} g_{0} ε^{\frac{5}{3}} {\bar{n}}_{F D} |_{0}^{\infty} + \frac{2}{5} g_{0} \int_{0}^{\infty} ε^{\frac{5}{2}} \frac{\partial {\bar{n}}_{F D}}{\partial ε} d ε$ (Let this equation be ( $1$ ))

If we substitute $ε = 0$ , the integral becomes zero due to the dependence of the term on $ε^{\frac{5}{3}}$ and the first term vanishes.

On simplifying, the equation ( $1$ ) becomes:

$U = \frac{2}{5} g_{0} \int_{0}^{\infty} ε^{\frac{5}{2}} \frac{\partial {\bar{n}}_{F D}}{\partial ε} d ε$ (Let this equation be ( $2$ ))

We know that ${\bar{n}}_{F D} = \frac{1}{e^{\frac{(ε - μ)}{k T}} + 1}$

Taking derivative with respect to $ε$ , we get:

$\frac{\partial {\bar{n}}_{F D}}{\partial ε} = \frac{\partial [\frac{1}{e^{\frac{(ε - μ)}{k T}} + 1}]}{\partial ε}$

On simplifying, we get:

$\frac{\partial {\bar{n}}_{F D}}{\partial ε} = \frac{1}{k T} \frac{e^{\frac{(ε - μ)}{k T}}}{{(e^{\frac{(ε - μ)}{k T}} + 1)}^{2}}$

Let $\frac{(ε - μ)}{k T = x}$ , then we can write as :

$d x = \frac{d ε}{k T}$

Substitute $d x, x, \frac{\partial {\bar{n}}_{F D}}{\partial ε}$ in equation ( $2$ ), we get:
$U = \frac{2}{5} g_{0} \int_{0}^{\infty} ε^{\frac{5}{2}} \frac{1}{k T} \frac{e^{x}}{{(e^{x} + 1)}^{2}} k T d x$

$U = \frac{2}{5} g_{0} \int_{0}^{\infty} ε^{\frac{5}{2}} \frac{e^{x}}{{(e^{x} + 1)}^{2}} d x$

3Step 3: Changing the limits of integration

We need to change the boundaries of integration, so:

$ε \to \infty x \to \infty ε \to 0 x \to - \frac{μ}{k T}$

As $k T ≪ μ$ , so we can put $- \infty$ as the lower limit of integral, so the integral will become:

$U = \frac{2}{5} g_{0} \int_{- \infty}^{\infty} ε^{\frac{5}{2}} \frac{e^{x}}{{(e^{x} + 1)}^{2}} d x$ (Let this equation be ( $3$ ))

Now by expanding the term $ε^{\frac{5}{2}}$ about $μ$ using Taylor series, we get:

$ε^{\frac{5}{2}} = μ^{\frac{5}{2}} + \frac{5}{2} (ε - μ) μ^{\frac{3}{2}} + \frac{5}{16} {(ε - μ)}^{2} μ^{\frac{1}{2}} + \frac{5}{128} {(ε - μ)}^{3} μ^{\frac{- 1}{2}} + \frac{15}{8} {(ε - μ)}^{4} μ^{\frac{- 3}{2}} . . .$

Substitute $ε - μ = k T x$ , we get:

$ε^{\frac{5}{2}} = μ^{\frac{5}{2}} + \frac{5}{2} (k T x) μ^{\frac{3}{2}} + \frac{15}{8} {(k T x)}^{2} μ^{\frac{1}{2}} + \frac{5}{16} {(k T x)}^{2} μ^{\frac{- 1}{2}} + \frac{5}{128} {(k T x)}^{2} μ^{\frac{- 3}{2}} . . .$

4Step 4: Finding the values of integrals

Substitute the value of $ε^{\frac{5}{2}}$ in equation ( $3$ ), we get three integral say $I_{1}, I_{2}, I_{3}$ :

$U = \frac{2}{5} g_{0} (I_{1} + I_{2} + I_{3} + I_{4} + I_{5})$ (Let this equation be ( $4$ ))

where, $I_{1} = μ^{\frac{5}{2}} \int_{- \infty}^{\infty} \frac{e^{x}}{{(e^{x} + 1)}^{2}} d x$

$I_{2} = \frac{5}{2} k T μ^{\frac{3}{2}} \int_{- \infty}^{\infty} \frac{x e^{x}}{{(e^{x} + 1)}^{2}} d x$

$I_{3} = \frac{15}{8} {(k T)}^{2} μ^{\frac{1}{2}} \int_{- \infty}^{\infty} \frac{x^{2} e^{x}}{{(e^{x} + 1)}^{2}} d x$

$I_{4} = \frac{5}{16} {(k T)}^{3} μ^{\frac{- 1}{2}} \int_{- \infty}^{\infty} \frac{x^{3} e^{x}}{{(e^{x} + 1)}^{2}} d x$

$I_{5} = \frac{5}{128} {(k T)}^{4} μ^{\frac{- 3}{2}} \int_{- \infty}^{\infty} \frac{x^{4} e^{x}}{{(e^{x} + 1)}^{2}} d x$

On simplifying $I_{1}, I_{2}, I_{3}, I_{4}, I_{5}$ , we get:

$I_{1} = μ^{\frac{5}{2}} I_{2} = 0 I_{3} = \frac{5 π^{2}}{8} {(k T)}^{2} μ^{\frac{1}{2}} I_{4} = 0 I_{5} = \frac{7 π^{4}}{384} {(k T)}^{4} μ^{\frac{- 3}{2}}$

Substituting $I_{1}, I_{2}, I_{3}, I_{4}, I_{5}$ in equation ( $4$ ), we get:

$U \approx \frac{2}{5} g_{0} [μ^{\frac{5}{2}} + \frac{5 π^{2}}{8} {(k T)}^{2} μ^{\frac{1}{2}} + \frac{7 π^{2}}{384} {(k T)}^{4} μ^{\frac{- 3}{2}}]$

We have proved that the proportional term to $T^{3}$ is zero. We can also evaluate the further terms using computer algebra program, if wanted.

Q. 7.31

Q. 7.33

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