PHYS6562 W3 Daily Question

## Entropy maximization and temperature

First of all, *Ω*_{1}(*E*_{1})*δ**E*_{1} and *Ω*_{2}(*E*_{2})*δ**E*_{2} are the phase space volume for these two subsystems. Since most of the volume in phase space is near the surface, we can instead write *Ω*_{1}(*E*_{1}) and *Ω*_{2}(*E*_{2}) to denote the two volumes in phase space. Second, it’s clear that the probability density of subsystem 1 having energy *E*_{1} is proportional to the ’allowed’ volume product of the two subsystems.

\begin{equation}
\rho_1(E_1) \propto \Omega_1(E_1)\Omega_2(E_2)
\end{equation} where *E*_{2} = *E* − *E*_{1}. And finally, we impose the normalization condition:

\begin{equation}
\int \rho_1(E_1)dE_1 = A \int \Omega_1(E_1)\Omega_2(E-E_1)dE_1 = A \Omega(E) = 1
\end{equation} A is the proportional constant and equal to 1/*Ω*(*E*). Therefore,

\begin{equation} \rho_1(E_1) = \frac{\Omega_1(E_1)\Omega_2(E_2)}{\Omega(E)} \end{equation}

If we maximize the probability at *E*_{1}:

\begin{equation} \frac{d\rho_1}{dE_1} = 0 \end{equation}

\begin{equation} \frac{1}{\Omega_1}\frac{d\Omega_1}{dE_1} - \frac{1}{\Omega_2}\frac{d\Omega_2}{dE_2} = 0 \end{equation}

By plugging the definition of entropy *S* = *k*_{B}*l**o**g*(*Ω*),

\begin{equation} \frac{1}{k_B}\frac{dS_1}{dE_1} - \frac{1}{k_B}\frac{dS_2}{dE_2} = 0 \end{equation}

\begin{equation} \frac{dS_1}{dE_1} + \frac{dS_2}{dE_1} = \frac{d}{dE_1}(S(E_1)+S(E_2)) = 0 \end{equation} ,which implies the maximization of the sum of entropies of both subsystems.

For (6), we see that if the probability density *ρ*(*E*_{1}) is maximized,

\begin{equation} \frac{1}{T_1} = \frac{1}{T_2} \end{equation}

PHYS6562 F3 Daily Quaetion

## Undistinguished particles

For N distinguishable particles, if at certain instant they occupy N states in the phase space, there are *N*! different choices of such configuration. The total volume for this system in phase space is the product of the volume for each particle in its own sub phase space.

\begin{equation} \Omega_D = N! \space \omega_1 \omega_2 \omega_3 ... \omega_N \end{equation}

If the particles are now undistinguished, there is only one choice of such configuration without permutation.

\begin{equation} \Omega_U = \omega_1 \omega_2 \omega_3 ... \omega_N \end{equation}

Therefore, $\Omega_D = N!\space \Omega_U $.

For the mixing problem of black and white particles of the same pressure on both sides, if you cannot tell the difference between black and white ones, you cannot extract work from the mixing. Actully, you won’t know if they’ve mixed already or not. However, suppose the black and white particles are of different pressure, and you still cannot tell them apart. If there were a membrane which allows particles of lower pressure to penetrate through, this membrane then can tell the difference between the two sides without telling the particles apart. Then work can be extracted.

Now if a door can distinguish the black from the white particles and let through white ones, after very long time, the white particles will distribute themselves equally in the two sides, i.e. there will be $\frac{1}{4} N\space white + 0 \space black$ in the left and $\frac{1}{4} N \space white + \frac{1}{2} N \space black$ in the right. And we can extract work from it due to the pressure difference between the two sides.

PHYS6562 W4 Daily Question

## No attractors in Hamiltonian system

It’s not true that neighboring initial conditions will converge to the fixed state after long time. As we saw from the Jupiter problem, there are some regions where the initial conditions for chaotic and periodic trajectories lies infinitesimally close but lead to different results. The chaotic initial conditions will not make the system converge to certain state. Therefore, there is no attractor in this 3-body problem.

Stable Hamiltonian systems can withstand a minor perturbation *in the process* and the system will still end up where it’s supposed to.

Unstable Hamiltonian system will diverge if perturbed *in the process* and the system will not come to the original equilibrium regardless of whether the system becomes chaotic or not.

