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## 物理代写|统计力学代写Statistical mechanics代考|Heat capacity

Using Eq. (5.104) we can calculate the heat capacity,
$$C_V=\left(\frac{\partial U}{\partial T}\right)_{V, N}=N k\left[\frac{\pi^2}{2} \frac{T}{T_F}-\frac{3 \pi^4}{20}\left(\frac{T}{T_F}\right)^3+\cdots\right] . \quad\left(T \ll T_F\right)$$
Clearly, $C_V$ vanishes as $T \rightarrow 0$ (compare with Eq. (5.82)). We also see a characteristic feature of the Fermi gas: $C_V$ is linear with $T$ at low temperature.

One can understand $C_V \propto T$ at low temperature qualitatively. ${ }^{42}$ As per the considerations leading to Eq. (5.96), the energy states that can participate in energy transfers between the system and its environment are those roughly within $k T$ of $\mu$, which at low temperature is ostensibly the Fermi energy, $E_F$. How many of those states are there? The number of energy states per energy range is the density of states, $g(E)$. Thus, $g\left(E_F\right) \times k T$ is approximately the number of states near $E_F$ available to participate in energy exchanges, and is therefore the number of states that contribute to the heat capacity. ${ }^{43}$ The excitation energy for each energy transfer is approximately $k T$. Thus, we can estimate the energy of the Fermi gas at low temperatures as
$$U \approx U_0+\left[g\left(E_F\right) \times k T\right] \times k T,$$
where $U_0=\frac{3}{5} N E_F$, Eq. (5.93). As shown in Exercise 5.19, $g\left(E_F\right) E_F=\frac{3}{2} N$. The heat capacity based on this line of reasoning would then be
$$C_V \approx 2 k^2 g\left(E_F\right) T=3 N k \frac{T}{T_F} .$$
Such an argument gets you on the same page with the exact result, Eq. (5.106); the two differ by a multiplicative factor of order unity, $\pi^2 / 6$.

## 物理代写|统计力学代写Statistical mechanics代考|DEGENERACY PRESSURE IN THE LIFE OF STARS

Degeneracy pressure, associated with the large zero-point energy of collections of identical fermions, accounts reasonably well for the bulk modulus of metals (Section 5.6) -that compressing an electron gas meets with a significant resisting force associated with the Pauli exclusion principle. In this section we consider an astrophysical application of degeneracy pressure.

Stars generate energy through nuclear fusion, “burning” nuclei in processes that are fairly well understood. Stars convert hydrogen to helium by a series of fusion reactions: $\mathrm{H}^1+\mathrm{H}^1 \rightarrow \mathrm{H}^2+\mathrm{e}^{+}+\nu$, $\mathrm{H}^2+\mathrm{H}^1 \rightarrow \mathrm{He}^3+\gamma, \mathrm{He}^3+\mathrm{He}^3 \rightarrow \mathrm{He}^4+\mathrm{H}^1+\mathrm{H}^1$. When all hydrogen has been converted to helium, this phase of the burning process stops. Gravitational contraction then compresses the helium until the temperature rises sufficiently that a new sequence of reactions can take place, $\mathrm{He}^4+\mathrm{He}^4 \rightarrow$ $\mathrm{Be}^8+\gamma, \mathrm{Be}^8+\mathrm{He}^4 \rightarrow \mathrm{C}^{12}+\gamma$. As helium is exhausted, gravitational contraction resumes, heating the star until new burning processes are initiated. The variety of nuclear processes gets larger as new rounds of nuclear burning commence (the subject of stellar astrophysics). Eventually burning stops when the star consists of iron, silicon, and other elements. The force of gravity, however, is ever present. Can gravitational collapse can be forestalled? We now show that the degeneracy pressure of electrons in stars is enough to balance the force of gravity under certain circumstances.

We assume temperatures are sufficiently high that all atoms in stars are completely ionized, i.e., stars consist of fully-ionized plasmas, gases of electrons and positively charged species. Let there be $N$ electrons (of mass $m$ ) and assume conditions of electrical neutrality, that $N=N_p$, the number of protons (of mass $m_p$ ). Assume that the number of neutrons $N_n$ (locked up in nuclei) is the same as the number of protons, an assumption valid only for elements up to $Z \approx 20$; for heavier nuclei there might be $\approx 1.5$ neutrons per proton. We take $N_n=N_p$ to simplify the analysis. The neutron mass is nearly equal to the proton mass. The mass $M$ of the star is then, approximately (because $\left.m_p \gg m\right)$,
$$M \approx N\left(m+2 m_p\right) \approx 2 N m_p .$$
We make another simplifying assumption that the mass density $\rho=M / V$ is uniform 44 throughout a star of volume $V$. With these assumptions, the average electron number density $n \equiv N / V$ is
$$n=\frac{N}{V}=\frac{M /\left(2 m_p\right)}{M / \rho}=\frac{\rho}{2 m_p} .$$
White dwarfs are a class of stars thought to be in the final evolutionary state wherein nuclear burning has ceased and the star has become considerably reduced in size through gravitational contraction, where a star of solar mass $M_{\odot}$ might have been compressed into the volume of Earth. Sirius B, for example, is a white dwarf of mass $1.018 M_{\odot}$ and radius $8.4 \times 10^{-3} R_{\odot}$, implying an average electron density $7.2 \times 10^{29} \mathrm{~cm}^{-3}$, some seven orders of magnitude larger than the electron density in metals (see Table 5.2). We can take $n=10^{30} \mathrm{~cm}^{-3}$ as characteristic of white dwarf stars. Such high densities necessitate a relativistic treatment of the electrons (see Exercise 5.20).

# 统计力学代考

## 物理代写|统计力学代写Statistical mechanics代考|Heat capacity

$$C_V=\left(\frac{\partial U}{\partial T}\right)_{V, N}=N k\left[\frac{\pi^2}{2} \frac{T}{T_F}-\frac{3 \pi^4}{20}\left(\frac{T}{T_F}\right)^3+\cdots\right] . \quad\left(T \ll T_F\right)$$

$$U \approx U_0+\left[g\left(E_F\right) \times k T\right] \times k T,$$

$$C_V \approx 2 k^2 g\left(E_F\right) T=3 N k \frac{T}{T_F} .$$

## 物理代写|统计力学代写Statistical mechanics代考|DEGENERACY PRESSURE IN THE LIFE OF STARS

$$M \approx N\left(m+2 m_p\right) \approx 2 N m_p .$$

$$n=\frac{N}{V}=\frac{M /\left(2 m_p\right)}{M / \rho}=\frac{\rho}{2 m_p} .$$

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