PH503 Homework
1.
(a)
Liouville's equation $\frac{\partial \rho}{\partial t} = -\{\rho, \mathcal{H}\}$, or
$$\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t} + \sum_i \left( \frac{\partial \rho}{\partial q_i}\dot{q}_i + \frac{\partial \rho}{\partial p_i}\dot{p}_i \right) = \frac{\partial \rho}{\partial t} + \{\rho, \mathcal{H}\}$$
Which is $\frac{d\rho}{dt} = 0$, volume element $d\Gamma = dpdq$ is time invariant.
Entropy is $S = - \int \rho \ln \rho \, d\Gamma$.
Then $\rho$ invariant, $\rho \ln \rho$ invariant, $(\rho \ln \rho) \times d\Gamma$ invariant.
$\frac{dS}{dt} = 0$
(b)
Normalization: $C_1 = \int dpdq \, \rho(p, q) = 1$
Fixed average energy: $C_2 = \int dpdq \, \rho(p, q) \mathcal{H}(p, q) = E$
Lagrange multipliers:
$$L[\rho] = S - \alpha C_1 - \beta C_2$$
$$L[\rho] = - \int \rho \ln \rho \, dpdq - \alpha \left(\int \rho \, dpdq - 1\right) - \beta \left(\int \rho \mathcal{H} \, dpdq - E\right)$$
$$L[\rho] = \int (-\rho \ln \rho - \alpha \rho - \beta \rho \mathcal{H}) \, dpdq + \alpha + \beta E$$
To find the function $\rho$ that maximizes $L$,
$$\frac{\partial}{\partial \rho} (-\rho \ln \rho - \alpha \rho - \beta \rho \mathcal{H}) = 0$$
$$\ln \rho = -1 - \alpha - \beta \mathcal{H}$$
Then
$$\rho_{\text{max}} = e^{-1-\alpha} e^{-\beta \mathcal{H}}$$
$$\int \rho_{\text{max}} \, dpdq = \int e^{-1-\alpha} e^{-\beta \mathcal{H}} \, dpdq = e^{-1-\alpha} \int e^{-\beta \mathcal{H}} \, dpdq = 1$$
Partition function $Z = \int e^{-\beta \mathcal{H}} \, dpdq$.
$$e^{-1-\alpha} Z = 1 \quad \implies \quad e^{-1-\alpha} = \frac{1}{Z}$$
Finally,
$$\rho_{\text{max}}(p, q) = \frac{1}{Z} e^{-\beta \mathcal{H}(p, q)}$$
This is the probability distribution of the canonical ensemble.
(c)
A density function is stationary if its partial time derivative is zero $\frac{\partial \rho}{\partial t} = 0$.
Use Liouville's equation($\frac{\partial \rho}{\partial t} = -\{\rho, \mathcal{H}\}$).
Goal is $\{\rho_{\text{max}}, \mathcal{H}\}=0$.
$$\rho_{\text{max}} = f(\mathcal{H}) = \frac{1}{Z} e^{-\beta \mathcal{H}}$$
$$\{f(\mathcal{H}), \mathcal{H}\} = \sum_i \left( \frac{\partial f(\mathcal{H})}{\partial q_i} \frac{\partial \mathcal{H}}{\partial p_i} - \frac{\partial f(\mathcal{H})}{\partial p_i} \frac{\partial \mathcal{H}}{\partial q_i} \right)$$
Using the chain rule:
$$\frac{\partial f(\mathcal{H})}{\partial q_i} = \frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial q_i} \quad \text{and} \quad \frac{\partial f(\mathcal{H})}{\partial p_i} = \frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial p_i}$$
$$\{f(\mathcal{H}), \mathcal{H}\} = \sum_i \left( \left(\frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial q_i}\right) \frac{\partial \mathcal{H}}{\partial p_i} - \left(\frac{df}{d\mathcal{H}} \frac{\partial \mathcal{H}}{\partial p_i}\right) \frac{\partial \mathcal{H}}{\partial q_i} \right)$$
$$\{f(\mathcal{H}), \mathcal{H}\} = \frac{df}{d\mathcal{H}} \sum_i \left( \frac{\partial \mathcal{H}}{\partial q_i} \frac{\partial \mathcal{H}}{\partial p_i} - \frac{\partial \mathcal{H}}{\partial p_i} \frac{\partial \mathcal{H}}{\partial q_i} \right)$$
$$\{\rho_{\text{max}}, \mathcal{H}\} = 0$$
Since $\frac{\partial \rho_{\text{max}}}{\partial t} = -\{\rho_{\text{max}}, \mathcal{H}\}$:
$$\frac{\partial \rho_{\text{max}}}{\partial t} = 0$$
2.
