Riemann surface










问题 1.

The pressure which a gas exerts on the walls of a vessel can be regarded as the time average of the impulses which the gas molecules impart on the wall when colliding with and recoiling from it. From this point of view, calculate the pressure $p$ and show that
p=\frac{2}{3} n \bar{\varepsilon} \quad \text { (Bernoulli’s formula), }
where $n$ is the average number of molecules per unit volume and $\bar{\varepsilon}$ the mean kinetic evergy per molecule.

Let the vessel be a cube with edges of length $l$. Suppose that the wall is perfectly smooth so that a molecule colliding with it is reflected in a completely elastic way. Let us take the axis of an orthogonal coordinate system $x$, $\boldsymbol{y}, \boldsymbol{z}$ parallel to the edges. We shall neglect collisions of molecules with each other. The momentum components of a given molecule do not change their magnitude as a result of collisions with the walls. Therefore, this molecule collides $\left|p_x\right| / 2 \mathrm{ml}$ times per unit time with one of the walls prependicular to the $x$-axis, where $m$ is the mass of the molecule. As a result of each collision the wall receives a momentum $2\left|p_x\right|$ directed along the outward normal of the wall (perfect reflection). Hence the time average of the momentum, in other

words, the sum of the momentum imparted to the wall per unit time, is equal to $2\left|p_x\right| \cdot\left|p_x\right| / 2 m l=p_x^2 / m l$. Since the sum of these momenta yields a force, the contribution of this molecule to the pressure is given by this sum divided by the area of the wall $l^2\left(V=l^3\right.$ is the volume of the vessel). Therefore, adding together the contributions from all the molecules, we obtain the pressure exerted by all the molecules:
p=\frac{1}{3 V} \sum_{i=1}^N \frac{p_{i x}^2+p_{i z}^2+p_{i z}^2}{m}=\frac{2}{3 V} \sum_{i=1}^N \frac{p_i^2}{2 m}=\frac{2}{3} \frac{N}{V} \bar{\varepsilon}=\frac{2}{3} n \bar{\varepsilon} .

问题 2.

Show that a system in contact with a heat and particle source has particle number $N$ and energy $E$ with the probability given by $(1.73 \mathrm{a}, \mathrm{b})$.

Let $\Omega\left(N_{\mathrm{s}}, E_{\mathrm{s}}\right)$ be the density of states of the heat-particle source. We shall denote by $N_{\mathrm{t}}$ and $E_{\mathrm{t}}$ the total number of particles and the total energy of the compound system composed by the system under consideration and this heat-particle source, respectively. The probability $\operatorname{Pr}(N, E)$ that the system under consideration is in a microscopic state with $N$ and $E$ is proportional to the thermodynamic weight of a state of the heat-particle source with $N_{\mathrm{t}}-N$ and $E_{\mathrm{t}}-E$ :
\operatorname{Pr}(N, E) & \propto \Omega\left(N_{\mathrm{t}}-N, E_{\mathrm{t}}-E\right) \delta E_{\mathrm{t}} \
& \propto \exp \frac{1}{k}\left{S\left(N_{\mathrm{t}}-N, E_{\mathrm{t}}-E\right)-S\left(N_{\mathrm{t}} E_{\mathrm{t}}\right)\right}

Since the heat-particle source is a very large system, we may put $N_{\mathrm{t}} \gg N$ and $E_{\mathrm{t}} \gg E$. Hence
& S\left(N_{\mathrm{t}}-N, E_{\mathrm{t}}-E\right)-S\left(N_{\mathrm{t}}, E_{\mathrm{t}}\right)= \
& =-N\left(\frac{\partial S}{\partial N_{\mathrm{t}}}\right){E{\mathrm{t}}}-E\left(\frac{\partial S}{\partial E_{\mathrm{t}}}\right){N{\mathrm{t}}}+\frac{1}{2}\left{N^2 \frac{\partial^2 S}{\partial N_{\mathrm{t}}^2}+2 N E \frac{\partial^2 S}{\partial N_{\mathrm{t}} \partial E_{\mathrm{t}}}+E^2 \frac{\partial^2 S}{\partial E_{\mathrm{t}}^2}\right}+\ldots \
& =\frac{\mu}{T} N-\frac{E}{T}+\frac{1}{2}\left{N \Delta\left(-\frac{\mu}{T}\right)+E \Delta\left(\frac{1}{T}\right)\right}+\ldots
Here $\mu$ and $T$ are the chemical potential and the temperature of the heatparticle source with $N_t$ and $E_t$, and $\Delta$ indicates the variations due to the deviation of $N$ and $E$ from $N_{\mathrm{t}}$ and $E_{\mathrm{t}}$. Since these changes are $\mathrm{O}\left(N / N_{\mathrm{t}}\right)$ and $\mathrm{O}\left(E / E_{\mathrm{b}}\right)$, we can neglect them as long as $N \ll N_{\mathrm{t}}, E \ll E_{\mathrm{t}}$. Since the probability of having $N \sim N_{\mathrm{t}}, E \sim E_{\mathrm{t}}$ is, at any rate, extremely small, we can use this approximation. Therefore, from (1) we get
\operatorname{Pr}(N, E) \propto \exp \left(\frac{\mu N-E}{k T}\right) .
Normalizing this, we obtain (1.73a). (The same argument applies to the case where many kinds of particles are present.)

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