PHYS6562 F4 Daily Quaetion

## Ergodicity

The system of classical particles bouncing back and forth is not ergodic since each particles can only occupy certain close volume in phase space, with fixed *x*, *y* position, *P*_{x}, *P*_{y} momenta and ±*p*_{z}. Therefore, the given volume in phase space of the ensemble will not evolve with time.

## Heat engine

(a) True.

(b) True, since the work in done on the gas.

(c) False, it’s a refrigerator.

(d) True.

\begin{equation}
W = (P_0)(3V_0) + \int_{4V_0}^{V_0} \frac{4P_0V_0}{V} dV = 3P_0V_0 - 4P_0V_0\log4
\end{equation} (e) True, a part does no work so *Δ**U* = *Q*. \begin{equation}
\Delta U = \frac{3}{2} nRT_c - \frac{3}{2} nRT_h = -\frac{3}{2} (3P_0V_0) = -\frac{9}{2}P_0V_0
\end{equation} where n is the number of particles in mole and *R* = 8.317 is the ideal gas constant. (f) True. The whole cycle *Δ**U* = 0, the heat input and the net work done *onto* the gas must equal the heat output. \begin{equation}
Q_h + W = Q_c
\end{equation}

PHYS6562 M5 Daily Question

## Undistinguished particles and the Gibbs factor

(a) \begin{equation} \Omega = \frac{V^N}{N!} \end{equation}

\begin{equation}
S = k_B \space \log(\Omega) = k_B(N\log(V) - B\log(N) + N) = Nk_B\left(1+\log\left(\frac{V}{N}\right) \right)
\end{equation} The average volume of a region for single particle is *v* = *V*/*N* with the entropy of *s* = *k*_{B}log(*v*). Therefore, the configuration entropy is related to N times the entropy of a particle in its region with average volume.

(b) \begin{equation}
S = Nk_B\left(\frac{5}{2}-\log(\rho\lambda^3)\right) = Nk_B\left(\frac{5}{2}+\log\left(\frac{V}{N}\lambda^3\right)\right)
\end{equation} The configuration entropy for an ideal gas particle is also related to the small average volume *v*. \begin{equation}
S_c = Nk_B\log(v)
\end{equation}

But the ideal gas particles also occupy the momentum space and each particle has the average length of *λ* in each dimension. So

\begin{equation} S_p = 3Nk_B\log(\lambda) = Nk_B\log(\lambda^3) \end{equation}

Combine the configuration and momentum entropy for one particle, \begin{equation} S = S_c+S_p = Nk_B\log(v*\lambda^3) = Nk_B\log\left(\frac{V}{N}*\lambda^3\right) \end{equation} The total entropy is related to the N times the space occupied both in configuration and momentum space.

PHYS6562 W5 Daily Question

PHYS6562 F5 Daily Quaetion

## Loaded dice, and sticky spheres

(a)

For discrete system, the entropy \begin{equation}
S_{discrete} = -k_B \sum p_i\log p_i
\end{equation} \begin{equation}
S_{discrete} = \frac{3}{2}k_B \log 2
\end{equation}

(b)

Entropy for being hit by a comet with $\rho_c(\theta,\phi) = \frac{1}{4\pi}$: \begin{equation}
S_c = -k_B \iint \rho_s \log (\rho_s) d\Omega = k_B \log (4\pi)
\end{equation} Entropy for being hit by an asteroid with $\rho_a(\theta,\phi) = \frac{cos(\theta)}{\pi^2}$: \begin{equation}
S_a = -k_B \iint \rho_a \log (\rho_a) d\Omega = -k_B \int \left( \frac{2\cos^2\theta}{\pi}\log(\cos\theta) - \frac{4\cos^2\theta}{\pi}\log\pi \right)d\theta
\end{equation} \begin{equation}
S_a = -k_B \left(\frac{1}{2}(1-\log 4) - 2\log(\pi) \right) = k_B \log \left(\frac{2\pi^2}{\sqrt{e}} \right)
\end{equation}

The entropy difference \begin{equation} \Delta S = S_c - S_a = k_B\log\left(\frac{2\sqrt{e}}{\pi} \right) \end{equation}

AEP 4830 HW3 Root Finding and Special Functions

AEP 4830 HW2 Numerical Integration