(a)
Define the Systems:
- Subsystem 'a' has energy $E_a$, particle number $N_a$, and its number of accessible microstates is given by the phase space volume $\Gamma_a(E_a)$.
- Subsystem 'b' has energy $E_b$, particle number $N_b$, and its number of accessible microstates is $\Gamma_b(E_b)$.
- The total system has energy $E = E_a + E_b$ and particle number $N = N_a + N_b$.
Calculate the Total Entropy:
$$S_{total} = k_B \log \Gamma(E)$$
If system is statistically independent,
$$\Gamma(E) = \Gamma_a(E_a) \cdot \Gamma_b(E_b)$$
$$S_{total} = k_B \log(\Gamma_a(E_a) \cdot \Gamma_b(E_b))$$
Using $\log(xy) = \log(x) + \log(y)$:
$$S_{total} = k_B (\log \Gamma_a(E_a) + \log \Gamma_b(E_b))$$
$$S_{total} = k_B \log \Gamma_a(E_a) + k_B \log \Gamma_b(E_b)$$
$$S_{total} = S_a + S_b$$
(b)
$$\Gamma(E) = \Sigma(E+\Delta) - \Sigma(E)$$
Since $\Delta \ll E$, first-order Taylor expansion:
$$\Sigma(E+\Delta) \approx \Sigma(E) + \frac{\partial \Sigma(E)}{\partial E} \Delta$$
$$\Gamma(E) \approx \left(\Sigma(E) + \frac{\partial \Sigma(E)}{\partial E} \Delta\right) - \Sigma(E) = \frac{\partial \Sigma(E)}{\partial E} \Delta$$
Also,
$$\Gamma(E) \approx g(E) \Delta$$
Compare $S_1$ and $S_3$:
$$S_1 = k_B \log \Gamma(E) \approx k_B \log(g(E) \Delta)$$
$$S_1 \approx k_B (\log g(E) + \log \Delta) = S_3 + k_B \log \Delta$$
Since the entropy $S$ is extensive ($S \propto N$), this constant difference is negligible for large $N$. Then, $S_1 \approx S_3$.
Compare $S_2$ and $S_3$:
For large DoF $f$ system, let
$$\Sigma(E) \propto E^f$$
$$g(E) = \frac{\partial \Sigma(E)}{\partial E} \propto \frac{\partial}{\partial E}(E^f) = f E^{f-1} = \frac{f}{E} E^f \propto \frac{f}{E} \Sigma(E)$$
$$S_3 = k_B \log g(E) \approx k_B \log \left(C \frac{f}{E} \Sigma(E)\right) $$
$$S_3 \approx k_B \log \Sigma(E) + k_B \log\left(C\frac{f}{E}\right) = S_2 + k_B \log\left(C\frac{f}{E}\right)$$
Both the number of degrees of freedom ($f \propto N$) and the energy ($E \propto N$) are extensive. Then second term $O(\log N)$ at most. Since $S_2$ is of order $O(N)$, the difference is negligible for large $N$. Then, $S_2 \approx S_3$.
3.
The entropy of a monatomic ideal gas is:
$$S = N k_B \ln \left[ V \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{3}{2} N k_B$$
The entropy with the Gibbs correction factor ($S = S_{dist} - k_B \ln(N!)$) is given by the Sackur-Tetrode equation:
$$S(E, V, N) = N k_B \ln \left[ \frac{V}{N} \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{5}{2} N k_B$$
(a)
Let's scale the variables $E \to \lambda E$, $V \to \lambda V$, $N \to \lambda N$:
$S(\lambda E, \lambda V, \lambda N) = (\lambda N) k_B \ln \left[ \frac{(\lambda V)}{(\lambda N)} \left( \frac{4\pi m (\lambda E)}{3(\lambda N) h^2} \right)^{3/2} \right] + \frac{5}{2} (\lambda N) k_B$
$S(\lambda E, \lambda V, \lambda N) = \lambda N k_B \ln \left[ \frac{V}{N} \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{5}{2} \lambda N k_B$
$S(\lambda E, \lambda V, \lambda N) = \lambda \left( N k_B \ln \left[ \frac{V}{N} \left( \frac{4\pi m E}{3N h^2} \right)^{3/2} \right] + \frac{5}{2} N k_B \right)$
$$S(\lambda E, \lambda V, \lambda N) = \lambda S(E, V, N)$$
(b)
The Gibbs correction is a term that depends only on $N$ ($S_{corr} = -k_B \ln(N!)$).
Pressure is defined as $p = T \left( \frac{\partial S}{\partial V} \right)_{E,N}$.
Since the correction term, $-k_B \ln(N!)$, does not depend on volume $V$, its derivative with respect to $V$ is zero, pressure is unchanged.
Heat capacity is defined as $C_V = \left( \frac{\partial E}{\partial T} \right)_{V,N}$. Monatomic ideal gas, $E = \frac{3}{2} N k_B T$.
Since the relationship between $E$ and $T$ is unchanged, the heat capacity is also unchanged.
The exponent $\gamma$ is the ratio of specific heats, $\gamma = C_p/C_V$. Since $C_V$ is unchanged and Mayer's relation for an ideal gas ($C_p = C_V + N k_B$) still holds, $C_p$ is also unchanged. Consequently, $\gamma$ is unchanged, and the adiabatic relation remains the same.
(c)
Let's use the final Sackur-Tetrode equation:
$S = N k_B \left( \ln\left[\frac{V}{N}\left(\frac{4\pi mE}{3Nh^2}\right)^{3/2}\right] + \frac{5}{2} \right) = N k_B \left( \ln(V E^{3/2} C) - \frac{5}{2}\ln N + \frac{5}{2} \right)$ where C is a constant.
$\frac{\partial S}{\partial N} = k_B \left( \ln(V E^{3/2} C) - \frac{5}{2}\ln N + \frac{5}{2} \right) + N k_B \left(-\frac{5}{2N}\right)$
$\frac{\partial S}{\partial N} = k_B \ln\left[\frac{V}{N^{5/2}} (\dots)\right] + \frac{5}{2}k_B - \frac{5}{2}k_B = k_B \ln\left[\frac{V}{N}\left(\frac{4\pi mE}{3Nh^2}\right)^{3/2}\right]$
$\mu = -T \frac{\partial S}{\partial N} = -k_B T \ln\left[\frac{V}{N}\left(\frac{4\pi mE}{3Nh^2}\right)^{3/2}\right]$
Since the entropy $S$ changes, the Helmholtz free energy $F$ must also change.
$F_{new} = E - T S_{new} = E - T(S_{old} - k_B \ln N!) = (E - T S_{old}) + k_B T \ln N!$
$$F_{new} = F_{old} + k_B T \ln(N!)$$
4.
(a)
Use the entropy formula for an ideal gas, $S(N,V) = N k_B \ln(V/N) + \text{constant}$.
Initial Entropy: $S_i = S(N_1, V_1) + S(N_2, V_2) = N_1 k_B \ln(V_1/N_1) + N_2 k_B \ln(V_2/N_2)$.
Final Entropy: $S_f = S(N_1+N_2, V_1+V_2) = (N_1+N_2) k_B \ln\left(\frac{V_1+V_2}{N_1+N_2}\right)$.
$$(\Delta S)_{1=2} = S_f - S_i = k_B \left[ (N_1+N_2) \ln\left(\frac{V_1+V_2}{N_1+N_2}\right) - N_1 \ln\left(\frac{V_1}{N_1}\right) - N_2 \ln\left(\frac{V_2}{N_2}\right) \right]$$
Set volume $v_1 = V_1/N_1$ and $v_2 = V_2/N_2$:
$$v_f = \frac{V_1+V_2}{N_1+N_2} = \frac{N_1 v_1 + N_2 v_2}{N_1+N_2}$$
Then it changes with $N = N_1+N_2$, $x_1 = N_1/N$, and $x_2 = N_2/N$:
$$\ln(v_f) - x_1 \ln(v_1) - x_2 \ln(v_2) \ge 0$$
$$\ln(x_1 v_1 + x_2 v_2) \ge x_1 \ln(v_1) + x_2 \ln(v_2)$$
Use Jensen's inequality ($f(x) = \ln(x)$, is strictly concave), it's true.
Equal at
$$v_1 = v_2 \implies \frac{V_1}{N_1} = \frac{V_2}{N_2}$$
(b)
$$(\Delta S)^* = -k_B (N_1 \ln x_1 + N_2 \ln x_2)$$
where $N = N_1 + N_2$ and $x_1 = N_1/N$ and $x_2 = N_2/N$.
Let $N = N_1+N_2$. then $x = x_1 = N_1/N$, so that $x_2 = 1-x$,
$$(\Delta S)^*(x) = -N k_B [x \ln x + (1-x) \ln(1-x)]$$
To find the maximum,
$$\frac{d(\Delta S)^*}{dx} = -N k_B \frac{d}{dx}[x \ln x + (1-x) \ln(1-x)]$$ $$\frac{d(\Delta S)^*}{dx} = -N k_B \left[ (\ln x + 1) + (-\ln(1-x) - 1) \right] = -N k_B [\ln x - \ln(1-x)]$$
$$\ln x - \ln(1-x) = 0 \implies \ln x = \ln(1-x) \implies x = 1-x \implies 2x=1$$
at $x = 1/2$.
The second derivative is $\frac{d^2(\Delta S)^*}{dx^2} = -N k_B (\frac{1}{x} + \frac{1}{1-x})$, which is negative for $x \in (0,1)$, then $x=1/2$ is a maximum.
Maximum when $x_1 = x_2 = 1/2$, which means $N_1 = N_2$.
$$(\Delta S)^*_{max} = -N k_B \left[ \frac{1}{2} \ln\left(\frac{1}{2}\right) + \frac{1}{2} \ln\left(\frac{1}{2}\right) \right]$$ $$(\Delta S)^*_{max} = -N k_B \left[ \ln\left(\frac{1}{2}\right) \right] = -N k_B (-\ln 2) = N k_B \ln 2$$
$$(\Delta S)^* \le (N_1+N_2)k_B \ln 2$$
5.
For ideal gas, the adiabatic exponent $\gamma$ and the molar heat capacities $C_{P,m}$ and $C_{V,m}$ are related by:
$$\gamma = \frac{C_{P,m}}{C_{V,m}}$$
$$C_{P,m} - C_{V,m} = R$$
$$\gamma = \frac{C_{V,m} + R}{C_{V,m}} = 1 + \frac{R}{C_{V,m}}$$
$$\gamma - 1 = \frac{R}{C_{V,m}} \quad \implies \quad C_{V,m} = \frac{R}{\gamma - 1}$$
For a mixture of ideal gases,
$$C_{V,mix} = f_1 C_{V,m1} + f_2 C_{V,m2}$$
where $f_1$ and $f_2$ are the mole fractions, and $C_{V,m1}$ and $C_{V,m2}$ are the molar heat capacities of gas 1 and gas 2.
Use $C_{V,mix} = \frac{R}{\gamma - 1}$, $C_{V,m1} = \frac{R}{\gamma_1 - 1}$, $C_{V,m2} = \frac{R}{\gamma_2 - 1}$
$$\frac{R}{\gamma - 1} = f_1 \left(\frac{R}{\gamma_1 - 1}\right) + f_2 \left(\frac{R}{\gamma_2 - 1}\right)$$
$$\frac{1}{\gamma - 1} = \frac{f_1}{\gamma_1 - 1} + \frac{f_2}{\gamma_2 - 1}$$